case study based questions class 12 physics chapter 4

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CBSE Class 12 Physics Case Study Questions PDF Download

Case Study questions for the Class 12 Physics board exams are available here. You can read the Class 12 Physics Case Study Questions broken down by chapter. Subject matter specialists and seasoned teachers created these quizzes. You can verify the right response to each question by referring to the answer key, which is also provided. To achieve success on your final exams, practice the following questions.

CBSE (Central Board of Secondary Education) is a renowned educational board in India that designs the curriculum for Class 12 Physics with the goal of promoting a scientific temperament and nurturing critical thinking among students. As part of their Physics examination, CBSE includes case study questions to assess students’ ability to apply theoretical knowledge to real-world scenarios effectively.

Chapter-wise Solved Case Study Questions for Class 12 Physics

Before the exams, students in class 12 should review crucial Physics Case Study issues. They will gain a better understanding of the kinds of Case Study questions that may be asked in Physics exams for Grade 12. These questions were created by our highly qualified standard 12 Physics staff based on the questions that appeared most frequently in last year’s exams. The solutions have been written in a way that will make them simple to grasp and will aid students in grade 12 in understanding the topics.

Best Books for Class 12 Physics Exams

Strictly in accordance with the new term-by-term curriculum for the class 12 board exams to be held in the academic year 2023–2024, which will include multiple choice questions based on new board typologies including stand-alone MCQs and case-based MCQs with an assertion–reason. Included are inquiries from the official CBSE Question Bank that was released in April 2024. What changes have been made to the book: strictly in accordance with the term-by-term syllabus for the board exams that will be held during the 2024 academic year? Chapter- and topic-based Questions with multiple choices that are based on the unique evaluation method used for the Class 12th Physics Final Board Exams.

Key Benefits of Solving CBSE Class 12 Physics Case Study Questions

• Application of Concepts: Case study questions demand the application of theoretical knowledge in practical scenarios, preparing students for real-world challenges and professional pursuits.
• Critical Thinking: By evaluating and analyzing case studies, students develop critical thinking abilities, enabling them to approach complex problems with a logical mindset.
• In-Depth Understanding: Addressing case study questions necessitates a thorough understanding of physics concepts, leading to a more profound comprehension of the subject matter.
• Holistic Evaluation: CBSE adopts case study questions as they provide a holistic evaluation of a student’s aptitude and proficiency in physics, moving beyond rote memorization.
• Preparation for Competitive Exams: Since competitive exams often include similar application-based questions, practicing case study questions equips students for various entrance tests.

How to Approach CBSE Class 12 Physics Case Study Questions

• Read and Analyze Thoroughly: Carefully read the case study to grasp its context and identify the underlying physics principles involved.
• Identify Relevant Concepts: Highlight the physics theories and concepts applicable to the given scenario.
• Create a Systematic Solution: Formulate a step-by-step solution using the identified concepts, explaining each step with clarity.
• Include Diagrams and Charts: If relevant, incorporate diagrams, charts, or graphs to visually represent the situation, aiding better comprehension.
• Double-Check Answers: Always review your answers for accuracy, ensuring that they align with the principles of physics.

Tips for Excelling in CBSE Class 12 Physics

• Conceptual Clarity: Focus on building a strong foundation of physics concepts, as this will enable you to apply them effectively to case study questions.
• Practice Regularly: Dedicate time to solving case study questions regularly, enhancing your proficiency in handling real-world scenarios.
• Seek Guidance: Don’t hesitate to seek guidance from teachers, peers, or online resources to gain additional insights into challenging concepts.
• Time Management: During exams, practice efficient time management to ensure you allocate enough time to each case study question without rushing.
• Stay Positive: Approach case study questions with a positive mindset, embracing them as opportunities to showcase your skills and knowledge.

CBSE Class 12 Physics case study questions play a pivotal role in promoting practical understanding and critical thinking among students. By embracing these questions as opportunities for growth, students can excel in their physics examinations and beyond.

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So, you’re in Class 12 and the concept of Physics case study questions is beginning to loom large. Are you feeling a little lost? Fear not! This article will guide you through everything you need to know to conquer these challenging yet rewarding Class 12 Physics Case Study Questions.

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

We have provided here Case Study questions for the Class 12 Physics for board exams. You can read these chapter-wise Case Study questions. These questions are prepared by subject experts and experienced teachers. The answer and solutions are also provided so that you can check the correct answer for each question. Practice these questions to score excellent marks in your final exams.

Physics is more than just formulas and equations, right? It’s a way to interpret the natural world, and case studies provide a perfect opportunity for students to apply theoretical knowledge to real-world situations. So, what’s the importance of case studies in class 12 Physics? Well, they help you to solidify your understanding and enhance your analytical skills, which are invaluable for exams and beyond.

Table of Contents

Case Study-Based Questions for Class 12 Physics

There are total of 14 chapters Electric Charges And Fields, Electrostatic Potential And Capacitance, Current Electricity, Moving Charges And Magnetism, Magnetism And Matter, Electromagnetic Induction, Alternating Current, Electromagnetic Waves, Ray Optics and Optical Instruments, Wave Optics, Dual Nature Of Radiation And Matter, Atoms, Nuclei, Semiconductor Electronics Materials Devices, And Simple Circuits.

• Case Study Based Questions on class 12 Physics Chapter 1 Electric Charges And Fields
• Case Study Based Questions on Class 12 Physics Chapter 2 Electrostatic Potential And Capacitance
• Case Study Based Questions on Class 12 Physics Chapter 3 Current Electricity
• Case Study Based Questions on Class 12 Physics Chapter 4 Moving Charges And Magnetism
• Case Study Based Questions on Class 12 Physics Chapter 5 Magnetism And Matter
• Case Study Based Questions on Class 12 Physics Chapter 6 Electromagnetic Induction
• Case Study Based Questions on Class 12 Physics Chapter 7 Alternating Current
• Case Study Based Questions on Class 12 Physics Chapter 8 Electromagnetic waves
• Case Study Based Questions on Class 12 Physics Chapter 9 Ray Optics and Optical Instruments
• Case Study Based Questions on class 12 Physics Chapter 10 Wave Optics
• Case Study Based Questions on class 12 Physics Chapter 11 Dual Nature of Matter and Radiation
• Case Study Based Questions on class 12 Physics Chapter 12 Atoms
• Case Study Based Questions on class 12 Physics Chapter 13 Nuclei
• Case Study Based Questions on class 12 Physics Chapter 14 Semiconductor Electronics
• Class 12 Chemistry Case Study Questions
• Class 12 Biology Case Study Questions
• Class 12 Maths Case Study Questions

Class 12 students should go through important Case Study problems for Physics before the exams. This will help them to understand the type of Case Study questions that can be asked in Grade 12 Physics examinations. Our expert faculty for standard 12 Physics have designed these questions based on the trend of questions that have been asked in last year’s exams. The solutions have been designed in a manner to help the grade 12 students understand the concepts and also easy-to-learn solutions.

How to Approach Physics Case Study Questions

Ah, the million-dollar question! How do you tackle these daunting case study questions? Let’s dive in.

Effective Reading Techniques

A vital part of cracking case study questions is understanding the problem correctly. Don’t rush through the question, take your time to fully grasp the scenario, and don’t skip over the details – they can be critical!

Conceptual Analysis

Once you’ve understood the problem, it’s time to figure out which physics principles apply. Remember, your foundational knowledge is your greatest weapon here.

Solving Strategies

Next, put pen to paper and start solving! Follow a systematic approach, check your calculations, and make sure your answer makes sense in the context of the problem.

Best Books for Class 12 Physics Exams

Class 12 Physics Syllabus 2024

Chapter–1: Electric Charges and Fields

Electric charges, Conservation of charge, Coulomb’s law-force between two point charges, forces between multiple charges; superposition principle, and continuous charge distribution.

Electric field, electric field due to a point charge, electric field lines, electric dipole, electric field due to a dipole, torque on a dipole in a uniform electric field.

Electric flux, statement of Gauss’s theorem and its applications to find field due to infinitely long straight wire, uniformly charged infinite plane sheet, and uniformly charged thin spherical shell (field inside and outside).

Chapter–2: Electrostatic Potential and Capacitance

Electric potential, potential difference, electric potential due to a point charge, a dipole and system of charges; equipotential surfaces, electrical potential energy of a system of two-point charges and of electric dipole in an electrostatic field.

Conductors and insulators, free charges and bound charges inside a conductor. Dielectrics and electric polarization, capacitors and capacitance, combination of capacitors in series and in parallel, capacitance of a parallel plate capacitor with and without dielectric medium between the plates, energy stored in a capacitor (no derivation, formulae only).

Chapter–3: Current Electricity

Electric current, flow of electric charges in a metallic conductor, drift velocity, mobility and their relation with electric current; Ohm’s law, V-I characteristics (linear and non-linear), electrical energy and power, electrical resistivity and conductivity, temperature dependence of resistance, Internal resistance of a cell, potential difference and emf of a cell, combination of cells in series and in parallel, Kirchhoff’s rules, Wheatstone bridge.

Chapter–4: Moving Charges and Magnetism

Concept of magnetic field, Oersted’s experiment.

Biot – Savart law and its application to current carrying circular loop.

Ampere’s law and its applications to infinitely long straight wire. Straight solenoid (only qualitative treatment), force on a moving charge in uniform magnetic and electric fields.

Force on a current-carrying conductor in a uniform magnetic field, force between two parallel current-carrying conductors-definition of ampere, torque experienced by a current loop in uniform magnetic field; Current loop as a magnetic dipole and its magnetic dipole moment, moving coil galvanometerits current sensitivity and conversion to ammeter and voltmeter.

Chapter–5: Magnetism and Matter

Bar magnet, bar magnet as an equivalent solenoid (qualitative treatment only), magnetic field intensity due to a magnetic dipole (bar magnet) along its axis and perpendicular to its axis (qualitative treatment only), torque on a magnetic dipole (bar magnet) in a uniform magnetic field (qualitative treatment only), magnetic field lines.

Magnetic properties of materials- Para-, dia- and ferro – magnetic substances with examples, Magnetization of materials, effect of temperature on magnetic properties.

Chapter–6: Electromagnetic Induction

Electromagnetic induction; Faraday’s laws, induced EMF and current; Lenz’s Law, Self and mutual induction.

Chapter–7: Alternating Current

Alternating currents, peak and RMS value of alternating current/voltage; reactance and impedance; LCR series circuit (phasors only), resonance, power in AC circuits, power factor, wattless current. AC generator, Transformer.

Chapter–8: Electromagnetic Waves

Basic idea of displacement current, Electromagnetic waves, their characteristics, their transverse nature (qualitative idea only). Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays) including elementary facts about their uses.

Chapter–9: Ray Optics and Optical Instruments

Ray Optics:  Reflection of light, spherical mirrors, mirror formula, refraction of light, total internal reflection and optical fibers, refraction at spherical surfaces, lenses, thin lens formula, lens maker’s formula, magnification, power of a lens, combination of thin lenses in contact, refraction of light through a prism.

Optical instruments:  Microscopes and astronomical telescopes (reflecting and refracting) and their magnifying powers.

Chapter–10: Wave Optics

Wave optics: Wave front and Huygen’s principle, reflection and refraction of plane wave at a plane surface using wave fronts. Proof of laws of reflection and refraction using Huygen’s principle. Interference, Young’s double slit experiment and expression for fringe width (No derivation final expression only), coherent sources and sustained interference of light, diffraction due to a single slit, width of central maxima (qualitative treatment only).

Chapter–11: Dual Nature of Radiation and Matter

Dual nature of radiation, Photoelectric effect, Hertz and Lenard’s observations; Einstein’s photoelectric equation-particle nature of light.

Experimental study of photoelectric effect

Matter waves-wave nature of particles, de-Broglie relation.

Chapter–12: Atoms

Alpha-particle scattering experiment; Rutherford’s model of atom; Bohr model of hydrogen atom, Expression for radius of nth possible orbit, velocity and energy of electron in his orbit, of hydrogen line spectra (qualitative treatment only).

Chapter–13: Nuclei

Composition and size of nucleus, nuclear force

Mass-energy relation, mass defect; binding energy per nucleon and its variation with mass number; nuclear fission, nuclear fusion.

Chapter–14: Semiconductor Electronics: Materials, Devices and Simple Circuits

Energy bands in conductors, semiconductors and insulators (qualitative ideas only) Intrinsic and extrinsic semiconductors- p and n type, p-n junction

Semiconductor diode – I-V characteristics in forward and reverse bias, application of junction diode -diode as a rectifier.

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You can click on the chapter-wise Case Study links given above and read the Class 12th Physics Case Study questions and answers provided by us. If you face any issues or have any questions please put your questions in the comments section below. Our teachers will be happy to provide you with answers.

FAQs Class 12 Physics Case Studies Questions

What is the best website for a  case   study  of physics  class   12 .

studyrate.in is the best website for Class 12 Physics Case Study Questions for Board Exams. Here you can find various types of Study Materials, Ebooks, Notes, and much more free of cost.

What are Physics case study questions?

Physics case study questions involve applying physics principles to solve real-world scenarios or problems.

How can I prepare for Physics case study questions?

Strengthen your basic concepts, practice solving a variety of problems, and make use of resources like the Class 12 Physics Case Study Questions PDF.

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CBSE 12th Standard Physics Subject Moving Charges And Magnetism Chapter Case Study Questions 2021

By QB365 on 21 May, 2021

QB365 Provides the updated CASE Study Questions for Class 12 , and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams 

QB365 - Question Bank Software

Cbse 12th standard physics subject moving charges and magnetism case study questions 2021.

12th Standard CBSE

Final Semester - June 2015

(ii) Radius of particle in second magnetic field B o is

(iii) Which of the following will trace a circular trajectory wit largest radius?

(iv) Mass of the particle in terms q, B o,  B,r and E is

(v) The particle comes out of velocity selector along a straight line, because

(ii) If v o = 2v o  then the time required for one revolution of the electron will change to

(iii) A charged particles is projected in a magnetic field  \(\vec{B}=(2 \hat{i}+4 \hat{j}) \times 10^{2} \mathrm{~T}\)  The acceleration of the particle is found to be  \(\vec{a}=(x \hat{i}+2 \hat{j}) \mathrm{m} \mathrm{s}^{-2}\) . Find the value of x.

(iv) If the given electron has a velocity not perpendicular to B, then trajectory of the electron is

(v) If this electron of charge (e) is moving parallel to uniform magnetic field with constant velocity v, the force acting on the electron is

(ii) To make the field radial in a moving coil galvanometer

(iii) The deflection in a moving coil galvanometer is

(iv) In a moving coil galvanometer, having a coil of N-turns of area A and carrying current I is placed in a radial field of strength B. The torque acting on the coil is

(v) To increase the current sensitivity of a moving coil galvanometer, we should decrease

(ii) There are 3 voltmeter A, B, C having the same range but their resistance are  \(15,000 \Omega, 10,000 \Omega\)  and  \(5,000 \Omega\)  respectively. The best voltmeter amongst them is the one whose resistance is

(iii) A milliammeter of range 0 to 25 mA and resistance of  \(10 \Omega\)  is to be converted into a voltmeter with a range of 0 to 25 V. The resistance that should be connected in series will be

(iv) To convert a moving coil galvanometer (MCG) into a voltmeter

(v) The resistance of an ideal voltmeter is

(ii) A proton is projected with a uniform velocity v along the axis of a current carrying solenoid, then

(iii) A charged particle experiences magnetic force in the presence of magnetic field. Which of the following statement is correct?

(iv) A charge q moves with a velocity 2 ms -1 along x-axis in a uniform magnetic field  \(\vec{B}=(\hat{i}+2 \hat{j}+3 \hat{k}) \mathrm{T}\)  then charge will experience a force

(v) Moving charge will produce

*****************************************

Cbse 12th standard physics subject moving charges and magnetism case study questions 2021 answer keys.

(i) (c): In mass spectrometer, the ions are sorted out by accelerating them through electric and magnetic field. (ii) (c):  As  \(\frac{m v^{2}}{r}=q v B_{0} \therefore r=\frac{m v}{q B_{0}}\) (iii) (b): As radius  \(r \propto \frac{m}{q}\) \(\therefore\)  r will be maximum for  \(\alpha\)  - particle. (iv) (b) : Here,  \(r=\frac{m v}{q B_{0}} \text { or } m=\frac{r q B_{0}}{v}\) As  \(v=\frac{E}{B}, \therefore m=\frac{q B_{0} B r}{E}\) (v) (c): From the relation v = E/B, it is clear electric and magnetic force balance each other.

(i) (b) : As  \(r_{0}=\frac{m v}{q B} \Rightarrow r^{\prime}=\frac{m\left(2 v_{0}\right)}{q B}=2 r_{0}\) (ii) (c): As,  \(T=\frac{2 \pi m}{q B}\) Thus, it remains same as it is in dependent of velocity (iii) (b) : As  \(F \perp B\) Hence,  \(a \perp B\) \(\therefore \vec{a} \cdot \vec{B}=0\) \(\Rightarrow \quad(x \hat{i}+2 \hat{j}) \cdot(2 \hat{i}+4 \hat{j})=0\) \(2 x+8=0 \Rightarrow x=-4 \mathrm{~m} \mathrm{~s}^{-2}\) (iv) (c): If the charged particle has a velocity not perpendicular to  \(\vec{B},\)  then component of velocity along  \(\vec{B}\)  remains unchanged as the motion along the  \(\vec{B}\)  will not be affected by  \(\vec{B}\) . Then, the motion of the particle in a plane perpendicular to  \(\vec{B}\)  is as before circular one. Thereby, producing helical motion. (v) (d): The force on electron  \(F=q v B \sin \theta\) As the electron is moving parallel to B So, \(\theta=0^{\circ} \Rightarrow q v B \sin 0^{\circ}=0\)

(I) (d): A moving coil galvanometer is a sensitive instrument which is used to measure a deflection when a current flows through its coil. (ii) (d) : Uniform field is made radial by cutting pole pieces cylindrically. (iii) (b): The deflection in a moving coil galvanometer  \(\phi=\frac{N A B}{k} \cdot I \text { or } \phi \propto N\)  where Nis number of turns in a coil, B is magnetic field and A is area of cross-section. (iv) (d): The deflecting torque acting on the coil  \(\tau_{\text {deflection }}=N I A B\) (v) (b): Current sensitivity of galvanometer \(\frac{\phi}{I}=S_{i}=\frac{N B A}{k}\) Hence, to increase (current sensitivity) S i  (torsional constant of spring) k must be decrease.

(i) (d) : A galvanometer can be converted into a voltmeter of given range by connecting a suitable high resistance R in series of galvanometer, which is given by  \(R=\frac{V}{I_{g}}-G=\frac{100}{10 \times 10^{-3}}-25=10000-25=9975 \Omega\) (ii) (c): An ideal voltmeter should have a very high resistance. (iii) (c): Resistance of voltmeter  \(=\frac{25}{25 \times 10^{-3}}=1000 \Omega\) \(\therefore \quad X=1000-10=990 \Omega\) (iv) (d) : To convert a moving coil galvanometer into a voltmeter, it is connected with a high resistance in series. The voltmeter is connected in parallel to measure the potential difference. As the resistance is high, the voltmeter itself does not consume current. (v) (d): The resistance of an ideal voltmeter is infinity.

(i) (a): For stationary electron, \(\vec{v}=0\) \(\therefore\)  Force on the electron is  \(\vec{F}_{m}=-e(\vec{v} \times \vec{B})=0\) (ii) (d) : Force on the proton  \(\vec{F}_{B}=e(\vec{v} \times \vec{B})\) Since,  \(\vec{v}\)  is parallel to  \(\vec{B}\) \(\therefore \quad \vec{F}_{B} \doteq 0\) Hence proton will continue to move with velocity v along the axis of solenoid. (iii) (b): Magnetic force on the charged particle q is  \(\vec{F}_{m}=q(\vec{v} \times \vec{B}) \text { or } F_{m}=q v B \sin \theta\) where  \(\theta\)  is the angle between  \(\vec{v} \text { and } \vec{B}\) Out of the given cases, only in case (b) it will experience the force while in other cases it will experience no force (iv) (a) :  \(\vec{F}=q(\vec{v} \times \vec{B})\) \(=q[(2 \hat{i} \times(\hat{i}+2 \hat{j}+3 \hat{k})]=(4 q) \hat{k}-(6 q) \hat{j}\) (v) (c): When an electric charge is moving both electric and magnetic fields are produced, whereas a static charge produces only electric field.

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Case Study Questions Class 12 Physics

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Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

CBSE will ask two Case Study Questions in the CBSE class 12 Physics questions paper. Question numbers 15 and 16 are case-based questions where 5 MCQs will be asked based on a paragraph. Each theme will have five questions and students will have a choice to attempt any four of them. You can download CBSE Class 12 Physics case study questions from the myCBSEguide App and our free student dashboard .

You can Score High

CBSE class 12 Physics question paper will carry questions for 70 marks. Certainly, the question paper is a bit easier this year. It is because the syllabus is already reduced and there are more internal choices. Besides this, the case study questions are a plus to winning the game with good marks.

In simple words, all circumstances are in favour of the sincere students who are working hard to score high this year. Although it has been a difficult time for students as they were not getting the personal attention of the teachers. We know that online classes have their own limits, but we still expect better scores, especially from students who are putting extra effort into their studies.

Class 12 Physics Case Study Questions

CBSE class 12 Physics question paper will have case study questions too. These case-based questions will be objective type in nature. So, class 12 Physics students must prepare themselves for such questions. First of all, you should study NCERT Textbooks line by line and then you should practice as many questions as possible.

Case Study Syllabus

We know that CBSE has reduced the syllabus. Hence, practice only relevant questions. don’t waste time on case study questions from deleted portion. It is of no use. You can download the latest class 12 Physics case study questions from the myCBSEguide App.

Physics Case Studies

Class 12 Physics has many chapters but all chapters are not important for case studies. As we know case studies are not exactly given from NCERT books but these may be extracted from some newspaper articles, magazines, journals or other books. So, it is very much important that you are studying only the most relevant case studies. Here, the myCBSEguide app helps you a lot. We have case study questions that are prepared by a team of expert teachers. These experts exactly know what types of questions can come in exams.

Case Study Questions

There are a number of study apps available over the internet. But if you are a CBSE student and willing to get an app for the CBSE curriculum, you have very limited options. And if you want an app that is specifically designed for CBSE students, your search will definitely end on finding myCBSEguide. Case study questions are the latest updates in CBSE syllabus. It is exclusively available in the myCBSEguide app.

Here are some example questions. For more questions, you can download the myCBSEguide App.

Physics Case Study -1

Read  the following source and answer any four out of the following questions: Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. There are two types of charges positive and negative charges. Also, like charges repel each other whereas unlike charges attract each other.

• -3.2  × ×  10 -18  C
• 3.2  × ×  10  18  C
• -3.2  × ×  10 -17  C
• 3.2  × ×  10  -17   C
• -1.6  × ×  10 -18  C
• 1.6  × ×  10  -18  C
• 2.6  × ×  10 -18  C
• 1.6  × ×  10 -21  C
• 9.1  × ×  10 -31  kg
• 9.1  × ×  10 -31  g
• 1.6  × ×  10 -19  kg
• 1.6  × ×  10 -19  g
• there is only a positive charge in the body
• there is positive as well as negative charge in the body but the positive charge is more than the negative charge
• there is equally positive and negative charge in the body but the positive charge lies in the outer regions
• the negative charge is displaced from its position
• valence electrons only
• electrons of inner shells
• both valence electrons and electrons of the inner shell.
• none of the above

Physics Case Study -2

Read the following source and answer any four out of the following questions: Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms. Also, Resistivity is the electrical resistance of a conductor of unit cross-sectional area, and unit length. … A characteristic property of each material, resistivity is useful in comparing various materials on the basis of their ability to conduct electric currents.

• nature of material
• temperature
• dimensions of material
• cross-sectional area
• length of wire
• wire’s nature
• all of the above
• more resistance
• less resistance
• same resistance

Physics Case Study -3

Read the source given below and answer any four out of the following questions: The Bohr model of the atom was proposed by Neil Bohr in 1915. It came into existence with the modification of Rutherford’s model of an atom. Rutherford’s model introduced the nuclear model of an atom, in which he explained that a nucleus (positively charged) is surrounded by negatively charged electrons.

• The energy of the electrons in the orbit is quantized
• The electron in the orbit nearest the nucleus has the lowest energy
• Electrons revolve in different orbits around the nucleus
• The position and velocity of the electrons in the orbit cannot be determined simultaneously
• Single proton
• Multiple electrons
• emitted only
• absorbed only
• both (a) and (b)
• none of these
•  It must emit a continuous spectrum
•  It loses its energy
• Gaining its energy
• A discrete spectrum
• dequantized

Physics Case Study & myCBSEguide App

We at myCBSEguide provide the best case study questions for CBSE class 12 Physics. We have Physics case study questions for every chapter in 12th class Physics. Students can access the Physics case study questions with answers on the myCBSEguide App or on the student dashboard . Here are some features that make myCBSEguide the best learning app for CBSE students:

• Updated syllabus
• Up to date question bank
• Model papers and 10-year questions
• NCERT and Exemplar sulutions
• Best quality learning videos
• Detailed revision notes

12 Physics Question Paper Design

Here is the question paper design for CBSE class 12 Physics. It shows the typology of the questions and their weightage in CBSE board exams.

QUESTION PAPER DESIGN Theory (Class: 12)

Maximum Marks: 70 Duration: 3 hrs

Note:  The above template is only a sample. Suitable internal variations may be made for generating similar templates keeping the overall weightage to different forms of questions and typology of questions the same.

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NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions

December 6, 2019 by Sastry CBSE

Moving Charges and Magnetism Class 12 Important Questions Very Short Answer Type

Question 2. Why should the spring/suspension wire in a moving coil galvanometer have low torsional constant? (All India 2008) Answer: Low torsional constant is basically required to increase the current/charge sensitivity in a moving coil ballistic galvanometer.

Question 3. Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid. Why? (Delhi 2008) Answer: At the edges of the solenoid, the field lines get diverged due to other fields and/or non-availability of dipole loops, while in toroids the dipoles (in loops) orient continuously.

Question 14. Why do the electrostatic field lines not form closed loops? (All India 2015) Answer: Electric field lines do not form closed loops because the direction of an electric field is from positive to negative charge. So one can regard a line of force starting from a positive charge and ending on a negative charge. This indicates that electric field lines do not form closed loops.

Question 15. A particle of mass ‘m’ and charge ‘q’ moving with velocity V enters the region of uniform magnetic field at right angle to the direction of its motion. How does its kinetic energy get affected? (Comptt. Delhi 2015) Answer: Kinetic energy will NOT be affected. *(When \(\vec{v}\) is perpendicular to \(\vec{B}\), then magnetic field provides necessary centripetal force)

Question 17. A coil, of area A, carrying a steady current I, has a magnetic moment, \(\vec{m}\), associated with it. Write the relation between \(\vec{m}\), I and A in vector form. (Comptt Delhi 2015) Answer: Relation for magnetic moment = \(\vec{m}\) = I\(\vec{A}\)

Moving Charges and Magnetism Class 12 Important Questions Short Answer Type SA II

Question 34. (a) How is a toroid different from a solenoid? (b) Use Ampere’s circuital law to obtain the magnetic field inside a toroid. (c) Show that in an ideal toroid, the magnetic field (i) inside the toroid and (ii) outside the toroid at any point in the open space is zero. (Comptt. All India 2014) Answer: (a) A toroid is essentially a solenoid which has been bent into a circular shape to close on itself. (b)

Question 42. Find the condition under which the charged particles moving with different speeds in the presence of electric and magnetic field vectors can be used to select charged particles of a particular speed. (All India 2015) Answer: Condition: The velocity \(\vec{v}\) of the charged particles, and the \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{B}}\) vectors, should be mutually perpendicular

Moving Charges and Magnetism Class 12 Important Questions Short Answer Type SA III

Direction : Towards the conductor/Attractive Net force on the loop will act towards the long conductor (attractive) if the current in its closer side is in the same direction as the current in the long conductor, otherwise it will be repulsive.

Justification : Component of velocity \(\vec{v}\), parallel to magnetic field, will make the particle move along the field.

Perpendicular component of velocity \(\vec{v}\) will cause the particle to move along a circular path in the plane perpendicular to the magnetic field Hence, the particle will follow a helical path, as shown above.

Question 67. A uniform magnetic field \(\overrightarrow{\mathrm{B}}\) is set up along the positive x-axis. A particle of charge ‘q’ and mass ‘m’ moving with a velocity v enters the field at the origin in X-Y plane such that it has velocity components both along and perpendicular to the magnetic field \(\overrightarrow{\mathrm{B}}\). Trace, giving reason, the trajectory followed by the particle. Find out the expression for the distance moved by the particle along the magnetic field in one rotation. (All India 2014) Answer: Since the velocity of the particle is inclined to x-axis, thererfore, the velocity has a component along B, this component remains unchanged as the motion along the magnetic field will not be affected by the magnetic field. The motion in a plane perpendicular to B is as before a circular one, thereby producing a helical motion, which is its trajectory.

Working. When current is passed, say along ABCD, the couple acts on it. Since the plane remains always parallel to the magnetic field in all positions of the coil (radial field), the force on the vertical arms always remains perpendicular to the place of the coil.

Let, I be the current flowing through coil, B be magnetic field supposed to be uniform and always parallel to the coil, A be area of the coil Deflecting torque acting on the coil is, τ = nI BA sin 900

Moving Charges and Magnetism Class 12 Important Questions Long Answer Type

(b) For radial magnetic field, sin θ = 1, so torque τ = NIAB. Thus when radial magnetic field is used, the deflection of the coil is proportional to the current flowing through it. Hence a linear scale can be used to determine the deflection of the coil.

(c) A high resistance is joined in series with a galvanometer so that when the arrangement (voltmeter) is used in parallel with the selected section of the circuit, it should draw least amount of current. In case voltmeter draws appreciable amount of current, it will disturb the original value of potential difference by a good amount.

To convert a galvanometer into ammeter, a shunt is used in parallel with it so that when the arrangement is joined in series, the maximum current flows through the shunt, and thus the galvanometer is saved from its damage, when the current is passed through ammeter.

Yes, there is an upper limit. The increase in the kinetic energy of particles is qv. Therefore, the radius of their path goes on increasing each time, their kinetic energy increases. The lines are repeatedly accelerated across the dees, untill they have the required energy to have a radius approximately that of the dees. Hence, this is the upper limit on the energy required by the particles due to definite size of dees.

(b) Since \(\mathrm{v}_{\mathrm{s}}=\frac{\mathrm{I}_{\mathrm{s}}}{\mathrm{R}}\) increase in current sensitivity may not necessarily increase the voltage sensitivity. It may be affected by the resistance used.

Question 99. (a) Write the expression for the force, \(\overrightarrow{\mathbf{F}}\), acting on a charged particle of charge ‘q’, moving with a velocity latex]\overrightarrow{\mathbf{v}}[/latex] in the presence of both electric field \(\overrightarrow{\mathrm{E}}\) and magnetic field \(\overrightarrow{\mathrm{B}}\). Obtain the condition under which the particle moves undeflected through the fields. (b) A rectangular loop of size l × b carrying a steady current I is placed in a uniform magnetic field \(\overrightarrow{\mathrm{B}}\). Prove that the torque \(\vec{\tau}\) acting on the loop is given by \(\vec{\tau}=\vec{m} \times \overrightarrow{\mathrm{B}}\), where \(\overrightarrow{\mathrm{m}}\) is the magnetic moment of the loop. (All India 2011) Answer: (a) A charge q in an electric field \(\overrightarrow{\mathrm{E}}\) experiences the electric force, \(\overrightarrow{\mathrm{F}}_{e}=q \overrightarrow{\mathrm{E}}\)

This force acts in the direction of field \(\overrightarrow{\mathrm{E}}\) and is independent of the velocity of the charge.

On the other hand, dB sin θ component of magnetic field due to each element of the coil or loop is directed in the same direction.

Important Questions for Class 12 Physics

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NCERT Solutions For Class 12 Physics Chapter 4 Moving Charges and Magnetism

Access comprehensive NCERT Solutions for Class 12 Physics Chapter 4 - Moving Charges and Magnetism. Explore detailed answers and explanations to enhance understanding in moving charges and magnetism concepts. Simplify learning with step-by-step solutions for CBSE Class 12 Physics.

February 9, 2024

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NCERT Solutions for class 12 Physics  Chapter 4 Moving Charges and Magnetism is prepared by our senior and renowned teachers of Physics Wallah primary focus while solving these questions of class-12 in NCERT textbook, also do read theory of this Chapter 4 Moving Charges and Magnetism while going before solving the NCERT questions. You can download and share  NCERT Solutions  of Class 12 Physics from Physics Wallah.

NCERT Solutions Class 12 Physics Chapter 4 Overview

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The instructors at Physics Wallah have dedicatedly created these solutions to enhance comprehension of the chapter’s ideas. The intention is for students to effortlessly achieve high exam scores after reviewing and practicing these solutions.

NCERT Solutions Class 12 Physics Science Chapter 4 PDF

Our team of experts at Physics Wallah has created comprehensive solutions for NCERT Class 12 Physics, Chapter 4, aiming to assist students in learning and practicing chapter concepts effectively. These questions are designed to simplify explanations, enhancing the learning process for students.

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NCERT Solutions For Class 12 Physics Chapter 4

Answer The Following Question Answer NCERT Solutions For Class 12 Physics Chapter 4 Moving Charges and Magnetism:

Question 1. A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Solution : Given:

Number of turns, n = 100

Radius of coil, r = 8 cm

Current through the coil, I = 0.40 A

Magnitude of magnetic field at centre of coil, B = ?

⇒ |B| = 3.14 × 10 -4 T

∴ Magnitude of magnetic field at the centre of the coil is 3.14 × 10 4 T. Question 2. A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?

Current through the wire, I = 35 A

Distance of point P from the wire, d = 20 cm

Question 3. A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.

Current through the wire, I = 50 A    (North to South)

Distance of point P East of the wire, d = 2.5 m

Direction of magnetic field,

The point is in a plane normal to the wire and the wire carries current in north to south. Using Right hand thumb rule we can conclude that the direction of magnetic field is vertically upwards, or out of the paper.

The magnitude of the magnetic field is 4 × 10 -6 T and its direction is upwards or out of paper.

Question 4. A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

Current through the wire, I = 90 A   (East to West)

Distance of point P below the wire, d = 1.5 m

We know that wire carries current in east to west direction. Using Right hand thumb rule, we can conclude that the direction of magnetic field is from north to south as indicated in the figure.

The magnitude of the magnetic field is 1.2 × 10 -5 T and its direction is from north to south.

Question 5. What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30º with the direction of a uniform magnetic field of 0.15 T?

Solution : Current in the wire, I = 8 A

Magnitude of the uniform magnetic field, B = 0.15 T

Angle between the wire and magnetic field, θ = 30°.

Magnetic force per unit length on the wire is given as:

f = BI sinθ

= 0.15 × 8 ×1 × sin30°

= 0.6 N m–1

Hence, the magnetic force per unit length on the wire is 0.6 N m–1.

Question 6. A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

Solution : Length of the wire, l = 3 cm = 0.03 m

Current flowing in the wire, I = 10 A

Magnetic field, B = 0.27 T

Angle between the current and magnetic field, θ = 90°

Magnetic force exerted on the wire is given as:

F = BIlsinθ

= 0.27 × 10 × 0.03 sin90°

= 8.1 × 10–2 N

Hence, the magnetic force on the wire is 8.1 × 10–2 N. The direction of the force can be obtained from Fleming’s left hand rule.

Question 7. Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

Current in wire A, I A  = 8.0 A

Current in wire B, I B  = 5.0 A

Distance between the conductors A and B, d = 4 cm

Length of conductor on which we have to calculate force, L = 10cm

So, the force on the 10 cm section on wire A is 2 × 10 -5 N. Since the current is flowing in the same direction the force will be attractive in nature.

Note: The force will be same on both the wires, we can use Newton’s third law of motion to such conclusion.

Question 8. A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.

Length of solenoid, L = 80cm

Number of turns = number of layers × number of turns per layer

Number of turns, n = 5 × 400 = 2000

Radius of solenoid, r = Diameter/2 = 0.9 cm

Current through the solenoid = 8.0A

Hence the magnetic field strength at the centre of the solenoid is 2.512 × 10 -2 T.

Question 9. A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30º with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Length of side of square, L = 10 cm

Number of turns, n = 20

Current through the square coil, I = 12 A

Angle between the normal to the coil and uniform magnetic field, θ = 30°

Magnitude of magnetic field, B = 0.80 T

Question 10. Two moving coil meters, M1 and M2 have the following particulars:

R1 = 10 Ω, N1 = 30,

A1 = 3.6 × 10–3 m2, B1 = 0.25 T

R2 = 14 Ω, N2 = 42,

A2 = 1.8 × 10–3 m2, B2 = 0.50 T

(The spring constants are identical for the two meters).

Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.

For moving coil meter M 1

Resistance of wire, R 1  = 10Ω

Number of turns, N 1  = 30

Area of cross-section, A 1  = 3.6 × 10 -3  m 2

Magnetic field strength, B 1  = 0.25 T

For moving coil meter M 2

Resistance of wire, R 2  = 14Ω

Number of turns, N 2  = 42

Area of cross-section, A 2  = 1.8 × 10 -3  m 2

Magnetic field strength, B 2  = 0.50 T\

Spring constant, K 1  = K 2  = K

Current sensitivity is given by,

Hence, the ratio of current sensitivities is 1.4.

Hence, the ratio of voltage sensitivity of M 1  and M 2  is 1.

Question 11. In a chamber, a uniform magnetic field of 6.5 G (1 G = 10–4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 106 m s–1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.6 × 10–19 C, me= 9.1×10–31 kg)

Magnetic field strength, B = 6.5 G = 6.5 × 10 -4 T

Initial velocity of electron = 4.8 × 10 6  ms -1

Angle between the initial velocity of electron and magnetic field, θ = 90 0

⇒ F e  = 1.6 × 10 -19  C × 4.8 × 10 6  ms -1  × 6.5 × 10 -4 T × sin 90

⇒ F e  = 4.99 × 10 -16 N

This force serves as the centripetal force, which explains the circular trajectory of the electron.

Centripetal force F c  = mv 2 /r       …(2)

By equating equation (1) and equation (2) we get,

Question 12. In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

We can relate the velocity of the electron to its angular frequency by the relation,

V = rω              …(1)

V = velocity of electron

r = radius of path

ω = angular frequency

Question 13. (a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60º with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

Number of turns in the coil, n = 30

Current through the coil, I = 6.0 A

Strength of magnetic field = 1.0 T

Angle between the direction of field and normal to coil, θ = 60°

We can understand that the counter torque required to prevent the coil from rotating is equal to the torque being applied by the magnetic field.

Torque on the coil due to magnetic field is given by,

T = n × B × I × A × sinθ           …(1)

n = number of turns

B = Strength of magnetic field

I = Current through the coil

A = Area of cross-section of coil

A = πr 2  = 3.14 × (0.08 × 0.08) = 0.0201m 2               …(2)

θ = Angle between normal to cross-section of coil and magnetic field

Now, by putting the values in equation (1) we get,

⇒ T = 30 × 6.0T × 1A × 0.0201m 2  × sin60°

T = 3.133 Nm

Hence, the counter torque required to prevent the coil from rotating is 3.133 Nm.

b) From equation (1) we can understand that, torques depends on the total area of cross-section and has no relation with the geometry of cross-section. Hence, the answer will remain unaltered if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area.

ADDITIONAL EXERCISES

Question 14. Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.

Solution : Here we have to find total magnetic field produced by the system so we will first find magnetic field due to each coil with direction and then add them in accordance with vector addition. Using the Right-hand thumb rule we can predict the direction of induced magnetic field in both the coils.

The orientation of both the coils is shown below in the figure.

Question 15. A magnetic field of 100 G (1 G = 10−4 T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10−3 m2. The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m−1. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.

Solution : Here, we have a particular value of No. of turns per unit Length and Current in the coil in order to obtain the given magnetic field.

The Required Magnetic field B = 100 G = 100 × 10 –4  = 10 –2  T

Maximum Number of turns per unit length, n = 1000/m

Maximum Current flowing in the coil, I = 15 A

Permeability of free space, μ 0  = 4π × 10 –4  TmA -1

We know magnetic field for a solenoid is given by

NCERT Solutions for Class 12 Physics Chapter 4 FAQs

NCERT Solutions for Chapter 4 of Class 12 Physics provide comprehensive explanations, solving methods, and clarifications for complex concepts, aiding students in understanding the fundamentals of moving charges and magnetism.

These solutions provide in-depth explanations, step-by-step solutions to problems, and clarify doubts regarding the chapter, aiding in better comprehension and exam preparation.

Yes, these solutions are designed in adherence to the NCERT textbook, ensuring accuracy and relevance to the prescribed syllabus.

The detailed explanations and step-by-step solutions in NCERT Solutions help in resolving doubts and provide clarity on intricate concepts related to moving charges and magnetism.

Yes, NCERT Solutions are a valuable resource for self-study, enabling students to revise and practice questions independently, reinforcing their understanding of the chapter.

NCERT Solutions For Class 12 Physics Chapter 9 Ray Optics and Opticals Instrument

NCERT Solutions For Class 12 Physics Chapter 7 Alternating Current

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NCERT Solutions for Class 12 Physics Chapter 4 - Moving Charges And Magnetism

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• Chapter 4 Moving Charges And Magnetism

NCERT Solutions for Class 12 Physics Chapter 4 - Moving Charges and Magnetism

The NCERT Solutions for Class 12 Physics Chapter 4 - Moving Charges and Magnetism provided by Vedantu is a comprehensive and detailed explanation of the concepts covered in the PDF which is FREE to Download! This topic deals with the interaction between moving charges and magnetic fields. In the given PDF the chapter begins by introducing the concept of the magnetic field and its properties. It then discusses the Lorentz force, which is the force exerted on a moving charge by a magnetic field. The chapter also covers the Biot-Savart law, which gives the magnetic field produced by a current element. Ampere's circuital law is also discussed, which gives the relationship between the magnetic field and the current enclosed by a loop. The solutions are also accompanied by solved examples and practice questions, which help students to understand the concepts better.

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Moving Charges and Magnetism Chapter at a Glance - Class 12 NCERT Solutions

The total force on a charge q moving with velocity v in the presence of magnetic and electric fields B and E, respectively is called the Lorentz force. It is given by the expression: F = q (v × B + E). The magnetic force q (v × B) is normal to v and work done by it is zero.

A straight conductor of length l and carrying a steady current I experiences a force F in a uniform external magnetic field B, $F=I\left ( l\times B \right )$ 

Where |l| = l and the direction of l is given by the direction of the current.

In a uniform magnetic field B, a charge q executes a circular orbit in a plane normal to B. Its frequency of uniform circular motion is called the cyclotron frequency and is given by: 

$v_e=\frac{qB}{2\pi m}$

This frequency is independent of the particle’s speed and radius. This fact is exploited in a machine, the cyclotron, which is used to accelerate charged particles. 

Magnetic force does not work when the charged particle is displaced while electric force does work in displacing the charged particle.

Cyclotrons cannot accelerate electrons because they have very small mass.

The Biot-Savart law asserts that the magnetic field dB due to an element dl carrying a steady current I at a point P at a distance r from the current element is:

$dB=\frac{\mu_0}{4\pi}\mathbb{I}\frac{\vec{dl}\times \vec{r}}{r^3}$

To obtain the total field at P, we must integrate this vector expression over the entire length of the conductor.

Magnetic Field at Centre O in different conditions  

Ampere’s Circuital Law:  Let an open surface S be bounded by a loop C. Then the Ampere’s law states that $\underset{C}{\oint }N/d=\mu_0I$ where Ι refers to the current passing through S. The sign of I is determined from the right-hand rule. If B is directed along the tangent to every point on the perimeter L of a closed curve and is constant in magnitude along perimeter then:       $ BL=\mu_0I_e$ , where I e is the net current enclosed by the closed circuit

The magnitude of the field B inside a long solenoid carrying a current I is: B = $\mu$ 0 nl

where n is the number of turns per unit length. For a toroid one obtains, $B=\frac{\mu_0NI}{2\pi R}$ , where N is the total number of turns and r is the average radius.

If a current carrying circular loop (n = 1) is turned into a coil having n identical turns then magnetic field at the centre of the coil becomes n 2 times the previous field i.e. B (n turn) = n 2 B (single turn) .

Parallel currents attract and anti-parallel currents repel.

A planar loop carrying a current I, having N closely wound turns, and an area A possesses a magnetic moment M where, M = N I A and the direction of M is given by the right-hand thumb rule.

When this loop is placed in a uniform magnetic field B, the force F on it is:  F = 0

And the torque on it is: $\tau =M\times B$ 

In a moving coil galvanometer, this torque is balanced by a counter- torque due to a spring, yielding: $k\varphi =NIAB$

An electron moving around the central nucleus has a magnetic moment M given by: $M=\frac{e}{2M}L$ 

where L is the magnitude of the angular momentum of the circulating electron about the central nucleus and m is the mass. The smallest value of M is called the Bohr magneton M B and it is M B = 9.27×10 –24 J/T.

Access NCERT Solutions for Class 12 Physics Chapter 4 – Moving Charges and Magnetism

1. A Circular Coil of Wire Consisting of $100$ Turns, Each of Radius $8.0cm$ Carries a Current of $0.40A$. What is the Magnitude of the Magnetic Field B at the Centre of the Coil?

Ans: We are given:

Number of turns on the circular coil, $n=100$

Radius of each turn, $r=8.0cm=0.08m$

Current flowing in the coil is given to be, $I=0.4A$

We know the expression for magnetic field at the centre of the coil as, 

$\left| B \right|=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi nI}{r}$

Where, ${{\mu }_{0}}=4\pi \times {{10}^{-4}}Tm{{A}^{-1}}$is the permeability of free space. 

On substituting the given values we get, 

$\left| B \right|=\frac{4\pi \times {{10}^{-7}}\times 2\pi \times 100\times 0.4}{4\pi \times 0.08}$

$\Rightarrow \left| B \right|=3.14\times {{10}^{-4}}T$

Clearly, the magnitude of the magnetic field is found to be $3.14\times {{10}^{-4}}T$. 

2. A Long Straight Wire Carries a Current of $35A$. What Is the Magnitude of Field B at a Point 20cm from the Wire?

Ans: We are given the following:

Current in the wire, $I=35A$

Distance of the given point from the wire, $r=20cm=0.2m$

We know the expression for magnetic field as,

$B=\frac{{{\mu }_{0}}}{4\pi }\frac{2I}{r}$

On substituting the given values, we get, 

$B=\frac{4\pi \times {{10}^{-7}}\times 2\times 35}{4\pi \times 0.2}$

$\Rightarrow B=3.5\times {{10}^{-5}}T$

Thus, we found the magnitude of the magnetic field at the given point to be $3.5\times {{10}^{-5}}T$.

3. A Long Straight Wire in the Horizontal Plane Carries a Current of $50A$ in North to South Direction. Give the Magnitude and Direction of B at a Point $2.5m$ East of the Wire. 

The current in the wire, $I=50A$

The distance of the given point from the wire, $r=2.5m$

We have the expression for magnetic field as, 

$B=\frac{2{{\mu }_{0}}I}{4\pi r}$

Substituting the given values, we get,

$B=\frac{4\pi \times {{10}^{-7}}\times 2\times 50}{4\pi \times 2.5}$

$\Rightarrow B=4\times {{10}^{-6}}T$

Now from Maxwell’s right hand thumb rule, we have the direction of the magnetic field at the given point B to be vertically upward. 

4. A Horizontal Overhead Power Line Carries a Current of $90A$ in East to West Direction. What is the Magnitude and Direction of the Magnetic Field Due to the Current $1.5m$ Below the Line? 

Current in the power line, $I=90A$

Distance of the mentioned point below the power line, $r=1.5m$

Now, we have the expression for magnetic field as, 

$B=\frac{4\pi \times {{10}^{-7}}\times 2\times 90}{4\pi \times 1.5}$

$\Rightarrow B=1.2\times {{10}^{-5}}T$

We found the magnitude of the magnetic field to be $1.2\times {{10}^{-5}}T$and it will be directed towards south as per Maxwell’s right hand thumb rule. 

5. What is the Magnitude of Magnetic Force Per Unit Length on a Wire Carrying a Current of 8 and Making an Angle of $30{}^\circ $ with the Direction of a Uniform Magnetic Field of $0.15T$? 

Ans: Given that,

Current in the wire, $I=8A$

Magnitude of the uniform magnetic field, $B=0.15T$

Angle between the wire and magnetic field, $\theta =30{}^\circ $

We have the expression for magnetic force per unit length on the wire as, 

$F=BI\sin \theta $

$F=0.15\times 8\times 1\times \sin 30{}^\circ $

$\Rightarrow F=0.6N{{m}^{-1}}$

Thus, the magnetic force per unit length on the wire is found to be $0.6N{{m}^{-1}}$

6. A $3.0cm$ Wire Carrying a Current of $10A$ is Placed Inside a Solenoid Perpendicular to Its Axis. The Magnetic Field Inside the Solenoid Is Given to Be $0.27T$. What is the Magnetic Force on the Wire?

Ans: We are given the following, 

Length of the wire, $l=3cm=0.03m$

Current flowing in the wire, $I=10A$

Magnetic field, $B=0.27T$

Angle between the current and magnetic field, $\theta =90{}^\circ $

(Since the magnetic field produced by a solenoid is along its axis and current carrying wire is kept perpendicular to the axis) 

The magnetic force exerted on the wire is given as, 

$F=BIl\sin \theta $

Substituting the given values, 

$F=0.27\times 10\times 0.03\sin 90{}^\circ $

$\Rightarrow F=8.1\times {{10}^{-2}}N$

Clearly, the magnetic force on the wire is found to be $8.1\times {{10}^{-2}}N$. The direction of the force can be obtained from Fleming’s left-hand rule. 

7. Two Long and Parallel Straight Wires A and B Carrying Currents of $8.0A$and $5.0A$ in the Same Direction are Separated by a Distance of $4.0cm$. Estimate the Force on a $10cm$ Section of Wire A. 

Current flowing in wire A, ${{I}_{A}}=8.0A$ 

Current flowing in wire B, ${{I}_{B}}=5.0A$

Distance between the two wires, $r=4.0cm=0.04m$

Length of a section of wire A, $l=10cm=0.1m$

Force exerted on length $l$ due to the magnetic field is given as,

$B=\frac{2{{\mu }_{0}}{{I}_{A}}{{I}_{B}}l}{4\pi r}$ 

$B=\frac{4\pi \times {{10}^{-7}}\times 2\times 8\times 5\times 0.1}{4\pi \times 0.04}$

$\Rightarrow B=2\times {{10}^{-5}}N$

The magnitude of force is $2\times {{10}^{-5}}N$. This is an attractive force that is normal to A towards B because the direction of the currents in the wires is the same.

8. A Closely Wound Solenoid $80cm$ Long has $5$ Layers of Windings of $400$ Turns Each. The Diameter of the Solenoid is $1.8cm$. If the Current Carried is $8.0A$, Estimate the Magnitude of B Inside the Solenoid Near its Centre. 

Length of the solenoid, $l=80cm=0.8m$

Since there are five layers of windings of 400 turns each on the solenoid. 

Total number of turns on the solenoid would be, $N=5\times 400=2000$

Diameter of the solenoid, $D=1.8cm=0.018m$

Current carried by the solenoid, $I=8.0A$

We have the magnitude of the magnetic field inside the solenoid near its centre given by the relation, 

$B=\frac{{{\mu }_{0}}NI}{l}$

$B=\frac{4\pi \times {{10}^{-7}}\times 2000\times 8}{0.8}$

$\Rightarrow B=2.512\times {{10}^{-2}}T$

Clearly, the magnitude of the magnetic field inside the solenoid near its centre is found to be $2.512\times {{10}^{-2}}T$.

9. A Square Coil of Side $10cm$ Consists of 20 Turns and Carries a Current of $12A$. The Coil Is Suspended Vertically and the Normal to the Plane of the Coil Makes an Angle of $30{}^\circ $ with the Direction of a Uniform Horizontal Magnetic Field of Magnitude $0.80T$. What is the Magnitude of Torque Experienced by the Coil?

Length of a side of the square coil, $l=10cm=0.1m$

Area of the square, $A={{l}^{2}}={{\left( 0.1 \right)}^{2}}=0.01{{m}^{2}}$

Current flowing in the coil, $I=12A$

Number of turns on the coil, $n=20$

Angle made by the plane of the coil with magnetic field, $\theta =30{}^\circ $

Strength of magnetic field, $B=0.80T$

Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,

$\tau =nIAB\sin \theta $

$\tau =20\times 0.8\times 12\times 0.01\times \sin 30{}^\circ $

$\Rightarrow \tau =0.96Nm$

Thus, the magnitude of the torque experienced by the coil is 0.96 N m.

10. Two Moving Coil Meters, ${{M}_{1}}$ and ${{M}_{2}}$ Have the Following Particulars:

${{R}_{1}}=10\Omega $  , ${{N}_{1}}=30$, ${{A}_{1}}=3.6\times {{10}^{-3}}{{m}^{2}}$ , ${{B}_{1}}=0.25T$,${{R}_{2}}=14\Omega $ ,${{N}_{2}}=42$${{A}_{2}}=1.8\times {{10}^{-3}}{{m}^{2}}$ , ${{B}_{2}}=0.50T$

(The spring constants are identical for the meters).

Determine the Ratio of:

a) Current Sensitivity of ${{M}_{2}}\text{ and }{{\text{M}}_{1}}$

For moving coil meter ${{M}_{1}}$,

Resistance, ${{R}_{1}}=10\Omega $

Number of turns, ${{N}_{1}}=30$

Area of cross-section, ${{A}_{1}}=3.6\times {{10}^{-3}}{{m}^{2}}$

Magnetic field strength, ${{B}_{1}}=0.25T$

Spring constant, ${{K}_{1}}=K$

For moving coil meter ${{M}_{2}}$:

Resistance, ${{R}_{2}}=14\Omega $

Number of turns, ${{N}_{2}}=42$

Area of cross-section, ${{A}_{2}}=1.8\times {{10}^{-3}}{{m}^{2}}$

Magnetic field strength, ${{B}_{2}}=0.50T$

Spring constant, ${{K}_{2}}=K$

Current sensitivity of ${{M}_{1}}$ is given as:

${{I}_{S1}}=\frac{{{N}_{1}}{{B}_{1}}{{A}_{1}}}{{{K}_{1}}}$

And, current sensitivity of ${{M}_{2}}$ is given as:

${{I}_{S2}}=\frac{{{N}_{2}}{{B}_{2}}{{A}_{2}}}{{{K}_{2}}}$

On taking the ratio, we get, 

$\Rightarrow \frac{{{I}_{S2}}}{{{I}_{S1}}}=\frac{\frac{{{N}_{2}}{{B}_{2}}{{A}_{2}}}{{{K}_{2}}}}{\frac{{{N}_{1}}{{B}_{1}}{{A}_{1}}}{{{K}_{1}}}}$

Substituting the values we get,

$\Rightarrow \frac{{{I}_{S2}}}{{{I}_{S1}}}=\frac{42\times 0.5\times 1.8\times {{10}^{-3}}\times 10\times K}{14\times 30\times 0.25\times 3.6\times {{10}^{-3}}\times K}$

$\Rightarrow \frac{{{I}_{S2}}}{{{I}_{S1}}}=1.4$

Therefore, the ratio of current sensitivity of ${{M}_{2}}\text{ and }{{\text{M}}_{1}}$ is 1.4.

b) Voltage Sensitivity of ${{M}_{2}}\text{ and }{{\text{M}}_{1}}$

Ans: Voltage sensitivity for ${{M}_{2}}$is given is:

${{V}_{S2}}=\frac{{{N}_{2}}{{B}_{2}}{{A}_{2}}}{{{K}_{2}}{{R}_{2}}}$

And, voltage sensitivity for ${{M}_{1}}$is given as:

${{V}_{S1}}=\frac{{{N}_{1}}{{B}_{1}}{{A}_{1}}}{{{K}_{1}}{{R}_{1}}}$

On taking the ratio we get, 

$\Rightarrow \frac{{{V}_{S2}}}{{{V}_{S1}}}=\frac{{{N}_{2}}{{B}_{2}}{{A}_{2}}{{K}_{1}}{{R}_{1}}}{{{K}_{2}}{{R}_{2}}{{N}_{1}}{{B}_{1}}{{A}_{1}}}$

Substituting the given values, we get, 

$\Rightarrow \frac{{{V}_{S2}}}{{{V}_{S1}}}=\frac{42\times 0.5\times 1.8\times {{10}^{-3}}\times 10\times K}{K\times 14\times 30\times 0.25\times 3.6\times {{10}^{-3}}}=1$

Thus, the ratio of voltage sensitivity of ${{M}_{2}}\text{ and }{{\text{M}}_{1}}$is 1. 

11. In a Chamber, a Uniform Magnetic Field of $6.5G\left( 1G={{10}^{-4}}T \right)$is Maintained. An Electron Is Shot Into the Field With a Speed Of $4.8\times {{10}^{6}}m{{s}^{-1}}$ Normal to the Field. Explain Why the Path of the Electron is a Circle. Determine the Radius of the Circular Orbit.$\left( e=1.6\times {{10}^{-19}}C,{{m}_{e}}=9.1\times {{10}^{-31}}kg \right)$

Ans: Magnetic field strength, $B=6.5G=6.5\times {{10}^{-4}}T$

Speed of the electron, $V=4.8\times {{10}^{6}}m/s$

Charge on the electron, $e=1.6\times {{10}^{-19}}C$

Mass of the electron, ${{m}_{e}}=9.1\times {{10}^{-31}}kg$

Angle between the shot electron and magnetic field, $\theta =90{}^\circ $

Magnetic force exerted on the electron in the magnetic field could be given as:

$F=evB\sin \theta $

This force provides centripetal force to the moving electron and hence, the electron starts moving in a circular path of radius r.

Hence, centripetal force exerted on the electron would be,

${{F}_{C}}=\frac{m{{v}^{2}}}{r}$

However, we know that in equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,

${{F}_{C}}=F$

$\Rightarrow \frac{m{{v}^{2}}}{r}=evB\sin \theta $

$\Rightarrow r=\frac{mv}{Be\sin \theta }$

Substituting the given values we get, 

$\Rightarrow r=\frac{9.1\times {{10}^{-31}}\times 4.8\times {{10}^{6}}}{6.5\times {{10}^{-4}}\times 1.6\times {{10}^{-19}}\times \sin 90{}^\circ }$

$\Rightarrow r=4.2cm$

Clearly, we found the radius of the circular orbit to be 4.2cm.

12. In Exercise 4.11 Obtain the Frequency of Revolution of the Electron in Its Circular Orbit. Does the Answer Depend on the Speed of the Electron? Explain.

Magnetic field strength, $B=6.5\times {{10}^{-4}}T$

Charge of the electron, $e=1.6\times {{10}^{-19}}C$

Velocity of the electron, $v=4.8\times {{10}^{6}}m/s$

Radius of the orbit, $r=4.2cm=0.042m$

Frequency of revolution of the electron $\nu $

Angular frequency of the electron $\omega =2\pi \theta $

Velocity of the electron is related to the angular frequency as:

$v=r\omega $

In the circular orbit, the magnetic force on the electron provides the centripetal force. Hence,

$evB=\frac{m{{v}^{2}}}{r}$

$\Rightarrow eB=\frac{m}{r}\left( r\omega  \right)=\frac{m}{r}\left( r2\pi \nu  \right)$

$\Rightarrow \nu =\frac{6.5\times {{10}^{-4}}\times 1.6\times {{10}^{-19}}}{2\times 3.14\times 9.1\times {{10}^{-31}}}$$\therefore \nu =18.2\times {{10}^{6}}Hz\approx 18MHz$

Thus, the frequency of the electron is around 18 MHz and is independent of the speed of the electron.

a) A Circular Coil of 30 Turns and Radius $8.0cm$ Carrying a Current of $6.0A$ is Suspended Vertically in a Uniform Horizontal Magnetic Field of Magnitude $1.0T$ The Field Lines Make an Angle of $60{}^\circ $ with the Normal of the Coil. Calculate the Magnitude of the Counter Torque That Must Be Applied to Prevent the Coil from Turning.

Ans: Number of turns on the circular coil, $n=30$

Radius of the coil, $r=8.0cm=0.08m$

Area of the coil, $A=\pi {{r}^{2}}=\pi {{\left( 0.08 \right)}^{2}}=0.0201{{m}^{2}}$

Current flowing in the coil is given to be,  $I=6.0A$

Magnetic field strength, $B=1T$

Angle between the field lines and normal with the coil surface, $\theta =60{}^\circ $

The coil will turn when it experiences a torque in the magnetic field. The counter torque applied to prevent the coil from turning is given by the relation, $\tau =nIAB\sin \theta $

$\Rightarrow \tau =30\times 6\times 1\times 0.0201\times \sin 60{}^\circ $

$\Rightarrow \tau =3.133Nm$

b) Would Your Answer Change, If the Circular Coil in (a) Were Replaced by a Planar Coil of Some Irregular Shape that Encloses the Same Area? (all Other Particulars Are Also Unaltered).

Ans: From the part(a) we could infer that the magnitude of the applied torque is not dependent on the shape of the coil. 

On the other hand, it is dependent on the area of the coil. 

Thus, we could say that the answer would not change if the circular coil in the above case is replaced by a planar coil of some irregular shape that encloses the same area.

14. Two Concentric Circular Coils X and Y Radii 16 cm and 10 cm, Respectively, Lie in the Same Vertical Plane Containing the North to South Direction. Coil X has 20 Turns and Carries a Current of 16 A; Coil Y has 25 Turns and Carries a Current Of 18 A. The Sense of the Current in X is Anticlockwise, and Clockwise in Y, for an Observer Looking at the Coils Facing West. Give the Magnitude and Direction of the Net Magnetic Field Due to the Coils at Their Centre.

Ans: We are given,

Radius of coil X, ${{r}_{1}}=16cm=0.16m$

Radius of coil Y, ${{r}_{2}}=10cm=0.1m$

Number of turns of on coil X,${{n}_{1}}=20$

Number of turns of on coil Y,${{n}_{2}}=25$

Current in coil X, ${{I}_{1}}=16A$

Current in coil Y,${{I}_{2}}=18A$

Magnetic field due to coil X at their centre is given by the relation,

${{B}_{1}}=\frac{{{\mu }_{0}}{{n}_{1}}{{I}_{1}}}{2{{r}_{1}}}$

Where, Permeability of free space, ${{\mu }_{0}}=4\pi \times {{10}^{-4}}Tm{{A}^{-1}}$

${{B}_{1}}=\frac{4\pi \times {{10}^{-7}}\times 20\times 16}{2\times 0.16}$

$\Rightarrow {{B}_{1}}=4\pi \times {{10}^{-4}}T$(towards East)

Magnetic field due to coil Y at their centre is given by the relation,

${{B}_{2}}=\frac{{{\mu }_{0}}{{n}_{2}}{{I}_{2}}}{2{{r}_{2}}}$

$\Rightarrow {{B}_{2}}=\frac{4\pi \times {{10}^{-7}}\times 25\times 18}{2\times 0.10}$

$\Rightarrow {{B}_{2}}=9\pi \times {{10}^{-4}}T$(towards West)

Clearly, net magnetic field could be obtained as:

$B={{B}_{2}}-{{B}_{1}}=9\pi \times {{10}^{-4}}-4\pi \times {{10}^{-4}}$

$\Rightarrow B=1.57\times {{10}^{-3}}T$(towards West)

15. A magnetic field of $100G$$\left( where,\text{ }1G={{10}^{-4}}T \right)$ is required which is uniform in a region of linear dimension about $10cm$ and area of cross-section about${{10}^{-3}}{{m}^{2}}$. The maximum current carrying capacity of a given coil of wire is $15A$ and the number of turns per unit length that can be wound a core is at most $1000\text{ turns per m}$. Suggest some appropriate design particulars to a solenoid for the required purpose. Assume the core is not ferromagnetic.

Magnetic field strength,$B=100G=100\times {{10}^{-4}}T$

Number of turns per unit length,$n=1000turns\text{ per m}$

Current flowing in the coil,$I=15A$

Permeability of free space, ${{\mu }_{0}}=4\pi \times {{10}^{-7}}Tm{{A}^{-1}}$

Magnetic field is given the relation,

$B={{\mu }_{0}}nI$

$\Rightarrow nI=\frac{B}{{{\mu }_{0}}}=\frac{100\times {{10}^{-4}}}{4\pi \times {{10}^{-7}}}$

$\Rightarrow nI\approx 8000A/m$

If the length of the coil is taken as 50 cm, radius 4 cm, number of turns 400, and current 10 A, then these values are not unique for the given purpose. There is always a possibility of some adjustments with limits.

16. For a Circular Coil of Radius R and N Turns Carrying Current I, the Magnitude of the Magnetic Field at a Point on Its Axis at a Distance X from Its Centre Is Given By,

a) Show that this reduces to the familiar result for the field at the centre of the coil.

Radius of circular coil = R

Number of turns on the coil = N

Current in the coil = I

Magnetic field at a point on its axis at distance x is given by the relation,

Where,${{\mu }_{0}}=4\pi \times {{10}^{-4}}Tm{{A}^{-1}}$Permeability of free space

If the magnetic field at the centre of the coil is considered, then $x=0$

$B=\frac{{{\mu }_{0}}I{{R}^{2}}N}{2{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\frac{3}{2}}}}$

This is the familiar result for the magnetic field at the centre of the coil.

b) Consider Two Parallel Co-axial Circular Coils of Equal Radius R, and the Number of Turns N, Carrying Equal Currents in the Same Direction, and Separated by a Distance R. Show That the Field on the Axis Around the Mid-point Between the Coils Is Uniform Over a Distance That Is Small as Compared to R, and Is Given By, Approximately, (such an Arrangement to Produce a Nearly Uniform Magnetic Field Over a Small Region Is Known as Helmholtz Coils.)

Radii of two parallel co-axial circular coils = R

Number of turns on each coil = N

Current in both coils = I

Distance between both the coils = R

Let us consider point Q at distance d from the centre.

Then, one coil is at a distance of $\frac{R}{2}+d$from point Q.

Magnetic field at point Q could be given as:

Also, the other coil is at a distance of  $\frac{R}{2}+d$from point Q.

Magnetic field due to this coil is given as:

${{B}_{2}}=\frac{{{\mu }_{0}}NI{{R}^{2}}}{2{{\left[ {{\left( \frac{R}{2}-d \right)}^{2}}+{{R}^{2}} \right]}^{\frac{3}{2}}}}$

Now we have the total magnetic field as, 

$B={{B}_{1}}+{{B}_{2}}$

$\Rightarrow B=\frac{{{\mu }_{0}}I{{R}^{2}}}{2}\left[ \left\{ {{\left\{ \left. {{\left( \frac{R}{2}-d \right)}^{2}}+{{R}^{2}} \right\} \right.}^{\frac{-3}{2}}}+{{\left. \left\{ {{\left( \frac{R}{2}+d \right)}^{2}}+{{R}^{2}} \right. \right\}}^{\frac{-3}{2}}}\times N \right. \right]$

$\Rightarrow B=\frac{{{\mu }_{0}}I{{R}^{2}}}{2}\left[ \left\{ {{\left\{ \left. \frac{5{{R}^{2}}}{4}+{{d}^{2}}-Rd \right\} \right.}^{\frac{-3}{2}}}+{{\left. \left\{ \frac{5{{R}^{2}}}{4}+{{d}^{2}}+Rd \right. \right\}}^{\frac{-3}{2}}}\times N \right. \right]$

$\Rightarrow B=\frac{{{\mu }_{0}}I{{R}^{2}}}{2}\left[ \left\{ {{\left\{ \left. 1+\frac{4{{d}^{2}}}{5{{R}^{2}}}-\frac{4d}{5R} \right\} \right.}^{\frac{-3}{2}}}+{{\left\{ 1+\frac{4{{d}^{2}}}{5{{R}^{2}}}+\frac{4d}{5R} \right\}}^{\frac{-3}{2}}}\times N \right. \right]$

Now for $d\ll R$, we could neglect the factor $\frac{{{d}^{2}}}{{{R}^{2}}}$, we get, 

$B\approx \frac{{{\mu }_{0}}I{{R}^{2}}}{2}\times {{\left( \frac{5{{R}^{2}}}{4} \right)}^{\frac{-3}{2}}}\left[ {{\left( 1-\frac{4d}{5R} \right)}^{\frac{-3}{2}}}+{{\left( 1+\frac{4d}{5R} \right)}^{\frac{-3}{2}}} \right]\times N$

$\Rightarrow B\approx \frac{{{\mu }_{0}}I{{R}^{2}}}{2}\times {{\left( \frac{5{{R}^{2}}}{4} \right)}^{\frac{-3}{2}}}\left[ 1-\frac{6d}{5R}+1+\frac{6d}{5R} \right]$

$\Rightarrow B\approx {{\left( \frac{4}{5} \right)}^{\frac{3}{2}}}\frac{{{\mu }_{0}}IN}{R}=0.72\left( \frac{{{\mu }_{0}}IN}{R} \right)$

Clearly, we proved that the field along the axis around the mid-point between the coils is uniform. 

17. A Toroid Has a Core (non-Ferromagnetic) of Inner Radius 25 Cm and Outer Radius 26 Cm, Around Which 3500 Turns of a Wire Are Wound. If the Current in the Wire is 11 A, What is the Magnetic Field 

a) Outside the Toroid 

Inner radius of the toroid, ${{r}_{1}}=25cm=0.25m$

Outer radius of the toroid, ${{r}_{2}}=26cm=0.26m$

Number of turns on the coil, $N=3500$

Current in the coil, $I=11A$

Magnetic field outside a toroid is zero. It is non-zero only inside the core of a toroid.

b) Inside the Core of the Toroid.

Ans: Magnetic field inside the core of a toroid is given by the relation, $B=\frac{{{\mu }_{0}}NI}{l}$

Where, Permeability of free space ${{\mu }_{0}}=4\pi \times {{10}^{-7}}Tm{{A}^{-1}}$

$l$ is the length of toroid, given by

$l=2\pi \left( \frac{{{r}_{1}}+{{r}_{2}}}{2} \right)=\pi \left( 0.25+0.26 \right)=0.51\pi $

$\Rightarrow B=\frac{4\pi \times {{10}^{-7}}\times 3500\times 11}{0.51\pi }\approx 3.0\times {{10}^{-2}}T$

Thus, magnetic field inside the core of the toroid is $3.0\times {{10}^{-2}}T$.

c) In the Empty Space Surrounded by the Toroid.

Ans: The empty space that is surrounding the toroid has a magnetic field equal to zero. 

18. Answer the Following Questions:

a) A Magnetic Field That Varies in Magnitude from Point to Point but Has a Constant Direction (east to West) Is Set up in a Chamber. A Charged Particle Enters the Chamber and Travels Undeflected Along a Straight Path With Constant Speed. What Can You Say About the Initial Velocity of the Particle?

Ans: The initial velocity of the particle could either be parallel or be anti-parallel to the magnetic field. So, it travels along a straight path without undergoing any deflection in the field.

b) A Charged Particle Enters an Environment of a Strong and Non-Uniform Magnetic Field Varying from Point to Point Both in Magnitude and Direction, and Comes Out of it Following a Complicated Trajectory. Would Its Final Speed Equal the Initial Speed If it Suffered No Collisions With the Environment? 

Ans: Yes, the final speed of the charged particle would be equal to its initial speed as the magnetic force has the potential to change the direction of velocity even though not its magnitude.

c) An Electron Travelling West to East Enters a Chamber Having a Uniform Electrostatic Field in the North to South Direction. Specify the Direction in Which a Uniform Magnetic Field Should Be Set up to Prevent the Electron from Deflecting from Its Straight-Line Path.

Ans: An electron travelling from West to East enters a chamber having a uniform electrostatic field along the North-South direction. 

This moving electron stays undeflected when the electric force acting on it is equal and opposite of the magnetic field. 

The magnetic force would stay directed towards the South. Also, according to Fleming’s left-hand rule, the magnetic field must be applied in a vertically downward direction.

19. An Electron Emitted by a Heated Cathode and Accelerated Through a Potential Difference of $2.0kV$, Enters a Region With Uniform Magnetic Field of $0.15T$. Determine the Trajectory of the Electron If the Field 

a) Is Transverse to Its Initial Velocity. 

Ans: We are given, 

Magnetic field strength, $B=0.15T$

Mass of the electron, $m=9.1\times {{10}^{-31}}kg$

Potential difference, $V=2.0kV=2\times {{10}^{3}}V$

Now we have the kinetic energy of the electron given by, 

$eV=\frac{1}{2}m{{v}^{2}}$

$\Rightarrow v=\sqrt{\frac{2eV}{m}}$…………………….. (1)

Where, $v$is the velocity of the electron

Since the magnetic force on the electron provides the required centripetal force of the electron, the electron traces a circular path of radius $r$.

Now, the magnetic force on the electron is given by the relation, 

Centripetal force, 

$\Rightarrow Bev=\frac{m{{v}^{2}}}{r}$

$\Rightarrow r=\frac{mv}{Be}$………………….. (2)

From the equations (1) and (2), we get, 

$r=\frac{m}{Be}{{\left[ \frac{2eV}{m} \right]}^{\frac{1}{2}}}$

$\Rightarrow r=\frac{9.1\times {{10}^{-31}}}{0.15\times 1.6\times {{10}^{-19}}}{{\left( \frac{2\times 1.6\times {{10}^{-19}}\times 2\times {{10}^{3}}}{9.1\times {{10}^{-31}}} \right)}^{\frac{1}{2}}}$

$\Rightarrow r=100.55\times {{10}^{-5}}$

$\Rightarrow r=1mm$

Thus, we found that the electron has a circular trajectory of radius 1.0 mm normal to the magnetic field.

b) Makes an Angle of $30{}^\circ $ with the Initial Velocity.

Ans: When the field makes an angle $\theta $ of $30{}^\circ $with initial velocity, the initial velocity will be,

${{v}_{1}}=v\sin \theta $

From equation (2), we can write the following expression:

${{r}_{1}}=\frac{m{{v}_{1}}}{Be}$

$\Rightarrow {{r}_{1}}=\frac{mv\sin \theta }{Be}$

$\Rightarrow {{r}_{1}}=\frac{9.1\times {{10}^{-31}}}{0.15\times 1.6\times {{10}^{-19}}}\left[ \frac{2\times 1.6\times {{10}^{-19}}\times 2\times {{10}^{3}}}{9\times {{10}^{-31}}} \right]\sin 30{}^\circ $

$\Rightarrow r=0.5\times {{10}^{-3}}m=0.5mm$

Clearly, we found that the electron has a helical trajectory of radius$0.5mm$, with axis of the solenoid along the magnetic field direction.

20. A Magnetic Field Set up Using Helmholtz Coils (Described in Exercise $4.16$) is Uniform in a Small Region and Has a Magnitude of $0.75T$. In the Same Region, a Uniform Electrostatic Field Is Maintained in a Direction Normal to the Common Axis of the Coils. A Narrow Beam of (single Species) Charged Particles All Accelerated Through $15kV$ Enters This Region in a Direction Perpendicular to Both the Axis of the Coils and the Electrostatic Field. If the Beam Remains Undeflected When the Electrostatic Field Is $9.0\times {{10}^{-5}}V{{m}^{-1}}$ Make a Simple Guess as to What the Beam Contains. Why Is the Answer Not Unique?

Magnetic field, $B=0.75T$

Accelerating voltage, $V=15kV=15\times {{10}^{3}}V$

Electrostatic field, $E=9\times {{10}^{5}}V{{m}^{-1}}$

Mass of the electron$=m$

Charge of the electron $=e$

Velocity of the electron $=v$

Kinetic energy of the electron $=eV$

Thus, 

$\frac{1}{2}m{{v}^{2}}=eV$

$\Rightarrow \frac{e}{m}=\frac{{{v}^{2}}}{2V}$…………………….. (1)

Since the particle remains undeflected by electric and magnetic fields, we could infer that the electric field is balancing the magnetic field.

$\Rightarrow v=\frac{E}{B}$……………………….. (2)

Now we could substitute equation (2) in equation (1) to get,

$\frac{e}{m}=\frac{1}{2}\frac{{{\left( \frac{E}{B} \right)}^{2}}}{V}=\frac{{{E}^{2}}}{2V{{B}^{2}}}$

$\Rightarrow \frac{e}{m}=\frac{{{\left( 9.0\times {{10}^{5}} \right)}^{2}}}{2\times 15000\times {{\left( 0.75 \right)}^{2}}}=4.8\times {{10}^{7}}C/kg$

This value of specific charge $\left( \frac{e}{m} \right)$ is equal to the value of deuteron or deuterium ions. This is not a unique answer. Other possible answers are $H{{e}^{++}}$, $L{{i}^{+++}}$

21. A Straight Horizontal Conducting Rod of Length $0.45m$ and Mass $60g$ is Suspended by Two Vertical Wires at Its Ends. A Current of $5.0A$ is Set up in the Rod Through the Wires.

a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?

Length of the rod, $l=0.45m$

Mass suspended by the wires, $m=60g=60\times {{10}^{-3}}kg$

Acceleration due to gravity, $g=9.8m{{s}^{-2}}$

Current in the rod flowing through the wire, $I=5A$

We could say that magnetic field (B) is equal and opposite to the weight of the wire i.e.,

$\Rightarrow B=\frac{mg}{Il}=\frac{60\times {{10}^{-3}}\times 9.8}{5\times 0.45}$

$\Rightarrow B=0.26T$

Clearly, a horizontal magnetic field of 0.26 T normal to the length of the conductor should be set up.

b) What will be the Total Tension in the Wires If the Direction of Current is Reversed Keeping the Magnetic Field Same as Before? (Ignore the Mass of the Wires.) $g=9.8m{{s}^{-2}}$

Ans: When the direction of the current is reversed, $BIl$ and $mg$ will act downwards. Clearly, the effective tension in the wires is found to be, 

$T=0.26\times 5\times 0.45+\left( 60\times {{10}^{-3}} \right)\times 9.8$

$\Rightarrow T=1.176N$

22. The Wires Which Connect the Battery of an Automobile to Its Starting Motor Carry a Current of $300A$(for a short time). What Is the Force Per Unit Length Between the Wires if they are $70cm$ Long and $1.5cm$ Apart? Is the Force Attractive or Repulsive?

Current in both wires, $I=300A$

Distance between the wires, $r=1.5cm=0.015m$

Length of the two wires, $l=70cm=0.7m$

We know that, Force between the two wires is given by the relation,

$F=\frac{{{\mu }_{0}}{{I}^{2}}}{2\pi r}$

Where, Permeability of free space${{\mu }_{0}}=4\pi \times 10Tm{{A}^{-1}}$

As the direction of the current in the wires is found to be opposite, a repulsive force exists between them.

23. A Uniform Magnetic Field of $1.5T$ Exists in a Cylindrical Region of Radius $10.0cm$, its Direction Parallel to the Axis Along East to West. A Wire Carrying Current of $7.0A$ in the North to South Direction Passes Through This Region. What is the Magnitude and Direction of the Force on the Wire If,

a) The Wire Intersects the Axis,

Magnetic field strength, $B=1.5T$

Radius of the cylindrical region, $r=10cm=0.1m$

Current in the wire passing through the cylindrical region, $I=7A$

If the wire intersects the axis, then the length of the wire is the diameter of the cylindrical region. Thus, $l=2r=0.2m$

Angle between magnetic field and current, $\theta =90{}^\circ $

We know that, Magnetic force acting on the wire is given by the relation,

$\Rightarrow F=1.5\times 7\times 0.2\times \sin 90{}^\circ $

$\Rightarrow F=2.1N$

Clearly, a force of 2.1 N acts on the wire in a vertically downward direction.

b) The Wire is Turned from N-S to northeast-northwest Direction,

Ans: The new length of the wire after turning it to the northeast-northwest direction can be given as:

${{l}_{1}}=\frac{l}{\sin \theta }$

Angle between magnetic field and current, $\theta =45{}^\circ $

Force on the wire,

$F=BI{{l}_{1}}\sin \theta =BIl=1.5\times 7\times 0.2$

Thus, a force of 2.1 N acts vertically downward on the wire. This is independent of angle $\theta $ as $l\sin \theta $ is fixed.

c) The Wire in the N-S Direction is Lowered from the Axis by a Distance of $6.0cm$?

Ans: The wire is lowered from the axis by distance, $d=6.0cm$

Let ${{l}_{2}}$be the new length of the wire, 

${{\left( \frac{{{l}_{2}}}{2} \right)}^{2}}=4\left( d+r \right)=4\left( 10+6 \right)=4\times 16$

$\Rightarrow {{l}_{2}}=8\times 2=16cm=0.16m$

Magnetic force that is exerted on the wire is, 

${{F}_{2}}=BI{{l}_{2}}=1.5\times 7\times 0.16$

$\Rightarrow F=1.68N$

Clearly, a force of $1.68N$acts in a vertically downward direction on the wire.

24. A Uniform Magnetic Field of $3000G$ is Established Along the Positive Z-Direction. A Rectangular Loop of Sides $10cm$ and $5cm$ Carries a Current of $12A$. What is the Torque on the Loop in the Different Cases Shown in Figure? What Is the Force on Each Case? Which Case Corresponds to Stable Equilibrium?

a)                              

Magnetic field strength, $B=3000G=3000\times {{10}^{-4}}T=0.3T$

Length of the rectangular loop, $l=10cm$

Width of the rectangular loop, $b=5cm$

Area of the loop, $A=l\times b=\left( 10\times 5 \right)c{{m}^{2}}=50\times {{10}^{-4}}{{m}^{2}}$

Current in the loop, $I=12A$

Now, we could take the anti-clockwise direction of the current as positive and vice-versa, 

We have the expression for torque given as,

$\vec{\tau }=I\vec{A}\times \vec{B}$

We could see from the given figure that A is normal to the y-z plane and B is directed along the z-axis. Substituting the given values, 

$\tau =12\times \left( 50\times {{10}^{-4}} \right)\hat{i}\times 0.3\hat{k}$

$\Rightarrow \tau =-1.8\times {{10}^{-2}}\hat{j}Nm$

Now, the torque is found to be directed along negative y-direction. Since the external magnetic field is uniform, the force on the loop would be zero. 

Ans: This case is very similar to case (a), and hence, the answer here would be same as (a). 

Ans: Torque here would be, 

$\tau =I\vec{A}\times \vec{B}$

$\Rightarrow \tau =-12\left( 50\times {{10}^{-4}} \right)\hat{j}\times 0.3\hat{k}$

$\Rightarrow \tau =-1.8\times {{10}^{-2}}\hat{i}Nm$

The direction here is along the negative x direction and the force is zero. 

$\left| \tau  \right|=IAB$

$\Rightarrow \tau =12\times \left( 50\times {{10}^{-4}} \right)\times 0.3$

$\Rightarrow \left| \tau  \right|=1.8\times {{10}^{-2}}Nm$

Here, the direction is found to be at $240{}^\circ $with positive x-direction and the force is zero.

Ans: Torque, 

$\Rightarrow \tau =\left( 50\times {{10}^{-4}}\times 12 \right)\hat{k}\times 0.3\hat{k}$

$\Rightarrow \tau =0$

Here, both torque and force are found to be zero. 

f)  

(Image will be Uploaded Soon)

Ans: Torque is given by, 

Here also both torque and force are found to be zero. 

For the case (e) the direction $I\vec{A}$and $\vec{B}$ is the same and the angle between them is zero. They would come back to equilibrium on being displaced and so the equilibrium is stable. 

For the case (f), the direction of $I\vec{A}$and $\vec{B}$ are opposite and the angle between them is $180{}^\circ $. Here it doesn’t come back to its original position on being disturbed and hence, the equilibrium is unstable. 

25. A Circular Coil of 20 Turns and Radius 10 Cm is Placed in a Uniform Magnetic Field of 0.10 T Normal to the Plane of the Coil. If the Current in the Coil Is 5.0 A, What is the: (The Coil is Made of Copper Wire of Cross-Sectional Area ${{10}^{-5}}{{m}^{2}}$, and the Free Electron Density in Copper Is Given to be about${{10}^{29}}{{m}^{-3}}$).

a) Total Torque on the Coil?

Number of turns on the circular coil, $n=20$

Radius of the coil, $r=10cm=0.1m$

Magnetic field strength, $B=0.10T$

Current in the coil, $I=5.0A$

As the angle between force and the normal to the loop is zero, the total torque on the coil is zero. 

So, $\tau =NIAB\sin \theta $is zero.

b) Total Force on the Coil, 

Ans: There is no total force on the coil because the field is uniform.

C) Average Force on Each Electron in the Coil Due to the Magnetic Field?

Cross-sectional area of copper coil, $A={{10}^{-5}}{{m}^{2}}$

Number of free electrons per cubic meter in copper, $N={{10}^{29}}/{{m}^{3}}$

Charge on the electron would be, $e=1.6\times {{10}^{-19}}C$

Magnetic force, $F=Be{{v}_{d}}$

Where, ${{v}_{d}}$ is drift velocity of electrons given by $\frac{I}{NeA}$

$\Rightarrow F=\frac{BeI}{NeA}=\frac{0.10\times 5.0}{{{10}^{29}}\times {{10}^{-5}}}=5\times {{10}^{-25}}N$

Clearly, the average force on each electron is $5\times {{10}^{-25}}N$.

26. A Solenoid $60cm$ Long and Radius $4.0cm$has 3 Layers of Windings of 300turns Each. A $2.0cm$ Long Wire of Mass $2.5g$ Lies Inside the Solenoid (near Its Centre) Normal to Its Axis; Both the Wire and the Axis of the Solenoid Are in the Horizontal Plane. the Wire Is Connected Through Two Leads Parallel to the Axis of the Solenoid to an External Battery Which Supplies a Current of $6.0a$ in the Wire. What Value of Current (with Appropriate Sense of Circulation) in the Windings of the Solenoid Can Support the Weight of the Wire? $g=9.8m{{s}^{-2}}$

Length of the solenoid, $L=60cm=0.6m$

Radius of the solenoid, $r=4.0cm=0.04m$

It is given that there are 3 layers of windings of 300 turns each.

Total number of turns, $n=3\times 300=900$

Length of the wire, $l=2cm=0.02m$

Mass of the wire, $m=2.5g=2.5\times {{10}^{-3}}kg$

Current flowing through the wire, $i=6A$

Magnetic field produced inside the solenoid, $B=\frac{{{\mu }_{0}}nI}{L}$

Current flowing through the windings of the solenoid, $I$

Magnetic force is given by the relation,

$F=Bil=\frac{{{\mu }_{0}}nI}{L}il$

Now, we have the force on the wire equal to the weight of the wire.

$mg=\frac{{{\mu }_{0}}nIil}{L}$

$\Rightarrow I=\frac{mgL}{{{\mu }_{0}}nil}=\frac{2.5\times {{10}^{-3}}\times 9.8\times 0.6}{4\pi \times {{10}^{-7}}\times 900\times 0.02\times 6}$

$\Rightarrow I=108A$

Clearly, the current flowing through the solenoid is 108 A.

27. A Galvanometer Coil Has a Resistance of $12\Omega $and the Metre Shows Full Scale Deflection for a Current of $3mA$. How will you Convert the Metre Into a Voltmeter of Range $0$ to $18V$?

Resistance of the galvanometer coil, $G=12\Omega $

Current for which there is full scale deflection, ${{I}_{g}}=3mA=3\times {{10}^{-3}}A$

Range of the voltmeter needs to be converted to $18V$.

Let a resistor of resistance R be connected in series with the galvanometer to convert it into a voltmeter. This resistance can be given as:

$R=\frac{V}{{{I}_{g}}}-G$

$R=\frac{18}{3\times {{10}^{-3}}}-12=6000-12$

$\Rightarrow R=5988\Omega $

Clearly, we found that a resistor of resistance $5988\Omega $is to be connected in series with the given galvanometer. 

28. A Galvanometer Coil Has a Resistance of $15\Omega $ and the Metre Shows Full Scale Deflection for a Current of $4mA$. How will you Convert the Metre Into an Ammeter of Range $0$ to $6A$?

Resistance of the galvanometer coil, $G=15\Omega $

Current for which the galvanometer shows full scale deflection, ${{I}_{g}}=4mA=4\times {{10}^{-3}}A$

We said that, Range of the ammeter needs to be $6A$.

In order to convert the given galvanometer into an ammeter, a shunt resistor of resistance S is to be connected in parallel with the galvanometer. 

The value of S could be given as:

$S=\frac{{{I}_{g}}G}{I-{{I}_{g}}}$

$S=\frac{4\times {{10}^{-3}}\times 15}{6-4\times {{10}^{-3}}}=\frac{0.06}{5.996}\approx 0.01\Omega $

$\Rightarrow S=10m\Omega $

Clearly, we found that a \[10m\Omega \] shunt resistor is to be connected in parallel with the galvanometer.

NCERT Solutions for Class 12 Physics Chapter 4 PDF

Students preparing for their class 12 boards must focus on the fourth chapter, which is crucial for the exam. Moreover, it is also significant for other competitive exams. There are multiple subtopics in chapter 4 Physics class 12, and they demand a thorough understanding. While studying this topic, students should also refer to a solutions book for learning these topics quickly.

These solutions are readily available online in PDF formats. Several dedicated E-learning platforms offer these solutions free of cost. Students can easily download these and study it as per their convenience. The PDFs of chapter 4 Physics class 12 NCERT solutions are curated by experts with years of experience. Thus, these solutions are accurate and highly reliable.

Magnetism is caused by moving charges or the flow of charge. Our revision notes on Moving Charges and Magnetism elaborate on the fact that magnetic fields exert further forces on the flow of charge. This, in turn, exerts a force on other magnets as well. This phenomenon occurs due to the presence of constant moving charges. For the benefits of these NCERT Solutions students should consider referring to the PDF for class 12 Physics chapter 4. This will enable them to acquire in-depth knowledge about these topics and thereby will help them to score well.

Chapter 4 Physics Class 12: Subtopics and Exercises

Moving charges and magnetism is divided into eleven subtopics, each of which is important for CBSE class 12 as well as JEE and NEET. To do well on any of these tests, students must focus on gaining in-depth knowledge in all of these areas. They may use NCERT solutions for class 12 Physics chapter 4 to help them in this procedure. These solutions have been specifically created to make this subject more comprehensible. Also, it will make it easier for them to complete the workouts.

There are approximately twenty-eight questions in the exercise of this chapter. Mostly all the questions are based on calculations. To solve these questions, students must have a clear concept of all the sections of the chapter.

Students who face difficulty in solving these sums or the other questions can consider referring to NCERT solutions for class 12 Physics Chapter 4 Moving Charges and Magnetism. As most of the questions are mathematical, everyone should cross-check it from the solutions book. This will help them to identify their mistakes and rectify them.

Topics Covered in Ch 4 Physics Class 12 NCERT Solutions

The answers given in the solutions are elaborately explained. The calculations are shown step-by-step which will be beneficial for the students to understand the logic behind each step. Moreover, students studying NCERT solutions for class 12 Physics Moving Charges and Magnetism will get a clear idea about the following:

Magnetic force and its sources.

Magnetic field (shown in the image below).

(image will be uploaded soon)

Lorentz Field

Magnetic force

Motion in a magnetic field

Velocity Selector

Biot-Savart Law

Ampere’s Circuital Law

The Solenoid

Magnetic dipole

The moving coil galvanometer

Moving charges and Magnetism has a share of 8 marks in the exam. The distribution of marks are as follows:

Moving Charges and Magnetism Class 12: Marks Distribution

Benefits of NCERT Solutions for Class 12 Physics Chapter 4

To ensure good grades in Physics, students should make sure that their concept regarding every topic is clear. However, studying only from the textbooks is not sufficient to achieve this, they must refer to class 12 NCERT solutions Physics ch 4. Learning from the solutions has many benefits, some of which are as follows:

It is written in an easy language.

The solutions are designed by faculty having years of knowledge; thereby it is accurate.

It has illustrations which make learning easier.

The calculations are explained in detailed steps.

Conclusion 

NCERT Solutions for Class 12 Physics Chapter 4 - Moving Charges and Magnetism offer comprehensive and invaluable learning resources for students. By addressing key concepts such as Lorentz force, Biot-Savart law, and Ampere's circuital law, these solutions enable a deeper understanding of the intricate relationship between moving charges and magnetism. The step-by-step explanations and solved examples provided in these solutions aid students in tackling complex problems with confidence. Additionally, the emphasis on real-life applications fosters an appreciation for the practical relevance of the subject. Overall, NCERT Solutions for Class 12 Physics Chapter 4 empower learners to grasp the fundamentals and excel in their academic journey.

FAQs on NCERT Solutions for Class 12 Physics Chapter 4 - Moving Charges And Magnetism

1. How Should I Prepare Magnetism and Moving Charges for my Boards?

While preparing this chapter, students should first go through the previous year papers. After finding the most relevant questions, they should start their preparation. Moreover, students should emphasise on practising sums as these have a majority of share in the question paper. They should revise the formulas regularly.

The questions are designed to check the analytical and quantitative aptitudes of the students. Therefore, they should focus on understanding the concepts well. To do this, they can take the help of Moving Charges and Magnetism class 12 NCERT solutions. This will help them to gain comprehensive knowledge about the topics. Besides, students while solving sums can refer to these solutions to identify their mistakes and rectify them.

2. Why Should I Refer to the NCERT Solution of Physics Class 12 Chapter 4?

Physics demands an in-depth understanding of the concepts. Without understanding the topics, students will not be able to score well in their exams. Moreover, this chapter helps students to prepare for other competitive exams like JEE and NEET as well.

Considering the importance of this chapter, students should study it thoroughly. NCERT solutions are designed to help them in the process. All the answers are explained elaborately, which allows students to understand the logic behind it. Besides, these are curated by an experienced faculty, and therefore the answers are highly reliable. The solutions strictly adhere to the CBSE pattern and help students to score well in their exams.

3. What is a Magnetic Field?

Magnetic field refers to an area around a magnet which shows its magnetic force. The magnetic field has direction as well as magnitude. This is an essential topic of class 12 Physics. Students while studying this should refer to Physics class 12, chapter 4, NCERT solutions. The solutions will help them to understand this topic better. While referring to the guide, they will also come across other related concepts like a magnetic force, motion in a magnetic field, velocity selector and others. After understanding these concepts, students will be able to solve the questions quickly.

4.  Why should I choose Vedantu’s NCERT Solutions as my first option for preparation?

Our science masters have rendered these solutions in an easily understandable letter. Methodological flow has been followed in this chapter to prepare yourself for the exam. You can use the techniques given for topics and subtopics of this chapter to score more marks. These solutions will assist you in attaining a better understanding of various basic concepts composed in this chapter with expertise. You can score well if you have secured 100% confidence to answer any question asked from this chapter. Our faculty made sure that CBSE and NCERT guidelines are strictly followed while drafting these solutions.

5. What are the key features of NCERT Solutions for Class 12 Physics Chapter 4?

The experienced team at Vedantu has diligently prepared all of the questions in NCERT Solutions for Class 12 Physics Chapter 4 Moving Charge and Magnetism, and the major benefits of these solutions are:

These solutions are provided free of cost. Thus, anyone can access them.

The solutions are provided in a PDF format that can be downloaded and printed for revision.

The professional team of Vedantu makes sure to structure all the solutions based only on the CBSE curriculum.

6. Is it necessary to use the NCERT Solutions for Class 12 Physics Chapter 4 PDF?

Class 12 Physics is important not only for the board exams but also for other competitive exams such as JEE, NEET, etc. Thus, one should clear all the concepts to score good grades in CBSE boards and other competitive exams. In order to accomplish this, it is advisable to refer to NCERT Books and NCERT Solutions for a comprehensive understanding of the concepts. The solutions to each question are thoroughly discussed on Vedantu .

7. What is Permittivity and Permeability Class 12 Physics?

A physical property that specifies how an electric field influences and is affected by a medium is known as electric permittivity. It is determined by a material's ability to polarise in response to an applied field. Similarly, magnetic permeability is an ability of a substance to acquire magnetisation in magnetic fields. It is a measure of how far a magnetic field can penetrate matter.

8. What are the concepts explained in Class 12 Physics Chapter 4?

Following are the concepts that are explained in Class 12 Physics Chapter 4:

Magnetic Force

Sources and Fields

Magnetic field and Lorentz force

Motion In Magnetic Field

Motion In Combined Electric and Magnetic Fields

Solenoid and Toroid

Torque On Current Loop, Magnetic Dipole

Circular Current Loop as A Magnetic Dipole

The Magnetic Dipole Moment of a Revolving Electron

The Moving Coil Galvanometer

9. Where can I find PDFs for NCERT Solutions Class 12 Physics?

Ans: You can find PDFs of NCERT Solutions Class 12 Physics on Vedantu. You can check out the app or the website for the same. Vedantu provides students with well-curated NCERT Solutions prepared by the Subject-Matter Experts and are totally based on the curriculum presented by CBSE. This free and downloadable study material will not only help to score good grades in Class 12 Boards but also in other competitive exams. Students can also access the study material from Vedantu’s App. All the resources are available free of cost.

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NCERT Solutions for Class 12 Physics Chapter 4 PDF Download

Some of the students often waste their time searching for the right study material; for them, the NCERT Solutions for Class 12 Physics Chapter 4 are the right ones. It is considered to be the right one as it provides accurate answers for the Class 12 Physics Chapter 4; accordingly, students can improve their accuracy level. 

NCERT Solutions for Class 12 Physics Chapter 4 PDF

To have success in Class 12 Physics Chapter 4, students need to take some effort or steps. One of the important steps is to access the NCERT Solutions for Class 12 Physics Chapter 4 PDF and need to start solving questions. Students can access the Class 12 Physics Chapter 4 questions through the Selfstudys website from their comfort zone. 

How to Download the NCERT Solutions for Class 12 Physics Chapter 4? 

It is important for students to download the NCERT Solutions for Class 12 Physics Chapter 4 so that they can easily refer to the answers; steps to download are: 

• Open the Selfstudys website. 

• Click the navigation button and select NCERT Books & Solutions. 

• A drop down menu will appear, select NCERT Solutions from the list. 

• Select Class 12th from the list 

• And select Physics from the list of subjects. 
• A new page will appear, now click Chapter 4 from the list of Chapters. 

Everything Students Need to Know About NCERT Solutions for Class 12 Physics Chapter 4 PDF

The NCERT Solutions for Class 12 Physics Chapter 4 PDF are considered to be important study material for students as it provides answers to all the textbook questions; here is that students need to know: 

• Keywords or Formulas are Presented: Formulas provide a way to present the substance using the symbol of elements; these keywords or formulas are given in the NCERT Solutions for Class 12 Physics Chapter 4. 
• Explained in Step Wise Manner: The answers in the NCERT Solutions for Class 12 Physics Chapter 4 revision are explained in a stepwise manner so that students can understand each point and step. 
• Explained in an Easy Language: Easy language is considered to be a plain language which is easier for students to read and comprehend; the answers in the NCERT Solutions for Class 12 Physics Chapter 4 theory are explained in easy language. 
• Provided in a Eye Catching Format: The eye catching format contains colours in a presentable format which can attract many; the same format is followed in the Chapter 4 questions of Class 12 Physics NCERT Solutions. 
• Tables are Included: Inside the Class 12 Physics NCERT Solutions, tables are included for Chapter 4 so that students can have organised information about the important themes. 
• Available in the PDF: The Class 12 Physics questions of Chapter 4 from the NCERT Solutions are available in the PDF so that students can access accurate answers from their comfort zone.

How Can NCERT Solutions for Class 12 Physics Chapter 4 Help Students?  

The NCERT Solutions for Class 12 Physics Chapter 4 can help students in various ways; those ways are discussed below: 

• Helps to Enhance the Learning Skills: Learning skills means the term that describes the tasks involved in the learning process; students can enhance their learning skills with the help of NCERT Solutions for Class 12 Physics Chapter 4 revision.  
• Offers Ample Questions to Practise: Students can only excel in the Class 12 Physics Chapter 4 only if they practise questions. Students can practise vast questions with the help of the NCERT Solutions for Class 12 Physics Chapter 4 theory. 
• Offers In Depth Knowledge: The NCERT Solutions for Class 12 Physics Chapter 4 PDF offers in depth knowledge about the concepts and topics so that students can build a strong and firm foundation. 
• Helps in Progress: Progress generally considered to be movement or improvement towards achieving something; students can judge their progress in Chapter 4 questions with the help of Class 12 Physics NCERT Solutions. 
• Can Get a Brief Idea About Important Questions: Through the Class 12 NCERT Physics Solutions, students can get a brief idea about the Chapter 4 questions; accordingly students can prepare for that particular Chapter. 
• Are Free of Cost: The Class 12 Physics NCERT Solutions are free of cost; accordingly students can access the Chapter 4 questions without paying a single penny. 

How to Manage Time While Solving Questions from NCERT Solutions for Class 12 Physics Chapter 4 PDF? 

Managing of time is considered to be a process in which organising and planning to divide time efficiently; there are tips to manage time while solving questions from the NCERT Solutions for Class 12 Physics Chapter 4 PDF: 

• Create a Plan: A study plan is defined as a time management plan that helps students in achieving their learning goals; in the same manner they need to make a study plan in order to manage time while solving questions from the NCERT Solutions for Class 12 Physics Chapter 4. 
• Start from the Easier Ones: Students need to start solving the easy questions for the NCERT Solutions for Class 12 Physics Chapter 4 revision so that they don’t waste much time on difficult questions and can manage time. 
• Set a Timer: It is advisable for students to set a timer to solve each question from the NCERT Solutions for Class 12 Physics Chapter 4 theory so that they can have enough time for each question. 
• Try to Avoid Distractions: Distractions are considered to be those activities which can take the attention of students; to manage time while solving Class 12 Physics Chapter 4 questions from the NCERT Solutions, students need to avoid the distractions. 
• Use Shortcuts and Tricks: Some Chapter 4 questions of the Class 12 Physics from the NCERT Solutions can be solved through shortcuts and tricks which can save a lot of time. 
• Take Breaks: To manage time while solving Class 12 Physics Chapter 4 questions from the NCERT Solutions, students can take breaks at equal intervals which can save much time in doing unnecessary activities. 

What are the Challenges Faced While Solving Questions from the NCERT Solutions for Class 12 Physics Chapter 4 PDF? 

Challenges are generally defined as something new or difficult that forces students to make efforts; students may face challenges while solving questions from the NCERT Solutions for Class 12 Physics Chapter 4 PDF: 

• Understanding of Concepts: Students may face challenges in solving questions from the NCERT Solutions for Class 12 Physics Chapter 4 if they are not able to understand the concepts properly. 
• Memorising Formulas or Keywords: They are formulas or keywords that students need to memorise, otherwise they may face challenges while solving questions from the NCERT Solutions for Class 12 Physics Chapter 4 revision. 
• Management of Time: Solving questions of NCERT Solutions for Class 12 Physics Chapter 4 theory within the given time frame is a big task for students; if they are not able to do then it may lead to challenges. 
• Confusing Questions: The Class 12 Physics NCERT Solutions may contain confusing questions of Chapter 4 due to difficult words; this may lead to a big challenge for students to understand what is being asked. 
• Fear of Failure: The fear of failure among students can lead to the challenge while solving Class 12 Physics Chapter 4 questions from the NCERT Solutions. 
• Lack of Clarity: Lack of clarity in the Class 12 Physics concepts can lead to big challenges in solving Chapter 4 questions from the NCERT Solutions. 

How to Analyse the Mistakes After Solving Questions from NCERT Solutions for Class 12 Physics Chapter 4? 

After solving questions from the NCERT Solutions for Class 12 Physics Chapter 4, it is necessary for students to analyse the mistakes so that they can assess their progress; steps to analyse are discussed below: 

• Identify the Patterns of Mistakes: Patterns of mistakes are considered to be those errors which are done by most students; in that case, they need to identify the pattern of mistakes while solving questions from the NCERT Solutions for Class 12 Physics Chapter 4 revision. 
• Keep a Record of Mistakes: In order to analyse the mistakes while solving questions from the NCERT Solutions for Class 12 Physics Chapter 4 theory, they need to keep a proper record so that they can judge the progress. 
• Review the Correct Questions: Students need to go through the correct Chapter 4 questions and answers in the Class 12 Physics NCERT Solutions so that they can get an idea about the pattern of correct answers; through this students can easily analyse the mistakes. 
• Seek Help: By seeking help from the concerned teachers, students can analyse the mistakes after solving Class 12 Physics Chapter 4 questions from the NCERT Solutions; by seeking help students can solve all their doubts. 
• Stay Motivated: In the process of analysing the mistakes after solving Chapter 4 questions from the Class 12 Physics NCERT Solutions, students need to stay motivated and focused. 
• Use Different Approaches: To analyse the mistakes of Chapter 4 questions from the Class 12 Physics NCERT Solutions, one needs to try on different approaches or strategies; through this students can get an idea of wrong approaches. 

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• Physics Important Questions
• Class 12 Physics
• Chapter 4: Moving Charges Magnetism

Important Questions for Class 12 Physics Chapter 4 – Moving Charges and Magnetism

Magnetism is a physical characteristic caused by magnetic fields that produce attractive and repulsive effects on other objects. A magnetic field is generated by electric currents and magnetic moments of particles and influences other currents and magnetic moments. Ferromagnetic materials, which are strongly drawn to magnetic fields and can become permanent magnets themselves, are the most well-known materials that exhibit magnetism. Only a limited number of substances are ferromagnetic, including iron, cobalt, nickel and their alloys, which are the most common. The less common ferromagnetic substances include the rare-earth metals neodymium and samarium. Ferro- refers to iron as the permanent magnetism that was first found in lodestone, a type of iron ore called magnetite (Fe3O4).

A magnetic field is a field of vectors representing magnetism’s impact on moving electric charges, electric currents, and magnetic materials. A force perpendicular to both the velocity and magnetic field is experienced by a moving charge in a magnetic field. A permanent magnet’s magnetic field attracts ferromagnetic materials like iron and other magnets. Nonuniform magnetic fields can also apply tiny forces to “nonmagnetic” materials through paramagnetism, diamagnetism, and antiferromagnetism, but these forces are usually so weak that they can only be measured with laboratory equipment. Magnetised materials are surrounded by magnetic fields generated by electric currents in electromagnets and changing electric fields. As the strength and direction of magnetic fields can vary in different locations, they are described mathematically with a vector field function assigning a vector to each point in space.

Magnetic fields are produced by the movement of electric charges and the inherent magnetic moments of particles connected to their spin, a fundamental quantum property. Magnetic and electric fields are related, and both are part of the electromagnetic force, one of nature’s four fundamental forces. Magnetic fields are widely utilised in modern technology, especially in electrical engineering and electromechanics. Rotating magnetic fields are applied in electric motors and generators. The study of magnetic fields in electrical devices like transformers is called magnetic circuits. The Hall effect shows the presence of charge carriers in a material through magnetic forces. The Earth generates its own magnetic field, protecting the ozone layer from solar wind and crucial for compass navigation.

Important Questions For Class 12 – Physics – Chapter 4 – Moving Charges and Magnetism are provided here. Students must go through these questions and solve them to prepare for their Physics papers. They can also refer to these questions for quick revision. These questions are more likely to be asked in the exam; hence, students must practise them thoroughly.

Very Short Answer Type Questions

1. How does the radius of the path of a charged particle moving in a cyclotron change when the radio frequency field’s frequency is doubled?

(a) It will be halved.

(b) It will be doubled.

(c) It will remain unchanged.

(d) It will be increased by four times.

Answer: (c) It will remain unchanged.

2. Which of the following statements about a cyclotron is not true?

(a) A cyclotron is a device that uses high energies to accelerate ions or charged particles.

(b) Cyclotron combines electric and magnetic fields to accelerate charged particles and increase their energy.

(c) The cyclotron works by utilising the principle that the time it takes for an ion to complete one revolution is not affected by its speed or the radius of its orbit.

(d) In a cyclotron, the path of charged particles and ions is not fixed and can be arbitrary.

Answer: (d) In a cyclotron, the path of charged particles and ions is not fixed and can be arbitrary.

3. If an electron is travelling with velocity ν, it generates a magnetic field B, then

(a) the direction of field B will align with the direction of velocity ν.

(b) the direction of field B will be in the opposite direction to the direction of velocity ν.

(c) the direction of field B will be at a right angle to the direction of velocity ν.

(d) the direction of field B is independent of the direction of velocity ν.

Answer: (c) the direction of field B will be at a right angle to the direction of velocity ν.

4. If the magnetic compass needle is brought close to a straight wire bearing current, then

(I) the straight wire results in a significant deviation of the compass needle.

(II) The needle tangentially aligns to an imaginary circle whose centre is the straight wire and whose plane is perpendicular to the wire.

(a) (II) is correct

(b) (I) is correct

(c) neither (I) nor (II) is correct

(d) both (I) and (II) are correct

Answer: (d) both (I) and (II) are correct

5. Which statement about magnetic forces is correct?

(a) Magnetic forces always obey Newton’s third law.

(b) Magnetic forces do not follow Newton’s third law.

(c) Magnetic forces adhere to Newton’s third law when the current is very high.

(d) Magnetic forces comply with Newton’s third law in a low magnetic field environment.

Answer: (b) Magnetic forces do not follow Newton’s third law.

6. Converting a moving coil galvanometer into a voltmeter is achieved by

(a) placing a high resistance in series.

(b) placing a low resistance in parallel.

(c) placing a high resistance in parallel.

(d) placing a low resistance in series.

Answer: (a) placing a high resistance in series.

7. Converting a moving coil galvanometer into an ammeter is done by

(a) placing a high shunt resistance in series.

(b) placing a low shunt resistance in parallel.

(c) placing a low resistance in series.

(d) placing a high resistance in parallel.

Answer: (b) placing a low shunt resistance in parallel.

Short Answer Type Questions

1. Verify that the cyclotron frequency = eB/m has the correct dimensions of [T] -1

When a charged particle travels normal to the magnetic field, then

\(\begin{array}{l}\frac{ mv^2 }{ r } = qvB\end{array} \)

\(\begin{array}{l}\textup{Therefore, dimensions of } \omega = \frac{qB}{m} = \frac{v}{r} = \frac{LT^{-1}}{L} = [T^{-1}]\end{array} \)

2. Show that a force that does no work and is a velocity-dependent force.

Work is a scalar product of displacement and force. Hence work done at any instant is given by the following formula:

\(\begin{array}{l}dW = \vec{F}. \vec{ds}\end{array} \)

Where displacement is given as:

\(\begin{array}{l}\textup{Displacement, } \vec{ds} = \vec{v} dt \end{array} \)

According to the question, the work done by the force is:

\(\begin{array}{l}dW = 0\end{array} \)

\(\begin{array}{l}\Rightarrow \vec{ F }. \vec{v}dt = 0\end{array} \)

\(\begin{array}{l}\vec{F}.\vec{v} = 0\end{array} \)

This shows that the angle between force and velocity will be 90 o . If the direction of velocity changes, then the direction of the force will also vary.

3. The magnetic force depends on v, which depends on the inertial frame of reference. Does then the magnetic force differ from inertial frame to frame? Is it reasonable that the net acceleration has a different value in different frames of reference?

The magnetic force relies on vector v, which depends on the inertial reference frame. Magnetic force on a charged particle,

\(\begin{array}{l}F = q ( \vec{v} \times \vec{B} ) \end{array} \)

\(\begin{array}{l} F = qvBsin \theta \end{array} \)

Thus, the magnetic force is velocity-dependent. It varies from one inertial frame to another inertial frame.

4. Describe the motion of a charged particle in a cyclotron if the frequency of the radio (rf) field was doubled.

When the frequency of the radio frequency (rf) field is doubled, then the resonance condition is violated, and the time period of the radio frequency (rf) field is halved.

Therefore, the duration in which a particle completes half revolution inside the dees, radio frequency completes the cycle. As a result, particles will accelerate and decelerate alternatively.

5. Two long wires carrying current I 1 and 1 2 are arranged as shown in the figure. The one carrying current I 1 is along the x-axis. The other carrying current, I 2 is along a line parallel to the y-axis given by x = 0 and z =d. Find the force exerted at 0 2 because of the wire along the x-axis.

Magnetic field at the point O 2 due to current I 1 in long wire along the x-axis is

\(\begin{array}{l} B_{O_{2}} = \frac{ \mu_0 }{ 4 \pi } \frac{ 2I }{ d } \end{array} \)

It will be along the y-axis. Since the second wire is along the y-axis, the angle between the direction of

\(\begin{array}{l} I_2 \vec{l} \textup{ and } \vec{B}_{O_{2}} \textup{ is zero.} \end{array} \)

\(\begin{array}{l} \theta = 0^o \end{array} \)

Force on the wire existing along the y-axis is,

\(\begin{array}{l} f = I_2 | \vec{ l } \times \vec{B}_{O_{2}} | = I_2 l B_{O_{2}} sin 0^o = 0 \end{array} \)

Long Answer Type Questions

1. A current carrying loop consists of 3 identical quarter circles of radius R, lying in the positive quadrants of the x-y, y-z and z-x planes with their centres at the origin, joined together. Find the direction and magnitude of B at the origin.

Solution: Draw a rough diagram.

The equation used is

\(\begin{array}{l}B = \frac{ \mu_0 I \theta }{ 4 \pi R }\end{array} \)

It is given that

\(\begin{array}{l} \textup{Quarter Circles } \theta = \frac{ \pi }{ 2 } \end{array} \)

The electric current in circular arc = I, Radius = R

The magnetic field at the origin point due to the electric current carrying arc existing in x-y plane:

\(\begin{array}{l} \vec{ B } _ { xy } = \frac{ \mu _ { 0 } I ( \pi / 2 ) }{ 4 \pi R } \hat{ k } \end{array} \)

\(\begin{array}{l} = \frac{ \mu _ { 0 } I }{ 8 R } \hat{ k } \end{array} \)

Likewise, the Magnetic field at the origin point due to the electric current circular arc existing in the y-z plane:

\(\begin{array}{l} \vec{ B } _ { xy } = \frac{ \mu _ { 0 } I ( \pi /2 ) } { 4 \pi R } \hat{ i } \end{array} \)

\(\begin{array}{l} = \frac{ \mu _ { 0 } I }{ 8 R } \hat{ i } \end{array} \)

Likewise, the magnetic field at origin due to the electric current carrying a circular arc existing in the y-x plane:

\(\begin{array}{l} \vec{ B } _ { zx } = \frac{ \mu _ { 0 }I ( \pi / 2 ) }{ 4 \pi R } \hat{ j } \end{array} \)

\(\begin{array}{l} = \frac{ \mu _ { 0 } I }{ 8R } \hat{ j } \end{array} \)

The total vector sum of the magnetic field at the origin point due to each quarter circle,

\(\begin{array}{l} \vec{ B } = \vec{ B } _ { xy } + \vec{ B } _ { yz } + \vec{ B } _ { zx } \end{array} \)

\(\begin{array}{l} \vec{ B } _ { net } = \left ( \frac{ \mu I }{ 8 R } \right ) ( \hat{ i } + \hat{ j } + \hat{ k } ) \end{array} \)

2. A charged particle of charge e and mass m is moving in an electric field E, and magnetic field B. Construct dimensionless quantities and quantities of dimension [T] -1 .

No dimensionless quantity made using the given quantities. In the case of a charged particle travelling normally to the magnetic field, the magnetic Lorentz forces lend the required circular or centripetal force for revolution.

\(\begin{array}{l} \frac{ mv ^ { 2 } }{ R } = qvB \end{array} \)

By simplifying the equation, we get,

\(\begin{array}{l} \frac{ q B }{ m } = \frac{ v }{ R } = \omega \end{array} \)

The dimensional form of angular frequency

\(\begin{array}{l} [\omega] = \left [ \frac{ qB }{ m } \right ] = \frac{ v } { R } = [T] ^ { -1 } \end{array} \)

3. An electron enters with a velocity v = V o i into a cubical region (faces parallel to coordinate planes) in which there are uniform electric and magnetic fields. The orbit of the is found to spiral down inside the cube in the plane parallel to the x-y plane. Suggest a configuration of fields E and B that can lead to it.

If a charged particle is positioned in an electric field and magnetic field, then it begins moving in a spiral trajectory where magnetic field B and electric field E is perpendicular to the direction of motion. The electric field applies force on the charged particle, which results in the particle’s deceleration or acceleration and the centripetal force works due to the magnetic force. Assume that a magnetic field B = B 0 exists in the area, and an electron enters with a velocity of into the cubical region.

Using magnetic Lorentz force, the net force on the electron is represented by

\(\begin{array}{l} \vec{ F} _ { m } = – e ( v _ { 0 } \hat{ i } \times B _ { 0 } \hat{ k } ) \end{array} \)

\(\begin{array}{l} = e v _ { 0 } B _ { 0 } \hat{ j } \end{array} \)

The orbit of the electron circles around down inside the cube in the plane parallel to the x-y plane. So, E should be along the x-axis, and magnetic field B should be along the +z direction.

\(\begin{array}{l} \vec{ B } = B _ { 0 } \hat{ k } \textup{ and } \vec{ E } = E _ { 0 } \hat{ i } \end{array} \)

(where E 0 > 0)

4. Do magnetic forces obey Newton’s third law? Verify for two current elements dl 1 – dl i (unit vector) located at the origin and dl 2 = dl j (unit vector) located at (O, R, O). Both carry current I.

Here, it is required to find the magnetic field’s direction due to one wire at the point on a different wire, then the magnetic force on the current bearing wire.

As per Biot-Savart’s law, the magnetic field is parallel to idl × r, and idl is the current bearing element which has its direction along the current flow’s direction.

In the case of the magnetic field direction, at dl 2 , positioned at (0, R, 0) due to wire dl s represented by B || idl × r or i × j (point (0, r, 0) lies on the y-axis) and i × j = k. Therefore, the magnetic field at dl 2 is along the z-direction.

The magnetic force direction exerted at dl 2 due to the magnetic field of the first wire is along the x-axis.

F -i(I × ), i.e., F || (i × k) or along the -j direction.

Thus, the force due to dl 1 and dl 2 is non-zero.

Hence, magnetic forces do not follow Newton’s third law. But they follow Newton’s third law if the current carrying sections are placed parallel to each other.

5. A multirange voltmeter can be constructed by using a galvanometer circuit, as shown in the figure. We want to construct a voltmeter that can measure 2V, 20V and 200V using a galvanometer of resistance 10Ω and produce maximum deflection for a current of 1 mA. Find R 1 , R 2 , and R 3 that have to be used.

In this scenario,

\(\begin{array}{l} G = 10\Omega \end{array} \)

\(\begin{array}{l} I _ { g } = 1 m A = 10 ^ { -3 } A \end{array} \)

\(\begin{array}{l} \textup{ Case (i): } V = 2V \end{array} \)

\(\begin{array}{l}R _ { 1 } = \frac{ V }{ I _ g } – G \end{array} \)

\(\begin{array}{l} \frac{ 2 }{ 10 ^ { -3 } } -10 \end{array} \)

\(\begin{array}{l} 1990 \Omega \approx 2\Omega \end{array} \)

\(\begin{array}{l} \textup{ Case (b): } V = 20V \end{array} \)

\(\begin{array}{l} ( R _ { 1 } + R _ { 2 } ) = \frac{ 20 }{ 10 ^ { -3 } } – 10 \end{array} \)

\(\begin{array}{l} 20, 000 – 10 \approx 20k\Omega \end{array} \)

\(\begin{array}{l} R_2 = 20k \Omega – 2 k \Omega = 18 k \Omega \end{array} \)

\(\begin{array}{l} \textup{ Case(c): } V = 200V \end{array} \)

\(\begin{array}{l} R _ { 1 } + R _ { 2 } + R _ { 3 } = \frac{ 200 }{ 10 ^ { -3 } } – 10 \end{array} \)

\(\begin{array}{l} \approx 200 k \Omega \end{array} \)

\(\begin{array}{l} \therefore R _ { 3 } = 200 k \Omega – 20 k \Omega \approx 180 k \Omega \end{array} \)

6. A long straight wire carrying a current of 25A rests on a table, as shown in the figure. Another wire PQ of length 1m, mass 2.5g, carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will PQ rise?

In this case,

\(\begin{array}{l} I _ { 1 } = 25 A \end{array} \)

\(\begin{array}{l} I _ { 2 } = 25 A \end{array} \)

\(\begin{array}{l} l = 1m \end{array} \)

\(\begin{array}{l} m = 2.5g = 2.5 \times 10 ^ { -3 } kg \end{array} \)

In the equilibrium position,

\(\begin{array}{l} m g = \frac{ \mu _ { 0 } }{ 4 \pi } \frac{ 2 l _ { 1 } l _ { 2 } l }{ h } \end{array} \)

\(\begin{array}{l} h = \frac{ \mu _ { 0 } }{ 4 \pi } \frac{ 2 l _ { 1 } l _ { 2 } l }{ mg } \end{array} \)

\(\begin{array}{l} = \frac{ 10 ^ { -7 } \times 2 \times 25 \times 25 \times 1 }{ (2.5 \times 10 ^ { -3 }) \times 9.8 } = 51 \times 10 ^ { -4 } m = 0.51 cm \end{array} \)

7. A 100-turn rectangular coil, ABCD (in XY plane), is hung from one arm of a balance. A mass of 500g is added to the other arm to balance the weight of the coil. A current 4.9 A passes through the coil, and a constant magnetic field of 0.2 T acting inward (in xz plane) is switched on such that only arm CD of length 1 cm lies in the field. How much additional mass ‘m’ must be added to regain the balance?

If the magnetic field is zero and the weight is added to one balance pan, balance the rectangular coil in the other balance pan. Then

\(\begin{array}{l} M g l = W _ { coil } l \textup{ or } W _ { coil } = M g = 500 \times 9.8 N \end{array} \)

When the current (I) is passed through the coil and the magnetic field is switched on, let ‘m’ be the mass added in a pan to counterbalance the beam. Then

\(\begin{array}{l} M g l + mgl + = W _ { coil } l + I B L \, sin 90 ^ { 0 } \, l \end{array} \)

\(\begin{array}{l} m g l = I B L I \textup{ or } m = \frac{ I B L }{ 9 } \end{array} \)

\(\begin{array}{l} \therefore m = \frac{ 4.9 \times 0.2 \times ( 1 \times 10 ^ {-2} ) }{ 9.8 } = 10 ^ {-3} kg = 1g \end{array} \)

8. A rectangular conducting loop consists of two wires on two opposite sides of length ‘I’ joined together by a rod of length ‘d’. The wires are each of the same material but with cross-sections differing by a factor of 2. The thicker wire has a resistance E, and the rods are of low resistance, which in turn are connected to a constant voltage source V 0 . The loop is placed in uniform a magnetic field B at 45 o to its plane. Find τ, the torque exerted by the magnetic field on the loop about an axis through the centres of rods.

Equation used: \(\begin{array}{l}B = B_0 \hat{i} \end{array} \)

\(\begin{array}{l} \vec{ \tau } = \vec{ r } + \vec{ F } \end{array} \)

Force on current bearing wire placed in a magnetic field,

\(\begin{array}{l} \vec{ F } = i ( \vec{ I } \times \vec{ B } ) \end{array} \)

The force on the thicker wire (wire 1) is

\(\begin{array}{l} F _ { 1 } = i _ { 1 } I B = \left ( \frac{ V _ { O } }{ R } \right ) I B \end{array} \)

\(\begin{array}{l} \textup{Since resistance of wire } R = \frac{ \rho I}{ A } \end{array} \)

\(\begin{array}{l} \textup{Each wire is of the same material } R \propto \frac{ 1 }{ A} \end{array} \)

\(\begin{array}{l} \textup{ Force on another wire } \vec{ F } _ { 2 } = ( \vec{ I } \times \vec{ B } ) \end{array} \)

\(\begin{array}{l} F _ { 2 } = i _ { 2 } I B = \left ( \frac{ V _ 0 }{ 2 R } \right ) I B \end{array} \)

Torque on wire on wire 1 due to F 1 ,

\(\begin{array}{l} \vec{ \tau } _ { 1 }\left ( \frac{d}{2} \right ) \left ( \frac{ V _ {0} }{ R } \right ) I B \, sin 45 ^ { o } = \left ( \frac{ V _ { 0 } d }{ 2 \sqrt{ 2 } } \right ) I B \end{array} \)

Torque on the wire 2 due to F 2 ,

\(\begin{array}{l} \vec{ \tau } _ { 2 } = \vec{ r } \times \vec{ F } _ { 2 } = \left ( \frac{ d }{ 2 } \right ) F _ { 2 } \, sin \theta \end{array} \)

\(\begin{array}{l} \tau _ { 2 } = \left ( \frac{ d }{ 2 } \right ) \left ( \frac{ V _ { 0 } }{ 2 R } \right ) I B sin 45 ^ { o } = \left ( \frac{ V _ { 0 } d }{ 4 \sqrt{ 2 } } \right ) I B \end{array} \)

\(\begin{array}{l} \textup{Net torque } \tau = \tau _ 1 – \tau _ 2 \end{array} \)

\(\begin{array}{l} \tau = \frac{ 1 }{ 4 \sqrt{ 2 } } \frac{ V _ { 0 } I d B }{ R } \end{array} \)

(A = Id = Area of loop)

9. An electron and a positron are released from (0, 0, 0) and (0, 0, 1.5R), respectively, in a uniform magnetic field, each with an equal momentum of magnitude p = e BR. Under what conditions on the directions of momentum will the orbits be non-intersecting circles?

Since B is along the x-axis, for a circular orbit, the memento of the two particles is in the y-z plane. Assume p 1 and p 2 as the moment of the electron and positron, respectively. Both of them are defined as a circle of radius R. Assume p 1 makes an angle θ with the y-axis, and p 2 should make the same angle. The centre of the circles should be normal to the momenta and at a distance of R.

The circles of the two shall not overlap if the length between the two centres is more than 2R.

Let d be the length between Ce and Cp.

\(\begin{array}{l} d ^ { 2 } = [ R sin \theta – ( – R sin \theta ) ^ 2 ] \left [ R cos \theta \left ( \frac{ 3 }{ 2 } – R cos \theta \right ) \right ] ^ { 2 } \end{array} \)

\(\begin{array}{l} d ^ { 2 } = ( 2 R sin \theta ) ^ { 2 } + \left ( \frac{ 3 }{ 2 } R – 2 R cos \theta \right ) ^ { 2 } \end{array} \)

\(\begin{array}{l} = 4 R ^ { 2 } sin ^ { 2 } \theta + \frac{ 9 ^ { 2 } }{ 4 } r – 6 R ^ { 2 } cos \theta + 4 R ^ 2 cos ^ 2 \theta \end{array} \)

\(\begin{array}{l} = 4 R ^ { 2 } + \frac{ 9 }{ 4 } R ^ { 2 } – 6 R ^ { 2 } cos \theta \end{array} \)

As ‘d’ has to be larger than 2R

\(\begin{array}{l} d ^ { 2 } > 4 R ^ { 2 } \end{array} \)

\(\begin{array}{l} \Rightarrow 4 R ^ { 2 } + \frac{ 9 }{ 4 } R ^ { 2 } – 6 R ^ { 2 } cos \theta > 4 R ^ { 2 } \end{array} \)

\(\begin{array}{l} \frac{ 9 }{ 4 } > 6 cos \theta \textup{ or } cos \theta < \frac{ 3 }{ 8 } \end{array} \)

10. A uniform conducting wire of length 12a and resistance R is wound as a current-carrying coil in the shape of (i) an equilateral triangle of side a; (ii) a square of sides a, and (iii) a regular hexagon of sides a. The coil is connected to a voltage source V 0 . Find the magnetic moment of the coil in each case.

(i) No. of turns, n = 4

\(\begin{array}{l}\textup{ Area of triangle A} = \frac{ 1 }{ 2 } a \times \alpha \, sin 60 ^ { o } \end{array} \)

\(\begin{array}{l} = a ^ { 2 } \sqrt{ 3 } / 4 \end{array} \)

(ii) \(\begin{array}{l} n = 3, \, \textup{area A} = a \times a = a ^ { 2 } \end{array} \)

(iii) \(\begin{array}{l} n = 2, \textup{ area A} = 6 [\frac{ a }{ 2 } \times \frac{ a }{ 2 } tan 60 ^ { o }] = \frac{ 3 \sqrt{ 3 } }{ 2 } a ^ { 2 } \end{array} \)

Magnetic moment, M = nIA

\(\begin{array}{l} \textup{For (i), } M _ { 1 } = 4 \times I \times a ^ { 2 } \sqrt{ 3 } / 4 = \sqrt{ 3 } a ^ { 2 } I \end{array} \)

\(\begin{array}{l} \textup{For (ii), } M _2 = 3 \times I \times a ^ { a } = 3 a ^ { 2 } I \end{array} \)

\(\begin{array}{l} \textup{For (iii), } M _ { 3 } = 2 \times \frac{ 3 \sqrt{ 3 } }{ 2 } a ^ { 2 } = 3 \sqrt{ 3 } a ^ { 2 } I \end{array} \)

(a) Show that ζ(L) monotonically increases with L.

(b) Use an appropriate Amperian loop to show that ζ(∞)=μ 0 I, where I is the current in the wire.

(c) Verify directly the above result.

(d) Suppose we replace the circular coil with a square coil of sides R carrying the same current I. What can you say about ζ(L) and ζ(∞)?

Since the circular loop carrying an electric current is lying in the x-y plane, the magnetic field is along the z-axis, in the line integral direction,

\(\begin{array}{l} \xi ( L ) = \left| \int _ { – L } ^ { + L } \vec{ B } . \vec{ dl } \right| \end{array} \)

\(\begin{array}{l} = \int _ { – L } ^ { + L } B d l \cos 0 ^ { o } \end{array} \)

\(\begin{array}{l} = \int _ { -L } ^ { +L } B d l = 2 B L \end{array} \)

So, ζ(L) is monotonically rising with L.

(b) Take a closed amperian path PQRP as represented in the figure.

As per the Ampere circuital law, the integral of B over the closed trajectory PQRP is

\(\begin{array}{l} \oint _ { P Q R P } \vec{ B } . \vec{ dl } = \int _ { P Q R } \vec{ B } . \vec{d l} + \int _ { R P } \vec{ B } . \vec{ d l } \end{array} \)

\(\begin{array}{l} = \int _ { P Q R } \vec{ B } . \vec{ d l } + \int _ { – L } ^ { + L } \vec{ B } . \vec{ d l } = \mu _ { 0 } I \end{array} \)

\(\begin{array}{l} \int _ { P Q R } \vec{ B } . \vec{ d l } + \zeta ( L ) = \mu _ { 0 } I \end{array} \)

If L → ∞, then the value of B → 0 (as B ∝ 1 / r3)

\(\begin{array}{l} \therefore \int _ { P Q R } \vec{ B } . \vec{ d l } = 0 \textup{, from (i), } 0 + \zeta ( \infty ) = \mu _ { 0 } I \textup{ or } \zeta ( \infty ) = \mu _ { 0 } I \end{array} \)

(c) Magnet field B at a point on the circular coil axis at a distance z from the centre of a circular coil of radius R bearing current I is

\(\begin{array}{l} B _ { z } = \frac{ \mu _ { 0 } I R ^ { 2 } }{ 2 ( z ^ { 2 } + R ^ { 2 } ) ^ { 3/2} } \end{array} \)

\(\begin{array}{l} \therefore \int _ { – \infty } ^ { + \infty } B _ { z } dz = \int _ { -\infty } ^ { + \infty } \frac{ \mu _ { 0 } I R ^ { 2 } }{ 2 ( z ^ { 2 } + R ^ { 2 } ) ^ { 3/2 } } dz \end{array} \)

\(\begin{array}{l} \textup{Put }z = R \tan \theta, \,\, dz = R \sec ^ { 2 } \theta d \theta \end{array} \)

\(\begin{array}{l} \therefore \int _ { – \infty } ^ { + \infty } B _ { z } dz = \frac{ \mu _ { 0 } I }{ 2 } \int _ { \pi / 2 } ^ { + \pi / 2 } \cos \theta d \theta = \mu _ { 0 } I \end{array} \)

(d) If the circular coil is replaced by a square coil of side R, bearing the same current I, then

\(\begin{array}{l} B ( z ) _ { \textup{square} } < B ( z ) _ { \textup{circular coil } } \end{array} \)

\(\begin{array}{l} \therefore \zeta ( L ) _ { \textup{square }} < \zeta ( L ) _ { \textup{circular coil } } \end{array} \)

Utilising arguments as in (b)

\(\begin{array}{l} \zeta ( \infty ) _ { \textup{square } } = \zeta ( \infty ) _ { \textup{ circular } } = \mu _ { 0 } I \end{array} \)

12. A multi-range current metre can be constructed by using a galvanometer circuit, as shown in the figure. We want a current metre that can measure 10mA, 100mA and 1A using a galvanometer of resistance 10Ω and that produces maximum deflection for a current of 1mA. Find S 1 , S 2 and S 3 that have to be used.

\(\begin{array}{l} G = 10 \Omega \end{array} \)

In order to convert a galvanometer into an ammeter of the given current range 0 to 1, the shunt resistance needed is

\(\begin{array}{l} S = \frac{ I _ { g } }{ I – I _ { g } } \end{array} \)

\(\begin{array}{l} I = 10 m A = 10 \times 10 ^ { -3 } A \end{array} \)

S = S _ { 1 } + S _ { 2 } + S _ { 3 }

\(\begin{array}{l} \therefore S _ { 1 } + S _ { 2 } + S _ { 3 } = \frac{ 1 m A \times 10 \Omega }{ (10 – 1) m A } = \frac{ 10 }{ 9 } \Omega \end{array} \)

\(\begin{array}{l} \textup{When } I = 100 m A \end{array} \)

\(\begin{array}{l} S = S _ { 2 } + S _ { 3 } \end{array} \)

\(\begin{array}{l} \textup{Galvanometer Resistance } = G + S _ { 1 } \end{array} \)

\(\begin{array}{l} \therefore S _ { 2 } + S _ { 3 } = \frac{ I _ { g } (G + S _ { 1 }) }{ I – { l _ { g } } } \end{array} \)

\(\begin{array}{l} = \frac{ 1 m A \times ( 10 + S _ { 1 } ) }{ (100 – 1) m A } = \frac{ 10 + S _ { 1 } }{ 99 } \textup{ …. (ii)} \end{array} \)

Case (iii) If I – 1A, then S = S 3

Galvanometer Resistance = (G + S 1 + S 2 )

\(\begin{array}{l} \therefore S _ { 3 } = \frac{ I _ { g ( G + S + S _ { 2 } ) } }{ I – I _ { g } } \end{array} \)

\(\begin{array}{l} = \frac{ 1 m A [ 10 + S _ { 1 } + S _ { 2 } ] }{ ( 1000 – 1) m A } \end{array} \)

\(\begin{array}{l} \frac{ 10 + S _ { 1 } + S _ { 2 } }{ 999 } \textup{ …. (iii)} \end{array} \)

By inserting the value (ii) in (i), we get

\(\begin{array}{l} S _ { 1 } + \frac{ 10 + S _ { 1 } }{ 99 } = \frac{ 10 }{ 9 } \end{array} \)

\(\begin{array}{l} \textup{or } S _ { 1 } \left ( 1 + \frac{1}{99} \right ) = \frac{10}{9} – \frac{ 10 }{ 99 } = \frac{ 100 }{ 99 } \end{array} \)

\(\begin{array}{l} \therefore S _ { 1 } \times \frac{ 100 }{ 99 } = \frac{ 100 }{ 99 } \textup{ or } S _ { 1 } = 1 \Omega \end{array} \)

From equation (ii)

\(\begin{array}{l} S _ { 2 } + S _ { 3 } = \frac{ 1 }{ 99 } ( 10 + 1 ) = \frac{ 1 }{ 9 } \textup{ ….(iv)} \end{array} \)

From equation (iii)

\(\begin{array}{l} S _ { 3 } = \frac{ 10 + 1 + S _ { 2 } }{ 999 } = \frac{ 11 + S _ { 2 } }{ 999 } \end{array} \)

\(\begin{array}{l} \textup{Or } S _ { 3 } – \frac{ S _ { 2 } }{ 999 } = \frac{ 11 }{ 999 } \textup{ ….(iv)} \end{array} \)

Minusing equation (v) from equation (iv), we get

\(\begin{array}{l} S _ { 2 } = \frac{ S _ { 2 } }{ 999 } = \frac{ 1 }{ 9 } – \frac{ 11 }{ 999 } = \frac{ 1000 }{ 999 } \end{array} \)

\(\begin{array}{l} \textup{Or } \frac{ S _ { 2 } \times 1000}{ 999 } = \frac{ 1000 }{ 999 } or S _ { 2 } = \frac{1}{10} = 0.1 \Omega \end{array} \)

From equation (iv)

\(\begin{array}{l} \frac{1}{10} + s _ { 3 } = \frac{ 1 }{ 9 } \end{array} \)

\(\begin{array}{l} \textup{Or } \frac{ 1 }{ 9 } – \frac{ 1 }{ 10 } = \frac{ 1 }{ 90 } \end{array} \)

\(\begin{array}{l} \textup{Or } S _ { 3 } = 0.011 \omega \end{array} \)

13. Five long wires, A, B, C, D, and E, each carrying current I, are arranged to form the edges of a pentagonal prism, as shown in the figure. Each carries current out of the plane of the paper.

(a) What will be magnetic induction at a point on the axis O? Axis is at a distance R from each wire.

(b) What will be the field if the current in one of the wires (say A) is switched off?

(c) What if the current in one of the wires (say) A is reversed?

(a) The five wires A, B, C, D and E are positioned perpendicular to the plane of the paper, as shown in the figure. The total magnetic field induction at point O resulting from the current flowing through all five wires is zero, as it can be represented by the sides of a closed pentagon in the plane of the paper.

(b) The overall magnetic field induction at point O caused by the currents flowing through wires A, B, C, D, and E is the same as, but opposite in direction, to the magnetic field induction at O due to the current through wire A alone.

Therefore, the magnetic field induction at O due to the electric current flowing through wire A is

\(\begin{array}{l} B = \frac{ \mu _ { 0 } }{ 4 \pi } \frac{ 2 I }{ R } \textup{ acting }\perp \, \, r \, \, \textup{to AO towards right. } \end{array} \)

Therefore, the magnetic field induction at point O due to currents through the wires E, D, C, and B is

\(\begin{array}{l} \frac{ \mu _ { 0 } }{ 4 \pi } \frac{ 2 I }{ R } \textup{, acting} \perp \, \, r \textup{ to AO towards left. } \end{array} \)

(c) When the electric current flowing wire A is reversed, then the net magnetic field induction at point O is equal to mag. The magnetic field induction due to wires E, D, C, and B is

\(\begin{array}{l} \frac{ \mu _ { 0 } }{ 4 \pi } \frac{ 2 I }{ R } \,\, (\textup{ acting } \perp \, \, r \,\, \textup{AO towards left}) + \frac{ \mu _ { 0 } }{ 4 \pi } \frac{ 2 I }{ R } (\textup{acting} \, \, \perp r \, \, \textup{AO towards left} ) \end{array} \)

\(\begin{array}{l} = \frac{ \mu _ { 0 } I }{ \pi R } \textup{ acting } \perp \,\, r \textup{ AO towards left.} \end{array} \)

14. What are the applications of magnetism?

• Although we frequently use computers in our daily lives, we may not have considered the presence of a magnet inside them. Magnets play a crucial role in storing data on a hard disk and allow the computer to retrieve information.
• Magnets are also utilised in various electronic devices such as TVs, speakers, and radios. The combination of a small coil of wire and a magnet in a speaker converts electronic signals into sound waves.
• Magnets are used within generators to convert mechanical energy into electrical energy. In contrast, certain motors also use magnets to transform electrical energy into mechanical energy.
• Additionally, a powerful magnet is used by cranes to move large metal objects.
• Magnet technology is utilised in various industries, including ore processing, for separating metallic materials from other substances.
• Magnets also play a crucial role in medical imaging, such as MRI machines, to create detailed images of the body’s internal structures.
• Magnet usage can be found in everyday household items such as refrigerator magnets and magnetic bottle openers.
• A compass, which uses a magnetic needle to indicate a direction, is commonly used in outdoor activities.
• Additionally, magnets are used in technology, such as credit and debit cards, as the dark strip on the back is magnetic and stores data.

15. Write an essay on Galvanometer.

A galvanometer is an instrument that measures electric current and the direction of the current. The early versions of this instrument were not precise, but later models, known as ammeters, were calibrated and able to give more accurate measurements of the current flow. A galvanometer is an instrument that uses the interaction between an electric current and a magnetic field to measure current. It deflects a pointer in reaction to the current moving through a conducting coil in a fixed magnetic field.

Galvanometers are considered a type of actuator and were the first to measure small amounts of electric current. They were discovered by Hans Christian Ørsted in 1820 and were named after Luigi Galvani, who, in 1791, observed that electric current causes the legs of a dead frog to move. André-Marie Ampère mathematically described Ørsted’s discovery and named the instrument after Galvani’s discovery. Galvanometers have been crucial in advancing science and technology across various fields. In the 19th century, they were used to establish long-distance communication via submarine cables, such as the first transatlantic telegraph cables. They also played a vital role in discovering electrical activity in the heart and brain through their precise current measurement.

Galvanometers have been utilised as display elements in other types of analogue metres, such as light and VU metres, by measuring the output of their sensors. The most commonly used galvanometer type today is the D’Arsonval/Weston type. It operates by deflecting a pointer in response to an electric current flowing through a coil in a fixed magnetic field and can be considered a type of actuator. It is important to note that ammeters, calibrated and accurate instruments for measuring current, cannot be compared with values from a galvanometer.

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Important Questions Class 12 Physics Chapter 4

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Important Questions for CBSE Class 12 Physics Chapter 4 – Moving Charges and Magnetism 

Since more than a century ago, magnetism and moving charges, or electricity, have been explored. The occurrences that were seen with the alignment of a needle served as the basis for the link between the two.

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With these Class 12 Physics Chapter 4 Important Questions , students will come to know that it was observed that the alignment of a needle remained tangent to a hypothetical circle whose plane was perpendicular to the straight line at its centre. However, the needle’s orientation changes as the current is passed. A magnetic field is thought to be created as a result of the movement of charges.

Moreover, with these Class 12 Physics Chapter 4 Important Questions , Extramarks will provide students with detailed and authentic solutions to important questions according to CBSE past years’ question papers so that students can prepare for their examination according to the CBSE syllabus.

CBSE Class 12 Physics Chapter 4 Important Questions

Study important questions for class 12 physics chapter 4 – moving charges and magnetism .

This part of the Class 12 Physics Chapter 4 Important Questions , includes important questions that may come in your examination. Additionally, here we all also see how formulas are applied to a given question.

1. Give two details about the wire’s composition that was utilised to suspend the coil in a moving coil galvanometer.

Ans. The wire used to suspend the coil in a moving coil galvanometer has the following two characteristics.

• Non-brittle conductor
• Restoring small torque should be used per unit of twist

2. Two equal-length wires are bent into two loop shapes. One of the loops is circular, while the other is square in shape. The same current flows through them while they are hung in a constant magnetic field. Which loop will be subject to more torque? Explain.

Ans. We know that a circular loop has a larger area than a square loop, the torque experienced by the circular loop would be greater than that of the square loop because torque is directly proportional to area. A circular loop hence experiences more torque.

3. What would a charged particle travelling in the direction of a constant magnetic field take?

Ans. As there would be no forces acting on the charged particle, its route would be a straight line going in the direction of a uniform magnetic field.

4. Define a radial magnetic field. What is it? How is it measured in a galvanometer with a moving coil?

Ans. The magnetic field in which the coil’s plane constantly faces the magnetic field’s direction is known as a radial magnetic field. The following methods can be used to obtain it.

• Concavely cutting the pole sections as instructed
• Inserting a cylindrical core made of soft iron between the pole pieces

5. A long, straight solenoid carrying current I, on an axis along which an electron is travelling at velocity v. What will the magnetic field of the solenoid’s magnetic field be doing to the electron?

Ans. F = Bqv sin is the formula for zero as a force acting on a charged particle travelling in a magnetic field.

Since v and B are both along the -solenoid’s axis in this instance, the angle between them is 0°. Consequently, F= qvB sin = 0.

6. Would your response alter if the circular coil in (a) were swapped out for a planar coil that encloses the same space but has a different shape? (All other information remains the same.)

Ans. From relation (1), it may be concluded that the coil’s shape has no bearing on the applied torque’s size. The size of the coil will determine this. Therefore, the answer would remain the same if the circular coil in the example above were swapped out for a planar coil that encloses the same space but has an irregular shape.

7. A charged particle enters a region of a powerful and irregular magnetic field that varies from point to point in both strength and direction, and it exits the region via a convoluted track. If it didn’t collide with anything, would its final speed be the same as its starting speed?

Ans. Yes, the charged particle’s end speed would be the same as its starting speed since the magnetic force can only change the direction of a particle’s velocity, not its magnitude.

8. An electron moving from west to east enters a chamber with a consistent electrostatic field that runs from north to south. Give instructions on how to create a uniform magnetic field in that direction to stop the electron from veering off course.

Ans. In a chamber with a homogeneous electric field running north to south, an electron that is travelling from west to east enters. If the electric force exerted on it is equal to and opposite to the magnetic field, the travelling electron is not deflected. Southward would be the direction of the magnetic force. Fleming’s left-hand rule also states that the magnetic field should be applied vertically downward.

9. Define cyclotron. Explain in detail.

Ans. Protons, deuterons, and other charged – particles, among others, are accelerated by cyclotrons.

It operates on the tenet that a charged particle can be accelerated to extremely high energies by repeatedly passing through a moderate electric field while being subjected to a strong magnetic field.

10. In a chamber, a magnetic field is created that changes in strength from location to location but always faces east to west. A charged particle enters the chamber and moves with constant speed in a straight line without being deflected. What can you say about the particle’s starting velocity?

Ans. The particle’s starting velocity may be parallel or antiparallel to the magnetic field. As a result, it follows a straight course without experiencing any field deflection.

Important Questions Of Chapter 4 Physics Class 12

Class 12 physics chapter 4 important questions.

Here are some CBSE extra questions that may come in your exams. Go through these important questions in Class 12 Physics Chapter 4 to prepare yourself for your upcoming physics examination. You can use these CBSE revision notes as your guide before your examination to test your grasp of important concepts. These questions are inspired by the Class 12 Syllabus and NCERT books prescribed by CBSE .

Q.11. In an area where a uniform magnetic field B is directed normally to the plane of the paper, an alpha particle and a proton are travelling in that plane. What will the ratio of the radii of the trajectories in the field of two particles with equal linear momenta be?

Ans. The radius of the path can be represented by the following equation.

RaRp=qpqa=e2e=12

Here Ra is the radii of – particle and Rp is its proton, and their respective charge are qa and qp.

∴ Ra: Rp= 1:2

Hence, the ratio will be 1:2.

Q.12. In the presence of a magnetic field B, calculate the force that will act on a charged particle of charge q travelling at a velocity. Demonstrate that when this force is present,

(a) the particle’s K.E. remains constant.

(b) its instantaneous power amounts to zero.

Ans. We already know that

The magnetic force can be expressed as F= q(vB)

Since the force’s direction is parallel to the plane containing (vB)

F = qvBsin 90°=qvB

Here, the displacement and the force will be perpendicular to each other.

W = FScos90°=0

Hence, the kinetic energy will be constant in this particular condition.

(b) the expression for instantaneous power is

When velocity and force are perpendicular to each other.

p = FScos90° = 0

Hence, the instantaneous power will be zero.

Q.13. Answer the following.

(a) What distinguishes a toroid from a solenoid? Create a diagram of the magnetic field lines in the two scenarios and compare them.

(b) How is a given solenoid’s magnetic field made strong?

Ans. (a) The toroid is a hollow circular ring on which a substantial number of tightly wound wire turns are placed, in contrast to the solenoid, which is made up of a long wire twisted in the shape of a helix with the neighbouring turns closely spaced.

(b) A soft iron core is inserted within a particular solenoid to create a powerful magnetic field inside of it. By running more current through it, it becomes stronger.

Q.14.  A current I is carried by a spherical coil with a radius of r and N tightly wrapped turns. Create the expressions for the following.

(a) A magnetic field that is present at its centre

(b) The coil’s magnetic moment

Ans. (a) A circular coil with N turns and radius r carrying a current, I, has a magnetic field at its centre that is

(b) The magnetic moment, M = NIA = NIr²

Q.15. A current of 10A travels down a straight wire. A 2.0 cm gap separates an electron travelling at 107 m/s from the wire. Find the force on the electron that is being affected by its velocity approaching the wire.

Ans. We know that the current running through a straight wire is I=10A,

The speed of the electron is v=107 m/s

The distance between the wire and the electron is  R=20cm= 210-2 m

The force that is acting on the travelling electron can be written as

The expression for magnetic field is

If we substitute the given values,

B = 10-7210210-2=10-4T, and it is known to be to the plane

The force that is acting on the electron can be represented by,

F = 1.610-16 N

Hence, the force will be F = 1.610-16 N.

Q.16. A wire coil of 100 turns, each with a radius of 8.0 cm, is used to transport a current of 0.40. How strong is the magnetic field B in the centre within the coil?

Ans. We are given that,

Number of turns on the coils, n = 100

Each turn’s radius, r = 8.0 cm = 0.08 m

The current that is flowing within the coil is I = 0.4 A

The relationship could be used to determine the strength of the magnetic field at the coil’s centre.

B = 04 2nlr

Here, the permeability of free space, 0=410-7T m A-1

B = 410-7421000.40.08

B = 3.1410-4 T

Hence, the magnetic field’s magnitude will be 3.1410-4 T.

Q.17. A 35A current is carried along a long, straight wire. How big is field B at a location 20 cm from the wire?

Ans. The following is given.

The current present in the wire is I = 35 A

The distance between the point and the wire, r = 20 cm = 0.2 m

The magnetic field at a given point is

Here 0= Permeability of free space = 410-7 T m A-1

B = 410-723540.2

B = 3.510-5 T

Hence, the magnetic field’s magnitude at a distance of 20 cm from the wire will be 3.510-5 T.

Q.18. A 50 A current travels north to south via a straight wire in the horizontal plane. Give  B’s magnitude and direction at a location 2.5 metres east of the wire.

Ans. The following information is given.

The current in the wire, I = 50 A

The point is 2.5 m away from the wire’s east.

The magnitude of the distance between the point and the wire is r = 2.5 m.

The magnetic field at a given point can be represented by the following relation.

Here, 0 = Permeability of free space = 410-7 T m A-1

B = 410-725042.5

B = 410-6 T

Since the direction of the wire’s current is vertically downward, and the point is 2.5 metres away from the wire length, we may use Maxwell’s right-hand thumb rule to determine the magnetic field’s direction at the given point, which is vertically upward.

Q.19. Inside a solenoid, a 3.0 cm wire with a 10 A current is inserted perpendicular to the axis. It is stated that there is a 0.27 T magnetic field inside the solenoid. What is the wire’s magnetic field strength?

Ans. The following information is given,

The wire’s length, I = 3 cm = 0.03 m

The current that flows through the wire, I = 10 A

Magnetic field, B = 0.27 T

The angle between the current and the magnetic field = 90°

The magnetic force that is exerted on the wire can be represented as

If we substitute the values, we get,

F = 0.27100.03 sin90°

F = 8.110-2 N

Hence, the magnetic force that will be exerted on the wire will be 8.110-2 N and the force’s direction can be derived using Flemming’s left-hand rule.

Q.20. A tightly wound solenoid measuring 80 cm long has five layers with 400 turns each. The solenoid’s diameter is 1.8 cm. Calculate the magnitude of B inside the solenoid located near its centre if the current carried is 8.0A.

The solenoid length is given as I = 80 cm = 0.8 m

Since the solenoid has five layers of windings with a total of 400 turns each.

The total turns of the solenoid, D = 1.8 cm = 0.018 m

The current that is carried by the solenoid, I = 8.0 A

The magnetic field’s magnitude present inside and near the centre of the solenoid can be represented by the following equation.

Here, 0= 410-4 TmA-1 will be the permeability of free space.

When we substitute the values given, we get

B = 410-7200080.8

B = 2.51210-2 T

Hence, the magnitude of the magnetic field present inside and at its centre will be 2.51210-2 T.

Q.21. Answer the following.

(a) A 6.0A current-carrying circular coil with 30 turns and an 8.0cm radius is hung vertically in a 1.0T uniform horizontal magnetic field. The field lines are at an angle to the coil’s normal. Determine how much counter-torque will need to be provided to stop the coil from turning.

(b) Would your response alter if the circular coil in (a) were swapped out for a planar coil that encloses the same space but has a different shape? (All other information remains the same.)

Ans. (a) The information given is as follows.

The number of turns that the circular coil does, n = 30

The radius of the coil, r = 8.0 cm = 0.08 m

The coil’s area= r² = (0.08)² = 0.0201m²

The current that is flowing through the coil, I = 6.0 A

The strength of the magnetic field, = B = 1 T

The angle existing between the normal with the coil surface and the field lines is, = 60°

The magnetic field produces a torque in the coil and hence it turns. The relation, which is used to calculate the counter torque used to keep the coil from spinning is

T = nIBA sin           ……………………….(1)

T = 30610.0201sin 60°

Hence, the counter torque that will be applied to keep the coil from turning is 3.133 Nm.

(b) From relation (1), it may be concluded that the coil’s shape has no bearing on the applied torque’s size. The size of the coil will determine this. Therefore, the answer would remain the same if the circular coil in the example above were swapped out for a planar coil that encloses the same space but has an irregular shape.

Q.22. A square plane coil with 200 turns and a surface area of 100 cm2 can carry 5A of constant current. It is positioned in a 0.2 T uniform magnetic field that is acting perpendicular to the coil’s plane. Calculate the coil’s torque when the field’s direction and its plane are at an angle of 60 degrees. What direction will the coil be in when it reaches a stable equilibrium?

Ans. A = 100 cm² = 100(10-4)m² = 10-2 m

N = 200 turns, I = 5A, B = 0.2 T

= 90°-60°=30°

= (200)(5)(10-2)(0.2)12

Hence, when the coil and the magnetic field are parallel, the coil’s equilibrium state will be stable.

Q.23. Answer the following. 

(a) In a chamber, a magnetic field is created that changes in strength from location to location but always faces east to west. A charged particle enters the chamber and moves with constant speed in a straight line without being deflected. What can you describe about the particle’s starting velocity?

(b) A charged particle enters a region of a powerful and irregular magnetic field that varies from point to point in both strength and direction, and it exits the region via a convoluted track. If it didn’t collide with anything, would its final speed be the same as its starting speed?

(c) An electron moving from west to east enters a chamber with a consistent electrostatic field that runs from north to south. Give instructions on creating a uniform magnetic field in that direction to stop the electron from veering off course.

Ans. (a) The particle’s starting velocity may be parallel or anti-parallel to the magnetic field. As a result, it follows a straight course without experiencing any field deflection.

(b) Yes, the charged particle’s end speed would be the same as its starting speed since the magnetic force can only change the direction of a particle’s velocity, not its magnitude.

(c) In a chamber with a homogeneous electric field running north to south, an electron travelling from west to east enters. If the electric force exerted on it is equal to and opposite to the magnetic field, the travelling electron is not deflected. Southward would be the direction of the magnetic force. Fleming’s left-hand rule also states that the magnetic field should be applied vertically downward.

Q.24. A 300A current flows via the cables connecting an automobile’s battery to its starting motor (for a short time). If the distance between the wires is 1.5 cm and they are 70 cm long, what is the force per unit length? Also, answer whether the force will be attractive or repulsive.

Ans. The information given is as follows.

Current present in both the wires, I = 300A

The distance between the two wires, r = 1.5 cm = 0.0015 m

The two wire’s lengths, I = 70 cm = 0.7 m

We already know that the force that exists between two wires can be denoted by the following relation.

Here, the permeability of free space 0 = 410TmA-1

If we substitute the values given, we get

F = 410-7300220.015

There is a repulsive force between the wires since the direction of the current in them is observed to be opposing.

Q.25. A constant magnetic field of 6.5G (1G=10 -4 T) is maintained inside a chamber.

A normal-to-the-field moving electron enters the field at a speed of 4.8 10 6 ms -1 . Explain why the electron’s route is a circle. Calculate the orbit’s circumference. (e=1.6 10 -19 C, m e = 9.1 10 -31 kg) 

Ans. It is given that that magnetic field strength is B = 6.5 G = 6.510-4 T

The electron’s speed, v = 4.8106 m/s

The charge present on the electron, e = 1.610-19C

The electron’s mass, me= 9.110-31 kg

The angle that the magnetic field and the shot electron make, = 90°

The magnetic force that is exerted in the magnetic field on the electron can be represented in the following way.

The electron in motion is given centripetal force by this force. As a result, the electron begins to travel in a circular path with a radius r.

Hence the centripetal force that is exerted on the electron is

In this equilibrium, the magnetic force and the centripetal force that is exerted on the electron will be equal

That is, Fc = F

mv2r = evBsin

r = mvBesin

r = 9.110-314.81066.510-41.610-19sin90°

r = 4.210-2 m

Hence, the circular orbit’s radius of the electron will be 4.2 cm.

Q.26. Imagine a cylindrical region having a radius of 10.0 cm containing a magnetic field of 1.5 T, which runs parallel to the axis running east to west. This area is traversed by a wire carrying 7.0A of current in the north-to-south direction. What size and direction is the force acting on the wire in the event that,

• the axis is intersected by the wire?
• the wire changes its direction from north to south to northeast to northwest direction?
• the wire, which runs from the north to south direction, is lowered 6.0 cm from the axis?

Ans. (a) The information given is that,

The strength of the magnetic field, B = 1.5T

The radius of the given cylindrical region, r = 10 cm = 0.1 m

The current present in the wire that is travelling through the cylindrical region, I = 7A

If the axis is intersected by the wire, then the cylindrical region’s diameter will be the same as the wire’s length. Hence, I = 2r = 0.2m

The angle formed by the current and the magnetic field = 90°

We already know that the magnetic force that is acting on the wire can be represented by the following equation.

F = 1.570.2sin90°

Hence, the force that is acting on the wire that is in a vertically downward direction will be 2.1 N.

(b) The new length of the wire after it is turned to the northeast to northwest direction can be represented as

The angle that exists between the current and the magnetic field, = 45°

The force remaining on the wire,

F = BIII sin = BII = 1.570.2

Hence, the force amounting to 2.1 N will act on the wire in a vertically downward manner. This will be independent of the angle because 1sin is fixed.

(c) d = 6.0 cm is the distance to which the wire is lowered from the axis.

Imagine the new wire’s length to be l2

l222= 4(d+r) = 4(10+6)=416

l2= 82 = 16 cm = 0.16 m

The magnetic force exerted on the wire will be

F2 = Bll2 = 1.570.16

Hence, a force that will act on the wire in a vertically downward manner will be 1.68N.

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Q1. Read the assertion and reason carefully to mark the correct option out of the options given below. Assertion: The kinetic energy of a moving proton placed in a uniform perpendicular magnetic field remains constant. Reason: The moving charge placed in a uniform perpendicular magnetic field does not experience force.

Assertion is true but reason is false.

Assertion and reason both are false.

Both assertion and reason are true and the reason is the correct explanation of the assertion.

Both assertion and reason are true but reason is not the correct explanation of the assertion.

Q2. A charged particle of charge ‘q’ moving with velocity ‘v’ enters along the axis of a current carrying solenoid . The magnetic force on the particle is

finite but different from qvB

Q3. In an ammeter, 0.5% of main current pass through galvanometer. If resistance of galvanometer is G, what will be the resistance of ammeter?

Q4. What is the force between two straight conductors if they carry current in opposite direction?

The force will be repulsive in nature and will be given by

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Faqs (frequently asked questions), 1. describe the fundamental principle behind a moving coil galvanometer..

A current-carrying coil experiences a torque in the presence of a magnetic field, which causes a corresponding deflection, or deflection () ∝ (torque).

2. A coil of area A that is carrying a constant current I is connected with a magnetic moment called m. In vector form, express the link between m, I, and A.

The relation can be expressed in the following way = m = ∣A

3. Give one distinction between ferromagnetic and diamagnetic materials. Give one illustration of each.

The substances that a magnet only faintly repels are the diamagnetic materials. For instance, gold. The materials that a magnet is most drawn to are ferromagnetic materials. For instance, iron.

4. Why should a moving coil galvanometer's spring/suspension wire have a low torsional constant?

To maximise the current/charge sensitivity in a moving coil ballistic galvanometer, a low torsional constant is necessary.

5. Why don't the lines of the electrostatic field form closed loops?

Since an electric field flows from a positive to a negative charge, it is impossible for field lines to form closed loops. Therefore, a line of force that begins on a positive charge and ends on a negative charge can be considered. This demonstrates that closed loops are not formed by electric field lines.

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