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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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1569+ Class 12 Physics Case Study Questions PDF Download

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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.

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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

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What are Physics case study questions?

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

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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 

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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

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

Ncert solutions for class 12 physics chapter 4 – free pdf download.

The NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism is given here for the benefit of CBSE Class 12 Science students. In Chapter 4, NCERT Solutions for Class 12 Physics , students will learn the concepts of Moving Charges and Magnetism. Also, this chapter covered several topics with significant marks allotment, like magnetic field and direction of a circular coil, magnetic force, etc.

The concepts present in this chapter of NCERT Solutions for Class 12 Physics are not only important for the CBSE Board exam but also are crucial for competitive exams like JEE and NEET. The solutions are created by subject experts following the latest CBSE Syllabus 2023-24 and its guidelines. Students can now download the PDF of the NCERT Solutions for Class 12 Physics Chapter 4 from the link given below.

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Class 12 Physics NCERT Solutions Moving Charges and Magnetism Important Questions

Q 4.1) 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?

Answer 4.1:

The number of turns on the coil (n) is 100

The radius of each turn (r) is  8 cm or 0.08 m

The magnitude of the current flowing in the coil (I) is 0.4 A

The magnitude of the magnetic field at the centre of the coil can be obtained by the following relation:

The magnitude of the magnetic field is \(\begin{array}{l}3.14 \times 10^{-4}\; T\end{array} \) .

Q 4.2) 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?

Answer 4.2:

The magnitude of the current flowing in the wire (I) is 35 A

The distance of the point from the wire (r) is 20 cm or 0.2 m

At this point, the magnitude of the magnetic field is given by the relation:

Substituting the values in the equation, we get

Hence, the magnitude of the magnetic field at a point 20 cm from the wire is \(\begin{array}{l}3.5 \times 10^{-5}\; T\end{array} \) .

Q 4.3) A long straight wire in the horizontal plane carries a current of 50 A in the north-to-south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.

Answer 4.3:

The magnitude of the current flowing in the wire is (I) = 50 A.

Point B is 2.5 m away from the east of the wire.

Therefore, the magnitude of the distance of the point from the wire (r) is 2.5 m

The magnitude of the magnetic field at that point is given by the relation:

The point is located normal to the wire length at a distance of 2.5 m. The direction of the current in the wire is vertically downward. Hence, according to Maxwell’s right-hand thumb rule, the direction of the magnetic field at the given point is vertically upward.

Q 4.4) A horizontal overhead power line carries a current of 90 A in the east-to-west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

Answer 4.4:

The magnitude of the current in the power line is (I) = 90 A

The point is located below the electrical cable at a distance (r) = 1.5 m

Hence, the magnetic field at that point can be calculated as follows,

Substituting values in the above equation, we get

The current flows from east to west. The point is below the electrical cable.

Hence, according to Maxwell’s right-hand thumb rule, the direction of the magnetic field is towards the south.

Q 4.5) 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?

Answer 4.5:

In the problem,

The current flowing in the wire is (I) = 8 A

The magnitude of the uniform magnetic field (B) is 0.15 T

The angle between the wire and the magnetic field, \(\begin{array}{l}\theta = 30°\end{array} \) .

Hence, the magnetic force per unit length on the wire is \(\begin{array}{l}0.6\; N\; m^{-1}\end{array} \) .

Q 4.6) 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?

Answer 4.6:

The length of the wire (l) is 3 cm or 0.03 m

The magnitude of the current flowing in the wire (I) is 10 A

The strength of the magnetic field (B) is 0.27 T

The angle between the current and the magnetic field is \(\begin{array}{l}\theta = 90°\end{array} \) .

The magnetic force exerted on the wire is calculated as follows:

Substituting the values in the above equation, we get

Hence, the magnetic force on the wire is \(\begin{array}{l}8.1\times 10^{-2}\; N\end{array} \) . The direction of the force can be obtained from Fleming’s left-hand rule.

Q 4.7) 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.

Answer 4.7:

The magnitude of the current flowing in the wire A ( \(\begin{array}{l}I_{A}\end{array} \) ) is 8 A

The magnitude of the current flowing in wire B ( \(\begin{array}{l}I_{B}\end{array} \) ) is 5 A

The distance between the two wires (r) is 4 cm or 0.04 m

The length of the section of wire A (L) = 10 cm = 0.1 m

The force exerted on the length L due to the magnetic field is calculated as follows:

Substituting the values, we get

The magnitude of the force is \(\begin{array}{l}2\times 10^{-5} \; N\end{array} \) . This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.

Q 4.8) 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.

Answer 4.8:

Solenoid length (l) = 80 cm = 0.8 m

Five layers of windings of 400 turns each on the solenoid.

Solenoid Diameter (D) = 1.8 cm = 0.018 m

The current carried by the solenoid (I) = 8.0 A

The relation that gives the magnitude of the magnetic field inside the solenoid near its centre is given below:

Hence, the magnitude of B inside the solenoid near its centre is \(\begin{array}{l}2.5\times 10^{-2}\; T\end{array} \) .

Q 4.9) 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?

Answer 4.9:

In the given problem,

The length of a side of the square coil (l) is 10 cm or 0.1 m

The magnitude of the current flowing in the coil (I) is 12 A

The number of turns on the coil (n) is 20

The strength of the magnetic field (B) is 0.8 T

The following relation gives the magnitude of the magnetic torque experienced by the coil in the magnetic field:

A = Area of the square coil

= 1 x 1 = 0.1 x 0.1 = 0.01 m 2

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

Q 4.10) Two moving coil meters, M 1 and M 2, have the following particulars:

\(\begin{array}{l}R_{1} = 10\; \Omega ,\; N_{1} = 30\end{array} \) ,

\(\begin{array}{l}A_{1} = 3.6\times 10^{-3}\; m^{2},\; B_{1} = 0.25\; T\end{array} \)

\(\begin{array}{l}R_{2} = 14\; \Omega ,\; N_{2} = 42\end{array} \) ,

\(\begin{array}{l}A_{2} = 1.8\times 10^{-3}\; m^{2},\; B_{2} = 0.5\; T\end{array} \)

(The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M 2 and M 1 . Answer 4.10: Given data:

(a) Current sensitivity of M 1 ­ is given as:

And, Current sensitivity of M 2 ­ is given as:

Hence, the ratio of the current sensitivity of M 2 to M 1 is 1.4.

(b) Voltage sensitivity for M 2 is given as:

And voltage sensitivity for M 1 is given as:

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

Q 4.11) 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 x 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. ( \(\begin{array}{l}e=1.6\times 10^{-19} \;C,\; m_{e}=9.1\times 10^{-31} \;kg\end{array} \) )

Answer 4.11:

The relation for Magnetic force exerted on the electron in the magnetic field is given as:

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

Hence, the centripetal force exerted on the electron,

In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force, i.e.,

Hence, 4.2 cm is the radius of the circular orbit of the electron.

Q 4.12) In Exercise 4.11, find the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

Answer 4.12:

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

Frequency of revolution of the electron = v

In the circular orbit, the magnetic force on the electron is balanced by the centripetal force.

Hence, we can write:

This expression for frequency is independent of the speed of the electron. On substituting the known values in this expression, we get the frequency as:

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

Q 4.13) (a) A circular coil having a radius of 8.0 cm, the number of turns as 30 and 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 60 0 with the normal of the coil. To prevent the coil from turning, determine the magnitude of the counter-torque that must be applied.

(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.)

Answer 4.13:

(a) Number of turns on the circular coil (n) = 30

Radius of the coil (r) = 8.0 cm = 0.08 m

Current flowing in the coil (I) = 6.0 A

Magnetic field strength, B = 1 T

The coil experiences a torque in the magnetic field. Hence, it turns. The counter torque applied to prevent the coil from turning is given by the  relation,

= 3.133 N m

(b) It can be inferred from the relation \(\begin{array}{l}\tau = nIBA\; sin\theta\end{array} \) that the magnitude of the applied torque is not dependent on the shape of the coil. It depends on the area of the coil. Hence, 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.

Q 4.14) 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.

Radius of the coil X, r 1 = 16 cm = 0.16 m

Number of turns in coil X, n 1 = 20

Current in the coil X, I 1 = 16 A

Radius of the coil Y, r 2 = 10 cm = 0.1 m

Number of turns in coil Y, n 2 = 25

Current in the coil Y, I 2 = 18 A

The magnetic field due to the coil X at the centre is given as

Here, μ 0 is the permeability of the free space = 4π x 10 -7 Tm/A

= 4π x 10 -4 T (Towards east)

The magnetic field due to the coil Y at the centre is given as

= 9π x 10 -4 T (Towards west)

Therefore, the net magnetic field is given as

B = B 2 – B 1 = 9π x 10 -4 T – 4π x 10 -4 T

= 5π x 10 -4 T

= 5 x 3.14 x  10 -4 

= 1.57 x 10 -3 T (Towards west)

Q 4.15) A magnetic field of 100 G (1 G = 10 –4 T) is required, which is uniform in a region of linear dimension of about 10 cm and an area of cross-section of about 10 –3 m 2 . 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 around 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.

Magnetic field strength, B = 100 G = 100 x 10 -4 T

Number of turns per unit length, N = 1000 turns/m

Current carrying capacity of the coil = 15 A

Permeability of free space, μ 0 = 4π x 10 -7 TmA -1

The magnetic field is given as

B = μ 0  NI/ l

⇒ NI/ l = B/μ 0 

= (100 x 10 -4 )/(4π x 10 -7 )

NI/ l = 7961

Now, we can consider a possible combination. Let the current, I = 10 A and the length of the solenoid, l = 0.5 m

(N x 10)/0.5 =  7961

N = 398 turns ≈ 400 turns

Length about 50 cm, number of turns about 400, current about 10 A. These particulars are not unique. Some adjustment with limits is possible.

Q 4.16) 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,

\(\begin{array}{l}B=\frac{\mu _{0}IR^{2}N}{2(x^{2}+R^{2})^{3/2}}\end{array} \)

(a) Show that this reduces to the familiar result for the field at the centre of the coil. (b) Consider two parallel co-axial circular coils of equal radius R and 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,

\(\begin{array}{l}B= 0.72\frac{\mu _{0}NI}{R}\end{array} \) , approximately. [Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]

At the centre of the coil, x = 0

(b) Let the small distance between the points P and O be d.

For the coil C, the distance O 1 P = x 1 = (R/2) +d

For the coil d, the distance O 2 P = x 2 = (R/2) – d

The magnetic field on the axis  at a distance x from the centre of the circular coil having a radius a, with n number of turns and having a current I is given by

The magnetic field at P due to coil C

d 2 can be neglected when compared to R 2

It acts along PO 2

The magnetic field at P is given by

The total Magnetic field at the point P due to the current through the coils B = B 1 + B 2

Q 4.17 ) 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, (b) inside the core of the toroid, and (c) in the empty space surrounded by the toroid.

The inner radius of the core, r 1 = 25 cm = 0.25 m

The outer radius of the core, r 2 = 26 cm = 0.26 m

Number of turns of the wire, N= 3500 turns

Current in the wire, I = 11 A

(a) The magnetic field outside the toroid is zero.

(b) Inside the core of the toroid, the magnetic field induction is

B = μ 0 NI/ l

Mean length of the toroid,

= π (r 1 + r 2 )  =  π (0.25 + 0.26) = π x 0.51

So, B = μ 0 NI/ l 

(c) The magnetic field in the empty space surrounded by the toroid is zero.

Q 4.18) 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? (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? (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.

(a) Initial velocity is either parallel or antiparallel to the magnetic field. There is no magnetic force acting on the particle when it is parallel or antiparallel, and it moves undeflected.

(b) Yes, because magnetic force can change the direction of velocity but not its magnitude.

(c) Magnetic field should be in a vertically downward direction.

Q 4.19) An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV enters a region with a uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30º with the initial velocity.

Magnetic field, B = 0.15 T

Potential difference, V = 2.0 kV

An electron gains kinetic energy, which is given by

E = (1/2) mv 2

eV = (1/2) mv 2

(a) When the field applied is transverse to the initial velocity, the force F 1 acting on the electron is perpendicular to the direction of the magnetic field and the direction of motion of the electron.  The perpendicular force provides the centripetal force F 2 to the electron and makes it move in a circular path.

The forces F 1 and F 2 are equal

F 2 = mv 2 /2

evB = mv 2 /2

⇒ r = mv/eB

= 10 -3 m = 1 mm

The electron will move in a circular path with a radius of 1 mm in the direction perpendicular to the magnetic field.

(b) The applied force is inclined to the initial velocity. The velocity of the electron is resolved into two components, V cos θ and V sin θ. V cos θ is along the direction of the field, and it causes the electron to move in a straight line. V sin θ is along the normal, and it causes the electron to move in a circular path. Therefore, the electron moves in a helical path. The radius of the helical path is given as,

r = mV sin θ/Be

= 50.25 x 10 -5 m

The helical trajectory of radius 0.5 mm along the direction of the magnetic field.

Q 4.20) A magnetic field set up using Helmholtz coils (described in Exercise  4.16) is uniform in a small region and has a magnitude of 0.75 T. 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 15 kV, 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 × 10 –5 V m –1 , make a simple guess  as to what the beam contains. Why is the answer not unique?

Magnetic field, B = 0.75 T

Accelerating voltage, V = 15 kV = 15 x 10 3 V

Electrostatic field, E = 9.0 × 10 –5 V m –1

Kinetic energy of the electron, E = (1/2) mv 2

Therefore, (e/m) = (v 2 /2V)

Mass of the electron = m

Charge of the electron = e

The velocity of the electron = v

The particles are not deflected by the electric field and the magnetic field. So the force on the particle due to the electric field is balanced by force due to the magnetic field.

Therefore, (1/2) m(E/B) 2   = eV

e/m = E 2 /2VB 2

The beam contains Deuterium ions or deuterons

The answer is not unique because only the ratio of charge to mass is determined. Other possible answers are He++, Li++, etc.

Q 4.21) A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A 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? (b) What will be the total tension in the wires if the direction of the current is reversed, keeping the magnetic field the same as before? (Ignore the mass of the wires.) g = 9.8 m s –2 .

Length of the rod, l = 0.45 m

Mass suspended, m = 60 g = 60 x 10 -3 Kg

Current, I = 5 A

(a)  Tension in the wire is zero if the force on the current-carrying wire due to the magnetic field is equal and opposite to the weight on the wire.

The magnetic field set up normal to the conductor is 0. 26133 T

(b) When the direction of the current is reversed, BI l and mg will act vertically downwards; the effective tension in the wire is

T = BI l +mg

= (0. 26133 x 5.0 x 0.45) + (60 x 10 -3 x 9.8) = 0.587 + (0.588) = 1.176 N

Total tension in the wire =  1.176 N

Q 4.22) The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?

Current in the wires, I = 300 A

Distance between the wires, d = 1.5 cm = 0.015 m

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

The force between the two wires, F = μ 0 I 2 /2πd

μ 0  = permeability of the free space = 4π x 10 -7 Tm/A

Since the direction of the wires is opposite, the force will be repulsive.

Q 4.23)  A uniform magnetic field of 1.5 T exists in a cylindrical region of a radius of 10.0 cm, its direction parallel to the axis along east to west. A wire carrying a current of 7.0 A 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, (b) the wire is turned from N-S to the northeast-northwest direction (c) the wire in the N-S direction is lowered from the axis by a distance  of 6.0 cm?

(a) Magnetic field, B = 1.5 T

Current in the wire, I = 7.0 A

Radius, r = 10 cm = 0.1 m

Diameter, l = 2 x r = 0.2 m

A force 2.1 N acts vertically downward on the wire

(a) The wire intersects the axis, θ = 90 0

F = BI l sinθ = 1.5 x 7 x 0.20 x sin 90 0

(b) The wire is turned from N-S to the northeast-northwest direction

F = BI l 1 Sin θ

The length of the wire after turning to the northeast-northwest direction is l 1 = l/sinθ

Therefore, l = l 1 sin θ

= 1.5 x 7 x 0.20= 2.1 N

(c) When the wire is lowered by 6 cm, then

2x = l 2 = 16 cm

F 2 = BI l 2 = 1.5 x 7 x 0.16 = 1. 68 N

The force is directed vertically downwards.

Q 4.24) A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in the figure? What is the force in each case? Which case corresponds to stable equilibrium?

Magnetic field strength, B = 3000 G = 0.3 T

Area of the loop, A = 10 x 5 = 50 cm 2  = 50 x 10 -4 m 2

Current in the loop, I = 12 A

A is normal to the y-z plane, and B is directed along the z-axis

The torque is 1.8 x 10 -2 N m along the negative y-direction. The net force on the loop is zero since the opposite forces acting on the two ends of the loop will cancel each other.

(b) This is the same as (a). Therefore, the torque is 1.8 x 10 -2 N m along the negative y-direction. The net force is zero.

(c) A is normal to the x-z plane, and B is along the z-axis

Therefore, the torque is 1.8 x 10 -2 N m along the negative x-direction. The net force is zero.

From the given figure, it can be observed that A is normal to the coil. Since the coil makes an angle of -30° with the y-axis, A will make an angle of 30 0  with the positive x-axis in the negative y-direction, and B will be directed along the z-axis. The angle between A and B is θ = 90 0 . Therefore,

Accordingly, the magnitude of torque will be,

τ = 12×50 × 10 -4 × 0.3

= 1.8×10 -2  N m

The direction of the torque is (90° + 30°) from the negative x-axis or 360°−120° = 240° from the positive x-axis. The net force on the loop is zero.

(e)  From the given figure, we can see that A is normal to the x-y plane in the positive z-direction and B is directed along the z-axis.

Accordingly,

Hence, the torque is zero. The force is also zero.

(f) From the given figure, we can see that A is normal to the x-y plane in the negative z-direction and B is directed along the z-axis. The angle between A and B is θ = 180°. Therefore,

Therefore, torque and force is zero.

Case (e) corresponds to stable, and case (f ) corresponds to unstable equilibrium.

Q 4.25) A circular coil of 20 turns and a radius of 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  (a) the total torque on the coil (b) the total force on the coil (c) the average force on each electron in the coil due to the magnetic field?  (The coil is made of copper wire of a cross-sectional area of 10 –5 m 2 , and the free electron density in copper is about 10 29 m –3 .)

Number of turns, n = 20 turns

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

Current in the coil, I = 5 A

Magnetic field strength, B = 0.10 T

The cross-sectional area of the wire, a = 10 -5 m 2

The angle between  \(\begin{array}{l}\vec{A}\end{array} \) and  \(\begin{array}{l}\vec{B}\end{array} \) is θ = 0

(a) Total torque on the coil

⇒ τ = NIAB sin 0 = 0

Therefore, the torque is zero.

(b) Equal and opposite forces act on opposite sides of the coil. Therefore, the total force on the coil is zero.

(c) Average force on the moving electron

F = B e v d

v d is the drift velocity of the electrons

v d = I/N e A

Therefore, F = B e I/ N e a

F = B I/n a

The average force on each electron is 5 x 10 -25 N

Q. 4.26) A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g 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 6.0 A in the wire. What value of current (with an appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire? g = 9.8 m s –2 ?

Length of the solenoid, l = 60 cm

Layers of windings = 3

Each layer has 300 turns

Number of turns per unit length, n = (3 x 300)/0.6 = 1500 m -1

The magnetic field inside the solenoid is given by, B = μ 0 nI

μ 0   = 4π x 10 -7 T

B =  (4π x 10 -7 ) x 1500 x I

= 6π x 10 -4  I———-(1)

Force due to magnetic field, F = I’B l

I’ = Current in the wire = 6 A

l = length of the wire = 2 cm

The windings of the solenoid would support the weight of the wire when the force due to the magnetic field inside the solenoid balances the weight of the wire.

I’B l = mg

m = mass of the wire = 2.5 g

B = mg/I’ l

From equation (1), we get

6π x 10 -4   I= mg/I’ l

⇒ I = mg/(6π x 10 -4   ) I’ l

A current of 108.37 A will support the wire.

Q 4.27) A galvanometer coil has a resistance of 12 Ω, and the metre shows full-scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?

Resistance of the galvanometer coil, G = 12 Ω

Current for which there is full-scale deflection, I = 3mA

A resistor with a resistance R is connected in series with the galvanometer to convert it into a voltmeter. The resistance R is given as

R = (V/Ig) – G

A galvanometer can be converted into a voltmeter by connecting a resistor of 5988 Ω.

Q 4.28) A galvanometer coil has a resistance of 15 Ω, and the metre shows full-scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?

Resistance of the galvanometer, G = 15 Ω

Current through the galvanometer, I g = 4 mA = 4 x 10 -3 A

Ammeter range = 0 to 6 A

= 10 x 10 -3 A

A 10 mA resistor must be connected in parallel to the galvanometer.

NCERT Class 12 Solutions PDF for Moving Charges and Magnetism provided here has answers to textbook questions and Class 12 Physics Chapter 4 important questions from previous year question papers, exercises and worksheets. These solutions will help you prepare Chapter 4 Physics Class 12 notes, which can be kept handy when the exam nears.

Class 12 Physics NCERT Solutions for Moving Charges and Magnetism

Chapter 4 Moving Charges and Magnetism of Class 12 Physics, is prepared as per the latest CBSE Syllabus 2023-24. The NCERT Solutions of Chapter 4 given here can help the students to clear all their doubts instantly. The NCERT Solutions of this chapter comprise diagrams, graphs, illustrations and day-to-day examples that make your study interesting and help you remember the concepts for a longer period.

The direct derivations of topics like forces between two current-carrying wires and ampere circuital law are frequently asked in exams. Numerical problems are also asked in this chapter. The working and figure of the cyclotron should be studied intensively.

Subtopics of NCERT Class 12 Physics Chapter 4 Moving Charges and Magnetism

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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

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Important Questions for CBSE Class 12 Physics Chapter 4 - Moving Charges and Magnetism 2024-25

• Class 12 Important Question

CBSE Class 12 Physics Chapter- 4 Important Questions - Free PDF Download

Important Questions of Chapter 4 Physics Class 12 PDF available on the Vedantu website will help students in their preparation for the board examinations and any entrance examinations. Among all the science subjects, Physics is always found to be difficult for most students. The only key to excelling in Physics is to study with dedication and with quality materials. Chapter 4 of Physics Class 12 deals with Moving Charges and Magnetism. The Class 12 Physics Chapter 4 important questions will be useful for students in their preparation.

Download CBSE Class 12 Physics Important Questions 2024-25 PDF

Also, check CBSE Class 12 Physics Important Questions for other chapters:

Important Topics Covered in Chapter 4

The following is a list of the important topics covered under the chapter on Moving Charges and Magnetism:

Ampere's Circuital Law

Biot-Savart Law, Magnetic Field Due to a Current Element 

The Moving Coil Galvanometer

The Toroid and the Solenoid

Magnetic Field and Magnetic Force

Magnetic Field and Motion in Combined Electric 

Magnetic Dipole, Torque on Current Loop

Related Chapters

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

Very Short Answer Questions  1 Mark

1. State two properties of the material of the wire used for suspension of the coil in a moving coil galvanometer.  

Ans: Two properties of the material of the wire used for suspension of the coil in a moving coil galvanometer are:

Non-Brittle conductor 

Restoring Torque per unit twist should be small. 

2. What will be the path of a charged particle moving along the direction of a uniform magnetic field? 

Ans: The path of a charged particle moving along the direction of a uniform magnetic field would be a straight line path as no force would act on the particle. 

3. Two wires of equal lengths are bent in the form of two loops. One of the loops is square shaped whereas the other loop is circular. These are suspended in a uniform magnetic field and the same current is passed through them. Which loop will experience greater torque? Give reasons.  

Ans : We know the expression for torque as, 

$\tau =\text{NIAB}$

\[\Rightarrow \tau \propto A\]

Since, we know that the area of circular loops is more than that of a square loop, torque which is directly proportional to area would experience greater torque than the square loop.  

Therefore, torque experienced by a circular loop is greater.

4. A cyclotron is not suitable to accelerate electron. Why? 

Ans: A cyclotron is not suitable to accelerate electron as its mass is known to be less due to which they gain speed and step out of the dee immediately.

Short Answer Questions 2 Marks

1. A steady current flows in the network shown in the figure. What will be the magnetic field at the center of the network?

Ans:   The magnetic field at the center of the network is zero. This is because, the magnetic field at the center of the loop would just be equal and opposite i.e., magnetic field due \[\text{PQR}\] is equal and opposite to that due to \[\text{PSR}\].

2. An $\alpha $ - particle and a proton are moving in the plane of paper in a region where there is uniform magnetic field B directed normal to the plane of paper. If two particles have equal linear momenta, what will be the ratio of the radii of their trajectories in the field? 

Ans : We know the radius of the path to be given by, 

 \[R=\frac{mv}{Bq}\]

\[\Rightarrow R\propto \frac{1}{q}\] 

\[\Rightarrow \frac{{{R}_{\alpha }}}{{{R}_{p}}}=\frac{{{q}_{p}}}{{{q}_{\alpha }}}=\frac{e}{2e}=\frac{1}{2}\] 

Where, ${{R}_{\alpha }}\text{ and }{{\text{R}}_{p}}$are radii of $\alpha $-particle and proton respectively and ${{q}_{\alpha }}\text{ and }{{\text{q}}_{p}}$ are their respective charges. 

$\therefore {{R}_{\alpha }}:{{R}_{p}}=1:2$ 

Therefore, we find the required ratio to be 1:2. 

3. Give one difference each between diamagnetic and ferromagnetic substances. Give one example of each. 

Ans: Diamagnetic substances are the ones that are weakly repelled by a magnet. For example, gold. Ferromagnetic materials are the ones that are strongly attracted by a magnet. For example, iron. 

4. Write the expression for the force acting on a charged particle of charge q moving with velocity is in the presence of magnetic field B. Show that in the presence of this force,

The K.E. of the particle does not change. 

Ans : We know the expression for magnetic force as, $F=q(\vec{v}\times \vec{B})$

Since direction of force is perpendicular to the plane containing $(\vec{v}\times \vec{B})$, 

$F=qvB\sin 90{}^\circ =qvB$

Here, we find the force and displacement to be perpendicular to each other. So, 

$W=FS\cos \theta $ 

$\Rightarrow W=FS\cos 90{}^\circ =0$ 

$\Rightarrow \text{KE}=0$ 

Therefore, we find the kinetic energy to be constant at the given condition. 

Its instantaneous power is zero. 

Ans:   We have the expression for instantaneous power given by,

$p=Fv\cos \theta $ 

When force and velocity are perpendicular to each other, 

$p=Fv\cos 90{}^\circ =0$

Therefore, we find the instantaneous power to be zero.

5. An electron of kinetic energy \[25\mathbf{KeV}\]moves perpendicular to the direction of a uniform magnetic field of \[0.2\mathbf{millitesla}\]. Calculate the time period of rotation of the electron in the magnetic field.  

Ans : We are given the magnetic field to be, $B=0.2T=0.2\times {{10}^{-3}}T$ 

We know the expression for Time Period to be, $T=\frac{2\pi M}{QB}$

Substituting the given values, 

$\Rightarrow T=\frac{2\times 3.14\times 9.1\times {{10}^{-31}}}{1.6\times {{10}^{-17}}\times 0.2\times {{10}^{-3}}}$ 

$\Rightarrow T=1.787\times {{10}^{-7}}\text{ second}$ 

We find the time period of rotation of the electron in the magnetic field to be $T=1.787\times {{10}^{-7}}\text{ second}$. 

6. It is desired to pass only \[10%\] of the current through a galvanometer of resistance $90\Omega $. How much shunt resistance should be connected across the galvanometer?   

Ans: Current through galvanometer, 

${{I}_{G}}=10%\text{ of I}=\frac{10}{100}\times I$  

Galvanometer resistance is given to be, $G=90\Omega $ 

Now, we could find the shunt resistance as, 

S $=\frac{\frac{9I}{10I-I}}{100}$ 

$\Rightarrow S=\frac{90I}{90I}=10$ 

$\Rightarrow S=10\Omega $ 

Therefore, we found the shunt resistance to be $S=10\Omega $. 

Short Answer Question 3 Marks

1. Derive an expression for the force acting on a current carrying conductor placed in a uniform magnetic field. Name the rule which gives the direction of the force. Write the condition for which this force will have (1) maximum (2) minimum value.  

Ans : Let us consider a conductor that is placed in a uniform magnetic field \[\vec{B}\]making an angle $\theta $ with \[\vec{B}\]. Let \[I\] be the current that flows through the conductor.

If $n$ is the number of electrons per unit volume of the conductor, then the total number of electrons in the small current element \[dl,\text{ N}=nAdl\].

We have, 

$\theta =Ne$ 

$\Rightarrow \theta =nAdle$ 

Let $\vec{f}$ be the force experienced by each electron, then, 

$\vec{f}=e({{\vec{v}}_{d}}\times \vec{B})$ 

Now, force experienced by a small current element would be, 

\[d\vec{f}=neAdl({{\vec{v}}_{d}}\times \vec{B})\] 

$d\vec{f}=neAdlB\sin \theta $ 

But we have, $I=neAvd$ 

$\Rightarrow d\vec{f}=IdlB\sin \theta $ 

Now, the total force experienced will be,

\[F=\int_{0}^{1}{df}=\int_{0}^{1}{IdlB\sin \theta }\] 

$\Rightarrow F=IBl\sin \theta $ 

In vector form total force could be given by, $\vec{F}=I(\vec{l}\times \vec{B})$ 

Force will be maximum when $\theta =90{}^\circ $ 

Force will be minimum when $\theta =0{}^\circ $

2. A straight wire carries a current of \[\mathbf{10}\]A. An electron moving at ${{10}^{7}}\text{ m/s}$is at distance \[\mathbf{2}.\mathbf{0}\] cm from the wire. Find the force acting on the electron if its velocity is directed towards the wire.

Ans : We are given the current through the straight wire to be, \[I\text{ }=\text{ }10A\] 

Speed of the electron, $v={{10}^{7}}\text{ m/s}$ 

Distance of electron from the wire, $R=2.0\text{ cm}=2\times {{10}^{-2}}\text{ m}$ 

Force acting on a moving electron would be,  \[F=qVB\sin \theta \] 

We have the expression for magnetic field as, 

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

$B=\frac{{{10}^{-7}}\times 2\times 10}{2\times {{10}^{-2}}}={{10}^{-4}}T$ and it is given to be $\bot $ to the plane of paper and directed towards.

Now, force acting on the electron could be given by, $\Rightarrow F=1.6\times {{10}^{-19}}\times {{10}^{7}}\times {{10}^{-4}}\sin 90{}^\circ $ 

$\Rightarrow F=1.6\times {{10}^{-16}}N$ 

Therefore, we find the force to be, $F=1.6\times {{10}^{-16}}N$. 

3. State Biot-Savart's law. Derive an expression for the magnetic field at the center of a circular coil of \[\mathbf{n}\] -turns carrying current – I. 

Ans: Biot – Savart law states that the magnetic field $dB$ due to a current element $\overrightarrow{dl}$ at any point would be as following:

$dB\propto I$ 

$dB\propto dl$ 

$dB\propto \sin \theta $ 

$dB\propto \frac{1}{{{r}^{2}}}$ 

Combining all the above conditions, we get, 

$dB\propto \frac{Idl\sin \theta }{{{r}^{2}}}$ 

\[\Rightarrow dB=\frac{{{\mu }_{0}}}{4\pi }\frac{Idl\sin \theta }{{{r}^{2}}}\]

Consider a circular loop of radius $r$ that is carrying a current $I$, 

Since $dl\bot \vec{r}$ , 

$\Rightarrow \theta =90{}^\circ $ 

Now, on applying Biot Savart law 

$dB=\frac{{{\mu }_{0}}}{4\pi }\frac{Idl\sin 90{}^\circ }{{{r}^{2}}}$ 

For entire closed circular loop, 

$B=\int_{0}^{2\pi r}{\frac{{{\mu }_{0}}}{4\pi }\frac{Idl\sin 90{}^\circ }{{{r}^{2}}}}$ 

$\Rightarrow B=\frac{{{\mu }_{0}}}{4\pi }\frac{I}{{{r}^{2}}}\int_{0}^{2\pi r}{dl}=\frac{{{\mu }_{0}}}{4\pi }\frac{I}{{{r}^{2}}}\times 2\pi r$ 

For $n$ turns of a coil we would get, 

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

Therefore, we find the expression for magnetic field at the center of a circular coil of n -turns carrying current – I to be, 

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

4. What is a radial magnetic field? How is it obtained in a moving coil galvanometer?

Ans : A radial magnetic field is the magnetic field in which the plane of the coil always lies in the direction of the magnetic field. It can be obtained by the following ways:

Properly cutting the pole pieces concave in shape. 

Placing soft iron cylindrical core between the pole pieces.

5. Two straight parallel current carrying conductors are kept at a distance r from each other in air. The direction for current in both the conductors is the same. Find the magnitude and direction of the force between them. Hence define one ampere. 

Ans: Let us consider two parallel conductors carrying current ${{I}_{1}}\,\text{ and }{{I}_{2}}$ and is separated by a distance \[d\], 

Magnetic field due to current ${{I}_{1}}$ at any point on conductor 2 could be given by, 

\[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2Il}{d}\,\] ……………………….. (1)

($\bot $ to the plane and downwards $(\times )$)

Since current carrying conductor is placed at right angles with the magnetic field, we get the magnetic force to be, 

$F=BIl\sin 90{}^\circ $ 

$\Rightarrow F=BIl$ ………………………… (2) 

This would be the Force experienced per unit length of conductor. 

Now, we have, 

${{F}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2{{I}_{1}}{{I}_{2}}}{d}$…………………………… (3) 

Fleming’s left hand Rule says ${{F}_{2}}$ is directed towards conductor 1.

Similarly, ${{F}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2{{I}_{1}}{{I}_{2}}}{d}$ (Directed Towards conductor $2$)

Since ${{F}_{1}}\text{ and }{{F}_{2}}$ are equal in magnitude and directed opposite, two parallel current carrying conductors would attract each other.

Since, $F=\frac{{{\mu }_{0}}}{4\pi }\left( \frac{2{{I}_{1}}{{I}_{2}}}{d} \right)$ 

If ${{I}_{1}}={{I}_{2}}=1\text{ A}$  and $d=1\text{m}$ , then, 

$F=2\times {{10}^{-7}}\,\text{m}$ 

Hence, we found that one ampere is that current which is flowing in two infinitely long parallel conductors that are separated by a distance of $1$ meter in vacuum and experiences a force of $F=2\times {{10}^{-2}}\,\text{m}$ on each meter of the other wire.

6. A circular coil of wire consisting of $100$ turns, each of radius $8.0$ cm carries a current of $0.40\,\text{A}$. 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 coils, $n=100$ 

Radius of each turn, $r=8.0\text{ cm}=0.08\text{ m}$ 

Current flowing in the coil, $I=0.4\,\text{A}$ 

Magnitude of the magnetic field at the centre of the coil could be given by the relation,

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

Where, Permeability of free space, \[{{\mu }_{0}}=4\pi \times {{10}^{-7}}\,\text{T m }{{\text{A}}^{-1}}\] 

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

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

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

7. A long straight wire carries a current of $35\,\text{A}$. What is the magnitude of the field B at a point $20$ cm from the wire?

Ans: We are given the following:

Current in the wire, $I=35\text{ A}$ 

Distance of a point from the wire, $r=20\text{ cm}=0.2\,\text{m}$ 

Magnitude of the magnetic field at this point could be given as:

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

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

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

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

Hence, the magnitude of the magnetic field at a point $20\,\text{cm}$ from the wire is found to be $3.5\times {{10}^{-5}}\,\text{T}$.

8. A long straight wire in the horizontal plane carries a current of \[50\text{ A}\] in the north to south direction. Give the magnitude and direction of B at a point \[2.5\]m east of the wire. 

Current in the wire, \[I\text{ }=\text{ }50\text{ A}\] 

A point is said to be \[2.5\]m away from the East of the wire. 

Magnitude of the distance of the point from the wire is given as, \[r\text{ }=\text{ }2.5\text{ m}\]. 

Magnitude of the magnetic field at that point could be given by the relation, \[B=\frac{{{\mu }_{0}}2I}{4\pi r}\] 

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

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

Since, the point is located normal to the wire length at a distance of $2.5\text{ m}$and the direction of the current in the wire is vertically downward, by using Maxwell’s right hand thumb rule we get the direction of the magnetic field at the given point as vertically upward.

9. A horizontal overhead power line carries a current of \[\mathbf{90}\text{ }\mathbf{A}\]in an east to west direction. What is the magnitude and direction of the magnetic field due to the current \[\mathbf{1}.\mathbf{5}\text{ }\mathbf{m}\] below the line? 

Ans: We are given the following: 

Current in the power line, \[I\text{ }=\text{ }90\,\text{A}\] 

A Point is located below the power line that is at distance, \[r\text{ }=\text{ }1.5\text{ m}\] 

Now, the magnetic field at that point could be given by the relation,

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

$\Rightarrow B=\frac{4\pi \times {{10}^{-7}}\times 2\times 50}{4\pi \times 1.5}$ 

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

Since, the current is flowing from East to West and the point is given to be below the power line, by using Maxwell’s right hand thumb rule we get the direction of the magnetic field to be towards the South.

10. What is the magnitude of magnetic force per unit length on a wire carrying a current of \[8\text{ A}\] and making an angle of \[30{}^\text{o}\]  with the direction of a uniform magnetic field of \[0.15\text{ T}\] ? 

Current in the wire, \[I\text{ }=\text{ }8\text{ A}\] 

Magnitude of the uniform magnetic field, \[B\text{ }=\text{ }0.15\text{ T}\] 

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

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

$f=BI\sin \theta $ 

\[\Rightarrow f=0.15\times 8\times 1\times \sin 30{}^\circ \] 

\[\Rightarrow f=0.6\,\text{N }{{\text{m}}^{-1}}\] 

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

11. A \[\mathbf{3}.\mathbf{0}\] cm wire carrying a current of \[\mathbf{10}\]A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be \[\mathbf{0}.\mathbf{27}\] T. What is the magnetic force on the wire? 

Length of the wire, \[l\text{ }=\text{ }3\text{ cm }=\text{ }0.03\text{ m}\] 

Current flowing in the wire, \[I\text{ }=\text{ }10\text{ A}\] 

Magnetic field, \[B\text{ }=\text{ }0.27\text{ T}\] 

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

Magnetic force exerted on the wire could be given as:

$F=BIl\sin \theta $ 

Substituting the given values, we get, 

\[\Rightarrow F=0.27\times 10\times 0.03\sin 90{}^\circ \] 

\[\Rightarrow F=8.1\times {{10}^{-2}}\,\text{N}\] 

Therefore, the magnetic force on the wire is found to be \[8.1\times {{10}^{-2}}\,\text{N}\] and the direction of the force can be obtained using Fleming’s left-hand rule.

12. Two long and parallel straight wires A and B carrying currents of \[\mathbf{8}.\mathbf{0}\] A and \[\mathbf{5}.\mathbf{0}\] A in the same direction are separated by a distance of \[\mathbf{4}.\mathbf{0}\] cm. Estimate the force on a \[\mathbf{10}\] cm section of wire A. 

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

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

Distance between the two wires, \[r\text{ }=\text{ }4.0\text{ cm }=\text{ }0.04\text{ m}\] 

Length of a section of wire A, \[l\text{ }=\text{ }10\text{ cm }=\text{ }0.1\text{ m}\] 

Force exerted on length l due to the magnetic field could be given as:

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

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

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

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

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

13. 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, 

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

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

On substituting the given values we get, 

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

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

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

14. 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?

Ans: We are given,

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 $

Substituting the given values, we get,

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

$\Rightarrow \tau =0.96Nm$

Therefore, the magnitude of the torque experienced by the coil is 0.96Nm.

A circular coil of \[\mathbf{30}\] turns and radius \[\mathbf{8}.\mathbf{0}cm\] carrying a current of \[\mathbf{6}.\mathbf{0}A\] is suspended vertically in a uniform horizontal magnetic field of magnitude \[\mathbf{1}.\mathbf{0}T\]. The field lines make an angle of with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

Ans: Given that, number of turns on the circular coil, \[n=30\] 

Radius of the coil, \[r=8.0\text{ cm}=0.08\text{ m}\] 

Area of the coil $=\pi {{r}^{2}}=\pi {{(0.08)}^{2}}=0.0201\text{ }{{\text{m}}^{2}}$ 

Current flowing in the coil, \[I=6.0\text{ A}\] 

Magnetic field strength, \[B=1\text{ T}\] 

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

The coil experiences a torque in the magnetic field. So, it turns. The counter torque applied to prevent the coil from turning is given by the relation,

$T=nIBA\sin \theta \,$ ……………………. (1)

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

$\Rightarrow T=3.133\text{Nm}$ 

Therefore, counter torque to be applied against coil turning is $3.133\text{Nm}$.

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: It can be inferred from relation (1) that the magnitude of the applied torque is not dependent on the shape of the coil. It depends on the area of the coil. Hence, 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.

16. 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,

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

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

$\Rightarrow nI\approx 8000A/m$

So, 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.

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 

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, $I=11A$

So, we know, the magnetic field outside a toroid is zero. It is non-zero only inside the core of a toroid.

inside the core of the toroid.

Ans: Magnetic field inside the core of a toroid is given by the relation,

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

$l$ is the length of toroid

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

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

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

in the empty space surrounded by the toroid.

Ans: Magnetic field in the empty space that is surrounded by the toroid is zero. 

18. Answer the following questions: 

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 suffering any deflection in the field.

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 because the magnetic force can change direction of velocity, but not its magnitude.

An electron travelling west to east enters a chamber having a uniform electrostatic field in 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 remains undeflected if the electric force acting on it is equal and opposite to the magnetic field. Magnetic force would be directed towards the South. Also, according to Fleming’s left hand rule, the magnetic field should be applied in a vertically downward direction.

19. 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.

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.,

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

$\Rightarrow B=0.26T$

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

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. So, the effective tension in the wires is found to be, 

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

$\Rightarrow T=1.176N$

Therefore, total tension in the wires is $1.176N$. 

20. 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?

Ans: Given that,

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${{\mu }_{0}}=4\pi \times 10Tm{{A}^{-1}}$

$\Rightarrow F=\frac{4\pi \times {{10}^{-7}}\times {{300}^{2}}}{2\pi \times 0.015}$

$\Rightarrow F=1.2N$

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

21. 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${{10}^{29}}{{m}^{-3}}$).

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, $I=5.0A$

Since, the angle between force and the normal to the loop, the total torque on the coil is zero. So, $\tau =NIAB\sin \theta $ is zero.

total force on the coil,  

Ans: The total force on the coil is zero as the field is uniform.

average force on each electron in the coil due to the magnetic field?

Ans: 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.

${{v}_{d}}=\frac{I}{neA}=\frac{5}{{{10}^{29}}\times 1.6\times {{10}^{-19}}\times {{10}^{-5}}}=3.125\times {{10}^{-5}}m/s$

$\Rightarrow F=0.10\times 1.6\times {{10}^{-19}}\times 3.125\times {{10}^{-5}}$

$\Rightarrow F=5\times {{10}^{-25}}N$

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

22. 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 $18V$?

Ans: Given that, 

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:

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

Substituting the given values we get, 

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

$\Rightarrow R=5988\Omega $

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

23. 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:

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

Substituting the values, 

$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 $

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

Long Answer Questions                                                                                   5 Marks

What is a cyclotron? Explain its working principle.

Ans: Cyclotron is a device used to accelerate charged particles like protons, deuterons, $\alpha $ - particles, etc. 

It works on the basis of the principle that a charged particle can be accelerated to very high energies by making it pass through a moderate electric field a number of times and applying a strong magnetic field at the same time. 

A cyclotron’s oscillator frequency is \[10\,\text{MHz}\], what should be the operating magnetic field for accelerating protons? If radius is \[20\,\text{cm}\], what is the K.E. of the proton beam produced by the accelerator? \[\left( e=1.6\times {{10}^{-19}}\text{ c},mp=1.6\times {{10}^{-27}}\text{ kg},1\text{ MeV}=1.602\times {{10}^{-13}}\text{ J} \right)\]

Ans: Given that, \[v=10\,\text{MHz}=10\times {{10}^{6}}\text{ Hz}\] 

\[e=1.6\times {{10}^{-19}}\text{ c}\]

\[mp=1.6\times {{10}^{-27}}\text{ kg}\]

\[r=20\,\text{cm}=20\times {{10}^{-2}}\,\text{m}\] 

We have the expression for kinetic energy, 

KE $=\frac{{{q}^{2}}{{B}^{2}}{{r}^{2}}}{2m}$ 

Using $v=\frac{qB}{2\pi }$ 

$B=\frac{2\pi mV}{q}$ 

$\Rightarrow B=\frac{2\times 3.14\times 1.6\times {{10}^{-27}}\times {{10}^{7}}}{1.6\times {{10}^{-19}}}$ 

$\Rightarrow B=0.628\text{ T}$ 

Therefore, the operating magnetic field for accelerating protons is $0.628\text{ T}$.

KE $=\frac{{{\left( 1.6\times {{10}^{-19}} \right)}^{2}}\times {{\left( 0.66 \right)}^{2}}\times {{\left( 0.2 \right)}^{2}}}{2\times 1.67\times {{10}^{-27}}}$ 

KE $=13.35\times {{10}^{-13}}\text{ J}$ 

But we have, $1.602\times {{10}^{-13}}\text{ Joules}=1\text{ MeV}$ 

Since $12.02\times {{10}^{-13}}\text{ J}$ has $\frac{12.02\times {{10}^{-13}}}{1.602\times {{10}^{-13}}}\text{ MeV}$ 

$\Rightarrow KE=8.3\text{ MeV}$

Therefore, the K.E. of the proton beam produced by the accelerator is $8.3\text{ MeV}$.

Draw a labelled diagram of a moving coil galvanometer. Prove that in a radial magnetic field, the deflection of the coil is directly proportional to the current flowing in the coil. 

Ans: When a current $I$ is passed through a coil two equal and opposite forces act on the arms of a coil to form a couple which exerts a Torque on the coil. 

\[\tau =NIAB\sin \theta \] 

If $\theta =90{}^\circ \left( \sin 90{}^\circ =1 \right)$ 

$\theta $ is the angle made by the normal to the plane of coil with B 

$\tau =NIAB\text{ }$ ……………………………….. (1)

This is called as deflecting torque 

As the coil deflected the spring is twisted and a restoring torque per unit twist then the restoring torque for the deflecting & is given by 

${\tau }'=k\phi $ ……………………….. (2)

In equilibrium 

\[\text{Deflecting Torque}=\text{Restoring Torque}\] 

\[NIAB=K\phi \] 

\[I=\frac{K\phi }{NAB}\phi \] 

\[I=G\phi \] where \[G=\frac{K}{NAB}\] (galvanometer constant) 

$\Rightarrow I\propto \phi $ 

Therefore, deflection of the coil is directly proportional to the current flowing in the coil.

A galvanometer can be converted into a voltmeter to measure upto 

$V$ volt by connecting a resistance ${{R}_{1}}$ series with the coil 

\[\frac{V}{2}\] volt by connecting a resistance ${{R}_{2}}$ in series with a coil. Find \[R\] in terms of ${{R}_{1}}\text{ and }{{R}_{2}}$ required to convert – it into a voltmeter that can read up to \['2V'\] volt. 

Ans: We know that, ${{I}_{g}}=\frac{V}{R+{{R}_{G}}}$ 

\[\Rightarrow {{I}_{g}}\text{=}\frac{V}{{{R}_{1}}+{{R}_{G}}}\] …………………………….. (1)

And ${{I}_{g}}=\frac{\frac{V}{2}}{{{R}_{2}}+{{R}_{G}}}$  ……………………………. (2)

Equating \[\left( 1 \right)\text{ }\And \text{ }\left( 2 \right)\] 

$\frac{V}{{{R}_{1}}+{{R}_{G}}}=\frac{\frac{V}{2}}{{{R}_{2}}+{{R}_{G}}}$ 

That is, ${{R}_{1}}+{{R}_{G}}=2\left( {{R}_{2}}+{{R}_{G}} \right)$

${{R}_{G}}=-2{{R}_{2}}+{{R}_{1}}$ 

For conversion ${{I}_{g}}=\frac{2V}{R+{{R}_{G}}}$ 

$\Rightarrow {{I}_{g}}\frac{V}{{{R}_{1}}+{{R}_{G}}}=\frac{2V}{R+{{R}_{G}}}$ 

$\Rightarrow {{I}_{g}}=2{{R}_{1}}+2{{R}_{G}}=R+{{R}_{G}}$ 

$\Rightarrow R=2{{R}_{1}}+{{R}_{G}}$ 

$\Rightarrow R=2{{R}_{1}}+{{R}_{1}}-2{{R}_{2}}$ 

$\Rightarrow R=3{{R}_{1}}-2{{R}_{2}}$ 

Therefore, R in the case can be written as, $R=3{{R}_{1}}-2{{R}_{2}}$.

3. 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}}\text{ }{{\text{m}}^{2}},$ ${{B}_{1}}=0.25\text{ T}$

\[{{R}_{2}}=14\,\Omega ,\] \[{{N}_{2}}=42\],${{A}_{2}}=1.8\times {{10}^{-3}}\text{ }{{\text{m}}^{2}},$ ${{B}_{2}}=0.50\text{ T}$

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

Determine the ratio of (a) current sensitivity and (b) Voltage sensitivity of ${{M}_{1}}$ and  ${{M}_{2}}$.

a) current sensitivity of ${{M}_{1}}$ and  ${{M}_{2}}$:

Ans: From given data, moving coil meter ${{M}_{1}}$, 

Resistance, ${{R}_{1}}=10\,\Omega $ 

Number of turns, ${{N}_{1}}=30$ 

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

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

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, ${{A}_{2}}=1.8\times {{10}^{-3}}\text{ }{{\text{m}}^{2}}$

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

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

${{I}_{s2}}=\frac{{{N}_{2}}{{B}_{2}}{{A}_{2}}}{{{K}_{2}}}$ 

$\therefore $ Ratio $\frac{{{I}_{s2}}}{{{I}_{s1}}}=\frac{{{N}_{2}}{{B}_{2}}{{A}_{2}}{{K}_{1}}}{{{N}_{1}}{{B}_{1}}{{A}_{1}}{{K}_{2}}}$ $=\frac{42\times 0.5\times 1.8\times {{10}^{-3}}\times K}{K\times 30\times 0.25\times 3.6\times {{10}^{-3}}}=1.4$

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

(b) voltage sensitivity of ${{M}_{2}}\text{ and }{{M}_{1}}$ 

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

${{V}_{s2}}=\frac{{{N}_{2}}{{B}_{2}}{{A}_{2}}}{{{K}_{2}}{{R}_{2}}}$ 

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, 

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

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

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

4. 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: Given that, magnetic field strength, \[B=6.5\text{ G}=6.5\times {{10}^{-4}}\text{ T}\]

Speed of the electron, \[v=4.8\times {{10}^{6}}\text{ m/s}\]

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

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

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

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

\[F=evB\sin \theta \] 

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

Therefore, centripetal force exerted on the electron, 

${{F}_{c}}=\frac{m{{v}^{2}}}{r}$ 

In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force, 

That is, ${{F}_{c}}=F$ 

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

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

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

$\Rightarrow r=4.2\times {{10}^{-2}}\text{ m}$ 

\[\Rightarrow r=4.2\text{ cm}\] 

Therefore, the radius of the circular orbit of the electron is \[4.2\text{ cm}\].

5. 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. 

Ans: Given that, magnetic field strength, $B=6.5\times {{10}^{-4}}\text{ T}$ 

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

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

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

Radius of the orbit, \[r=4.2\text{ cm}=0.042\text{ m}\] 

Frequency of revolution of the electron $=\upsilon $ 

Angular frequency of the electron, $\omega =2n\upsilon $  

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

 $v=r\omega $ 

In the circular orbit, the magnetic force on the electron is balanced by the centripetal force. So, 

$evB=\frac{m{{v}^{2}}}{R}$ 

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

$\Rightarrow \upsilon =\frac{Be}{2\pi m}$ 

This expression for frequency is independent of the speed of the electron. 

On substituting, frequency, 

$\Rightarrow \upsilon =\frac{6.5\times {{10}^{-4}}\times 1.6\times {{10}^{-19}}}{2\times 3.14\times 9.1\times {{10}^{-31}}}$ 

$\Rightarrow \upsilon =18.2\times {{10}^{6}}\text{ Hz}$ 

$\Rightarrow \upsilon \approx 18\text{ MHz}$ 

Therefore, the frequency of the electron is around \[18\text{ MHz}\] and is independent of the speed of the electron.

6. 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.

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, ${{\mu }_{0}}=4\pi \times {{10}^{-4}}Tm{{A}^{-1}}$

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

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

So, 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)

Therefore, net magnetic field is $1.57\times {{10}^{-3}}T$ towards west.

7. 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,

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 $x=0$

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

Therefore, this is the familiar result for the magnetic field at the centre of the coil.

Consider two parallel coaxial circular coils of equal radius R, and 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 midpoint 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.)

Ans: Radii of two parallel coaxial 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:

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

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, 

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

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

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

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

8. An electron emitted by a heated cathode and accelerated through a potential difference of $2.0kV$, enters a region with a uniform magnetic field of $0.15T$. Determine the trajectory of the electron if the field 

is transverse to its initial velocity. 

Ans: We are given, 

Magnetic field strength, $B=0.15T$

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

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, 

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

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

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

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

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

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

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

$\Rightarrow r=1mm$

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

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 $

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

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

9. 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,

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

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

This value of the 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 $H{{e}^{++}}$, $L{{i}^{+++}}$

10. 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 a 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,

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,

$F=BIl\sin \theta $

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

$\Rightarrow F=2.1N$

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

The wire is turned from N-S to northeast-northwest direction,

Ans: 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,

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

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

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, 

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

$\Rightarrow F=1.68N$

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

11. 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?

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 a negative y-direction. Since the external magnetic field is uniform, the force on the loop would be zero. 

(Image will be uploaded soon)

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

Ans: Torque here would be, 

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

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

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, 

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

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

$\Rightarrow \tau =0$

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

(image will be uploaded soon)

Ans: Torque is given by, 

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

For this case, 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. 

12. A solenoid $60cm$ long and has a 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.

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

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

$\Rightarrow I=108A$

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

Download Important Questions Of Chapter 4 Physics Class 12 PDF

Class 12 physics chapter 4 important questions.

The Class 12 Physics Chapter 4 important questions PDF is highly beneficial for the students who are targeting good grades in competitive exams. The textbook prescribed by NCERT for Class 12 Physics consists of many topics. For elevating the grades and to improve their knowledge in the subject students are suggested to download chapter wise notes available on Vedantu for detailed explanation along with exercise problems. 

For thorough preparation, students should start with Chapter 4 Physics Class 12 important questions . The important questions of Chapter 4 Physics Class 12 are prepared with deep insight into the syllabus according to CBSE board exams , previous year’s questions. The Chapter 4 Physics Class 12 important questions included all the textbook exercises along with many Class 12 Physics Chapter 4 extra questions prepared by experts and the master teachers.

Chapter 4 Moving Charges and Magnetism has good marks allotment in the Board exams. The Moving charges and Magnetism include many interesting topics like magnetic forces, magnetic field, the direction of magnetic force in the current-carrying coil with different cases, etc. These concepts are not only important in the point of board exam but also for students who are aiming for IIT-JEE , NEET exams as well.

Students must start with Class 12 Physics Chapter 4 important questions after revising the chapter. Students must be thoroughly familiar with all the concepts of Chapter 4 for solving Class 12 Physics Chapter 4 extra questions. Let us have a look at some key points of Chapter 4:

The chapter consists of many direct derivations like ampere circuital law, the force between two parallel current-carrying wires, etc.

Chapter 4 mainly focused on explaining the idea that electric and magnetic fields are interrelated to one another. Because since the beginning it was assumed that electric fields and magnetic fields are independent of each other.

We come across the concepts like a magnetic force, Lorentz force which is a piece of evidence about electric and magnetic fields that are related to each other.

We study the effects of magnetic fields on current. We study a few devices that use the principle of magnetic induction.

We study cyclotrons, solenoids, moving coil galvanometers, etc.

These are a few key points to be remembered while jumping into the practice of Chapter 4 Physics Class 12 important questions .

Key Features of Vedantu’s CBSE Class 12 Important Questions for Physics Chapter 4 - Moving Charges and Magnetism

The following are the reasons why you should choose Vedantu’s Important Questions for Class 12 CBSE Physics Chapter 4 - Moving Charges and Magnetism:

Vedantu’s Important Questions for Class 12 CBSE Physics Chapter 4 - Moving Charges and Magnetism are aligned with the CBSE Board guidelines and answering pattern.

Our questions are prepared in a manner that will serve the purpose of revising the entire chapter to be able to solve any kind of question that is asked in the exam.

The questions adhere to the strict norms of the CBSE Class 12 Physics syllabus . The answers have also been framed in a straightforward manner using simple language. 

The set of important questions for CBSE Class 12 Physics Chapter 4 - Moving Charges and Magnetism for the academic year 2024-25 prove to be a valuable tool for students' exam preparation. These questions cover vital concepts such as magnetic fields, Ampere's law, and applications of moving charges in magnetic fields. By practicing these questions, students can strengthen their understanding of the topic and improve problem-solving abilities. The comprehensive nature of these questions ensures thorough coverage of the chapter, enabling students to excel in their examinations. With these resources at their disposal, students can confidently approach the subject and achieve academic success in the upcoming academic year.

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FAQs on Important Questions for CBSE Class 12 Physics Chapter 4 - Moving Charges and Magnetism 2024-25

1. What is Biot Savart’s law, according to Chapter 4 of Class 12 Physics?

This law is solely concerned with current sources and magnetic fields. The magnetic field is determined once the current involving vector products is spread. The distance between the field points and the current changes all the time, and this is the calculus problem. Aerodynamics and magnetic effects at the atomic and molecular levels are calculated. Visit Vedantu to learn more about this law.

2. What is Ampere's circuital law, according to Chapter 4 of Class 12 Physics?

Relation between the magnetic field and source current is discussed in ampere's circuital law. The density of a magnetic field in an imaginary path is studied in ampere's circuital law. For more information, visit Vedantu's official website. There you will find the important questions which will be helpful in your exams. These questions will cover all the important topics curated together from the exam's point of view. You can download these questions and save them on your PC for future reference free of cost from the Vedantu website and the Vedantu app.

3. What are the Different Applications of Ampere's Circuital Law, according to Chapter 4 of Class 12 Physics?

Different applications of ampere’s circuital law are:

Long current-carrying induced magnetism

Current carrying cylinder creating a magnetic field

Current carrying hollow cylinder creating magnetic fields

Vedantu has all the physics questions that are solved, so you do not have to take the pressure of solving them, especially when you have to submit the homework and have not read the chapter thoroughly. Also, in exams, you can go through these questions quickly and remember them till your exam day.

4. What is Understood by a Cyclotron, according to Chapter 4 of Class 12 Physics?

The first particle accelerator is responsible for accelerating charged particles to high energies. The charged particle accelerates outwards from the spiral routes alongside the spiral particles. Physics is a difficult topic that requires careful attention. All notes and content may be found on Vedantu. It also includes all of the practice examinations that will be useful in your board exams. Hence, if you want to get high grades, you should consult Vedantu.

5. What are the limitations to a cyclotron, according to Chapter 4 of Class 12 Physics?

The limitations of cyclotron are :

The electron cannot be accelerated by a cyclotron as they are of very small size.

Neutral particles cannot be accelerated.

Cannot accelerate large positive charged particles. 

Board exams are crucial for students. For these exams, a lot of practice is required. Vedantu has previous year's question papers which you can solve and then appear for the main exams. So, visit Vedantu for all such material.

CBSE Class 12 Physics Important Questions

Cbse study materials.

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

Table of Contents

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

These crucial subjects are covered in NCERT Solutions Class 12 Physics Chapter 4. To grasp the concepts presented in the chapter and effectively utilize the provided solutions, students are encouraged to carefully study each topic.

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.

To access the NCERT Solutions for Class 12 Physics, Chapter 4 Moving Charges and Magnetism PDF, you can download it from the link provided below:

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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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CBSE Class 12 Physics Case study Questions & Answers For Chapter Ch. 1, 2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14

Understudies can discover the chapter astute vital questions for course 12th Physics within the table underneath. These imperative questions incorporate questions that are regularly inquired in a long time. Moreover, arrangements are to give for these questions, with extraordinary accentuation on ease-of-study. Tap on the joins underneath to begin investigating.

CBSE Class 12 Case Study Question for Physics

To a distant observer, the light appears to be coming from somewhere below the ground. The observer naturally assumes that light is being reflected from the ground, say, by a pool of water near the tall object.

(i) Refraction, Total internal Reflection

a μω = 1/sin C

=>sin C = 1/a μω

(ii) n 1 = n 2

(ii) decrease

(iii) remain the same

(iv) 0< < 30°

Total internal reflection will occur if the angle i’ >i’ c , i.e. , if i’ > 59° or when r <r max’ where r max = 90° – 59° = 31°. Using Snell’s law,

Chapter-4: Moving Charges And Magnetism

Chapter – 10 Wave Optics

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

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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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CBSE Class 12th - PHYSICS : Chapterwise Case Study Question & Solution

CBSE will ask two Case Study Questions in the CBSE class 12 Physics questions paper. Question numbers 15 and 16 are cased-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.

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Case Study Question for Class 12 Physics Chapter 1 Electric Charges and Fields

• Last modified on: 2 years ago
• Reading Time: 19 Minutes

There is Case Study Questions in class 12 Physics in session 2020-21. The first two questions in the board exam question paper will be based on Case Study and Assertion & Reason. Case Study Questions will have 5 MCQs out of which students will have to attempt any 4 questions. Here are the questions based on case study.

Case Study Question 1:

Electric field strength is proportional to the density of lines of force i.e., electric field strength at a point is proportional to the number of lines of force cutting a unit area element placed normal to the field at that point. As illustrated in given figure, the electric field at P is stronger than at Q.

(i) Electric lines of force about a positive point charge are (a) radially outwards (b) circular clockwise (c) radially inwards (d) parallel straight lines

(ii) Which of the following is false for electric lines of force? (a) They always start from positive charge and terminate on negative charges. (b) They are always perpendicular to the surface of a charged conductor. (c) They always form closed loops. (d) They are parallel and equally spaced in a region of uniform electric field.

(iii) Which one of the following patterns of electric line of force is not possible in field due to stationary charges?

(iv) Electric field lines are curved (a) in the field of a single positive or negative charge (b) in the field of two equal and opposite charges. (c) in the field of two like charges. (d) both (b) and (c)

(v) The figure below shows the electric field lines due to two positive charges. The magnitudes E A , E B and E C of the electric fields at point A, B and C respectively are related as

(a) E A >E B >E C (b) E B >E A >E C (c) E A =E B >E C (d) E A >E B =E C

Case Study Question 2:

Smallest charge that can exist in nature is the charge of an electron. During friction it is only the transfer of electron which makes the body charged. Hence net charge on any body is an integral multiple of charge of an electron (1.6 x 10 -19 C) i.e., q=±ne where r= 1, 2, 3, 4 …. Hence no body can have a charge represented as 1.8e, 2.7e, 2e/5, etc. Recently, it has been discovered that elementary particles such as protons or neutrons are elemental units called quarks.

(i) Which of the following properties is not satisfied by an electric charge? (a) Total charge conservation. (b) Quantization of charge. (c) Two types of charge. (d) Circular line of force.

(ii) Which one of the following charges is possible? (a) 5.8 x 10 -18 C (b) 3.2 x 10 -18 C (c) 4.5 x 10 -19 C (d) 8.6 x 10 -19 C

(iii) If a charge on a body is 1 nC, then how many electrons are present on the body? (a) 6.25 x 10 27 (b) 1.6 x 10 19 (c) 6.25 X 10 28 (d) 6.25 X 10 9

(iv) If a body gives out 10 9 electrons every second, how much time is required to get a total charge of 1 from it? (a) 190.19 years (b) 150.12 years (c) 198.19 years (d) 188.21 years

(v) A polythene piece rubbed with wool is found to have a negative charge of 3.2 x 10 -7 C. Calculate the number of electrons transferred. (a) 2 x 10 12 (b) 3 x 10 12 (c) 2 x 10 14 (d) 3 x 10 14

Case Study Question 3:

When electric dipole is placed in uniform electric field, its two charges experience equal and opposite forces, which cancel each other and hence net force on electric dipole in uniform electric field is zero. However these forces are not collinear, so they give rise to some torque on the dipole. Since net force on electric dipole in uniform electric field is zero, so no work is done in moving the electric dipole in uniform electric field. However some work is done in rotating the dipole against the torque acting on it.

(i) The dipole moment of a dipole in a uniform external field Ē is B. Then the torque τ acting on the dipole is (a) τ=p x E (b) τ = P. Ē (c) τ = 2(p + Ē) (d) τ = (P + E)

(ii) An electric dipole consists of two opposite charges, each of magnitude 1.0 μC separated by a distance of 2.0 cm. The dipole is placed in an external field of 10 5 NC -1 . The maximum torque on the dipole is (a) 0.2 x 10 -3 Nm (b) 1x 10 -3 Nm (c) 2 x 10 -3 Nm (d) 4x 10 -3 Nm

(iii) Torque on a dipole in uniform electric field is minimum when θ is equal to (a) 0° (b) 90° (c) 180° (d) Both (a) and (c)

(iv) When an electric dipole is held at an angle in a uniform electric field, the net force F and torque τ on the dipole are (a) F= 0, τ = 0 (b) F≠0, τ≠0 (c) F=0, τ ≠ 0 (d) F≠0, τ=0

(v) An electric dipole of moment p is placed in an electric field of intensity E. The dipole acquires a position such that the axis of the dipole makes an angle with the direction of the field. Assuming that potential energy of the dipole to be zero when 0 = 90°, the torque and the potential energy of the dipole will respectively be (a) pEsinθ, -pEcosθ (b) pEsinθ, -2pEcosθ (c) pEsinθ, 2pEcosθ (d) pEcosθ, – pEsinθ

Case Study Question 4:

A charge is a property associated with the matter due to which it experiences and produces an electric and magnetic field. Charges are scalar in nature and they add up like real number. Also, the total charge of an isolated system is always conserved. When the objects rub against each other charges acquired by them must be equal and opposite.

(i) The cause of a charging is: (a) the actual transfer of protons. (b) the actual transfer of electrons. (c) the actual transfer of neutrons. (d) none the above

(ii) Pick the correct statement. (a) The glass rod gives protons to silk when they are rubbed against each other. (b) The glass rod gives electrons to silk when they are rubbed against each other. (c) The glass rod gains protons from silk when they are rubbed against each other. (d) The glass rod gains electrons when they are rubbed against each other.

(iii) If two electrons are each 1.5 × 10 –10 m from a proton, the magnitude of the net electric force they will exert on the proton is (a) 1.97 × 10 –8 N (b) 2.73 × 10 –8 N (c) 3.83 × 10 –8 N (d) 4.63 × 10 –8 N

(iv) A charge is a property associated with the matter due to which it produces and experiences: (a) electric effects only (b) magnetic effects only (c) both electric and magnetic effects (d) none of these.

(v) The cause of quantization of electric charges is: (a) Transfer of an integral number of neutrons. (b) Transfer of an integral number of protons. (c) Transfer of an integral number of electrons. (d) None of the above.

Case Study Question 5:

Surface Charge Density. Surface charge density is defined as the charge per unit surface area the surface (Arial) charge symmetric distribution and follow Gauss law of electro statics mathematical term of surface charge density σ=ΔQ/ΔS

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite sign (± s). Having magnitude 8.8 × 10 –12 cm –2 as shown here. The intensity of electrified at a point is E =σ/ε 0   and flux is Φ=E.ΔS, where ΔS = 1 m 2 (unit arial plate)

(i) E in the outer region (I) of the first (A) plate is (a) 1.7 × 10 –22 N/C (b) 1.1 × 10 –12 V/m (c) Zero (d) Insufficient data

(ii) E in the outer region (III) of the second plate (B) is (a) 1 N/C (b) 0.1 V/m (c) 0.5 N/C (d) zero

(iii) E between (II) the plate is (a) 1 N/C (b) 0.1 V/m (c) 0.5 N/C (d) None of these

(iv) The ratio of E from left side of plate A at distance 1 cm and 2 m respectively is (a) 1 : 2 (b) 10 : 2 (c) 1 : 1 (d) 20 : 1

(v) In order to estimate the electric field due to a thin finite plane metal plate the Gaussian surface considered is (a) Spherical (b) Linear (c) Cylindrical (d) Cybic

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