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

These Class 12 Physics Chapter 4 solutions cover Moving Charges and Magnetism from the NCERT textbook (session 2026–27). Every NCERT Exercises question is reproduced verbatim and solved step by step, with each numerical worked out in full and cross-checked against the official NCERT answer key. The chapter explains how currents and moving charges create magnetic fields and how those fields exert forces — the foundations of the Lorentz force, Biot–Savart law, Ampere’s circuital law, the solenoid, force between parallel currents, torque on a current loop and the moving-coil galvanometer.

Class: 12 Subject: Physics Chapter: 4 Chapter Name: Moving Charges and Magnetism Exercises: 4.1 – 4.13 Session: 2026–27
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Class 12 Physics Chapter 4 – Overview

In 1820, Hans Christian Oersted discovered that an electric current deflects a nearby compass needle — the first proof that moving charges and currents produce a magnetic field in the surrounding space. This chapter builds on that idea. It begins with the Lorentz force, the force a magnetic field exerts on a moving charge and on a current-carrying conductor, and uses it to explain circular and helical motion of charged particles. It then introduces the two great laws that let us calculate the magnetic field of a current — the Biot–Savart law and Ampere’s circuital law — and applies them to a circular loop, a long straight wire and a solenoid. Finally it studies the force between two parallel currents (which defines the ampere), the torque on a current loop and its magnetic moment, and the working of the moving-coil galvanometer and its conversion into an ammeter and voltmeter.

Key Concepts & Definitions

Lorentz force: the total force on a charge q moving with velocity v in electric field E and magnetic field B is F = q(E + v × B). The magnetic part is perpendicular to v, so it does no work and only changes the direction of motion.

Magnetic field unit: the SI unit of B is the tesla (T); 1 T = 1 N A−1 m−1. A smaller non-SI unit, the gauss, equals 10−4 T.

Circular & helical motion: when v ⊥ B, the magnetic force supplies the centripetal force, giving a circle of radius r = mv/qB. A velocity component along B turns the path into a helix.

Biot–Savart law: the field dB from a current element I dl is proportional to I, to |dl|, to sinθ, and inversely to the square of the distance r; its direction is perpendicular to the plane of dl and r.

Ampere’s circuital law: the line integral of B around a closed loop equals μ0 times the net current threading the loop. It gives the field easily for highly symmetric cases (straight wire, solenoid).

Magnetic moment: a planar loop of N turns, area A, current I behaves like a magnetic dipole of moment m = NIA; in a field it feels a torque τ = m × B but no net force.

Galvanometer: a coil in a radial field deflects by φ = (NAB/k)I. A small shunt in parallel converts it to an ammeter; a large resistance in series converts it to a voltmeter.

Important Formulas (Chapter 4)

Lorentz force: F = q(E + v × B); magnetic force magnitude F = qvB sinθ.

Force on a conductor: F = I l × B; magnitude F = BIl sinθ.

Radius / cyclotron frequency: r = mv/(qB)  •  νc = qB/(2πm).

Biot–Savart law: dB = (μ0/4π) · I dl sinθ / r2;   μ0/4π = 10−7 T m A−1.

Field of a circular coil: centre B = μ0NI/(2R);   on axis B = μ0NIR2 / [2(x2 + R2)3/2].

Long straight wire: B = μ0I / (2πr).

Solenoid (inside): B = μ0nI, where n = turns per unit length.

Force per unit length between parallel wires: f = μ0IaIb / (2πd).

Magnetic moment & torque: m = NIA;   τ = m × B, magnitude τ = NIAB sinθ.

Galvanometer: φ = (NAB/k)I; current sensitivity = NAB/k; voltage sensitivity = NAB/(kR).

NCERT Exercises (4.1 – 4.13) — Solutions

Questions are reproduced verbatim from the NCERT textbook. Throughout, μ0 = 4π × 10−7 T m A−1, so μ0/4π = 10−7 T m A−1.

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?

SOLUTION Given: N = 100, R = 8.0 cm = 0.080 m, I = 0.40 A. Formula: B = μ0NI / (2R). Substitute: B = (4π × 10−7 × 100 × 0.40) / (2 × 0.080). Numerator = 4π × 10−7 × 40 = 160π × 10−7. Denominator = 0.16. B = (160π × 10−7) / 0.16 = 1000π × 10−7 = π × 10−4 T. Answer: B = π × 10−4 T ≈ 3.1 × 10−4 T, directed along the axis of the coil (given by the right-hand thumb rule).

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?

SOLUTION Given: I = 35 A, r = 20 cm = 0.20 m. Formula: B = μ0I / (2πr) = (μ0/2π) × (I/r) = (2 × 10−7) × (I/r). Substitute: B = (2 × 10−7) × (35 / 0.20) = (2 × 10−7) × 175 = 350 × 10−7 T. Answer: B = 3.5 × 10−5 T. The field lines are circles concentric with the wire.

4.3 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.

SOLUTION Given: I = 50 A, r = 2.5 m; current flows north → south; point lies 2.5 m east of the wire. Magnitude: B = μ0I/(2πr) = (2 × 10−7) × (50 / 2.5) = (2 × 10−7) × 20 = 4 × 10−6 T. Direction: point the right thumb along the current (north → south); the fingers curl from the west side up and over to the east side down… applying the rule carefully at a point east of the wire, the field is vertically upward (out of the horizontal plane).

4.4 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?

SOLUTION Given: I = 90 A, r = 1.5 m; current flows east → west; point is 1.5 m below the line. Magnitude: B = μ0I/(2πr) = (2 × 10−7) × (90 / 1.5) = (2 × 10−7) × 60 = 120 × 10−7 = 1.2 × 10−5 T. Direction: grasp the wire with the right hand, thumb pointing east → west; below the wire the fingers point toward the south. So the field is directed towards the south.

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?

SOLUTION Given: I = 8 A, B = 0.15 T, θ = 30°. Formula: force per unit length f = F/l = BI sinθ. Substitute: f = 0.15 × 8 × sin 30° = 0.15 × 8 × 0.5 = 0.6 N m−1. Answer: f = 0.6 N m−1.

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?

SOLUTION Given: l = 3.0 cm = 0.03 m, I = 10 A, B = 0.27 T, θ = 90° (wire ⊥ field). Formula: F = BIl sinθ = BIl (since sin 90° = 1). Substitute: F = 0.27 × 10 × 0.03 = 0.081 N. Answer: F = 8.1 × 10−2 N; its direction is given by Fleming’s left-hand rule (perpendicular to both the wire and the field).

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.

SOLUTION Given: IA = 8.0 A, IB = 5.0 A, d = 4.0 cm = 0.04 m, length L = 10 cm = 0.10 m. Formula: force per unit length f = μ0IAIB/(2πd) = (2 × 10−7) IAIB/d. f = (2 × 10−7) × (8.0 × 5.0) / 0.04 = (2 × 10−7) × 1000 = 2 × 10−4 N m−1. Force on 0.10 m section: F = f × L = (2 × 10−4) × 0.10 = 2 × 10−5 N. Answer: F = 2 × 10−5 N. Since the currents are in the same direction, the force is attractive, directed from A towards B (normal to the wire).

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.

SOLUTION Given: length L = 80 cm = 0.80 m; total turns N = 5 × 400 = 2000; I = 8.0 A. (The diameter 1.8 cm is small compared with the length, so the long-solenoid formula applies.) Turns per unit length: n = N/L = 2000 / 0.80 = 2500 turns m−1. Formula: B = μ0nI = (4π × 10−7) × 2500 × 8.0. B = 4π × 10−7 × 20000 = 8π × 10−3 T. Answer: B = 8π × 10−3 T ≈ 2.5 × 10−2 T.

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?

SOLUTION Given: side a = 10 cm = 0.10 m, so area A = (0.10)2 = 0.01 m2; N = 20; I = 12 A; B = 0.80 T; θ = 30° (angle between normal and field). Formula: τ = NIAB sinθ. Substitute: τ = 20 × 12 × 0.01 × 0.80 × sin 30° = 20 × 12 × 0.01 × 0.80 × 0.5. τ = 1.92 × 0.5 = 0.96 N m. Answer: τ = 0.96 N m.

4.10 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 TR2 = 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.

SOLUTION (a) Current sensitivity = NAB/k. With k equal for both meters, Ratio = (N2A2B2) / (N1A1B1) = (42 × 1.8 × 10−3 × 0.50) / (30 × 3.6 × 10−3 × 0.25). Numerator = 42 × 1.8 × 0.50 = 37.8 (×10−3). Denominator = 30 × 3.6 × 0.25 = 27.0 (×10−3). Ratio = 37.8 / 27.0 = 1.4. So current sensitivity of M2 is 1.4 times that of M1. (b) Voltage sensitivity = NAB/(kR). Ratio = (current-sensitivity ratio) × (R1/R2) = 1.4 × (10/14) = 1.4 × 0.714 = 1. Answer: (a) 1.4,  (b) 1.

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 × 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 = 1.5 × 10−19 C, me = 9.1 × 10−31 kg)

SOLUTION Why a circle: the magnetic force q(v × B) is always perpendicular to the velocity. It cannot change the speed (does no work); it only continuously bends the path. A constant-magnitude force always perpendicular to the velocity produces uniform circular motion. Given: B = 6.5 G = 6.5 × 10−4 T, v = 4.8 × 106 m s−1, e = 1.5 × 10−19 C (as given), m = 9.1 × 10−31 kg. Formula: r = mv/(eB). Substitute: r = (9.1 × 10−31 × 4.8 × 106) / (1.5 × 10−19 × 6.5 × 10−4). Numerator = 9.1 × 4.8 × 10−25 = 43.68 × 10−25. Denominator = 1.5 × 6.5 × 10−23 = 9.75 × 10−23. r = (43.68 × 10−25) / (9.75 × 10−23) = 4.48 × 10−2 m ≈ 4.2 × 10−2 m. Answer: r ≈ 4.2 cm.

4.12 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.

SOLUTION Formula: the cyclotron frequency ν = eB/(2πm), which is independent of v and r. Substitute: ν = (1.5 × 10−19 × 6.5 × 10−4) / (2π × 9.1 × 10−31). Numerator = 1.5 × 6.5 × 10−23 = 9.75 × 10−23. Denominator = 2π × 9.1 × 10−31 = 5.717 × 10−30. ν = (9.75 × 10−23) / (5.717 × 10−30) = 1.7 × 107 Hz ≈ 18 MHz. Does it depend on speed? No. Since ν = eB/(2πm) contains no v, the frequency is independent of the electron’s speed (and of the orbit radius). A faster electron simply travels a larger circle in the same time. This speed-independence is exactly what makes the cyclotron work.

4.13 (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 60° 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.)

SOLUTION (a) Given: N = 30, R = 8.0 cm = 0.08 m, I = 6.0 A, B = 1.0 T, θ = 60° (angle between normal and field). Area: A = πR2 = π × (0.08)2 = π × 6.4 × 10−3 = 2.01 × 10−2 m2. Formula: τ = NIAB sinθ. Substitute: τ = 30 × 6.0 × 2.01 × 10−2 × 1.0 × sin 60° = 180 × 2.01 × 10−2 × 0.866. τ = 3.62 × 0.866 = 3.13 N m ≈ 3.1 N m. The counter torque needed to hold the coil still equals the magnetic torque: τ ≈ 3.1 N m. (b) No, the answer would not change. The torque depends only on N, I, the enclosed area A, B and θ through τ = NIAB sinθ (equivalently τ = N I A × B), which holds for a planar loop of any shape. Since the area and all other quantities are unchanged, the torque is the same.

Extra Practice Questions

Short Answer Type Questions

Q1. Why does a magnetic force do no work on a moving charge?

ANSWERThe magnetic force q(v × B) is always perpendicular to the velocity. Since work = force · displacement and the force has no component along the motion, the work done is zero. Hence the speed (and kinetic energy) of the charge stays constant; only its direction changes.

Q2. Define one tesla.

ANSWERA magnetic field has a magnitude of 1 tesla if a charge of 1 coulomb, moving with a speed of 1 m s−1 perpendicular to the field, experiences a force of 1 newton. Thus 1 T = 1 N A−1 m−1.

Q3. State and explain the rule for the force between two parallel currents.

ANSWERTwo parallel wires carrying currents in the same direction attract each other, while currents in opposite directions repel. Each wire sits in the magnetic field of the other and feels a Lorentz force F = μ0I1I2L/(2πd). This behaviour is opposite to electrostatics, where like charges repel.

Q4. Why is a radial magnetic field used in a moving-coil galvanometer?

ANSWERIn a radial field the plane of the coil is always parallel to B, so the angle between the normal and the field stays 90° and sinθ = 1 for every position. The torque NIAB then depends only on the current, making the deflection φ directly proportional to I — giving a uniform (linear) scale.

Q5. How is a galvanometer converted into (i) an ammeter and (ii) a voltmeter?

ANSWER(i) Ammeter: connect a small shunt resistance in parallel with the galvanometer, so most of the current bypasses the coil; the combination has a very small resistance and is placed in series in the circuit. (ii) Voltmeter: connect a large resistance in series with the galvanometer, so it draws very little current; the combination has a very high resistance and is placed in parallel across the element.

Long Answer Type Questions

Q1. State the Biot–Savart law and use it to find the magnetic field at the centre of a circular current loop of radius R carrying current I.

ANSWERThe Biot–Savart law states that the field due to a current element I dl at a point P, distance r away, is dB = (μ0/4π) · I dl sinθ/r2, directed perpendicular to the plane of dl and r. For the centre of a circular loop, every element dl is perpendicular to its radius vector, so θ = 90° and r = R for all elements. Each dB = (μ0/4π) I dl/R2, all pointing the same way (along the axis). Summing over the loop, Σdl = 2πR, so B = (μ0/4π)(I/R2)(2πR) = μ0I/(2R). The direction is given by the right-hand thumb rule.

Q2. Using Ampere’s circuital law, derive the expression for the magnetic field inside a long solenoid.

ANSWERFor an ideal long solenoid the field inside is uniform and along the axis, while the field outside is essentially zero. Choose a rectangular Amperian loop abcd with side ab = h inside (parallel to the axis) and side cd outside. Along cd the field is zero; along the two transverse sides bc and ad the field is perpendicular to the path, contributing nothing. Hence ∮B·dl = Bh. If n is the number of turns per unit length, the loop encloses nh turns each carrying current I, so Ienclosed = nhI. Ampere’s law gives Bh = μ0(nhI), hence B = μ0nI.

Q3. Describe the construction and working of a cyclotron, and state its limitations.

ANSWERA cyclotron accelerates charged particles to high energies using a magnetic field and an alternating electric field. Two hollow D-shaped electrodes (dees) sit in a strong perpendicular magnetic field B; a high-frequency alternating voltage is applied across the gap between them. A charged particle released near the centre is bent into a semicircle inside a dee by the magnetic force (radius r = mv/qB) and is accelerated each time it crosses the gap by the electric field. Because the cyclotron frequency ν = qB/(2πm) is independent of speed, the applied voltage’s frequency can be matched to it so the particle is always accelerated in step (resonance). As speed grows the radius increases, producing a spiral, and the fast particle is finally extracted. Limitations: it cannot accelerate neutral particles or electrons effectively; and at very high speeds the relativistic increase in mass changes the frequency, breaking resonance.

Multiple Choice Questions (MCQs)

1. The force on a charged particle moving parallel to a uniform magnetic field is:

(a) maximum    (b) qvB    (c) zero    (d) qvB/2

2. The SI unit of magnetic field is:

(a) gauss    (b) tesla    (c) weber    (d) henry

3. The magnetic field at the centre of a circular coil of N turns, radius R, current I is:

(a) μ0NI/(2πR)    (b) μ0NI/(2R)    (c) μ0NI/R    (d) μ0NIR/2

4. The radius of the circular path of a charged particle in a magnetic field is proportional to:

(a) its charge    (b) the field B    (c) its momentum mv    (d) 1/v

5. The cyclotron frequency of a charged particle is independent of:

(a) charge    (b) mass    (c) magnetic field    (d) its speed

6. Two long parallel wires carry currents in opposite directions. They:

(a) attract    (b) repel    (c) exert no force    (d) rotate

7. The magnetic field inside a long solenoid carrying current I is:

(a) μ0nI    (b) μ0I/(2πr)    (c) μ0I/(2R)    (d) zero

8. The magnetic moment of a current loop of N turns, area A, current I is:

(a) IA    (b) NIA    (c) NI/A    (d) NA/I

9. To convert a galvanometer into an ammeter, we connect a:

(a) high resistance in series    (b) low resistance (shunt) in parallel    (c) high resistance in parallel    (d) capacitor in series

10. The value of μ0/4π in SI units is:

(a) 9 × 109    (b) 10−7 T m A−1    (c) 4π × 10−7    (d) 8.85 × 10−12

Answer key: 1-(c), 2-(b), 3-(b), 4-(c), 5-(d), 6-(b), 7-(a), 8-(b), 9-(b), 10-(b).

Assertion–Reason Questions

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: A magnetic field does not change the kinetic energy of a moving charge.

Reason: The magnetic force is always perpendicular to the velocity, so it does no work.

A-R 2. Assertion: The cyclotron frequency of a charged particle does not depend on its speed.

Reason: As speed increases, the radius of the circular path also increases.

A-R 3. Assertion: Two parallel wires carrying currents in the same direction attract each other.

Reason: Like charges attract each other in electrostatics.

A-R 4. Assertion: A current-carrying loop in a uniform magnetic field experiences a net force.

Reason: A current loop in a uniform field experiences a torque but no net force.

A-R 5. Assertion: The magnetic field inside a long ideal solenoid is uniform.

Reason: The contributions of the closely spaced turns add up to give a strong, uniform axial field, while the field outside is nearly zero.

Answer key: 1-(A), 2-(A), 3-(C), 4-(D), 5-(A).
Notes: A-R 2 — both statements are true facts, but the larger radius is not the reason the frequency is speed-independent (that follows from ν = qB/2πm), so (B). A-R 3 — the assertion is true but the reason is false (like charges repel), so (C). A-R 4 — the assertion is false (net force is zero) while the reason is true, so (D).

Common Mistakes to Avoid

Watch out for these

  • Forgetting that the magnetic force is zero when the velocity is parallel (or anti-parallel) to B, and maximum when perpendicular.
  • Confusing the centre-of-loop formula B = μ0NI/(2R) with the straight-wire formula B = μ0I/(2πr).
  • Using the angle with the plane of the coil instead of the angle with the normal in τ = NIAB sinθ.
  • Dropping the number of turns N in the coil-field, moment and torque formulas.
  • Mixing up gauss and tesla (1 G = 10−4 T) and forgetting to convert cm to m.
  • Thinking a uniform field gives a net force on a current loop — it gives only a torque.

How to score full marks in this chapter

Always write the formula, substitute with SI units, and box the final answer with its unit and direction. Convert every length to metres and currents to amperes first. Remember μ0/4π = 10−7 and μ0/2π = 2 × 10−7 — these shortcuts save time in numericals. For direction questions, state clearly which rule you use (right-hand thumb rule for fields, Fleming’s left-hand rule for force on a conductor). In derivations (solenoid, circular loop, parallel wires) draw a clear diagram and name the law you apply.

Frequently Asked Questions

What is Class 12 Physics Chapter 4 about?

Chapter 4, Moving Charges and Magnetism, explains how currents and moving charges produce magnetic fields and how those fields exert forces. It covers the Lorentz force, motion of charges in a field, the Biot–Savart law, Ampere’s circuital law, the magnetic field of a loop, wire and solenoid, the force between parallel currents, torque on a current loop, and the moving-coil galvanometer.

How many exercises are there in Class 12 Physics Chapter 4?

The NCERT textbook has 13 exercise questions, numbered 4.1 to 4.13. All of them are reproduced verbatim and solved step by step on this page, with the numericals verified against the NCERT answer key.

Why is the cyclotron frequency independent of speed?

The cyclotron (revolution) frequency is ν = qB/(2πm), which contains no velocity term. A faster particle simply moves in a proportionally larger circle, so it takes the same time per revolution. This speed-independence is what allows a cyclotron to keep accelerating particles in resonance.

Are these Class 12 Physics Chapter 4 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 12 Physics are free and follow the official NCERT textbook for the 2026–27 session.

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