Class 12 Physics · Chapter 4

Moving Charges and Magnetism

Master Lorentz Force, Biot-Savart Law, Ampere's Law, Solenoid, Torque on current loops and the Galvanometer — all critical for AAI ATC.

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8 Subtopics
Full NCERT Coverage
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24+ MCQs
Numerical Practice
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All Key Formulas
Quick Reference
4.1
Magnetic Force — Lorentz Force F = q(v×B) · Direction by right-hand rule · Unit: Tesla

The Lorentz Force

When a charge q moves with velocity v in electric field E and magnetic field B, the total force is the Lorentz force: F = q[E + v×B]. The magnetic part q(v×B) is always perpendicular to velocity — so it does zero work and changes direction, not speed.

Key properties: (i) Force is zero if particle is stationary. (ii) Force is zero if v ∥ B or v ∥ (−B). (iii) Direction given by right-hand rule (or screw rule). (iv) Force on negative charge is opposite to positive charge.

Force on Current-Carrying Conductor

A straight conductor of length l carrying current I in field B: F = Il × B, magnitude F = BIl sinθ. For mid-air suspension: mg = BIl, so B = mg/(Il).

📐 Key Formulas
F = q[E + v×B]    — Lorentz force (total)
Fmag = qvB sinθ    — magnetic force magnitude
F = BIl sinθ    — force on current-carrying conductor
1 T = 1 N/(A·m)    — SI unit of B (Tesla)
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Insert Diagram: Right-hand rule for direction of F = q(v×B) — thumb along v, fingers along B, palm gives F for positive charge (Fig 4.2 NCERT)

Practice MCQs — Section 4.1
Q1 A wire of mass 200 g and length 1.5 m carries 2 A and is suspended mid-air by horizontal field B. The field magnitude is (g = 9.8 m/s²):
1.3 T
0.65 T
0.33 T
2.6 T
mg = BIl → B = mg/(Il) = 0.2×9.8/(2×1.5) = 1.96/3 = 0.65 T
Q2 Force per unit length on a wire carrying 8 A at 30° to a field of 0.15 T is:
1.2 N/m
0.6 N/m
0.3 N/m
1.04 N/m
f = BIsin30° = 0.15×8×0.5 = 0.6 N/m
Q3 A 3 cm wire carrying 10 A is placed perpendicular inside a solenoid (B = 0.27 T). Magnetic force on the wire is:
0.027 N
0.27 N
8.1 × 10⁻² N
2.7 × 10⁻³ N
F = BIl = 0.27×10×0.03 = 8.1×10⁻² N
4.2
Motion in a Magnetic Field Circular path · Radius r = mv/qB · Cyclotron frequency · Helical motion

Circular and Helical Motion

When v ⊥ B, the magnetic force acts as centripetal force, causing circular motion. Radius: r = mv/qB. The cyclotron frequency ν = qB/(2πm) is independent of velocity — the key principle behind a cyclotron.

If v has a component parallel to B, that component is unaffected (B exerts no force along itself), while the perpendicular component traces a circle. Combined: helical motion with pitch p = v∥ × T = 2πmv∥/(qB).

📐 Key Formulas
r = mv/(qB)    — radius of circular orbit
ω = qB/m    or    ν = qB/(2πm)    — cyclotron frequency (independent of v)
T = 2πm/(qB)    — time period of revolution
Pitch p = v∥ × T = 2πmv∥/(qB)    — pitch of helix
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Insert Diagram: Circular motion (v⊥B) and helical motion (v has component along B) — Figs 4.5, 4.6 NCERT

Practice MCQs — Section 4.2
Q4 An electron (m = 9×10⁻³¹ kg, q = 1.6×10⁻¹⁹ C) moves at 3×10⁷ m/s perpendicular to B = 6×10⁻⁴ T. Radius of path is:
56 cm
28 cm
14 cm
8 cm
r = mv/(qB) = 9×10⁻³¹×3×10⁷/(1.6×10⁻¹⁹×6×10⁻⁴) = 27×10⁻²⁴/9.6×10⁻²³ = 0.28 m = 28 cm
Q5 A proton (m = 1.67×10⁻²⁷ kg, q = 1.6×10⁻¹⁹ C) in B = 0.5 T. Cyclotron frequency is approximately:
76 MHz
7.6 MHz
0.76 MHz
760 MHz
ν = qB/(2πm) = 1.6×10⁻¹⁹×0.5/(2π×1.67×10⁻²⁷) = 8×10⁻²⁰/1.05×10⁻²⁶ ≈ 7.6×10⁶ Hz = 7.6 MHz
Q6 In a chamber B = 6.5 G (= 6.5×10⁻⁴ T), an electron (m = 9.1×10⁻³¹ kg, e = 1.6×10⁻¹⁹ C) moves at 4.8×10⁶ m/s normal to B. Its orbital radius is approximately:
0.042 m
0.042 m (≈4.2 cm)
0.42 m
4.2 m
r = mv/(eB) = 9.1×10⁻³¹×4.8×10⁶/(1.6×10⁻¹⁹×6.5×10⁻⁴) = 4.368×10⁻²⁴/1.04×10⁻²² ≈ 0.042 m = 4.2 cm
4.3
Biot-Savart Law dB due to current element · μ₀/4π · Field due to circular loop

Biot-Savart Law

The magnetic field dB at point P due to a current element I·dl at distance r is: dB = (μ₀/4π) × I dl sinθ / r². Direction is perpendicular to the plane containing dl and r (right-hand screw rule). μ₀ = 4π×10⁻⁷ T·m/A is the permeability of free space.

Contrast with Coulomb's law: Both ∝ 1/r². But B is produced by a vector source (I·dl) while E by a scalar source (charge); B is ⊥ to plane of dl and r.

📐 Key Formulas
|dB| = (μ₀/4π) × I·dl·sinθ/r²    — Biot-Savart law (scalar form)
Baxis = μ₀IR²/[2(x²+R²)^(3/2)]    — field on axis of circular loop of radius R
Bcentre = μ₀I/(2R)    — field at centre of circular loop (x=0)
BN-turn = μ₀NI/(2R)    — N-turn coil at centre
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Insert Diagram: Biot-Savart law — current element I·dl at origin, field dB at point P at distance r (Fig 4.7 NCERT) and circular loop field on axis (Fig 4.9)

Practice MCQs — Section 4.3
Q7 A circular coil of 100 turns, radius 8 cm, current 0.40 A. Field at centre is (μ₀ = 4π×10⁻⁷):
1.57×10⁻⁴ T
3.14×10⁻⁴ T
6.28×10⁻⁴ T
7.85×10⁻⁵ T
B = μ₀NI/(2R) = 4π×10⁻⁷×100×0.4/(2×0.08) = 4π×4×10⁻⁶/0.16 = 4π×2.5×10⁻⁵ = 3.14×10⁻⁴ T
Q8 A 100-turn coil of radius 10 cm carries 1 A. Field at centre is approximately:
3.14×10⁻³ T
6.28×10⁻⁴ T
1.57×10⁻³ T
2×10⁻³ T
B = μ₀NI/(2R) = 4π×10⁻⁷×100×1/(2×0.1) = 4π×10⁻⁵/0.2 = 2π×10⁻⁴ ≈ 6.28×10⁻⁴ T
Q9 A semicircular arc of radius 2 cm carries 12 A. Field at its centre is (μ₀ = 4π×10⁻⁷):
3.77×10⁻⁴ T
1.88×10⁻⁴ T
7.54×10⁻⁴ T
9.42×10⁻⁵ T
Full loop B = μ₀I/(2R) = 4π×10⁻⁷×12/(2×0.02) = 3.77×10⁻⁴ T; Semicircle = half = 1.88×10⁻⁴ T
4.4
Ampere's Circuital Law ∮B·dl = μ₀I · Infinite wire · Inside/outside cylinder

Ampere's Circuital Law

The line integral of magnetic field around any closed Amperian loop equals μ₀ times the total current enclosed: ∮B·dl = μ₀Ienc. For a symmetric case (tangential, constant B): BL = μ₀Ienc.

For an infinite straight wire at distance r: B = μ₀I/(2πr). Inside a wire of radius a: B ∝ r (B = μ₀Ir/2πa²). Outside: B ∝ 1/r.

AAI ATC Analogy: Ampere's law relates to Biot-Savart law the same way Gauss's law relates to Coulomb's law. Both express the same physics — Ampere's is easier for symmetric cases.
📐 Key Formulas
∮B·dl = μ₀Ienc    — Ampere's circuital law (general)
B = μ₀I/(2πr)    — field outside/at distance r from long wire
B = μ₀Ir/(2πa²)    — field inside wire of radius a (r < a)
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Insert Diagram: Circular Amperian loop around long straight wire; B vs r graph showing linear increase inside and 1/r decrease outside (Fig 4.13, 4.14 NCERT)

Practice MCQs — Section 4.4
Q10 A long straight wire carries 35 A. Field at 20 cm from the wire is (μ₀ = 4π×10⁻⁷):
7×10⁻⁵ T
3.5×10⁻⁵ T
1.75×10⁻⁵ T
1.1×10⁻⁴ T
B = μ₀I/(2πr) = 4π×10⁻⁷×35/(2π×0.2) = 2×10⁻⁷×35/0.2 = 7000×10⁻⁷/0.2 × 10⁻⁷ = 3.5×10⁻⁵ T
Q11 Wire A (Ia = 8 A) and wire B (Ib = 5 A) are 4 cm apart (parallel, same direction). Force per unit length between them is (μ₀ = 4π×10⁻⁷):
4×10⁻⁵ N/m (repulsive)
2×10⁻⁴ N/m (attractive)
1×10⁻⁴ N/m (repulsive)
4×10⁻⁴ N/m (attractive)
f = μ₀IaIb/(2πd) = 4π×10⁻⁷×8×5/(2π×0.04) = 2×10⁻⁷×40/0.04 = 2×10⁻⁴ N/m (attractive, parallel currents)
Q12 A long wire carries 50 A (north to south). Field at 2.5 m east of wire is:
8×10⁻⁶ T (north)
4×10⁻⁶ T (vertically downward)
4×10⁻⁶ T (vertically upward)
2×10⁻⁵ T (east)
B = μ₀I/(2πr) = 4π×10⁻⁷×50/(2π×2.5) = 2×10⁻⁷×50/2.5 = 4×10⁻⁶ T. Current S→N, point to east → by right-hand rule, B is directed vertically downward.
4.5
The Solenoid B = μ₀nI · Uniform interior field · n = turns per unit length

Magnetic Field of a Long Solenoid

A solenoid is a long helical coil. For an ideal long solenoid: field outside ≈ 0; field inside is uniform and along the axis: B = μ₀nI, where n = N/L is turns per unit length. The rectangular Amperian loop (abcd) is used to derive this via Ampere's law.

Inserting a soft iron core inside the solenoid greatly increases B. Solenoids are used as electromagnets, inductors, and MRI machines.

📐 Key Formulas
B = μ₀nI    — field inside long solenoid (n = turns/metre)
n = N/L    — number of turns per unit length
Bwith iron core = μ₀μrnI    — with relative permeability μr
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Insert Diagram: Long solenoid cross-section with uniform interior field, rectangular Amperian loop abcd, field lines (Figs 4.15, 4.16 NCERT)

Practice MCQs — Section 4.5
Q13 Solenoid: length = 0.5 m, N = 500 turns, I = 5 A. B inside is (μ₀ = 4π×10⁻⁷):
3.14×10⁻³ T
6.28×10⁻³ T
1.57×10⁻³ T
12.56×10⁻³ T
n = 500/0.5 = 1000/m; B = μ₀nI = 4π×10⁻⁷×1000×5 = 4π×5×10⁻⁴ = 20π×10⁻⁴ ≈ 6.28×10⁻³ T
Q14 A solenoid 80 cm long has 5 layers of 400 turns each. Current = 8 A. B at centre is approximately:
1×10⁻² T
2.51×10⁻² T
5×10⁻³ T
4×10⁻² T
Total turns N = 5×400 = 2000; n = 2000/0.8 = 2500/m; B = 4π×10⁻⁷×2500×8 = 4π×2×10⁻³ ≈ 2.51×10⁻² T
Q15 A solenoid needs B = 0.0628 T inside. If n = 500 turns/m, required current is:
50 A
0.1 A
100 A
10 A
I = B/(μ₀n) = 0.0628/(4π×10⁻⁷×500) = 0.0628/(6.28×10⁻⁴) = 100 A
4.6
Torque on Current Loop & Magnetic Dipole τ = m×B · m = NIA · Current loop as magnetic dipole

Torque on Rectangular Loop

A rectangular current loop (N turns, area A, current I) in uniform field B experiences zero net force but a torque: τ = NIAB sinθ, where θ is the angle between field and normal to loop.

The magnetic moment m = NIA (direction by right-hand thumb rule). In vector form: τ = m × B. This is analogous to electric dipole: τ = p × E. Stable equilibrium when m ∥ B (θ = 0); unstable when m antiparallel to B.

📐 Key Formulas
m = NIA    — magnetic moment (A·m²)
τ = mB sinθ = NIAB sinθ    — torque on current loop
τ = m × B    — vector form (analogous to p×E)
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Insert Diagram: Rectangular current loop in uniform B showing torque couple — side view, magnetic moment m direction (Figs 4.18, 4.19 NCERT)

Practice MCQs — Section 4.6
Q16 Square coil: side 10 cm, 20 turns, I = 12 A, B = 0.80 T, normal at 30° to B. Torque is:
1.92 N·m
0.96 N·m
3.84 N·m
0.48 N·m
τ = NIAB sinθ = 20×12×(0.1)²×0.80×sin30° = 20×12×0.01×0.8×0.5 = 0.96 N·m
Q17 A 100-turn coil (radius 10 cm, I = 3.2 A) is placed with axis along B = 2 T. After rotating 90°, final torque is:
0 N·m
10π N·m
20 N·m
6.28 N·m
m = NIπR² = 100×3.2×π×0.01 ≈ 10 A·m²; τ = mBsin90° = 10×2×1 = 20 N·m
Q18 A circular coil of 30 turns, radius 8 cm, I = 6 A in B = 1 T. Field lines at 60° to normal. Counter torque to prevent turning is:
1.92 N·m
3.13 N·m
6.28 N·m
0.96 N·m
τ = NIABsinθ = 30×6×π×(0.08)²×1×sin60° = 30×6×0.02011×0.866 ≈ 3.13 N·m
4.7
Moving Coil Galvanometer kφ = NIAB · Current sensitivity · Ammeter & Voltmeter conversion

Working Principle

The MCG uses torque on a current-carrying coil in a radial magnetic field. Magnetic torque NIAB is balanced by spring restoring torque kφ at equilibrium: kφ = NIAB → φ = (NAB/k) × I. Deflection φ is proportional to I.

Conversion to Ammeter and Voltmeter

Ammeter: Connect shunt resistance rs (small) in parallel with galvanometer. Most current bypasses the coil. Total resistance ≈ rs (small).

Voltmeter: Connect large resistance R in series. Total resistance = RG + R (large). Current sensitivity ∝ N but voltage sensitivity independent of N (when N doubled, RG also doubles).

📐 Key Formulas
kφ = NIAB    →    φ = (NAB/k)·I    — galvanometer deflection
Current sensitivity = NAB/k    — (rad/A)
Voltage sensitivity = NAB/(kR)    — (rad/V)
Ammeter: shunt rs in parallel    Voltmeter: R in series
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Insert Diagram: Moving coil galvanometer — coil between N-S poles, soft iron core, spring Sp, pointer and scale; conversion to ammeter and voltmeter (Figs 4.20, 4.21, 4.22 NCERT)

Practice MCQs — Section 4.7
Q19 Galvanometer: RG = 60 Ω, connected in circuit with 3 Ω external. EMF = 3 V. Current through it is:
1.0 A
0.99 A
0.048 A
0.5 A
I = V/(RG+R) = 3/(60+3) = 3/63 ≈ 0.048 A
Q20 Same galvanometer (RG = 60 Ω) converted to ammeter with shunt rs = 0.02 Ω. Circuit with 3 Ω external, 3 V EMF. Current is approximately:
0.048 A
1.0 A
0.99 A
3.0 A
Shunt equivalent = 60×0.02/60.02 ≈ 0.02 Ω; Total = 3.02 Ω; I = 3/3.02 ≈ 0.99 A
Q21 Two galvanometers M₁ (N₁=30, A₁=3.6×10⁻³ m², B₁=0.25 T, R₁=10 Ω) and M₂ (N₂=42, A₂=1.8×10⁻³ m², B₂=0.50 T, R₂=14 Ω), same spring constant. Ratio of current sensitivity M₂/M₁ is:
0.5
1.4
2.0
2.8
CS = NAB/k; Ratio = (N₂A₂B₂)/(N₁A₁B₁) = (42×1.8×10⁻³×0.50)/(30×3.6×10⁻³×0.25) = (0.0378)/(0.027) = 1.4
4.8
Force Between Parallel Currents — Definition of Ampere f = μ₀IaIb/(2πd) · Parallel attract · Antiparallel repel

Parallel and Antiparallel Currents

Two parallel current-carrying conductors exert magnetic forces on each other: parallel currents attract; antiparallel currents repel. This is opposite to the electrostatic rule (like charges repel).

Definition of Ampere: 1 A is that steady current which, when flowing in each of two very long parallel wires placed 1 m apart in vacuum, produces a force of exactly 2×10⁻⁷ N/m of length on each wire.

📐 Key Formula
fba = μ₀IaIb/(2πd)    — force per unit length between parallel wires
= 2×10⁻⁷ N/m when Ia=Ib=1A, d=1m    — defines 1 Ampere
Practice MCQs — Section 4.8
Q22 Force on 10 cm of wire A (8 A) due to wire B (5 A, 4 cm apart, parallel) is:
1×10⁻⁵ N (repulsive)
2×10⁻⁵ N (attractive)
4×10⁻⁵ N (attractive)
5×10⁻⁶ N (repulsive)
F = μ₀IaIbL/(2πd) = 4π×10⁻⁷×8×5×0.1/(2π×0.04) = 2×10⁻⁷×40×0.1/0.04 = 2×10⁻⁷×100 = 2×10⁻⁵ N (attractive)
Q23 Earth's horizontal field at a place is 3×10⁻⁵ T. A 1 A wire horizontal (east-west). Force per unit length on the wire is:
3×10⁻⁵ N/m (upward)
0 N/m
3×10⁻⁵ N/m (downward)
6×10⁻⁵ N/m
θ = 90° (current E-W, B points N-S); f = IBsin90° = 1×3×10⁻⁵ = 3×10⁻⁵ N/m (downward by right-hand rule)
Q24 Two long parallel wires carrying 1 A each, 1 m apart. Force per metre between them is:
4π×10⁻⁷ N/m
2×10⁻⁷ N/m
1×10⁻⁷ N/m
10⁻⁶ N/m
f = μ₀I²/(2πd) = 4π×10⁻⁷×1×1/(2π×1) = 2×10⁻⁷ N/m — this is the definition of 1 Ampere.

📋 Chapter Summary — Quick Revision

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