Moving Charges and Magnetism Class 12 Physics NCERT Solutions

Step-by-step solution

1. 1B = μ₀NI/(2r), with μ₀ = 4π × 10⁻⁷ T m A⁻¹, N = 100, I = 0.40 A, r = 0.08 m.
2. 2B = (4π × 10⁻⁷ × 100 × 0.40)/(2 × 0.08).
3. 3B = (1.6π × 10⁻⁵)/0.16 = π × 10⁻⁴ T.
4. 4B ≈ 3.14 × 10⁻⁴ T.

Step-by-step solution

1. 1B = μ₀I/(2πd) = 2 × 10⁻⁷ × I/d (in SI units).
2. 2B = 2 × 10⁻⁷ × 35/0.20 = 70 × 10⁻⁷/0.20.
3. 3B = 3.5 × 10⁻⁵ T.

Step-by-step solution

1. 1B = 2 × 10⁻⁷ × I/d = 2 × 10⁻⁷ × 50/2.5.
2. 2B = 1.0 × 10⁻⁵/2.5 = 4.0 × 10⁻⁶ T.
3. 3Direction by right-hand grip rule: thumb along the current (north → south), fingers curl; at a point due east of the wire the fingers point vertically upward.

Step-by-step solution

1. 1B = 2 × 10⁻⁷ × I/d = 2 × 10⁻⁷ × 90/1.5.
2. 2B = 1.8 × 10⁻⁵/1.5 = 1.2 × 10⁻⁵ T.
3. 3Direction by right-hand rule: current toward the west, observer directly below — the field at that point points toward the geographically south.

Step-by-step solution

1. 1F/l = B I sinθ.
2. 2F/l = 0.15 × 8 × sin30° = 0.15 × 8 × 0.5.
3. 3F/l = 0.6 N m⁻¹.

Step-by-step solution

1. 1Wire is perpendicular to B, so F = B I l (sin90° = 1).
2. 2l = 3.0 cm = 0.03 m.
3. 3F = 0.27 × 10 × 0.03 = 0.081 N.

Step-by-step solution

1. 1Force per unit length between parallel wires: F/l = 2 × 10⁻⁷ × I₁I₂/d.
2. 2F/l = 2 × 10⁻⁷ × (8.0 × 5.0)/0.04 = 2 × 10⁻⁷ × 100.
3. 3F/l = 2 × 10⁻⁵ N m⁻¹.
4. 4On a 10 cm (0.10 m) section: F = 2 × 10⁻⁵ × 0.10 = 2 × 10⁻⁶ N.
5. 5Currents flow in the same direction, so the force is attractive.

Step-by-step solution

1. 1Total turns N = 5 × 400 = 2000; length l = 0.80 m, so n = N/l = 2500 turns m⁻¹.
2. 2B = μ₀ n I = 4π × 10⁻⁷ × 2500 × 8.0.
3. 3B = 4π × 10⁻⁷ × 2.0 × 10⁴ = 8π × 10⁻³ T.
4. 4B ≈ 2.5 × 10⁻² T.
5. 5The diameter of the solenoid does not enter the estimate (long-solenoid approximation).

Step-by-step solution

1. 1Area A = (0.10)² = 1.0 × 10⁻² m².
2. 2τ = N I A B sinθ, with θ = 30° the angle between the normal and the field.
3. 3τ = 20 × 12 × 1.0 × 10⁻² × 0.80 × sin30°.
4. 4τ = 20 × 12 × 10⁻² × 0.80 × 0.5 = 0.96 N m.

Step-by-step solution

1. 1Current sensitivity S = NBA/k; since the spring constants k are identical, S₂/S₁ = (N₂B₂A₂)/(N₁B₁A₁).
2. 2(a) S₂/S₁ = (42 × 0.50 × 1.8 × 10⁻³)/(30 × 0.25 × 3.6 × 10⁻³) = 37.8/27 = 1.4.
3. 3Voltage sensitivity = S/R, so (V₂/V₁) = (S₂/S₁) × (R₁/R₂).
4. 4(b) V₂/V₁ = 1.4 × (10/14) = 1.4/1.4 = 1.

Step-by-step solution

1. 1B = 6.5 G = 6.5 × 10⁻⁴ T; the electron enters perpendicular to B, so F = evB acts perpendicular to v at every instant.
2. 2A force always perpendicular to the velocity changes only the direction, not the speed, and supplies exactly the centripetal force mv²/r — hence the path is a circle of constant radius.
3. 3mv²/r = evB → r = mv/(eB).
4. 4r = (9.1 × 10⁻³¹ × 4.8 × 10⁶)/(1.6 × 10⁻¹⁹ × 6.5 × 10⁻⁴).
5. 5r = (4.37 × 10⁻²⁴)/(1.04 × 10⁻²²) = 4.2 × 10⁻² m.

Step-by-step solution

1. 1T = 2πr/v and r = mv/eB, so T = 2π(mv/eB)/v = 2πm/eB — the speed cancels.
2. 2Hence ν = 1/T = eB/(2πm).
3. 3ν = (1.6 × 10⁻¹⁹ × 6.5 × 10⁻⁴)/(2π × 9.1 × 10⁻³¹).
4. 4ν = 1.04 × 10⁻²²/5.72 × 10⁻³⁰ = 1.82 × 10⁷ Hz ≈ 18 MHz.
5. 5The frequency does not depend on the speed: a faster electron simply moves in a proportionally larger circle, taking the same time per revolution.

Step-by-step solution

1. 1A = πr² = π(0.08)² = 2.01 × 10⁻² m².
2. 2(a) τ = N I A B sinθ with θ = 60° (angle between normal and field).
3. 3τ = 30 × 6.0 × 2.01 × 10⁻² × 1.0 × sin60°.
4. 4τ = 180 × 2.01 × 10⁻² × 0.866 ≈ 3.1 N m.
5. 5The counter-torque must be equal in magnitude and opposite in direction: 3.1 N m.
6. 6(b) The torque on a planar loop is τ = N I (A n̂) × B — it depends only on the enclosed area vector, not on the shape of the boundary. An irregular coil enclosing the same area has the same magnetic moment NIA, so the torque (and the required counter-torque) is unchanged.