Class 12 Physics Chapter 4: Moving Charges and Magnetism worksheet with answer key

Answer: (a) zero

The magnetic force is always perpendicular to the velocity, so it does no work along any part of the path; the speed and kinetic energy stay the same.

Answer: (c) N A⁻¹ m⁻¹

From F = qvB, B = F/(qv), with unit N s C⁻¹ m⁻¹. Since 1 C s⁻¹ = 1 A, this is N A⁻¹ m⁻¹.

Answer: (c) 1/c²

ε₀μ₀ = (4πε₀)(μ₀/4π) = [1/(9 × 10⁹)] × 10⁻⁷ = 1/(3 × 10⁸)² = 1/c².

Answer: (b) 0.79 T

ν_c = qB/2πm, so B = 2πmν_c/q = (2π × 1.67 × 10⁻²⁷ × 12 × 10⁶)/(1.6 × 10⁻¹⁹) = 0.79 T. Leaving out 2π gives 0.13 T.

Answer: (a) on the line of the element itself

|dB| = (μ₀/4π) I dl sin θ/r². On the line of the element θ = 0° or 180°, so sin θ = 0 and the field there is zero.

(a) F = IlB sin θ is largest at θ = 90°: F_max = 2.5 × 0.40 × 0.20 = 0.20 N, with the conductor perpendicular to the field. (b) sin θ = 0.10/0.20 = 0.5, so θ = 30° (or 150°).

• F_max = IlB = 0.20 N, conductor perpendicular to B (1)
• sin θ = 0.5, θ = 30° (1)

F/l = IB sin 90° = 5.0 × 3.5 × 10⁻⁵ = 1.75 × 10⁻⁴ N m⁻¹. With l upwards (ẑ) and B northwards (ŷ), l × B is along ẑ × ŷ = −x̂, that is, towards the west.

• F/l = IB = 1.75 × 10⁻⁴ N m⁻¹ (1)
• direction from l × B: towards the west (1)

The torque τ = m × B is zero when m is parallel or anti-parallel to B. (a) m parallel to B: a small rotation produces a torque that turns it back, so the equilibrium is stable. (b) m anti-parallel to B: a small rotation produces a torque that turns it further away, so it is unstable.

• stable with m parallel to B, restoring torque on small rotation (1)
• unstable with m anti-parallel to B, torque increases the displacement (1)

(a) Between the wires the two fields are opposite (right-hand rule). B = (μ₀/2π)(6.0 − 4.0)/0.050 = 2 × 10⁻⁷ × 40 = 8.0 × 10⁻⁶ T, in the sense of B's field. (b) The fields cancel between the wires where 4.0/x = 6.0/(0.10 − x): 0.40 − 4.0x = 6.0x, x = 0.040 m. The field is zero 4.0 cm from A (6.0 cm from B).

• fields opposite between the wires; uses B = μ₀I/2πr with superposition (1)
• B at mid-point = 8.0 × 10⁻⁶ T (1)
• zero 4.0 cm from A (1)

The sides of length a lie in the plane containing B and the normal; the forces on them are equal, opposite and collinear, so they cancel. The sides of length b each feel F = IbB, equal and opposite, so the net force is zero. These two forces are not collinear: their lines are separated by a sin θ, so they form a couple of moment τ = IbB × a sin θ = I(ab)B sin θ = IAB sin θ. In vector form τ = m × B with m = IA.

• forces on one pair of sides cancel and are collinear (1)
• IbB on the other pair: equal and opposite, net force zero (1)
• couple arm a sin θ, τ = IAB sin θ (1)

(a) The coil and the shunt are in parallel, so they share one potential difference, and the shunt carries I − I_g. With V = IR, I_g·R_G = (I − I_g)·r_s, so r_s = (4.0 × 10⁻³ × 30)/(1.0 − 0.004) = 0.12 Ω. (b) R_Gr_s/(R_G + r_s) = 30 × 0.12/30.12 ≈ 0.12 Ω, close to r_s since R_G ≫ r_s. (c) In series one current I_g flows through the coil and R, and their potential differences add: V = I_g(R_G + R), so R = V/I_g − R_G = 6.0/(4.0 × 10⁻³) − 30 = 1470 Ω.

• shunt r_s = 0.12 Ω (1)
• ammeter resistance ≈ 0.12 Ω (1)
• series resistance 1470 Ω (1)