Magnetism, Electromagnetic Induction and Alternating Current — Practice

Question 1

A charged particle moves parallel to a uniform magnetic field. The magnetic force on it is:
(a) zero (b) maximum (c) qvB (d) equal to its weight

Accepted Answer

(a) zero.

F = qvB sinθ, and for motion parallel to the field θ = 0, so sinθ = 0 and the force is zero.

Question 2

A 1 m long wire carrying 2 A lies perpendicular to a 0.5 T magnetic field. The force on it is:
(a) 1 N (b) 0.5 N (c) 2 N (d) 4 N

Accepted Answer

(a) 1 N.

F = BIL sinθ = 0.5 × 2 × 1 × sin90° = 1 N.

Question 3

The radius of the circular path of a charged particle of mass m, charge q, moving at speed v perpendicular to a magnetic field B is:
(a) mv/qB (b) qB/mv (c) mvB/q (d) qvB/m

Accepted Answer

(a) mv/qB.

The magnetic force provides the centripetal force: qvB = mv²/r, so r = mv/qB.

Question 4

The magnetic field inside a long solenoid with n turns per metre carrying current I is:
(a) μ₀nI (b) μ₀I/2πn (c) μ₀nI/2 (d) μ₀I/2n

Accepted Answer

(a) μ₀nI.

Inside a long solenoid the field is uniform and equal to μ₀nI, independent of the solenoid's radius.

Question 5

A rod of length 1 m moves at 2 m/s perpendicular to a 0.5 T magnetic field. The EMF induced across it is:
(a) 1 V (b) 2 V (c) 0.5 V (d) 4 V

Accepted Answer

(a) 1 V.

Motional EMF = BLv = 0.5 × 1 × 2 = 1 V.

Question 6

When the north pole of a magnet is pushed toward a coil, the induced current in the coil:
(a) opposes the magnet's approach (b) aids the magnet's approach (c) is zero (d) reverses the magnet's polarity

Accepted Answer

(a) opposes the magnet's approach.

By Lenz's law the induced current flows so as to oppose the change in flux, making the coil's near face a north pole that repels the approaching magnet — a consequence of energy conservation.

Question 7

The EMF induced in a coil is proportional to:
(a) the rate of change of magnetic flux (b) the magnetic flux itself (c) the current (d) the resistance

Accepted Answer

(a) the rate of change of magnetic flux.

Faraday's law: ε = −dΦ/dt. A steady flux, however large, induces no EMF; only a changing flux does.

Question 8

An alternating current has a peak value of 10 A. Its rms value is approximately:
(a) 7.07 A (b) 10 A (c) 14.1 A (d) 5 A

Accepted Answer

(a) 7.07 A.

I_rms = I₀/√2 = 10/1.414 ≈ 7.07 A.

Question 9

The Indian mains supply is rated at 220 V (rms). Its peak voltage is approximately:
(a) 311 V (b) 220 V (c) 156 V (d) 440 V

Accepted Answer

(a) 311 V.

V₀ = V_rms × √2 = 220 × 1.414 ≈ 311 V. The quoted 220 V is the rms (effective) value.

Question 10

The reactance of a 0.1 H inductor at 50 Hz is approximately:
(a) 31.4 Ω (b) 5 Ω (c) 0.1 Ω (d) 314 Ω

Accepted Answer

(a) 31.4 Ω.

X_L = 2πfL = 2π × 50 × 0.1 = 10π ≈ 31.4 Ω.

Question 11

The resonant frequency of a series LC circuit is given by:
(a) 1/(2π√(LC)) (b) 2π√(LC) (c) 1/(2πLC) (d) √(LC)/2π

Accepted Answer

(a) 1/(2π√(LC)).

At resonance X_L = X_C, i.e. ωL = 1/ωC, giving ω = 1/√(LC) and f₀ = 1/(2π√(LC)).

Question 12

A transformer has 100 turns in its primary and 1000 in its secondary. If 220 V AC is applied to the primary, the secondary voltage is:
(a) 2200 V (b) 22 V (c) 220 V (d) 1100 V

Accepted Answer

(a) 2200 V.

V_s = V_p × (N_s/N_p) = 220 × (1000/100) = 220 × 10 = 2200 V — a step-up transformer.

Question 13

The energy stored in a 2 H inductor carrying a current of 3 A is:
(a) 9 J (b) 6 J (c) 18 J (d) 3 J

Accepted Answer

(a) 9 J.

E = ½LI² = ½ × 2 × 3² = ½ × 2 × 9 = 9 J.

Question 14

The work done by the magnetic force on a charged particle moving in a uniform magnetic field is:
(a) zero (b) positive (c) negative (d) equal to its kinetic energy

Accepted Answer

(a) zero.

The magnetic force is always perpendicular to the velocity, so it does no work — the particle's speed (and kinetic energy) stays constant while only its direction changes.

Question 15

The magnetic field at distance r from a long straight wire carrying current I is:
(a) μ₀I/2πr (b) μ₀I/2r (c) μ₀nI (d) μ₀I/4πr²

Accepted Answer

(a) μ₀I/2πr.

For a long straight wire, B = μ₀I/2πr, falling off inversely with distance (not inverse-square).

Practice — Magnetism, Electromagnetic Induction and Alternating Current

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