JEE Physics Electromagnetic Induction MCQs Set 19

Practice MCQs for JEE Physics Electromagnetic Induction

Review structured MCQ sets for JEE Physics Electromagnetic Induction. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

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Access the complete set of multiple-choice questions for Electromagnetic Induction below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: Two coils of self-inductances L1 and L2 are placed so close together that the effective flux in one coil is completely linked with the other. If M is the mutual inductance between them, then
(a) \( M=L_1L_2 \)
(b) \( M=L_1/L_2 \)
(c) \( M=(L_1L_2)^2 \)
(d) \( M=\sqrt{L_1L_2} \)
Show Answer & Explanation

Answer: (d) \( M=\sqrt{L_1L_2} \)

Explanation:
1. For perfect coupling the coupling coefficient is 1: M = √(L1 L2).

Teacher's Note:
1. Recall the basic law.

Question 2: A coil of insulated wire is connected to a battery. If it is connected to a galvanometer, its pointer is deflected because
(a) an induced current is set up
(b) no induced current is set up
(c) the coil behaves as a magnet
(d) the number of turns is changed
Show Answer & Explanation

Answer: (a) an induced current is set up

Explanation:
1. When the battery is connected or disconnected the current changes.
2. The changing flux sets up an induced current in the galvanometer circuit.

Teacher's Note:
1. Recall the basic law.

Question 3: A conductor of length 0.4 m moves with a speed of 7 m/s perpendicular to a magnetic field of intensity 0.9 Wb/m². The induced emf across the conductor is
(a) 1.26 V
(b) 2.52 V
(c) 5.04 V
(d) 25.2 V
Show Answer & Explanation

Answer: (b) 2.52 V

Explanation:
1. e = B l v = 0.9 × 0.4 × 7 = 2.52 V.

Teacher's Note:
1. Recall the basic law.

Question 4: Eddy currents do not cause
(a) damping
(b) heating
(c) sparking
(d) loss of energy
Show Answer & Explanation

Answer: (c) sparking

Explanation:
1. Eddy currents cause damping, heating and loss of energy, but not sparking.

Teacher's Note:
1. Recall the basic law.

Question 5: Two pure inductors, each of self-inductance L and well separated from each other, are connected in series. The net inductance is
(a) L
(b) 2L
(c) L/2
(d) L/4
Show Answer & Explanation

Answer: (b) 2L

Explanation:
1. Inductors in series (with no mutual coupling) add: L + L = 2L.

Teacher's Note:
1. Recall the basic law.

Question 6: Two coaxial loops P and Q are separated by some distance. A clockwise current I_P (as seen from the end E on the far side of Q) is set up in P when a switch is closed, and an induced current I_Q1 flows in Q. When the switch is opened after a long time, a current I_Q2 flows in Q. The directions of I_Q1 and I_Q2 (as seen from E) are
(a) respectively clockwise and anticlockwise
(b) both clockwise
(c) both anticlockwise
(d) respectively anticlockwise and clockwise
Show Answer & Explanation

Answer: (d) respectively anticlockwise and clockwise

Explanation:
1. When the current in P grows, Q opposes the increase, so I_Q1 is opposite to I_P (anticlockwise).
2. When the current in P dies away, Q opposes the decrease, so I_Q2 is in the same sense as I_P (clockwise).

Teacher's Note:
1. Opposes the increase, supports the decrease.

Question 7: The henry is the SI unit of
(a) resistance
(b) capacity
(c) inductance
(d) current
Show Answer & Explanation

Answer: (c) inductance

Explanation:
1. Henry is the unit of inductance.

Teacher's Note:
1. Recall the basic law.

Question 8: A choke coil works on the principle of
(a) transient current
(b) self induction
(c) mutual induction
(d) wattless current
Show Answer & Explanation

Answer: (b) self induction

Explanation:
1. A choke coil reduces a.c. current by its large self-induced back emf.

Teacher's Note:
1. Recall the basic law.

Question 9: A small square loop of wire of side l is placed inside a large square loop of side L (L much greater than l). The loops are coplanar and their centres coincide. The mutual inductance of the system is proportional to
(a) \( \frac lL \)
(b) \( \frac{l^2}{L} \)
(c) \( \frac Ll \)
(d) \( \frac{L^2}{l} \)
Show Answer & Explanation

Answer: (b) \( \frac{l^2}{L} \)

Explanation:
1. The field at the centre of the large loop is proportional to 1/L and is nearly uniform over the small loop.
2. Flux = B l² is proportional to l²/L, so M is proportional to l²/L.

Teacher's Note:
1. Recall the basic law.

Question 10: A six-pole generator with fixed field excitation develops an emf of 100 V when operating at 1500 r.p.m. At what speed must it rotate to develop 120 V?
(a) 1200 r.p.m
(b) 1800 r.p.m
(c) 1500 r.p.m
(d) 400 r.p.m
Show Answer & Explanation

Answer: (b) 1800 r.p.m

Explanation:
1. The emf is proportional to the speed (ω) for a fixed field.
2. N = 1500 × 120/100 = 1800 r.p.m.

Teacher's Note:
1. Recall the basic law.

Question 11: Two coils A and B are linked such that an emf ε is induced in B when the current in A changes at the rate I (A per second). If the current i is now made to flow in B, the flux linked with A will be
(a) \( (\varepsilon/I)\,i \)
(b) \( \varepsilon\,i\,I \)
(c) \( (\varepsilon I)\,i \)
(d) \( iI/\varepsilon \)
Show Answer & Explanation

Answer: (a) \( (\varepsilon/I)\,i \)

Explanation:
1. M = ε/(dI/dt) = ε/I.
2. The flux linked with A when i flows in B is M i = (ε/I) i.

Teacher's Note:
1. φ = M i.

Question 12: A coil of 50 turns is pulled in 0.02 s between the poles of a magnet, where its area is changed so that the flux changes from 31 × 10⁻⁶ Wb to 1 × 10⁻⁶ Wb. The average emf is
(a) \( 7.5\times10^{-2}\ \text{V} \)
(b) \( 7.5\times10^{-3}\ \text{V} \)
(c) zero
(d) \( 7.5\times10^{-4}\ \text{V} \)
Show Answer & Explanation

Answer: (a) \( 7.5\times10^{-2}\ \text{V} \)

Explanation:
1. e = NΔφ/t = 50 × 30 × 10⁻⁶/0.02 = 7.5 × 10⁻² V.

Teacher's Note:
1. Recall the basic law.

Question 13: A coil has inductance 5 mH. If the current increases from zero to 1 A in 0.1 s, the magnitude of the induced emf is
(a) 0.005 V
(b) 0.5 V
(c) 0.05 V
(d) 5 V
Show Answer & Explanation

Answer: (c) 0.05 V

Explanation:
1. e = L ΔI/Δt = 5 × 10⁻³ × 1/0.1 = 0.05 V.

Teacher's Note:
1. Recall the basic law.

Question 14: A 100 millihenry coil carries a current of 1 ampere. The energy stored in its magnetic field is
(a) 0.5 J
(b) 1 J
(c) 0.05 J
(d) 0.1 J
Show Answer & Explanation

Answer: (c) 0.05 J

Explanation:
1. U = (1/2) L I² = 0.5 × 0.1 × 1 = 0.05 J.

Teacher's Note:
1. Recall the basic law.

Question 15: A coil of resistance 400 ohm is placed in a magnetic field. If the magnetic flux φ (Wb) linked with the coil varies with time t (s) as φ = 50t² + 4, the current in the coil at t = 2 s is
(a) 0.5 A
(b) 0.1 A
(c) 2 A
(d) 1 A
Show Answer & Explanation

Answer: (a) 0.5 A

Explanation:
1. e = dφ/dt = 100t = 200 V at t = 2 s.
2. I = e/R = 200/400 = 0.5 A.

Teacher's Note:
1. Recall the basic law.

Question 16: Two solenoids of the same cross-sectional area have lengths and numbers of turns in the ratio 1 : 2. The ratio of the self-inductances of the two solenoids is
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 4
Show Answer & Explanation

Answer: (b) 1 : 2

Explanation:
1. L = μ₀N²A/l.
2. L1/L2 = (N1²/l1)/(N2²/l2) = (1/1)/(4/2) = 1/2.

Teacher's Note:
1. Recall the basic law.

Question 17: When the plane of the armature of an a.c. generator is parallel to the field in which it is rotating,
(a) both the flux linked and induced emf in the coil are zero
(b) the flux linked with it is zero, while the induced emf is maximum
(c) the flux linked is maximum while the induced emf is zero
(d) both the flux and the emf have their respective maximum values
Show Answer & Explanation

Answer: (b) the flux linked with it is zero, while the induced emf is maximum

Explanation:
1. Flux = NBA cos(ωt) is zero when the plane is parallel to B.
2. The emf, which is proportional to dφ/dt, is maximum then.

Teacher's Note:
1. Recall the basic law.

Question 18: The emf is induced in a single isolated coil due to ____ A ____ of flux through the coil by means of varying the current through the same coil. This phenomenon is called ____ B ____. A and B refer to
(a) constancy, mutual induction
(b) change, self induction
(c) constancy, self induction
(d) change, mutual induction
Show Answer & Explanation

Answer: (b) change, self induction

Explanation:
1. A change of flux through a coil due to its own current is self induction.

Teacher's Note:
1. Recall the basic law.

Question 19: The laws of electromagnetic induction have been used in the construction of a
(a) galvanometer
(b) voltmeter
(c) electric motor
(d) generator
Show Answer & Explanation

Answer: (d) generator

Explanation:
1. A generator converts mechanical energy into electrical energy by electromagnetic induction.

Teacher's Note:
1. Recall the basic law.

Question 20: The mutual inductance of a pair of coils is 0.75 H. If the current in the primary changes from 0.5 A to zero in 0.01 s, find the average induced emf in the secondary coil.
(a) 25.5 V
(b) 12.5 V
(c) 22.5 V
(d) 37.5 V
Show Answer & Explanation

Answer: (d) 37.5 V

Explanation:
1. e = M ΔI/Δt = 0.75 × 0.5/0.01 = 37.5 V.

Teacher's Note:
1. Recall the basic law.

Question 21: The flux through a loop perpendicular to its plane varies with time as φ = 6t² + 7t + 1, where φ is in milliweber and t in seconds. The magnitude of the emf induced in the loop at t = 2 s is
(a) 19 mV
(b) 31 mV
(c) 43 mV
(d) 55 mV
Show Answer & Explanation

Answer: (b) 31 mV

Explanation:
1. e = dφ/dt = 12t + 7 = 12 × 2 + 7 = 31 mV.

Teacher's Note:
1. e = dφ/dt.

Question 22: The coils in resistance boxes are made of doubled insulated wire to nullify the effect of
(a) heating
(b) magnetism
(c) pressure
(d) none of these
Show Answer & Explanation

Answer: (b) magnetism

Explanation:
1. The two halves of the double winding carry currents in opposite directions, so their magnetic fields cancel and the coil is non-inductive.

Teacher's Note:
1. Opposite currents cancel the field.

Question 23: A metallic rod of length l is tied to a string of length 2l and made to rotate with angular speed ω on a horizontal table, with one end of the string fixed. There is a vertical magnetic field B in the region. The emf induced across the ends of the rod is
(a) \( \frac{2B\omega l^2}{2} \)
(b) \( \frac{3B\omega l^2}{2} \)
(c) \( \frac{4B\omega l^2}{2} \)
(d) \( \frac{5B\omega l^2}{2} \)
Show Answer & Explanation

Answer: (d) \( \frac{5B\omega l^2}{2} \)

Explanation:
1. The rod spans radial distances 2l to 3l from the fixed point.
2. emf = (1/2) B ω [(3l)² - (2l)²] = (1/2) B ω (5l²) = 5Bωl²/2.

Teacher's Note:
1. Recall the basic law.

Question 24: If the magnetic flux linked with a coil through which a current of x A is set up is y Wb, the coefficient of self-inductance of the coil is
(a) (x - y) henry
(b) \( \frac xy\ \text{henry} \)
(c) \( \frac yx\ \text{henry} \)
(d) x y henry
Show Answer & Explanation

Answer: (c) \( \frac yx\ \text{henry} \)

Explanation:
1. L = φ/I = y/x henry.

Teacher's Note:
1. Recall the basic law.

Question 25: A dynamo converts
(a) mechanical energy into thermal energy
(b) electrical energy into thermal energy
(c) thermal energy into electrical energy
(d) mechanical energy into electrical energy
Show Answer & Explanation

Answer: (d) mechanical energy into electrical energy

Explanation:
1. A dynamo is a generator that changes mechanical energy into electrical energy.

Teacher's Note:
1. Recall the basic law.

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