68 free MCQs on Electromagnetic Induction with worked answers and explanations. Faraday's and Lenz's laws, motional EMF, self/mutual inductance, eddy currents, and the AC generator - how a changing magnetic flux creates an electric current.
Take the timed Electromagnetic Induction chapterwise test →Below are 68 practice questions on Electromagnetic Induction, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Electromagnetic Induction notes.
As a magnet approaches a coil, flux rises steadily and a constant non-zero EMF is induced (Faraday's law). Once the magnet stops moving, flux stays constant and the induced EMF drops to zero - EMF only exists while flux is actively changing.
Q1.
Faraday's law of electromagnetic induction states that induced EMF is proportional to:
Answer: B. Rate of change of magnetic flux
Why: Faraday's law: EMF = -d(phi)/dt. Induced EMF equals the negative rate of change of magnetic flux.
Q2.
Lenz's law is a consequence of:
Answer: B. Conservation of energy
Why: Lenz's law: induced current opposes the change causing it: this is energy conservation (opposing change requires work).
Q3.
Electromagnetic induction was discovered by:
Answer: B. Faraday
Why: Michael Faraday discovered electromagnetic induction in 1831, showing that changing magnetic flux induces EMF.
Q4.
Mutual inductance is the property due to which change in current in one coil:
Answer: B. Induces EMF in a neighboring coil
Why: Mutual inductance M: changing current in one coil induces EMF in a nearby coil. M = phi_21/I<sub>1</sub>.
Q5.
Self-inductance of a coil depends on:
Answer: C. Geometry and material only
Why: Self-inductance L depends on coil geometry (number of turns, area, length) and core material, not on current.
Q6.
Magnetic flux is defined as:
Answer: B. B x A x cos(theta)
Why: Magnetic flux phi = B x A x cos(theta) where theta is angle between B and the area normal vector.
Q7.
SI unit of magnetic flux is:
Answer: B. Weber
Why: Magnetic flux is measured in Weber (Wb). 1 Wb = 1 V x s = 1 T x m<sup>2.</sup>
Q8.
An induced EMF is generated in a coil whenever there is a change in:
Answer: C. Magnetic flux linked with the coil
Why: According to Faraday's law, an EMF is induced in a coil only when the magnetic flux linked with it changes with time.
Q9.
A device that converts mechanical energy into electrical energy using the principle of electromagnetic induction is called a:
Answer: D. Generator
Why: A generator uses electromagnetic induction, typically by rotating a coil in a magnetic field, to convert mechanical energy into electrical energy.
Q10.
The direction of an induced current, as given by Lenz's law, is such that it:
Answer: A. Opposes the change in magnetic flux that produced it
Why: Lenz's law states that the induced current flows in a direction that opposes the change in magnetic flux causing it, consistent with the conservation of energy.
Q11.
Electromagnetic induction is the production of an EMF by a changing:
Answer: A. magnetic flux
Why: A changing magnetic flux through a circuit induces an EMF (Faraday’s law).
Q12.
Electromagnetic induction was discovered by:
Answer: A. Michael Faraday
Why: Faraday discovered electromagnetic induction in 1831.
Q13.
The induced EMF always opposes the change that produces it. This is:
Answer: A. Lenz's law
Why: Lenz's law expresses the opposition, reflected in the minus sign of Faraday's law.
Q14.
The SI unit of magnetic flux is the:
Answer: A. weber
Why: Magnetic flux is measured in weber (Wb).
Q15.
A current produced by a changing magnetic field is called an:
Answer: A. induced current
Why: Such a current, driven by an induced EMF, is called an induced current.
Q16.
Moving a magnet toward a coil ___ an EMF in the coil:
Answer: A. induces
Why: The changing flux as the magnet approaches induces an EMF in the coil.
Q17.
The SI unit of inductance is the:
Answer: A. henry
Why: Inductance is measured in henry (H).
Q18.
A device that uses electromagnetic induction to generate electricity is the:
Answer: A. generator
Why: An electric generator converts mechanical energy to electrical energy by induction.
Q19.
A transformer works on the principle of:
Answer: A. mutual induction
Why: A transformer transfers energy between coils by mutual induction.
Q20.
The faster the magnetic flux through a coil changes, the ___ the induced EMF:
Answer: A. greater
Why: Induced EMF is proportional to the rate of change of flux, so a faster change gives a larger EMF.
Q21.
Induced EMF in a loop rotating in magnetic field at angular velocity omega (area A, field B):
Answer: C. BAomega sin(omega t)
Why: Phi = BA cos(omega t). EMF = -d(phi)/dt = BA omega sin(omega t). Maximum EMF = BAomega.
Q22.
EMF induced in a straight conductor of length L moving at velocity v perpendicular to field B:
Answer: B. BLv
Why: Motional EMF = BLv (rod moving perpendicular to both L and B). This is called motional EMF.
Q23.
In an AC generator, the frequency of induced EMF depends on:
Answer: C. Angular velocity of coil rotation
Why: Frequency f = omega/(2 pi) where omega is angular velocity of rotation. f depends on rotation speed.
Q24.
A coil of self-inductance 100 mH carries current changing at 10 A/s. Induced EMF:
Answer: B. 1 V
Why: EMF = -L x dI/dt = 100 x 10<sup>-3</sup> x 10 = 1 V.
Q25.
Energy stored in an inductor L carrying current I:
Answer: C. (1/2)LI<sup>2</sup>
Why: Magnetic energy in inductor U = (1/2)LI<sup>2.</sup> Analogous to (1/2)mv<sup>2</sup> for kinetic energy.
Q26.
When a bar magnet is moved toward a coil, induced current creates a:
Answer: A. Magnetic north pole facing the magnet
Why: By Lenz's law, induced current opposes the magnet's approach. It creates a north pole (repulsion) facing the approaching north pole of magnet.
Q27.
A copper ring is dropped from rest above a strong magnet, oriented so it falls coaxially toward the magnet's pole. Compared to a similar non-conducting ring dropped from the same height, the copper ring:
Answer: A. Falls more slowly as it approaches the magnet, due to the retarding force from induced eddy currents opposing the change in flux
Why: As the copper ring approaches the magnet, the changing flux induces a current that, by Lenz's law, opposes the motion, producing a retarding force that the non-conducting ring does not experience.
Q28.
Two coils, P and Q, are placed near each other. When the current in coil P changes at a rate of 5 A/s, an EMF of 0.02 V is induced in coil Q. The mutual inductance between the coils is:
Answer: B. 4 mH
Why: EMF = M(dI/dt), so M = EMF/(dI/dt) = 0.02/5 = 4 × 10⁻³ H = 4 mH.
Q29.
A conducting rod slides on two parallel rails in a region with a uniform magnetic field perpendicular to the plane of the rails, generating a motional EMF. If the rod's speed is doubled while the field and rod length remain unchanged, the induced EMF:
Answer: C. Doubles
Why: Motional EMF is given by ε = Bvl, which is directly proportional to velocity v, so doubling v doubles the EMF.
Q30.
A solenoid has self-inductance L. If the number of turns per unit length is doubled while keeping the same length and cross-sectional area, the self-inductance becomes:
Answer: D. 4L
Why: Self-inductance of a solenoid L = μ0n²Al, which depends on the square of the number of turns per unit length, so doubling n makes L four times as large.
Q31.
Faraday’s law states that the induced EMF equals the rate of change of:
Answer: A. magnetic flux
Why: ε = −dΦ/dt, the negative rate of change of magnetic flux.
Q32.
Lenz’s law is a direct consequence of the conservation of:
Answer: A. energy
Why: The opposition described by Lenz’s law ensures energy is conserved.
Q33.
The EMF induced in a rod of length L moving at speed v perpendicular to a field B is:
Answer: A. BLv
Why: The motional EMF is ε = BLv.
Q34.
Self-inductance of a coil opposes any change in its:
Answer: A. current
Why: A changing current induces a back-EMF that opposes the change, the essence of self-inductance.
Q35.
A step-up transformer increases the:
Answer: A. voltage
Why: A step-up transformer raises voltage (while lowering current, conserving power).
Q36.
The energy stored in an inductor of inductance L carrying current I is:
Answer: A. ½LI²
Why: The magnetic energy stored is U = ½LI².
Q37.
Eddy currents are induced currents that circulate in:
Answer: A. bulk conductors
Why: A changing flux through a solid conductor sets up swirling eddy currents within it.
Q38.
For a coil of N turns, the induced EMF is −N times the rate of change of:
Answer: A. flux
Why: ε = −N dΦ/dt; the turns multiply the effect of the changing flux.
Q39.
A transformer operates only with ___ current:
Answer: A. alternating
Why: A transformer needs a changing flux, so it works only on alternating current.
Q40.
The mutual inductance between two coils depends on their:
Answer: A. their shapes and spacing
Why: Mutual inductance depends on the coils’ shapes, sizes and how they are placed relative to each other.
Q41.
A rectangular loop of area 0.02 m² lies in a field B = 0.5 T with its normal at 30° to B. The magnetic flux through the loop is:
Answer: A. 0.00866 Wb
Why: Φ = B·A·cosθ = 0.5 × 0.02 × cos30° = 0.5 × 0.02 × 0.866 ≈ 0.00866 Wb.
Q42.
A square coil (N=100, side 10 cm, R=10 ohm) rotates at 600 rpm in B=0.1 T. Peak induced EMF:
Answer: C. 6.28 V
Why: omega = 600/60 x 2pi = 20pi rad/s. E<sub>0</sub> = NBAomega = 100 x 0.1 x 0.01 x 20pi = 2pi = 6.28 V.
Q43.
Mutual inductance M of two coaxial solenoids (inner: n<sub>1</sub> turns/m, area A; outer: n<sub>2</sub> turns/m, length l):
Answer: A. mu<sub>0</sub> n<sub>1</sub> n<sub>2</sub> Al
Why: M = mu<sub>0</sub> n<sub>1</sub> n<sub>2</sub> A l (for closely wound coaxial solenoids). Flux linkage from inner to outer = n<sub>2</sub> l x mu<sub>0</sub> n<sub>1</sub> I x A.
Q44.
A conducting rod of length L rotates at angular velocity omega about one end in a perpendicular field B. EMF generated:
Answer: B. (1/2)BL<sup>2</sup> omega
Why: EMF of rotating rod = (1/2)BL<sup>2</sup> omega. Derived by integrating motional EMF: E = integral(0 to L) B x (omega r) dr = (1/2)B omega L<sup>2.</sup>
Q45.
Eddy currents in a transformer core are minimized by:
Answer: C. Using laminated iron core
Why: Eddy currents flow in loops in the core. Laminated core (thin insulated sheets) breaks these paths and reduces eddy current losses.
Q46.
A rectangular coil enters a uniform magnetic field region at constant velocity. During entry, the induced current:
Answer: D. Is constant during entry, zero when fully inside
Why: As coil enters: flux increases at constant rate (constant velocity), so constant EMF and constant current. When fully inside: no flux change, no current.
Q47.
AC generator operates at 50 Hz. Coil has 200 turns, area 0.01 m<sup>2</sup> in field 0.5 T. Peak EMF:
Answer: A. 100pi V
Why: E<sub>0</sub> = NBAomega = 200 x 0.5 x 0.01 x 2pi x 50 = 200 x 0.5 x 0.01 x 100pi = 100pi V.
Q48.
A metallic rod of length L rotates with angular velocity ω about a perpendicular axis through its centre (not through one end), in a uniform magnetic field B parallel to the axis. The potential difference between the two ends of the rod is:
Answer: C. Zero, since the EMFs generated in the two half-rods on either side of the centre are equal and opposite
Why: Each half of the rod (length L/2) generates an EMF of Bω(L/2)²/2 directed from the centre outward, so the two halves produce equal and opposite EMFs from the centre to each end, making the net potential difference between the two ends zero.
Q49.
A long solenoid of cross-sectional area A and n turns per unit length carries a current that varies as I = I<sub>0</sub> sin(ωt). A small circular coil of N turns and area a (a << A) is placed coaxially inside the solenoid. The peak EMF induced in the small coil is:
Answer: D. μ0nNaI0ω
Why: The field inside the solenoid is B = μ0nI, so flux through the small coil is φ = μ0nI×a×N, and peak EMF = N×μ0n×a×(dI/dt)_max = μ0nNaI0ω.
Q50.
A metal disc rotates with angular velocity ω about an axis through its centre, perpendicular to its plane, in a uniform magnetic field B parallel to the axis (Faraday disc dynamo). The EMF induced between the centre and the rim of the disc (radius R) is:
Answer: A. BωR²/2
Why: Treating the disc as made of rotating radial rods, the EMF between centre and rim works out to BωR²/2, the same formula as for a single rotating rod of length R about one end.
Q51.
In an ideal transformer, the ratio of secondary to primary voltage equals:
Answer: A. Ns/Np
Why: Vs/Vp = Ns/Np, the ratio of the numbers of turns.
Q52.
The direction of an induced current is found using:
Answer: A. Lenz's law or the right-hand rule
Why: Lenz's law, or equivalently Fleming's right-hand rule, gives the induced-current direction.
Q53.
The self-inductance of a long solenoid is L = μ₀n²Al, where n is the number of turns per unit:
Answer: A. length
Why: n is the turn density, i.e. turns per unit length of the solenoid.
Q54.
In a step-down transformer, the number of secondary turns is ___ the number of primary turns:
Answer: A. fewer than
Why: A step-down transformer has fewer secondary turns, giving a lower output voltage.
Q55.
Eddy-current losses in a transformer core are reduced by using:
Answer: A. thin laminated sheets
Why: Laminating the core with insulated sheets breaks up the eddy-current paths, cutting the losses.
Q56.
A metal plate swinging between the poles of a magnet is quickly damped because of:
Answer: A. eddy currents
Why: Induced eddy currents oppose the motion (Lenz’s law), damping the swing.
Q57.
The EMF induced in a coil rotating steadily in a uniform magnetic field is:
Answer: A. sinusoidal (alternating)
Why: A uniformly rotating coil produces a sinusoidally varying EMF - the basis of an AC generator.
Q58.
In the relation ε₂ = −M(dI₁/dt) between two coils, the quantity M is the:
Answer: A. mutual inductance
Why: M is the mutual inductance linking the changing current in one coil to the EMF in the other.
Q59.
An ideal transformer conserves ___, so that VpIp = VsIs:
Answer: A. power
Why: With no losses, input power equals output power, giving VpIp = VsIs.
Q60.
The unit henry is equivalent to:
Answer: A. volt-second per ampere
Why: From ε = −L dI/dt, 1 H = 1 V·s/A.
Q61.
The magnetic flux through a coil changes by 0.2 Wb in 0.1 s. The average induced emf is:
Answer: C. 2 V
Why: emf = ΔΦ/Δt = 0.2/0.1 = 2 V.
Q62.
A conducting rod of length 0.2 m moves at 10 m/s perpendicular to a 0.5 T field. The motional emf induced is:
Answer: C. 1 V
Why: emf = BLv = 0.5·0.2·10 = 1 V.
Q63.
The energy stored in an inductor of 2 H carrying a current of 3 A is:
Answer: C. 9 J
Why: U = (1/2)LI² = 0.5·2·9 = 9 J.
Q64.
According to Lenz law, the induced current in a circuit always flows so as to:
Answer: A. oppose the change in flux that produces it
Why: Lenz law states the induced current opposes the change in magnetic flux, consistent with energy conservation.
Q65.
Two coils have mutual inductance 0.5 H. If the current in one changes at 4 A/s, the emf induced in the other is:
Answer: B. 2 V
Why: emf = M dI/dt = 0.5·4 = 2 V.
Q66.
A coil of 100 turns links a flux of 0.01 Wb per turn when it carries 2 A. Its self-inductance is:
Answer: B. 0.5 H
Why: L = NΦ/I = 100·0.01/2 = 0.5 H.
Q67.
The peak emf of a coil rotating in a magnetic field is NBAω. If the angular speed ω is doubled, the peak emf:
Answer: B. doubles
Why: Peak emf ∝ ω, so doubling ω doubles the peak emf.
Q68.
Eddy currents induced in a metal plate moving through a magnetic field primarily cause:
Answer: A. damping and heating
Why: Eddy currents dissipate energy as heat and oppose the motion, producing electromagnetic damping.