JEE Physics Electromagnetic Induction MCQs Set 03

Physics Objective Questions and Answers: Electromagnetic Induction

Access targeted multiple-choice questions for Electromagnetic Induction designed to align with the latest JEE academic syllabus for JEE Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Multiple Choice Questions

Question 1: The SI unit of magnetic flux is
(a) ampere/metre²
(b) weber
(c) gauss
(d) oersted
Show Answer & Explanation

Answer: (b) weber

Explanation:
1. Flux \( \phi=BA \): tesla × metre² = weber (Wb).
2. Gauss is a CGS unit of field and oersted a CGS unit of H.

Teacher's Note:
1. 1 Wb = 1 T m².
2. Gauss is a unit of B, not flux.

Question 2: A coil of area 2 m² is placed in a magnetic field that changes uniformly from 1 Wb/m² to 4 Wb/m² in 2 s. The emf induced in the coil is
(a) 4 V
(b) 3 V
(c) 2 V
(d) 1 V
Show Answer & Explanation

Answer: (b) 3 V

Explanation:
1. \( \varepsilon=\frac{A\,\Delta B}{\Delta t}=\frac{2\times3}{2}=3 \) V.

Teacher's Note:
1. The field is perpendicular to the plane, so use A ΔB.
2. Rate of change is uniform, so emf is constant.

Question 3: A closed coil of 500 turns on a rectangular frame of area 4.0 cm² has resistance 50 Ω. It is kept with its plane perpendicular to a uniform field of 0.2 Wb/m². The charge that flows through the coil if it is turned over (rotated through 180°) is
(a) \( 1.6\times10^{-3} \) C
(b) \( 16\times10^{-3} \) C
(c) \( 0.16\times10^{-3} \) C
(d) \( 160\times10^{-3} \) C
Show Answer & Explanation

Answer: (a) \( 1.6\times10^{-3} \) C

Explanation:
1. Flux change \( \Delta\phi=2NBA=2\times500\times0.2\times4\times10^{-4}=0.08 \) Wb.
2. \( Q=\frac{\Delta\phi}{R}=\frac{0.08}{50}=1.6\times10^{-3} \) C.

Teacher's Note:
1. Turning over reverses the flux, so ΔΦ = 2NBA.
2. Charge does not depend on how fast it is done.

Question 4: A coil of 200 turns and area 70 cm² is placed in a field of 0.8 Wb/m² perpendicular to its plane. It is rotated through 180° in 0.1 s. The induced emf is
(a) 11.2 V
(b) 1.12 V
(c) 22.4 V
(d) 2.24 V
Show Answer & Explanation

Answer: (c) 22.4 V

Explanation:
1. \( \varepsilon=\frac{2NBA}{t}=\frac{2\times200\times0.8\times70\times10^{-4}}{0.1}=22.4 \) V.

Teacher's Note:
1. Convert cm² to m².
2. 180° rotation changes flux by 2NBA.

Question 5: A solenoid of inductance 10 H and resistance 2 Ω is connected to a 10 V battery. How long does it take for the magnetic energy to reach 1/4 of its maximum value?
(a) 3.5 s
(b) 2.5 s
(c) 5.5 s
(d) 7.5 s
Show Answer & Explanation

Answer: (a) 3.5 s

Explanation:
1. \( I_{max}=\frac{10}{2}=5 \) A and \( \tau=\frac LR=5 \) s.
2. Energy \( \propto I^2 \); one quarter of the energy means \( I=\frac{I_{max}}2 \).
3. \( 1-e^{-t/5}=\frac12\Rightarrow t=5\ln2\approx3.5 \) s.

Teacher's Note:
1. Quarter of the energy is half of the current.
2. Use I = I_max(1 − e^(−t/τ)).

Question 6: A current of 2 A gives rise to a magnetic flux of 5 × 10⁻⁵ Wb through each turn of a coil of 100 turns. The magnetic energy stored in the medium surrounding the coil is
(a) 5 J
(b) \( 5\times10^{-7} \) J
(c) \( 5\times10^{-3} \) J
(d) 0.5 J
Show Answer & Explanation

Answer: (c) \( 5\times10^{-3} \) J

Explanation:
1. Total linkage \( N\phi=100\times5\times10^{-5}=5\times10^{-3} \) Wb-turn, so \( L=\frac{N\phi}{I}=2.5\times10^{-3} \) H.
2. \( U=\tfrac12LI^2=\tfrac12\times2.5\times10^{-3}\times4=5\times10^{-3} \) J.

Teacher's Note:
1. U = ½ N φ I.
2. Do not forget the number of turns.

Question 7: The inductance of a solenoid is 5 H and its resistance 5 Ω. It is connected to a 10 V battery. The time taken for the current to reach 9/10 of its maximum value is
(a) 4.0 s
(b) 2.3 s
(c) 1.4 s
(d) 1.2 s
Show Answer & Explanation

Answer: (b) 2.3 s

Explanation:
1. \( \tau=\frac LR=1 \) s.
2. \( 1-e^{-t/\tau}=0.9\Rightarrow t=\tau\ln10=2.3 \) s.

Teacher's Note:
1. The battery voltage does not matter for the time.
2. ln 10 = 2.303.

Question 8: The time constant of an inductance coil is 2.0 × 10⁻³ s. When a 90 Ω resistor is joined in series, the time constant becomes 0.5 × 10⁻³ s. The inductance and resistance of the coil are
(a) 30 mH; 30 Ω
(b) 30 mH; 60 Ω
(c) 60 mH; 30 Ω
(d) 60 mH; 60 Ω
Show Answer & Explanation

Answer: (c) 60 mH; 30 Ω

Explanation:
1. \( \frac LR=2\times10^{-3} \) and \( \frac{L}{R+90}=0.5\times10^{-3} \).
2. Dividing: \( \frac{R+90}{R}=4\Rightarrow R=30\,\Omega \).
3. \( L=2\times10^{-3}\times30=60 \) mH.

Teacher's Note:
1. Take the ratio of the two time constants.
2. L = τ R.

Question 9: The energy of a choke coil is stored in the form of
(a) heat
(b) electric field
(c) magnetic field
(d) electromagnetic radiation
Show Answer & Explanation

Answer: (c) magnetic field

Explanation:
1. The energy \( \frac12LI^2 \) resides in the magnetic field of the coil.

Teacher's Note:
1. A choke is an inductor.
2. Heat is dissipated in its resistance, not stored.

Question 10: A coil of resistance 10 Ω and inductance 5 H is connected to a 100 V battery. The energy stored in the coil in the steady state is
(a) 125 erg
(b) 125 J
(c) 250 erg
(d) 250 J
Show Answer & Explanation

Answer: (d) 250 J

Explanation:
1. Steady current \( I=\frac{100}{10}=10 \) A.
2. \( U=\tfrac12LI^2=\tfrac12\times5\times100=250 \) J.

Teacher's Note:
1. In steady state the inductor has no back emf.
2. Use ½LI².

Question 11: When the current in a coil is reduced from 2 A to 1 A in 1 ms, the induced emf is 5 V. The inductance of the coil is
(a) 5 H
(b) 5000 H
(c) 5 mH
(d) 50 H
Show Answer & Explanation

Answer: (c) 5 mH

Explanation:
1. \( L=\frac{\varepsilon\Delta t}{\Delta I}=\frac{5\times10^{-3}}{1}=5\times10^{-3} \) H.

Teacher's Note:
1. 1 ms = 10⁻³ s.
2. The sign of emf is irrelevant here.

Question 12: A small square loop 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. Field at the centre due to the large loop \( B\propto\frac IL \).
2. Flux through the small loop \( \phi\approx B\,l^2\propto\frac{Il^2}{L} \), so \( M=\frac{\phi}{I}\propto\frac{l^2}{L} \).

Teacher's Note:
1. Treat the field as uniform over the small loop.
2. M depends only on geometry.

Question 13: Which type of loss does NOT occur in a transformer?
(a) mechanical loss
(b) copper loss
(c) hysteresis loss
(d) eddy current loss
Show Answer & Explanation

Answer: (a) mechanical loss

Explanation:
1. A transformer has no moving parts, so there is no mechanical (friction) loss.
2. Copper, hysteresis and eddy-current losses do occur (plus flux leakage).

Teacher's Note:
1. Static device, no friction.
2. Laminations reduce eddy loss.

Question 14: The primary and secondary windings of a transformer have 100 and 300 turns. If the input power is 60 W, the output power of an ideal transformer will be
(a) 240 W
(b) 180 W
(c) 60 W
(d) 20 W
Show Answer & Explanation

Answer: (c) 60 W

Explanation:
1. An ideal transformer has no losses: output power = input power.
2. The turns ratio changes voltage and current but not power.

Teacher's Note:
1. Do not multiply power by the turns ratio.
2. Efficiency 100% for ideal.

Question 15: A step-up transformer has a turn ratio of 10 : 1. A cell of emf 2 V is fed to the primary. The secondary voltage developed is
(a) 20 V
(b) 10 V
(c) 2 V
(d) zero
Show Answer & Explanation

Answer: (d) zero

Explanation:
1. A cell gives a steady current, so the flux in the core is constant.
2. No changing flux means no induced emf in the secondary: zero.

Teacher's Note:
1. Transformers work only on alternating (changing) current.
2. A steady current gives zero secondary emf.

Question 16: A conducting rod of length 2l rotates with constant angular speed ω about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The emf induced between the two ends of the rod is
(a) \( B\omega l^2 \)
(b) \( \tfrac12B\omega l^2 \)
(c) \( \tfrac18B\omega l^2 \)
(d) zero
Show Answer & Explanation

Answer: (d) zero

Explanation:
1. Each half gives \( \tfrac12B\omega l^2 \) between the centre and its end, with the centre at the same potential relative to both ends.
2. Both ends reach the same potential relative to the centre, so the emf between the ends is zero.

Teacher's Note:
1. Compare each half with the pivot.
2. Symmetrical halves cancel.

Question 17: A metallic disc of radius R rotates about its geometrical axis with constant angular speed ω in an external field B perpendicular to the disc. The induced emf between the centre and any peripheral point is
(a) \( \pi\omega BR^2 \)
(b) \( \omega BR^2 \)
(c) \( \frac{\pi\omega BR^2}{2} \)
(d) \( \frac{\omega BR^2}{2} \)
Show Answer & Explanation

Answer: (d) \( \frac{\omega BR^2}{2} \)

Explanation:
1. \( \varepsilon=\int_0^R B\omega r\,dr=\tfrac12B\omega R^2 \).

Teacher's Note:
1. The disc is a collection of radial rods.
2. Same result as for a rod rotating about its end.

Question 18: A transformer is used to light a 100 W, 110 V lamp from 220 V mains. If the main current is 0.5 A, the efficiency of the transformer is approximately
(a) 10%
(b) 30%
(c) 50%
(d) 90%
Show Answer & Explanation

Answer: (d) 90%

Explanation:
1. Input power \( =220\times0.5=110 \) W; output \( =100 \) W.
2. Efficiency \( =\frac{100}{110}\approx0.91\approx90\% \).

Teacher's Note:
1. Efficiency = output/input.
2. Lamp power is the useful output.

Question 19: A conducting circular loop in a uniform magnetic field of 0.04 T has its plane perpendicular to the field. The radius of the loop starts shrinking at 2 mm/s. The induced emf when the radius is 2 cm is
(a) \( 1.6\pi\,\mu V \)
(b) \( 3.2\pi\,\mu V \)
(c) \( 4.8\pi\,\mu V \)
(d) \( 0.8\pi\,\mu V \)
Show Answer & Explanation

Answer: (b) \( 3.2\pi\,\mu V \)

Explanation:
1. \( \phi=B\pi r^2\Rightarrow\varepsilon=B\,2\pi r\left|\frac{dr}{dt}\right| \).
2. \( \varepsilon=0.04\times2\pi\times0.02\times0.002=3.2\pi\times10^{-6} \) V.

Teacher's Note:
1. Differentiate the area with respect to r.
2. Convert mm/s to m/s.

Question 20: A transformer of efficiency 90% works on 200 V and 3 kW power supply. If the current in the secondary coil is 6 A, the voltage across the secondary and the current in the primary coil respectively are
(a) 300 V, 15 A
(b) 450 V, 15 A
(c) 450 V, 13.5 A
(d) 600 V, 15 A
Show Answer & Explanation

Answer: (b) 450 V, 15 A

Explanation:
1. Input power 3000 W gives \( I_p=\frac{3000}{200}=15 \) A.
2. Output power \( =0.9\times3000=2700 \) W, so \( V_s=\frac{2700}{6}=450 \) V.

Teacher's Note:
1. Use efficiency to find output power.
2. Input current uses the input voltage.

Question 21: The dimensional formula of inductance is
(a) \( ML^2T^{-2}A^{-2} \)
(b) \( ML^2T^{-1}A^{-1} \)
(c) \( MLT^{-1}A^{-1} \)
(d) \( MLTA^{-2} \)
Show Answer & Explanation

Answer: (a) \( ML^2T^{-2}A^{-2} \)

Explanation:
1. \( U=\tfrac12LI^2\Rightarrow L=\frac{U}{I^2} \).
2. \( [L]=\frac{ML^2T^{-2}}{A^2}=ML^2T^{-2}A^{-2} \).

Teacher's Note:
1. Use the energy formula.
2. Henry = kg m² s⁻² A⁻².

Question 22: A long solenoid has 1000 turns. When a current of 4 A flows through it, the flux linked with each turn is 4 × 10⁻³ Wb. The self-inductance is
(a) 4 H
(b) 3 H
(c) 2 H
(d) 1 H
Show Answer & Explanation

Answer: (d) 1 H

Explanation:
1. \( L=\frac{N\phi}{I}=\frac{1000\times4\times10^{-3}}{4}=1 \) H.

Teacher's Note:
1. Flux per turn × turns ÷ current.
2. Check units.

Question 23: A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced emf is
(a) six times per revolution
(b) once per revolution
(c) twice per revolution
(d) four times per revolution
Show Answer & Explanation

Answer: (c) twice per revolution

Explanation:
1. \( \varepsilon=NBA\omega\sin\omega t \) changes sign every half revolution.
2. So the direction reverses twice per revolution.

Teacher's Note:
1. Emf is zero when the plane is perpendicular to B.
2. Two zero crossings per cycle.

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Multiple Choice Questions (MCQs) for JEE Physics Electromagnetic Induction

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