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JEE • PHYSICS • ELECTRO MAGNETIC INDUCTION

Electro magnetic induction practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 121-150 of 276 questions

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121 Medium
121 of 276
Understanding +4 | -1
A cylindrical bar magnet is kept along the axis of a circular coil. If the magnet is rotated about its axis, then
View answer and explanation
Correct option: B
No current will be induced in a coil
Option B is correct because rotating a magnet about its axis along the axis of a coil does not change the magnetic flux linked with the coil, hence no current is induced.
122 Medium
122 of 276
Understanding +4 | -1
Three identical ring move with the same speed on a horizontal surface in a uniform horizontal magnetic field normal to the planes of the rings. The first (A) slips without rolling, the second (B) rolls without slipping, and the third rolls with slipping
View answer and explanation
Correct option: A
The same emf is induced in all the three rings
Option A (The same emf is induced in all the three rings) is correct because the induced EMF depends only on the linear motion of the rings and the magnetic field, not on their rolling or slipping.
123 Medium
123 of 276
Application +4 | -1
Two concentric circular coils have radius R1R_1 and R2R_2 such that R1≪R2R_1 \ll R_2. If a current is setup in the second coil then the mutual inductance of first coil with respect to second coil depends upon
View answer and explanation
Correct option: C
R1R2R_1 R_2
Option A (R12/R2R_1^2 / R_2) is correct because the mutual inductance depends on the area of the smaller coil and inversely on the radius of the larger coil.
124 Medium
124 of 276
Application +4 | -1
An electric motor operates on a 50  volt50\,\,volt supply and a current of 12 A12\,A. If the efficiency of the motor is 30  %\mathrm{30}\,\,\mathrm{\% }, what is the resistance of the winding of the motor
View answer and explanation
Correct option: C
2.9  Ω2.9\,\,\mathrm{\Omega }
Option C (2.9  Ω2.9\,\,\mathrm{\Omega }) is correct because it is calculated using the formula relating voltage, current, and resistance, considering the motor's efficiency.
125 Medium
125 of 276
Application +4 | -1
A long horizontal metallic rod with length along the east-west direction is falling under gravity. The potential difference between its two ends will
View answer and explanation
Correct option: C
Increase with time
Option C (Increase with time) is correct because the induced EMF is proportional to the velocity, which increases with time due to gravity.
126 Medium
126 of 276
Application +4 | -1
A coil has 1000 turns and 500  cm2500\,\,\mathrm{c}{\mathrm{m}^2} as its area. The plane of the coil is placed of 2×10−5  web/m2.2 \times {10^{ - 5}}\,\,\mathrm{web/}{\mathrm{m}^2}. The coil is rotated through 1800{180^0} in 0.2 seconds. The average emf induced in the coil, in milli volts is
View answer and explanation
Correct option: B
10
Option B (10) is correct because the average emf induced is calculated using Faraday's law of induction, considering the change in magnetic flux due to the coil's rotation.
127 Medium
127 of 276
Application +4 | -1
A circular conducting loop of radius 10 cm is placed perpendicular to a uniform magnetic field of 1 T. if the magnetic field is made zero in a time interval of 0.1 s the emf induced in the loop during this time is.
View answer and explanation
Correct option: B
0.314
Option B (0.314) is correct because the induced emf is calculated using Faraday's law of electromagnetic induction.
128 Medium
128 of 276
Application +4 | -1
A conducting rod of length ll is moving in a transverse magnetic field of strength BB with velocity vv. The resistance of the rod is RR. the current in the rod is
View answer and explanation
Correct option: C
Zero
Option C (Zero) is correct because if the rod is moving in a transverse magnetic field, but no external circuit is connected, there is no closed loop for current to flow.
129 Medium
129 of 276
Understanding +4 | -1
A magnet is allowed to move vertically through a resistance coil held horizontally. The
acceleration of the magnet is
View answer and explanation
Correct option: B
less than g
The correct answer is less than g because the induced emf opposes the motion of the magnet, resulting in a retarding force.
130 Medium
130 of 276
Analyzing +4 | -1
Consider two identical, coaxial circular loops carrying a current i each, circulating in the same direction. If the loops approach each other
View answer and explanation
Correct option: A
the current in each loop will decrease
Option A (the current in each loop will decrease) is correct because Lenz's law dictates that the induced current opposes the change in magnetic flux.
131 Medium
131 of 276
Understanding +4 | -1
If a magnet is plugged into a coil, then the magnitude of induced emf does not depend upon
View answer and explanation
Correct option: D
The resistance of the coil
Option D (The resistance of the coil) is correct because the induced emf depends on the rate of change of magnetic flux, number of turns, and magnetic field intensity, but not the resistance of the coil.
132 Medium
132 of 276
Understanding +4 | -1
The self inductance of a long solenoid cannot be increased by
View answer and explanation
Correct option: D
increasing the current through it
Option D (increasing the current through it) is correct because the self-inductance of a solenoid depends on its geometry and permeability, not the current flowing through it.
133 Medium
133 of 276
Understanding +4 | -1
The self-inductance LL of a solenoid of length ll and area of cross-section AA, with a fixed number of turns NN increases as
View answer and explanation
Correct option: B
ll decreases and AA increases
Option B (ll decreases and AA increases) is correct because the self-inductance LL is inversely proportional to the length ll and directly proportional to the area AA.
134 Medium
134 of 276
Application +4 | -1
A field of strength 5×104/π5 \times {10^4}/\pi Amp turns / meter acts at right angle to the coil of 50 turns of area 10−2 m2{10^{ - 2}}\,{m^2}. The coil is removed from the field in 0.1 second. Then the induced emf in the coil is
View answer and explanation
Correct option: A
0.1 V
Option A (0.1 V) is correct because the induced emf is calculated using Faraday's law of induction, considering the change in magnetic flux through the coil.
135 Medium
135 of 276
Remembering +4 | -1
Energy is stored in the choke coils in the form of
View answer and explanation
Correct option: C
Magnetic energy
Option C (Magnetic energy) is correct because choke coils store energy in the magnetic field created by the current flowing through them.
136 Medium
136 of 276
Application +4 | -1
A rectangular coil of wire has dimensions 0.2 m×0.1 m0.2 \mathrm{~m} \times 0.1 \mathrm{~m}. The coil has 2000 turns. The coil rotates in a magnetic field about an axis parallel to its length and perpendicular to the magnetic field of 0.02 Wb m−20.02 \mathrm{~Wb} \mathrm{~m}^{-2}. The speed of rotation of the coil is 4200rpm4200 \mathrm{rpm}. Calculate

(i) the maximum value of the induced emf in the coil

(ii) the instantaneous value of induced emf when the plane of the coil has rotated through an angle of 30∘30^{\circ} from the initial position.
View answer and explanation
Correct option: B
1760 V1760 \mathrm{~V}
Option B (1760 V) is correct because it represents the instantaneous induced EMF when the coil has rotated through 30 degrees, calculated using the formula ε = ε_0 sin(wt).
137 Medium
137 of 276
Understanding +4 | -1
An induced emf has
View answer and explanation
Correct option: B
A direction opposite to the field direction
Option B (A direction opposite to the field direction) is correct because Lenz's law states that the induced emf opposes the change in magnetic flux that produces it.
138 Medium
138 of 276
Understanding +4 | -1
In what form is the energy stored in an inductor or
A coil of inductance LL is carrying a steady current ii. What is the nature of its stored energy
View answer and explanation
Correct option: A
Magnetic
Option A (Magnetic) is correct because inductors store energy in the form of a magnetic field.
139 Medium
139 of 276
Application +4 | -1
A metal rod of length 1 m1 \mathrm{~m} is rotated about one of its ends in a plane right angles to a field of inductance 2.5×10−32.5 \times 10^{-3} Wb/m2\mathrm{Wb} / \mathrm{m}^2. If it makes 1800 revolutions/min. Calculate induced e.m.f. between its ends.
View answer and explanation
Correct option: A
0.471 V0.471 \mathrm{~V}
Option A (0.471 V0.471 \mathrm{~V}) is correct because it is the induced emf calculated using the formula for motional emf in a rotating rod.
140 Medium
140 of 276
Application +4 | -1
A conducting loop of area 5.0 cm25.0 \mathrm{~cm}^2 is placed in a magnetic field which varies sinusoidally with time as B=B0sin⁡ωtB=B_0 \sin \omega t where B0=0.20 TB_0=0.20 \mathrm{~T} and ω=300 s−1\omega=300 \mathrm{~s}^{-1}. The normal to the coil makes an angle of 60∘60^{\circ} with the field. Find

(a) the maximum emf induced in the coil,

(b) the emf induced at τ=(π/900) s\tau=(\pi / 900)\text{ s} and

(c) the emf induced at t=(π/600)st=(\pi / 600) \mathrm{s}.
View answer and explanation
Correct option: A
(a) 0.015 V;0.015 \mathrm{~V}; (b) 7.5×10−3 V;7.5 \times 10^{-3} \mathrm{~V}; (c) zero
Option A ( (a) 0.015 V;0.015 \mathrm{~V}; (b) 7.5×10−3 V;7.5 \times 10^{-3} \mathrm{~V}; (c) zero) is correct because it accurately calculates the maximum induced emf and the emf at the specified times using Faraday's law.
141 Medium
141 of 276
Understanding +4 | -1
According to Lenz’s law of electromagnetic induction
View answer and explanation
Correct option: B
The relative motion between the coil and magnet produces change in magnetic flux
Option B (The relative motion between the coil and magnet produces change in magnetic flux) is correct because Lenz's law is based on the principle that induced EMF opposes the change in magnetic flux, which is caused by relative motion.
142 Medium
142 of 276
Understanding +4 | -1
Certain galvanometers have a fixed core made of non magnetic metallic material. The function of this metallic material is
View answer and explanation
Correct option: B
To bring the coil to rest quickly
Option B is correct because the non-magnetic metallic core in galvanometers helps to quickly bring the coil to rest by inducing eddy currents.
143 Medium
143 of 276
Application +4 | -1
Over a solenoid of 50 cm50 \mathrm{~cm} length and 2 cm2 \mathrm{~cm} radius and having 500 turns, is wound another wire of 50 turns near the centre. Calculate the (i) mutual inductance of the two coils (ii) induced emf in the second coil when the current in the primary changes from 0 to 5 A5 \mathrm{~A} in 0.02 s0.02 \mathrm{~s}.
View answer and explanation
Correct option: D
78.96 μ H78.96 ~\mu \mathrm{~H} , −19.74 mV-19.74 \mathrm{~mV}
Option D (78.96 μ H78.96 ~\mu \mathrm{~H} , −19.74 mV-19.74 \mathrm{~mV}) is correct because it correctly calculates the mutual inductance and induced emf.
144 Medium
144 of 276
Analyzing +4 | -1
Assertion (A) : In parallel, current distributes in inverse ratio of inductance i∝1Li \propto \frac{1}{L}

Reason (R) : In electrical circuits, an inductor can be treated as a resistor.
View answer and explanation
Correct option: C
If Assertion is true, but the Reason is false.
Option C (If Assertion is true, but the Reason is false) is correct because in parallel, current distributes in the inverse ratio of inductance, but an inductor cannot be treated as a resistor.
145 Medium
145 of 276
Understanding +4 | -1
Pick out the false statement from the following
View answer and explanation
Correct option: A
The direction of eddy current is given by Fleming’s right hand rule
The direction of eddy current is given by Lenz's law, not Fleming's right-hand rule. Hence, option A is the correct answer.
146 Medium
146 of 276
Application +4 | -1
A very small circular loop of radius aa is initially (at t=0t=0 ) co-planar and concentric with a much larger fixed circular loop of radius bb. A constant current II flows in the larger loop. The smaller loop is rotated with a constant angular speed ω\omega about the common diameter. The emf induced in the smaller loop as a function of time tt is
View answer and explanation
Correct option: C
πa2μ0I2bωsin⁡ωt\frac{\pi a^2 \mu_0 I}{2 b} \omega \sin \omega t
Option C (πa2μ0I2bωsin⁡ωt\frac{\pi a^2 \mu_0 I}{2 b} \omega \sin \omega t) is correct because it is the induced emf calculated using the formula e=NBAωsin⁡ωte=N B A \omega \sin \omega t with appropriate substitutions.
147 Medium
147 of 276
Understanding +4 | -1
In an A.C. generator, when the plane of the armature is perpendicular to the magnetic field
View answer and explanation
Correct option: D
Magnetic flux is maximum and emf is zero
Option D (Magnetic flux is maximum and emf is zero) is correct because when the plane of the armature is perpendicular to the magnetic field, the flux is maximum and the induced emf is zero.
148 Medium
148 of 276
Understanding +4 | -1
A magnet is brought near a coil in two ways i. rapidly, ii. slowly. The induced charge will be:
View answer and explanation
Correct option: C
equal in both the cases
Option C (equal in both the cases) is correct because the induced charge depends only on the change in magnetic flux, not the rate of change.
149 Medium
149 of 276
Analyzing +4 | -1
Two identical coaxial coils P and Q carrying equal amount of current in the same direction are brought nearer. the current in
View answer and explanation
Correct option: D
Both P and Q decreases
Option D (Both P and Q decreases) is correct. Bringing the coils closer increases the magnetic flux, inducing a current that opposes the change, thus decreasing the current in both coils.
150 Medium
150 of 276
Application +4 | -1
An inductor of inductance 20mH20 \mathrm{mH} is connected across the terminals of a capacitor of capacitance 2μF2 \mu \mathrm{F} which is initially charged to a potential 102 V10 \sqrt{2} \mathrm{~V}. The charge on the capacitor, when the energy stored in the capacitor becomes equal to the energy stored in the inductor is
View answer and explanation
Correct option: A
20μC20 \mu \mathrm{C}
Option A (20μC20 \mu \mathrm{C}) is correct because it's the charge on the capacitor when the energy stored in the capacitor equals the energy stored in the inductor.
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