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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 61-90 of 276 questions

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61 Medium
61 of 276
Remembering +4 | -1
Changing magnetic fields can set up current loops
in nearby metal bodies and the currents are called
as
View answer and explanation
Correct option: A
eddy currents
Option A (eddy currents) is correct because changing magnetic fields induce circulating currents in conductive materials.
62 Medium
62 of 276
Remembering +4 | -1
If N\mathrm{N} is the number of turns in a coil, the value of self inductance varies as
View answer and explanation
Correct option: C
N2\mathrm{N}^2
Option C (N2\mathrm{N}^2) is correct because self-inductance is proportional to the square of the number of turns in the coil.
63 Medium
63 of 276
Application +4 | -1
Magnetic flux linked with a coil is ϕ=5 t2+2 t+3\phi = 5\,{t^2} + 2\,t + 3, where tt is second and ϕ\phi is in weber. At time t=1st = 1\mathrm{ }s, the value of induced emf in volt
View answer and explanation
Correct option: C
12
Option C (12) is correct because the induced emf is found by differentiating the magnetic flux with respect to time and evaluating at t=1s.
64 Medium
64 of 276
Application +4 | -1
A 12 V12 \, \mathrm{V} battery connected to a coil of resistance 6 Ω6 \, \Omega through a switch, drives a constant current in the circuit. The switch is opened in 1 ms1 \, \mathrm{ms}. The emf induced across the coil is 20 V20 \, \mathrm{V}. The inductance of the coil is :
View answer and explanation
Correct option: A
10 mH10 \, \mathrm{mH}
Option A (10 mH10 \, \mathrm{mH}) is correct because it is the inductance calculated using the induced emf and the rate of change of current.
65 Medium
65 of 276
Application +4 | -1
Two inductances each of inductance ‘L’ are joined in parallel their equivalent inductance is
View answer and explanation
Correct option: C
L/2
Option C (L/2) is correct because the equivalent inductance of two inductors in parallel is given by L2\frac{L}{2} when both inductances are equal.
66 Medium
66 of 276
Application +4 | -1
A train is running at a speed of 72 km hr−172 \mathrm{~km} \mathrm{~hr}^{-1} on the rails separated by a distance of 150 cm150 \mathrm{~cm}. If the vertical component of earth's magnetic field at the place is 4.0×10−5 T4.0 \times 10^{-5} \mathrm{~T}. The induced emf on the rails is
View answer and explanation
Correct option: A
1.2 mV1.2 \mathrm{~mV}
Option A (1.2 mV1.2 \mathrm{~mV}) is correct because it calculates the induced emf using the formula emf = Blv, where B is the magnetic field, l is the length, and v is the velocity.
67 Medium
67 of 276
Analyzing +4 | -1
The momentum in mechanics is expressed as m  ×  Vm\,\, \times \,\,V. The analogous expression in electricity is
View answer and explanation
Correct option: C
L×iL \times i
Option C is correct because inductance (L) is analogous to mass (m) and current (i) is analogous to velocity (v), thus momentum (mv) corresponds to Li.
68 Medium
68 of 276
Application +4 | -1
A 10-meter wire is kept in east-west direction. It is falling down with a speed of 5.0 meter/second5.0 \mathrm{~meter} / \mathrm{second}, perpendicular to the horizontal component of earth's magnetic field of 0.30×10−40.30 \times 10^{-4} weber // meter 2^2. The momentary potential difference induced between the ends of the wire will be
View answer and explanation
Correct option: A
0.0015 V0.0015 \mathrm{~V}
Option A (0.0015 V0.0015 \mathrm{~V}) is correct because it is calculated using the formula for induced EMF in a moving conductor in a magnetic field: V=BvlV = Bvl.
69 Medium
69 of 276
Application +4 | -1
An inductor coil stores 32 J32 \mathrm{~J} of magnetic field energy and dissipates energy as heat at the rate 320 W320 \mathrm{~W} when a current of 4 A4 \mathrm{~A} is passed through it. The time constant of the circuit when this coil is joined across an ideal battery will be
View answer and explanation
Correct option: A
0.2 s0.2 \mathrm{~s}
Option A (0.2 s) is correct because the time constant is calculated using the inductance and resistance of the coil.
70 Medium
70 of 276
Remembering +4 | -1
The energy stored in an inductor of self inductance LL henry carrying a current of 1 A\mathrm{1 }\,\mathrm{A} is
View answer and explanation
Correct option: C
12 LI2\frac{1}{2}\,L{I^2}
Option C (12 LI2\frac{1}{2}\,L{I^2}) is correct because it represents the formula for the energy stored in an inductor.
71 Medium
71 of 276
Application +4 | -1
The mutual inductance between two coils is 0.090.09 Henry. If the current in the primary coil changes from 0 to 20 A20 \mathrm{~A} in 0.006 s0.006 \mathrm{~s}, the e.m.f. induced in the secondary coil at that instant is
View answer and explanation
Correct option: D
300 V300 \mathrm{~V}
Option D (300 V300 \mathrm{~V}) is correct because it is the induced EMF calculated using the formula ε=MΔIΔt\varepsilon = M \frac{\Delta I}{\Delta t}.
72 Medium
72 of 276
Understanding +4 | -1
Two identical circular coils A and B are kept on a horizontal tube side by side without touching each other. If the current in the coil A increases with time, in response, the coil B :
View answer and explanation
Correct option: C
Is repelled
Option C (Is repelled) is correct because as the current in coil A increases, it induces a current in coil B in the same direction, causing them to repel each other.
73 Medium
73 of 276
Analyzing +4 | -1
The currents in two coils of self- inductance 5 mH5 \mathrm{~mH} and 1 mH1 \mathrm{~mH} are increasing at the same rate at a certain instant. The power supplied to the coils is also same. Then identify the false statement among the following
View answer and explanation
Correct option: D
The ratio of energy stored in the two coils at the instant is 9:259: 25
Option D (The ratio of energy stored in the two coils at the instant is 9:25) is correct because the ratio of energy stored is 1:5, not 9:25.
74 Medium
74 of 276
Application +4 | -1
A solenoidal coil has 50 turns per cm\mathrm{cm} along its length and a cross-sectional area of 4×10−4 m2.2004 \times 10^{-4} \mathrm{~m}^2 .200 turns of another wire is wound round the first solenoid co-axially. Both the two coils are electrically insulated from each other. Calculate the mutual inductance between the two coils.
View answer and explanation
Correct option: B
5.03×10−4H5.03 \times 10^{-4} \mathrm{H}
Option B (5.03×10−4H5.03 \times 10^{-4} \mathrm{H}) is correct because the mutual inductance is calculated using the formula involving the number of turns and area.
75 Medium
75 of 276
Understanding +4 | -1
Coefficient of coupling between two coils of self-inductances L1  and  L2{L_1}\,\,\mathrm{and}\,\,{L_2} is unity. It means
View answer and explanation
Correct option: B
100  %\mathrm{100}\,\,\mathrm{\% } flux of L1{L_1} is linked with L2{L_2}
Option B is correct because a coefficient of coupling of unity (1) means 100% of the flux from one coil links with the other.
76 Medium
76 of 276
Understanding +4 | -1
Assertion : A small magnet takes longer time in falling in a hollow metallic tube without touching the wall.

Reason : There is opposition of motion due to production of eddy currents in metallic tube.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.
Option A is correct because the magnet's slower fall is due to the opposition of motion caused by eddy currents induced in the metallic tube.
77 Medium
77 of 276
Remembering +4 | -1
Lenz's law provides a relation between
View answer and explanation
Correct option: C
Induced emf and rate of change of flux
Option C (Induced emf and rate of change of flux) is correct because Lenz's law relates the induced emf to the rate of change of magnetic flux.
78 Medium
78 of 276
Application +4 | -1
A short solenoid of radius aa, number of turns per Unit length n1n_1 and length LL is kept coaxially inside a very long solenoid of radius bb, the number of turns per Unit length n2n_2. What is the mutual inductance of the system?
View answer and explanation
Correct option: C
μ0πa2n1n2L\mu_0 \pi {a}^2 n_1 n_2 L
The correct answer is C (μ0πa2n1n2L\mu_0 \pi {a}^2 n_1 n_2 L) because the mutual inductance depends on the flux linkage with the inner solenoid due to the magnetic field of the outer solenoid.
79 Medium
79 of 276
Application +4 | -1
In 0.1s\mathrm{0}\mathrm{.1 s}, the current in a coil increases from 1 A\mathrm{1}\,\mathrm{A} to 1.5 A\mathrm{1.5}\,\mathrm{A}. If inductance of coil is 60  mH\mathrm{60}\,\,\mathrm{mH}, then induced current in external resistance of 3 Ω3\,\mathrm{\Omega } will be
View answer and explanation
Correct option: D
0.1 A\mathrm{0}\mathrm{.1}\,\mathrm{ A}
Option D (0.1 A\mathrm{0}\mathrm{.1}\,\mathrm{ A}) is correct because it is the induced current calculated using the induced emf and the external resistance.
80 Medium
80 of 276
Application +4 | -1
Two inductors A\mathrm{A} and B\mathrm{B} when connected in parallel are equivalent to a single inductor of inductance 1.5 H1.5 \mathrm{~H}, and when connected in series are equivalent to a single inductor of inductance 8 H8 \mathrm{~H}. Find the difference in the inductances of A\mathrm{A} and B\mathrm{B}.
View answer and explanation
Correct option: D
4 H4 \mathrm{~H}
Option D (4 H) is correct because it correctly solves the system of equations for inductors in series and parallel.
81 Medium
81 of 276
Understanding +4 | -1
Self induction of a solenoid is
View answer and explanation
Correct option: C
Directly proportional to area of cross-section
Option C is correct because the self-inductance of a solenoid is directly proportional to the area of cross-section.
82 Medium
82 of 276
Understanding +4 | -1
If a bar magnet is dropping through the copper ring, then its velocity
View answer and explanation
Correct option: A
Decreases
Option C (Positive and negative ions) is correct because in electrolysis, the current is carried by the movement of positive and negative ions towards the electrodes.
83 Medium
83 of 276
Application +4 | -1
A coil of cross-sectional area A having n turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity ω\omega, the maximum e.m.f. induced in the coil will be
View answer and explanation
Correct option: A
nBAωnBA\omega
Option A (nBAωnBA\omega) is correct because it is the formula for the maximum induced emf in a rotating coil in a magnetic field.
84 Medium
84 of 276
Understanding +4 | -1
The expression for the induced emf contains negative sign (E=−dϕdt)\left(\mathrm{E}=\frac{-\mathrm{d} \phi}{\text {dt}}\right). What is the significance of negative sign?
View answer and explanation
Correct option: B
The induced emf opposes the change in the magnetic flux
Option B (The induced emf opposes the change in the magnetic flux) is correct because Lenz's law states that the induced emf opposes the change in magnetic flux that produces it.
85 Medium
85 of 276
Application +4 | -1
A loop of area 1 m21 \mathrm{~m}^2 is placed in a magnetic field B=2 TB=2 \mathrm{~T}, such that plane of the loop is parallel to the magnetic field. If the loop is rotated by 180∘180^{\circ}, the amount of net charge passing through any point of loop, if its resistance is 10Ω10 \Omega, is
View answer and explanation
Correct option: A
0.4 C0.4 \mathrm{~C}
Option A (0.4 C) is correct because the net charge passing through the loop is calculated using the change in magnetic flux and the resistance of the loop.
86 Medium
86 of 276
Application +4 | -1
A planar loop of wire rotates in a uniform magnetic field. Initially at t=0t = 0, the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of 10 s about an axis in its plane, then the magnitude of induced emf will be maximum and minimum respectively at
View answer and explanation
Correct option: D
2.5 s and 5.0 s
Option D (2.5 s and 5.0 s) is correct because the induced emf is maximum when the plane of the loop is parallel to the magnetic field (t=2.5s) and minimum when it is perpendicular (t=5.0s).
87 Medium
87 of 276
Understanding +4 | -1
The inductance L\mathrm{L} of a solenoid depends upon its radius R\mathrm{R} as
View answer and explanation
Correct option: C
L∝R2\mathrm{L} \propto \mathrm{R}^2
Option C is correct because the inductance of a solenoid is proportional to the square of its radius, as shown in the formula L=μ0N2Al=μ0N2πR2lL = \frac{{\mu_0 N^2 A}}{l} = \frac{{\mu_0 N^2 \pi R^2}}{l}.
88 Medium
88 of 276
Application +4 | -1
In an inductor, the current I\mathrm{I} varies with time t as I=4+4t\mathrm{I=4+4 t}. If the emf induced in the inductor is 8 mV8 \mathrm{~mV}, what is self inductance? (Here current is in ampere and time is in second)
View answer and explanation
Correct option: D
2×10−3 H2 \times 10^{-3} \mathrm{~H}
Option D (2×10−3 H2 \times 10^{-3} \mathrm{~H}) is correct because it is the self-inductance calculated using the formula ∣e∣=LdIdt|e| = L \frac{dI}{dt}.
89 Medium
89 of 276
Remembering +4 | -1
Weber/m2Weber/{m^2} is equal to
View answer and explanation
Correct option: C
TeslaTesla
Option C (Tesla) is correct because Tesla is the SI unit of magnetic flux density (magnetic field strength), which is equivalent to Weber per square meter (Weber/m2Weber/{m^2}).
90 Medium
90 of 276
Application +4 | -1
If the magnetic flux through a coil at any time t\mathrm{t} is given by ϕ=(4t2+5t−3)\phi=\left(4 \mathrm{t^2+5 t}-3\right), then the increase in induced emf (in volt) 2 second after t=0\mathrm{t}=0 is
View answer and explanation
Correct option: B
16
Option B (16) is correct because it represents the difference in induced EMF calculated at t=2s and t=0s, based on the given flux equation.
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