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NEET • PHYSICS • ELECTRO MAGNETIC INDUCTION

Electro magnetic induction practice · Mixed difficulty

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121 Medium
121 of 129
Analyzing +4 | -1
Two identical coaxial circular loops carry a current ii each circulating in the same direction. As the loops are moved away from each other
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Correct option: B
The current in each will increase
Option B (The current in each will increase) is correct because as the loops move apart, the mutual inductance decreases, and to maintain the magnetic flux, the current must increase.
122 Easy
122 of 129
Understanding +4 | -1
When you walk through a metal detector carrying a metal object in your pocket, it raises an alarm. This phenomenon works on:
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Correct option: B
Resonance in ac circuits
Option B (Resonance in ac circuits) is correct because metal detectors use resonance to detect changes in the electromagnetic field caused by metal objects.
123 Medium
123 of 129
Application +4 | -1
A coil having 500\mathrm{500} square loops each of side 10  cm\mathrm{10}\,\,\mathrm{cm} is placed normal to a magnetic field which increases at the rate of 1   Wm−2\mathrm{1}\,\,\,\mathrm{W}{\mathrm{m}^{ - 2}}. The induced emf is
View answer and explanation
Correct option: B
5.0 V\mathrm{5}.\mathrm{0}\,\mathrm{V}
Option B (5.0 V\mathrm{5}.\mathrm{0}\,\mathrm{V}) is correct because the induced emf is calculated using Faraday's law, considering the number of loops, area of each loop, and the rate of change of the magnetic field.
124 Medium
124 of 129
Application +4 | -1
An axle of truck is 2.5m\mathrm{2}\mathrm{.5 m} long. If the truck is moving due north at 30  ms(−1)30\,\,m{s^{( - 1)}} at a place where the vertical component of the earth’s magnetic field is 90  μT90\,\,\mu T, the potential difference between the two ends of the axle is
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Correct option: A
6.75  mV\mathrm{6}.\mathrm{75}\,\,\mathrm{mV} with west end positive
Option A (6.75 mV\mathrm{6}.\mathrm{75}\,\mathrm{mV} with west end positive) is correct because the motional EMF is calculated using ε=vBL\varepsilon = vBL, and the direction is determined by the right-hand rule.
125 Hard
125 of 129
Application +4 | -1
A circular coil of area 0.01 m20.01 \mathrm{~m}^2 and 40 turns is rotated about its vertical diameter with an angular speed of 50 rad s−150 \mathrm{~rad~s}^{-1} in a uniform horizontal magnetic field 0.05 T0.05 \mathrm{~T}. If the average power loss due to joule heating is 25 mW25 \mathrm{~m W}. Then, the closed loop resistance of the coil is-
View answer and explanation
Correct option: D
20Ω20 \Omega
Option D (20Ω20 \Omega) is correct because it's the resistance value calculated from the average power loss due to Joule heating, considering the induced emf and angular speed.
126 Medium
126 of 129
Understanding +4 | -1
In a dc motor, induced e.m.f. will be maximum
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Correct option: A
When motor takes maximum speed
Option A (When motor takes maximum speed) is correct because the induced EMF in a DC motor is directly proportional to its speed, thus it's maximum at maximum speed.
127 Medium
127 of 129
Understanding +4 | -1
The coil of dynamo is rotating in a magnetic field. The developed induced e.m.f. changes and the number of magnetic lines of force also changes. Which of the following conditions is correct
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Correct option: B
Lines of force maximum but induced e.m.f. is zero
Option B (Lines of force maximum but induced e.m.f. is zero) is correct because when the coil is perpendicular to the magnetic field, the magnetic flux is maximum, but the rate of change of flux (and hence the induced EMF) is zero.
128 Medium
128 of 129
Analyzing +4 | -1
Assertion: An induced emf appears in any coil in which the current is constant.
Reason: Self induction phenomenon does not obey Faraday's law of induction.
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Correct option: D
If both Assertion and Reason are false
Option D (If both Assertion and Reason are false) is correct because an induced emf only appears when the current is changing, and self-induction does obey Faraday's law.
129 Hard
129 of 129
Application +4 | -1
In a uniform magnetic field of induction BB, a wire in the form of semicircle of radius rr rotates about the diameter of the circle with angular frequency ω\omega. If the total resistance of the circuit is RR, the mean power generated per period of rotation is
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Correct option: B
(Bπr2ω)25  Rt\frac{{{{\left( {B\pi {r^2}\omega } \right)}^2}}}{{5\,\,Rt}}
The correct answer is (Bπr2ω)25,,Rt\frac{{{{\left( {B\pi {r^2}\omega } \right)}^2}}}{{5\\,\\,Rt}}. The mean power generated per period of rotation involves calculating the induced EMF and then the power dissipated in the resistance.
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