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NEET UG PHYSICS MCQ: ELECTRO MAGNETIC INDUCTION
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In an induction coil, the coefficient of mutual inductance is 4H. If current of 5A in the primary coil is cut off 1/ 1500s, the emf at the terminals of the secondary coil will be
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Correct option: D
30 kV
Option D (30 kV) is correct because the induced emf is calculated using the formula emf = M * (change in current / change in time).
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Assertion (A): Whenever the magnetic flux linked with a closed coil changes there will be an induced emf as well as an induced current.
Reason (R): According to Faraday, the induced emf is inversely proportional to the rate of change of magnetic flux linked with a coil. The coefficient of mutual inductance of two coils is . If the current flowing in one is , then the induced e.m.f. in the second coil will be The induced emf cannot be produced by: Consider the statements:
(I)If magnetic field, , then magnetic flux is also zero.
(II)If magnetic flux, then magnetic field is also zero. The coefficient of self inductance of a solenoid is 0.18 mH. If a rod of soft iron of relative permeability 900 is inserted, then the coefficient of self inductance (in mH ) will become nearly: A uniformly wound solenoidal coil of self inductance H and resistance 6 Ω is broken up into two identical coils. These identical coils are then connected in parallel across a 12 V battery of negligible resistance. The time constant of the current in the circuit
Reason (R): According to Faraday, the induced emf is inversely proportional to the rate of change of magnetic flux linked with a coil. The coefficient of mutual inductance of two coils is . If the current flowing in one is , then the induced e.m.f. in the second coil will be The induced emf cannot be produced by: Consider the statements:
(I)If magnetic field, , then magnetic flux is also zero.
(II)If magnetic flux, then magnetic field is also zero. The coefficient of self inductance of a solenoid is 0.18 mH. If a rod of soft iron of relative permeability 900 is inserted, then the coefficient of self inductance (in mH ) will become nearly: A uniformly wound solenoidal coil of self inductance H and resistance 6 Ω is broken up into two identical coils. These identical coils are then connected in parallel across a 12 V battery of negligible resistance. The time constant of the current in the circuit