NEET UG PHYSICS MCQ: ELECTRO MAGNETIC INDUCTION

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A rectangular loop of sides

\mathrm{10}\,\mathrm{cm}

and

\mathrm{5}\,\mathrm{cm}

with a cut is stationary between the pole pieces of an electromagnet. The magnetic field of the magnet is normal to the loop. The current feeding the electromagnet is reduced so that the field decreased from its initial value of

\mathrm{0}\mathrm{.3 T}

at the rate of

0.02\,\,\mathrm{\Omega }

. If the cut is joined and the loop has a resistance of

2.0\,\,\mathrm{\Omega }

, the power dissipated by the loop as heat is

View answer and explanation

Correct option: A

\mathrm{5 nW}

Option A (

\mathrm{5 nW}

) is correct because the power dissipated is calculated from the induced EMF and the resistance of the loop, using Faraday's law and the power formula.

More Questions from ELECTRO MAGNETIC INDUCTION

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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

6\,\,mH

. If the current flowing in one is

2\,\,ampere

, 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,

\mathrm{B = 0}

, then magnetic flux is also zero.
(II)If magnetic flux,

\phi = 0,

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

1.8 \times {10^{ - 4}}

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 