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NEET UG PHYSICS MCQ: ELECTRO MAGNETIC INDUCTION

1 Hard

1 of 1

Application +4 | -1

A coil of '

\mathrm{n}

' turns and resistance '

\mathrm{R}

'

\Omega

is connected in series with a resistance

\frac{\mathrm{R}}{2}

. The combination is moved for time '

\mathrm{t}

' second through magnetic flux

\phi_1

to

\phi_2

. The induced current in the circuit is

View answer and explanation

Correct option: D

\frac{2 \mathrm{n}\left(\phi_1-\phi_2\right)}{3 \mathrm{Rt}}

Option D (

\frac{2 \mathrm{n}(\phi_1-\phi_2)}{3 \mathrm{Rt}}

) is correct because it's the induced current calculated using Faraday's law and Ohm's law, considering the total resistance in the circuit.

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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 