A coil is places perpendicular to a magnetic field of $5000 \mathrm{~T}$. When the field is changed to $3000 \mathrm{~T}$ in $2 \mathrm{~s}$, an induced emf of $22 \mathrm{~V}$ is produced in the coil. If the diameter of the coil is $0.02 \mathrm{~m}$, then the number of turns in the coil is

NEET UG PHYSICS MCQ: ELECTRO MAGNETIC INDUCTION

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A coil is places perpendicular to a magnetic field of

5000 \mathrm{~T}

. When the field is changed to

3000 \mathrm{~T}

2 \mathrm{~s}

, an induced emf of

22 \mathrm{~V}

is produced in the coil. If the diameter of the coil is

0.02 \mathrm{~m}

, then the number of turns in the coil is

View answer and explanation

Correct option: B

Option B (70) is correct because the number of turns is calculated using Faraday's law, considering the change in magnetic field and the area of the coil.

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 