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JEE PHYSICS MCQ: ELECTRO MAGNETIC INDUCTION

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Application +4 | -1

An e.m.f. of

5\,\,volt

is produced by a self inductance, when the current changes at a steady rate from

3\,\,A

to

2\,\,A

in

\mathrm{1\, milli\,second}

. The value of self inductance is

View answer and explanation

Correct option: D

5\,\,mH

Option D (

5\\,\text{{mH}}

) is correct because it is calculated directly from the formula relating induced EMF to self-inductance and rate of change of current.

More Questions from ELECTRO MAGNETIC INDUCTION

Practice All Chapter MCQs →

The current in a long solenoid of radius

R

n

turns per unit length is given by

i=i_0 \sin \omega t

. A coil having

N

turns is wound around it near the centre. Find the induced emf in the coil. The induced emf can be produced in a coil by

A. moving the coil with uniform speed inside
uniform magnetic field

B. moving the coil with non uniform speed inside
uniform magnetic field

C. rotating the coil inside the uniform magnetic
field

D. changing the area of the coil inside the uniform
magnetic field

Choose the correct answer from the options given
below: An inductance coil have the time constant

4 \mathrm{sec}

, if it is cut into two equal parts and connected parallel then new time constant of the circuit- If the coils of a transformer are made up of thick wire, then A loop of area

0.1\,\,{m^2}

rotates with a speed of

\mathrm{60}

rps perpendicular to a magnetic field of

\mathrm{0}\mathrm{.4}\,\,\mathrm{T}

\mathrm{100}

turns in the loop, maximum voltage induced in the loop is A

\mathrm{50}\,\,\mathrm{mH}

coil carries a current of

\mathrm{2 A}

, the energy stored in joule is