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JEE PHYSICS MCQ: ELECTRO MAGNETIC INDUCTION

1 Hard

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Application +4 | -1

A circular coil of area

0.01 \mathrm{~m}^2

and 40 turns is rotated about its vertical diameter with an angular speed of

50 \mathrm{~rad~s}^{-1}

in a uniform horizontal magnetic field

0.05 \mathrm{~T}

. If the average power loss due to joule heating is

25 \mathrm{~m W}

. Then, the closed loop resistance of the coil is-

View answer and explanation

Correct option: D

20 \Omega

Option D (20 Ω) is correct because the closed loop resistance is calculated from the induced EMF and average power loss.

More Questions from ELECTRO MAGNETIC INDUCTION

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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