A square loop of wire, side length ${\rm{10}}\,\,{\rm{cm}}$ is placed at angle of ${\rm{45}}^\circ $ with a magnetic field that changes uniformly from ${\rm{0}}{\rm{.1 T}}$ to zero in ${\rm{0}}{\rm{.7 s}}$. The induced current in the loop (its resistance is ${\rm{1}}\,\,{\rm{\Omega }})$ is

${\rm{1}}.{\rm{0}}\,\,{\rm{mA}}$

${\rm{2}}.{\rm{5}}\,\,{\rm{mA}}$

${\rm{3}}.{\rm{5}}\,\,{\rm{mA}}$

${\rm{4}}.{\rm{0}}\,\,{\rm{mA}}$

JEE PHYSICS MCQ: ELECTRO MAGNETIC INDUCTION

1 Medium

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Application +4 | -1

A square loop of wire, side length

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

is placed at angle of 

\mathrm{45}^\circ

with a magnetic field that changes uniformly from

\mathrm{0}\mathrm{.1 T}

to zero in

\mathrm{0}\mathrm{.7 s}

. The induced current in the loop (its resistance is

\mathrm{1}\,\,\mathrm{\Omega })

View answer and explanation

Correct option: A

\mathrm{1}.\mathrm{0}\,\,\mathrm{mA}

Option A (

\mathrm{1}.\mathrm{0}\,\,\mathrm{mA}

) is correct because the induced current is calculated using Faraday's law and Ohm's law, considering the change in magnetic flux.

Practice All Chapter MCQs →

R

n

i=i_0 \sin \omega t

N

4 \mathrm{sec}

0.1\,\,{m^2}

\mathrm{60}

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

\mathrm{100}

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

\mathrm{2 A}