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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Induction
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A conducting circular loop is placed in a uniform field of $B=0.05\;T$ with its plane perpendicular to the loop. The radius of loop is made to shrink at constant rate of $0.5\;mm$ per second. The induced emf when radius is $2 \;cm$ is

$(a)\;1 \times 10^{-6} \pi V \\ (b)\;2 \times 10^{-3} \pi V \\(c)\;1 \times 10^{-6} V \\(d)\;2 \times 10^{-3} V$

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1 Answer

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Induced emf $e= \large\frac{d\phi}{dt}$
Where $\phi$ is flux linked with the coil which is BA.
$\qquad= B \pi r^2$
$B= 0.05 \;T$
$\large\frac{dr}{dt} $$=0.5 \times 10^{-3}\;ms^{-1}$
$r= 2 \times 10^{-2}\;m$
$e= \large\frac{d \phi}{dt} =\frac{d}{dt} $$B \pi r^2$
$\qquad=B \pi 2r \large\frac{dr}{dt}$
$\qquad= 0.5 \pi \times 2 \times 2 \times 10^{-2} \times 0.5 \times 10^{-3}$
$\qquad= \pi \times 10^{-3} \times 10^{-2} \times 10^{-1}$
$\qquad= \pi \times 10^{-6}\;V$
Hence a is the correct answer.


answered Mar 21, 2014 by meena.p
edited Sep 24, 2014 by thagee.vedartham

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