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Consider a circular wire of radius R moving towards a region of constant and uniform magnetic field $\vec B$ perpendicular to the plane of circular wire . It is pushed into the region of magnetic field with a constant velocity v at t = 0 . The induced emf at time t , when the edge of field region subtends an angle $\theta$ at the centre
Option: 1
$BRv \sin \frac{\theta }{2}$

Option: 2
$BRv \sin {\theta }$

Option: 3
$2BRv \sin \frac{\theta }{2}$

Option: 4
$2BRv \sin {\theta }$

Answers (1)

best_answer

As we have learned

If the loop is free to move -

Relative motion can also cause the induced emf in the coil.

-

Let at any instant of time t , the edge of field region makes an angle $\theta$ at centre . So area , of circular wire which has gone inside the field region is

$dA = \frac{1}{2}R^2\theta -\frac{1}{2}R^2 \sin \theta$

NOw induced emf

$\varepsilon = \frac{-d\phi }{dt}$

$\varepsilon = -B\frac{d }{dt}\left ( \frac{1}{2}R^2 \theta -\frac{1}{2}R^2\sin \theta \right )$

$\varepsilon = -B\left [ \frac{R^2}{2}\frac{d\theta }{dt}-\frac{R^2}{2}\cos \theta \frac{d\theta }{dt} \right ]............ (1)$

Also

$v = \frac{d}{dt}(R \cos \frac{\theta }{2})$

$v =\frac{-R}{2}\sin \frac{\theta }{2} \frac{d\theta }{dt}$

$\varepsilon = B \frac{R^2}{2}(1- \cos \theta )\frac{2V}{R \sin \frac{\theta }{2}}$

$= \frac{BRv }{\sin \frac{\theta }{2}}2 \sin ^{2}\frac{\theta }{2}$

$\varepsilon = 2 BRv \sin \frac{\theta }{2}$

Posted by

Divya Prakash Singh

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