The figure shows a square loop of side 5 cm | vedclass

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(A) Given: Side length $l = 5 \ cm = 0.05 \ m$,speed $v = 1 \ cm/s = 0.01 \ m/s$,magnetic field $B = 0.6 \ T$,width of magnetic field region $w = 20 \

Vedclass · Accepted Answer

$(a) \ 3 	imes 10^{-4} \ V, (b) \ 0, (c) \ 3 	imes 10^{-4} \ V$

Vedclass · Answer

(A) Given: Side length $l = 5 \ cm = 0.05 \ m$,speed $v = 1 \ cm/s = 0.01 \ m/s$,magnetic field $B = 0.6 \ T$,width of magnetic field region $w = 20 \ cm = 0.2 \ m$.The induced $emf$ is given by $e = Bvl$ when the loop is entering or leaving the magnetic field,and $e = 0$ when the loop is completely inside the field (as magnetic flux is constant).Calculation of $emf$:$e = 0.6 	imes 0.01 	imes 0.05 = 0.0003 \ V = 3 	imes 10^{-4} \ V$.$(a)$ At $t = 2 \ s$,the loop is partially inside the field (distance moved = $2 \ cm $(b)$ At $t = 10 \ s$,the loop has moved $10 \ cm$. Since the front edge entered at $t = 0$,at $t = 10 \ s$,the entire loop is inside the magnetic field (as $5 \ cm $(c)$ At $t = 22 \ s$,the front edge has traveled $22 \ cm$. Since the magnetic field width is $20 \ cm$,the front edge has exited the field. The back edge is at $22 - 5 = 17 \ cm$ inside the field. The loop is leaving the field,so $e = 3 	imes 10^{-4} \ V$.

The figure shows a square loop of side $5 \ cm$ being moved towards the right at a constant speed of $1 \ cm/s$. The front edge enters the $20 \ cm$ wide magnetic field $(B = 0.6 \ T)$ at $t = 0$. Find the $emf$ induced in the loop at $(a) \ t = 2 \ s$,$(b) \ t = 10 \ s$,and $(c) \ t = 22 \ s$.

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