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(B) The induced electromotive force $(e.m.f.)$ in the moving wire is given by $\varepsilon = B \ell v$,where $B = 0.5 \ T$,$\ell = 0.25 \ m$,and $v$ i

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(B) The induced electromotive force $(e.m.f.)$ in the moving wire is given by $\varepsilon = B \ell v$,where $B = 0.5 \ T$,$\ell = 0.25 \ m$,and $v$ is the speed.So,$\varepsilon = 0.5 	imes 0.25 	imes v = 0.125v \ V$.The net current $I$ in the circuit is $I = \frac{E - \varepsilon}{R} = \frac{10 - 0.125v}{2} \ A$.The magnetic force acting on the wire is $F_m = B I \ell = 0.5 	imes I 	imes 0.25 = 0.125I \ N$.Since the wire moves at a constant speed,the external force must balance the magnetic force: $F_{ext} = F_m = 0.5 \ N$.Substituting $I$ into the force equation: $0.5 = 0.125 	imes \left( \frac{10 - 0.125v}{2} ight)$.$0.5 	imes 2 = 0.125 	imes (10 - 0.125v) \implies 1 = 1.25 - 0.015625v$.$0.015625v = 0.25 \implies v = \frac{0.25}{0.015625} = 16 \ m/s$.

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