A metal wire PQ slides on parallel metallic r | vedclass

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(B) The induced electromotive force (e.m.f.) in the moving wire is given by $\varepsilon = B \ell v = 0.5 	imes 0.25 	imes v = 0.125v \ V.$The curre

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(B) The induced electromotive force (e.m.f.) in the moving wire is given by $\varepsilon = B \ell v = 0.5 	imes 0.25 	imes v = 0.125v \ V.$The current $I$ in the circuit is $I = \frac{10 - \varepsilon}{R} = \frac{10 - 0.125v}{2} \ A.$The magnetic force acting on the wire is $F_m = B I \ell = 0.5 	imes \left( \frac{10 - 0.125v}{2} ight) 	imes 0.25 \ N.$Since the wire moves with constant speed,the external force applied to the left must balance the magnetic force to the right: $F_{ext} = F_m.$$0.5 = 0.5 	imes \left( \frac{10 - 0.125v}{2} ight) 	imes 0.25.$$1 = \frac{10 - 0.125v}{2} 	imes 0.25 \implies 8 = 10 - 0.125v.$$0.125v = 2 \implies v = \frac{2}{0.125} = 16 \ m/s.$

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