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 $\varepsilon = B \ell v$,where $B = 0.5\ T$,$\ell = 0.25\ m$,and $v$ is th

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.$\varepsilon = 0.5 	imes 0.25 	imes v = 0.125v\ V$.The current $I$ in the circuit is given by $I = \frac{E - \varepsilon}{R}$,where $E = 10\ V$ and $R = 2\,\Omega$.$I = \frac{10 - 0.125v}{2}\ A$.The magnetic force acting on the wire is $F = BI\ell$. Given $F = 0.5\ N$ to the left to maintain constant speed to the right.$0.5 = 0.5 	imes \left( \frac{10 - 0.125v}{2} ight) 	imes 0.25$.$1 = \frac{10 - 0.125v}{2} 	imes 0.25$.$8 = 10 - 0.125v$.$0.125v = 2$.$v = \frac{2}{0.125} = 16\ m/s$.

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