A wire having mass m and length l can freely | vedclass

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Question

(A) As the wire moves with an instantaneous velocity $v$,the motional electromotive force (e.m.f.) induced across the wire is $\varepsilon = Bvl$.Sinc

Vedclass · Accepted Answer

$a = \frac{F}{m + C B^2 l^2}$

Vedclass · Answer

(A) As the wire moves with an instantaneous velocity $v$,the motional electromotive force (e.m.f.) induced across the wire is $\varepsilon = Bvl$.Since the wire is connected to a capacitor $C$,the charge $q$ stored on the capacitor is $q = C\varepsilon = CBvl$.The current $i$ flowing through the circuit is the rate of change of charge on the capacitor:$i = \frac{dq}{dt} = CBl \frac{dv}{dt} = CBla$,where $a$ is the acceleration of the wire.The magnetic force $F_m$ acting on the wire due to this current is $F_m = Bil = (CBla)Bl = CB^2 l^2 a$.According to Newton's second law,the net force on the wire is $F_{net} = F - F_m = ma$.Substituting the expression for $F_m$:$F - CB^2 l^2 a = ma$$F = ma + CB^2 l^2 a = a(m + CB^2 l^2)$Therefore,the acceleration $a$ is:$a = \frac{F}{m + CB^2 l^2}$

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