A rod closing the circuit shown in the figure | vedclass

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(B) The rate of heat generation $Q$ in the circuit is equal to the power dissipated,which is given by $P = I^2 R = \frac{E^2}{R}$.Alternatively,the me

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$F = \frac{Q}{v}$

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(B) The rate of heat generation $Q$ in the circuit is equal to the power dissipated,which is given by $P = I^2 R = \frac{E^2}{R}$.Alternatively,the mechanical power supplied by the external force $F$ is $P = Fv$.Since the rod moves at a constant speed,the external force $F$ must be equal to the magnetic force $F_m = IlB$ acting on the rod.The induced electromotive force $(EMF)$ is $E = Blv$.The current in the circuit is $I = \frac{E}{R} = \frac{Blv}{R}$.The magnetic force is $F_m = IlB = (\frac{Blv}{R})lB = \frac{B^2 l^2 v}{R}$.The power supplied is $P = Fv = (\frac{B^2 l^2 v}{R})v = \frac{B^2 l^2 v^2}{R}$.Since $Q = P$,we have $Q = Fv$.Therefore,$F = \frac{Q}{v}$.

$A$ rod closing the circuit shown in the figure moves along a $U$-shaped wire at a constant speed $v$ under the action of a force $F$. The circuit is in a uniform magnetic field perpendicular to the plane. Calculate $F$ if the rate of heat generation in the circuit is $Q$.

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