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(C) The motional electromotive force $(EMF)$ induced in the conducting bar moving with speed $v$ in a magnetic field $B$ is given by $E = B \ell v$,wh

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(C) The motional electromotive force $(EMF)$ induced in the conducting bar moving with speed $v$ in a magnetic field $B$ is given by $E = B \ell v$,where $\ell$ is the length of the bar.The rate of heat dissipation (power) in the circuit with resistance $R$ is given by $P = \frac{E^2}{R}$.Substituting the expression for $E$,we get $P = \frac{(B \ell v)^2}{R} = \frac{B^2 \ell^2 v^2}{R}$.From this expression,it is clear that the power $P$ is proportional to the square of the speed,i.e.,$P \propto v^2$.If the speed of the bar is doubled $(v' = 2v)$,the new rate of heat dissipation $P'$ will be:$P' = \frac{B^2 \ell^2 (2v)^2}{R} = 4 	imes \frac{B^2 \ell^2 v^2}{R} = 4P$.Therefore,the rate of heat dissipation becomes four times the initial value.

$A$ conducting bar is pulled with a constant speed $v$ on a smooth conducting rail. The region has a steady magnetic field of induction $B$ as shown in the figure. If the speed of the bar is doubled then the rate of heat dissipation will

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