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

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(C) The induced electromotive force $(EMF)$ in a conducting bar moving in a magnetic field is given by $E = B \ell v$,where $B$ is the magnetic field,$\ell$ is the length of the bar,and $v$ is the velocity.The rate of heat dissipation (power) $P$ in the circuit with resistance $R$ is given by $P = \frac{E^2}{R}$.Substituting the expression for $E$,we get $P = \frac{(B \ell v)^2}{R} = \frac{B^2 \ell^2 v^2}{R}$.This shows that $P \propto v^2$.If the speed of the bar is doubled,i.e.,$v' = 2v$,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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