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(D) The motional $Emf$ induced in the rod is given by $e = B l v$,where $l$ is the length of the rod.The current $I$ flowing through the resistor $R$

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$4$

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(D) The motional $Emf$ induced in the rod is given by $e = B l v$,where $l$ is the length of the rod.The current $I$ flowing through the resistor $R$ is $I = \frac{e}{R} = \frac{B l v}{R}$.The rate of heat dissipation (power) $P$ across the resistor $R$ is given by $P = I^2 R$.Substituting the expression for $I$,we get $P = \left( \frac{B l v}{R} \right)^2 R = \frac{B^2 l^2 v^2}{R}$.From this expression,it is clear that the rate of heat dissipation is proportional to the square of the speed: $P \propto v^2$.If the speed is doubled $(v' = 2v)$,the new rate of heat dissipation $P'$ will be $P' \propto (2v)^2 = 4v^2$.Therefore,the ratio of the new rate of heat dissipation to the initial rate is $\frac{P'}{P} = \frac{4v^2}{v^2} = 4$.Thus,the rate of heat dissipation increases by a factor of $4$.

Similar Questions

$A$ circular disc of radius $0.1 \ m$ is rotating about its axis at a frequency of $10 \ rev/s$ in a uniform magnetic field of $0.1 \ T$ directed perpendicular to the disc. Calculate the induced $emf$ between the center and the rim of the disc.

$A$ conducting rod of length $l$ moves with velocity $\upsilon$ in a direction parallel to a long wire carrying a steady current $I$. The axis of the rod is maintained perpendicular to the wire with the near end at a distance $r$ away as shown in the figure. Find the emf induced in the rod.

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