• # question_answer A conducting bar is pulled with a constant speed U 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: A)  remain constant.B)  become quarter of the initial value.C)  become four fold.D)  get doubled.

$\varepsilon =Blv,\,I=\frac{\varepsilon }{r}=\frac{Blv}{R},P={{I}^{2}}R=\frac{{{B}^{2}}{{l}^{2}}{{v}^{2}}}{R}$ i.e., $P\propto {{v}^{2}}.$ Hence, the correction option is [c].