Question

A conducting bar 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 be

• A. Constant.
• B. Quarter of the initial value.
• C. Four fold.
• D. Doubled.

Answer 1

Answer: (C)

Four fold.

Answer 2

The induced emf between A and B = ε = B v

⇒ The induced current = i = $\frac{\varepsilon}{R}$ ⇒ i = $\frac{B\mathcal{\mspace{6mu} l\mspace{6mu}}v}{R}$

The electrical power = P = i²R = $\frac{B^{2}\mspace{6mu}\mathcal{l}^{2}\mspace{6mu} v^{2}}{R}$ Since v is doubled, the electrical power, becomes four times. Since heat dissipation per second is proportional to electrical power, it becomes four fold.

Hence (3) is correct.