Figure shows a metal rod PQ resting on the rails AB and positioned between the poles of a permanent magnet. The rails, the rod and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = \( 180.0\,\mathrm{m\Omega} \). Assume the field to be uniform.
The power dissipated as heat in the closed circuit is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 4.5 \times 10^{-4}\,\mathrm{W} \)
With K closed, the motional emf of the rod drives a current round the loop, and the electrical energy appears as Joule heat in the loop resistance. The rod moves at \( 12\,\mathrm{cm\,s^{-1}} \), the speed used in this set of questions.
\( l = 15\,\mathrm{cm} = 0.15\,\mathrm{m} \)
\( v = 12\,\mathrm{cm\,s^{-1}} = 0.12\,\mathrm{m\,s^{-1}} \)
\( R = 180.0\,\mathrm{m\Omega} = 0.180\,\Omega \)
\( \displaystyle \varepsilon = B\,v\,l = 0.50 \times 0.12 \times 0.15 = 9.0\times10^{-3}\,\mathrm{V} \)
\( \displaystyle P = \frac{\varepsilon^{2}}{R} = \frac{(9.0\times10^{-3})^{2}}{0.180} \)
\( \displaystyle P = \frac{8.1\times10^{-5}}{0.180} = 4.5\times10^{-4}\,\mathrm{W} \)