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 required (by an external agent) to keep the rod moving at the same speed (\( = 12\,\mathrm{cm\,s^{-1}} \)) when K is closed will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 4.5 \times 10^{-4}\,\mathrm{W} \)
With K closed a current \( I \) flows in the rod, and the field exerts a force \( F = BIl \) on it. By Lenz's law this force opposes the motion, so the external agent must push with an equal force to keep the speed constant. The power it supplies is \( P = Fv \).
\( l = 0.15\,\mathrm{m} \)
\( v = 0.12\,\mathrm{m\,s^{-1}} \)
\( R = 0.180\,\Omega \)
\( \displaystyle \varepsilon = B\,v\,l = 0.50 \times 0.12 \times 0.15 = 9.0\times10^{-3}\,\mathrm{V} \)
\( \displaystyle I = \frac{\varepsilon}{R} = \frac{9.0\times10^{-3}}{0.180} = 0.05\,\mathrm{A} \)
\( \displaystyle F = B\,I\,l = 0.50 \times 0.05 \times 0.15 = 3.75\times10^{-3}\,\mathrm{N} \)
\( \displaystyle P = F\,v = 3.75\times10^{-3} \times 0.12 = 4.5\times10^{-4}\,\mathrm{W} \)
This is equal to the Joule heat \( \varepsilon^{2}/R \) produced in the loop: the work done by the agent is converted into heat, as conservation of energy requires.