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 magnetic force experienced by the rod when K is closed will be (Assume \( v = 12\,\mathrm{cm\,s^{-1}} \)):
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 3.75 \times 10^{-3}\,\mathrm{N} \)
With K closed the induced current flows through the rod, and a current-carrying rod perpendicular to a magnetic field feels the magnetic force \( F = BIl \).
\( l = 15\,\mathrm{cm} = 0.15\,\mathrm{m} \)
\( v = 12\,\mathrm{cm\,s^{-1}} = 0.12\,\mathrm{m\,s^{-1}} \)
\( R = 180.0\,\mathrm{m\Omega} = 0.180\,\Omega \)
\( \displaystyle \varepsilon = B\,v\,l = 0.50 \times 0.12 \times 0.15 = 9.0\times10^{-3}\,\mathrm{V} \)
\( \displaystyle I = \frac{\varepsilon}{R} = \frac{9.0\times10^{-3}}{0.180} = 0.05\,\mathrm{A} \)
\( \displaystyle F = B\,I\,l = 0.50 \times 0.05 \times 0.15 = 3.75\times10^{-3}\,\mathrm{N} \)
By Lenz's law this force acts opposite to the velocity of the rod.