Find the current in the conductor PQ so that it remains in equilibrium under its own weight and the magnetic force. If direction of current is reversed then what would be initial acceleration of the rod? Mass of the rod is m. (Friction is assumed to be absent)

Explanation

Magnetic force on the conductor is equal to `F_(m) = Bil`. This force is along horizontal. Direction of I should be from Q to P. In this case its component `F_(m) cos theta` will balance mg `sin theta` to keep it in equilibrium.
Since the rod can not move along perpendicular to the
plane of the rails
Hence for equilibrium `F_(m) cos theta = mg sin theta`
`implies F_(m) = mg tan theta`
`implies BI l = mg tan theta implies i = mg tan theta // Bl`
On reversing the direction of current accleration of the rod will be
`a = (F_(m) cos theta + mg sin theta)/(m) = (BI l cos theta + mg sin theta)/(m) = 2g sin theta`