In the above diagram, a strong bar magnet is moving towards solenoid-2 from solenoid-1. The direction of induced current in solenoid-1 and that in solenoid-2, respectively, are through the directions:
Answer: (C) $AB$ and $DC$
The magnet moves to the right, with its N pole facing solenoid-1 and its S pole facing solenoid-2.
Solenoid-1: the N pole is moving away. By Lenz's law, the induced current opposes this by attracting the magnet, so the end of solenoid-1 facing the magnet becomes an S pole.
Solenoid-2: the S pole is approaching. The induced current opposes this by repelling the magnet, so the end of solenoid-2 facing the magnet also becomes an S pole.
In the figure, both solenoids are wound the same way. For this winding, current entering at the left terminal (A or C) makes the right-hand face of the solenoid an S pole.
- Solenoid-1 needs its right face to be S: current flows $A \to B$.
- Solenoid-2 needs its left face to be S (so its right face is N): current flows $D \to C$.
So the currents are along $AB$ and $DC$.
Solution by Sreeraj P, M.Sc Physics