Predict the direction of induced current in the situations described by the following figures (a) to (d).

Explanation

(a) Here south pole of bar magnet is approaching end q of coil. Hence, according to Lenz.s law, to oppose the motion of bar magnet, face of coil at end q in the figure must become induced south pole, for which induced current at end q must flow clockwise (as seen from right to left) i.e. in the direction `q to r to p to q`
(b) Here induced current in the left coil should flow in a direction `p to r to q to p` so that face of this coil at end q can be induced south pole to oppose the motion of south pole of bar magnet towards it.
Similarly induced current in the right coil should flow in a direction `x to y to zx` so that face of this coil at end x can be induced south pole to oppose the motion of north pole of bar magnet, receding from it.
(c )(c) Here when key is closed, right face of left coil becomes south pole and so left face of right coil must become induced south pole and so current in the right coil should flow in a direction `y to z to x to y` (which will be found clockwise as seen from left side and anticlockwise as seen from right side into this coil).
(d) Here when slider terminal is shifted towards fixed terminal, resistance decreases and so current through right loop increases in anticlockwise direction as we see from left to right. At this time left face of right loop behaves like a north pole. So according to Lenz.s law, right face of left loop should become induced north pole. Hence, its left face should become induced south pole. Hence direction of induced current in the left loop should be clockwise as we see from left to right. i.e. in the direction `y to x to z to y`.
(e) Here when the closed key is made open, current flowing through left solenoid (in anticlockwise direction, as seen from right to left) decreases. Hence induced current in the right solenoid should flow in such a direction which can increase the magnetic flux through left solenoid. For this induced current should flow in right solenoid in the direction `x to r to y to x`
Here magnetic field lines are in the form of coplanar, concentric circles with the centres on the current carrying extremely long and thin straight wire. Here magnetic field lies in horizontal planes which are parallel to plane of given circular ring. Hence for this circular ring, its area vector `vecA bot vecB` and so no magnetic flux is linked with the ring. Hence though the current through the wire is increased we will not get any induced current in the circular ring, shown in the diagram.