As shown in the figure, P and Q are two coaxial conducting loops separated by some distance. When the switch S is closed, a clockwise current $I_{P}$ flows in P (as seen by E) and an induced current $I_{Q 1}$ flows in Q. The switch remains closed for a long time. When S is opened, a current $I_{Q 2}$ flows in Q. Then the direction of $I_{Q 1}$ and $I_{Q 2}$ (as seen by E) are [2002]

Question

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(4)

When switch $S$ is closed, a magnetic field is set up in the space around $P$ . The field lines threading $Q$ increase in the direction from right to left. According to Lenz's law, $I_{Q_{1}}$ will flow so as to oppose the cause or change and flow in the anticlockwise direction as seen by $E$ . Opposite is the case when $S$ is opened. $I_{Q_{2}}$ will be clockwise.

Solution

1 respectively clockwise and anti-clockwise

2 both clockwise

3 both anti-clockwise

4 respectively anti-clockwise and clockwise

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