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)` (as seen by E) and an induced current `I_(Q1)` flows in Q. The switch remains closed for a long time. when S is opened, a current `I_(Q2)` flows in Q. Then the direction `IQ_(1) and IQ_(2)` (as seen by E) are

Options

A. respectively clockwise and anti-clockwise
B. both clockwise
C. both anti-clockwise
D. respectively anti-clockwise and clockwise

Answer

D

Explanation

Let the switch be closed at time `t=0`. The current `O_(p)` flowing in `P` grows for a time, say `t_(0)`, after which it becomes steady. During this time the magnetic field (du to `I_(p))` (from left to right) increases at the location of loop Q. According to Lenz's law, the induced current `(I_(Q))_(1)` should be such that it tries to decrease the magnetic field. Therefore, the magnetic field due to this current must be from right to left. Hence this induced current `(I_(Q))_(1)` should be anticlockwise (opposite to the direction of `I_(p))`. After the switch S is opened, the current `I_(P)` takes a finite time to decay to zero and the reverse of the above phenomenon is observed. Hence the induced current `(I_(Q))_(2)` should be clockwise.