A conducting bar PQ is free to slide on two parallel conducting rails as shown in figure. Two resistors $R_1$ and $R_2$ are connected across the ends of the rails. There is a uniform magnetic field B pointing into the page. An external agent pulls the bar to the left at a constant speed v. The correct statement about the directions of induced currents $I_1$ and $I_2$ flowing through $R_1$ and $R_2$ respectively is:

(A) $I_1$ is in anticlockwise direction and $I_2$ is in clockwise direction
(B) Both $I_1$ and $I_2$ are in anticlockwise direction
(C) Both $I_1$ and $I_2$ are in clockwise direction
(D) $I_1$ is in clockwise direction and $I_2$ is in anticlockwise direction

Solution

As PQ moves leftwards, the area of loop ABQP decreases causing reduction in flux. So, the direction of current through $R_1$ should be such that the flux can be increased. An upward current through $R_1$ helps to strengthen the magnetic field. Upward current through $R_1$ is the clockwise current through the left loop.

As PQ moves leftwards, the area of loop A’B’QP increases causing increase in flux. So, the direction of current through $R_2$ should be such that the flux can be decreased. An upward current through $R_2$ helps to weaken the magnetic field. Upward current through $R_2$ is the anticlockwise current through the right loop.

Hence, Option (D).