Question

A conducting bar of length L is free to slide on two parallel conducting rails as shown in the figure. Two resistors $R_1\&R_2$ are connected across the ends of the rails. There is a uniform magnetic field $B$ pointing to 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 $I_1\&I_2$ flowing through $R_1\&R_2$ respectively is

• A. $I_1$ is in the clockwise direction and $I_2$ is in the anticlockwise direction
• B. Both $I_1\&I_2$ are in a clockwise direction
• C. $I_1$ is in the anticlockwise direction and $I_2$ is in a clockwise direction
• D. Both $I_1\&I_2$ are in the anticlockwise direction

Answer 1

The correct option is A

$I_1$ is in the clockwise direction and $I_2$ is in the anticlockwise direction

Circuit diagram:

$I_1$= Current flowing in loop 1

$I_2$= current flowing in loop 2

$R_1$=Resistance in loop 1

$R_2$=Resistance in loop 2

$v$=speed of the rod

Magnetic field $B$ is perpendicular to bar

Applied rule:

Right-hand rule -

1. Thumb indicates the direction of motion
2. The index finger indicates the direction of the magnetic field
3. The middle finger indicates the direction of current

Explanation:

1. When the bar slides towards the 1st loop, the area of the loop decreases due to which magnetic flux in it decreases.
2. An EMF is induced. In loop one if we apply the right-hand rule, the direction of the current comes out to be clockwise which will induce the magnetic field in the same direction as the external magnetic field to increase the flux.
3. The area of the 2nd loop increases due to which magnetic flux increases
4. In loop 2 if we apply the right-hand rule, the direction of the current comes out to be anticlockwise which will induce the magnetic field opposite to the direction of the external magnetic field to decrease the flux.

Therefore, the correct option is A.