Question

The figure given below shows a square-shaped loop MNOP of side 25 cm placed horizontally in a uniform magnetic field `vec B` directed vertically downward. The position of the loop at t = 0 s is as shown in figure. The loop is pulled with a constant velocity of 25 cm/s till it goes out of the field.

1. What will be the direction of the induced current in the loop as it goes out of the field? For how long would the current in the loop persist?
2. Plot graphs showing the variation of magnetic flux and the magnitude of induced emf as a function of time.

Answer 1

a. Direction and Duration:

As the loop moves out to the right, the area inside the magnetic field decreases. Thus, the magnetic flux (directed downwards) decreases.

The generated current must provide a downward (within the loop) magnetic field to counteract this drop. The current must flow clockwise according to the right-hand rule (M → N → O → P).

The side of the square is a = 25 cm. The current persists as long as the loop is exiting the field.

Time taken for the loop to completely exit:

t = `"Length of side"/"Velocity"`

= `(25 cm)/(25 cm//s)`

= 1 s

b.