In the figure two parallel loops of wire having a common axis have been shown. The smaller loop (radius r) is above the larger loop (radius R), by a distance . Consequently the magnetic field, due to the current in the larger loop, is nearly constant throughout the smaller loop and equal to the value on the axis. Suppose that x is increasing at constant rate . (a) We have to determine the magnetic flux across the area bounded by the smaller loop as a function of x. (b) We have to compute the emf generated in the smaller loop. (c) We have to determine the direction of the induced current flowing in the smaller loop.

Solution:             Click For PDF Version

As the point P is on the axis of the loop of radius R and is at a distance x from the plane of the loop such that

,

and a current is flowing in the counter-clockwise direction, the magnetic field at P will be approximately given by

It is the magnetic field due to a dipole of moment at a point on its axis far away from the dipole at a distance x from it.

In this approximation we can assume that the magnetic field in the region enclosed by the loop of radius r will be uniform and equal to the value of the field at the centre of the circle. Therefore, the flux enclosed by the loop of radius r will be

As the smaller loop is moving in the direction of the axis with speed v, the induced emf in it will be given by the Faraday’s law,

Therefore, the induced emf in the loop of radius r will be

As the flux enclosed in the loop of radius r will decrease as it moves up along the axis, therefore, by the Lenz’ law the direction of the induced current will be counter-clockwise as seen from the top along the line joining the centres of the two loops.