Paragraph A special metal $S$ conducts electricity without any resistance. A closed wire loop, made of $S$,…

Paragraph
A special metal $S$ conducts electricity without any resistance. A closed wire loop, made of $S$, does not allow any change in flux through itself by inducing a suitable current to generate a compensating flux. The induced current in the loop cannot decay due to its zero resistance. This current gives rise to a magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of radius $a$, with its center at the origin. A magnetic dipole of moment $m$ is brought along the axis of this loop from infinity to a point at distance $r(\gg a)$ from the center of the loop with its north pole always facing the loop, as shown in the figure below.

The magnitude of magnetic field of a dipole $m$, at a point on its axis at distance $r$, is $\frac{\mu_{0}}{2 \pi} \frac{m}{r^{3}}$, where $\mu_{0}$ is the permeability of free space. The magnitude of the force between two magnetic dipoles with moments, $m_{1}$ and $m_{2}$, separated by a distance $r$ on the common axis, with their north poles facing each other, is $\frac{k m_{1} m_{2}}{r^{4}}$, where $k$ is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles.

QuestionWhen the dipole m is placed at a distance r from the centre of the loop (as shown in the figure), the current induced in the loop will be proportional to:
  1. mr3
  2. m2r2
  3. mr2
  4. m2r

Solution

For a super conducting loop, net flux passing through it will remain constant.

ϕtotal i=ϕtotal f

As area is same, therefore,

μ0m2πr3=μ0I2a

I=amπr3

Imr3

`

Asked in: JEE Advanced 2021 (Paper 2)

Practice more Electromagnetic Induction questions on Aicharya