JEE Advanced · Physics · 20. Magnetism & Current
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.

Question :
The work done in bringing the dipole from infinity to a distancefrom the centre of the loop by the given process is proportional to:
- A
- B
- C
- D
Answer & Solution
Correct Answer
(C)
Step-by-step Solution
Detailed explanation
For a super conducting loop, net flux passing through it will remain constant.
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