Question

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 centre at the origin. A magnetic dipole of moment $m$ is brought along the axis of this loop from infinity to a point at distance $r(>>$ a) from the centre 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 $\frac{ km _{1} m _{2}}{ r ^{4}}$, where $k$ is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles.

The work done in bringing the dipole from infinity to a distance $r$ from the center of the loop by the given process is proportional to

• A. $m / r ^{5}$
• B. $m ^{2} / r ^{5}$
• C. $m ^{2} / r ^{6}$
• D. $m ^{2} / r ^{7}$

Answer 1

Answer: (C)

$dW = F . d x =\frac{- Km \left( i _{1} \pi a ^{2}\right)}{ r ^{4}}( dr )$
$i _{1}=\frac{ ma }{\pi r ^{3}}$
$W =\int\limits_{\infty}^{r} \frac{ km \left(\frac{ ma }{\pi r ^{3}} \times \pi a ^{2}\right) dr }{ r ^{3}}= km ^{2} a ^{3} \int \frac{ dr }{ r ^{7}}$
$W \alpha \frac{ m ^{2}}{ r ^{6}}$