ParagraphA special metal $S$ conducts electricity without any resistance. A closed wire loop,...

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.

QuestionThe work done in bringing the dipole from infinity to a distance

r

from the centre of the loop by the given process is proportional to:

\frac{m}{r^{5}}

\frac{m^{2}}{r^{5}}

\frac{m^{2}}{r^{6}}

\frac{m^{2}}{r^{7}}

Solution:

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

${(ϕ_{total})}_{i} = {(ϕ_{total})}_{f}$

As area is same, therefore,

$\frac{μ_{0} m}{2 {πr}^{3}} = \frac{μ_{0} I}{2 a}$

$\Rightarrow I = \frac{a m}{{πr}^{3}}$

$\Rightarrow I \propto \frac{m}{r^{3}}$

The current carrying superconducting loop will also behave like a magnet, whose magnetic dipole moment is given by, $m_{1} = N I A = (1) (\frac{a m}{{πr}^{3}}) \times π a^{2} = \frac{m a^{3}}{r^{3}}$

$\Rightarrow m_{1} \propto \frac{m}{r^{3}}$

The repulsive force felt on the magnet will be

$F = \frac{K m_{1} m_{2}}{r^{4}} = \frac{(K^{'}) (\frac{m}{r^{3}}) (m)}{r^{4}} \propto \frac{m^{2}}{r^{7}}$

$W_{ext} = - \int F d r = - K' \int \frac{m^{2}}{r^{7}} d r$

Using $\int x^{n} d x = \frac{x^{n + 1}}{n + 1}$

We get,

$W_{ext} \propto \frac{m^{2}}{r^{6}}$

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ParagraphA special metal \(S\) conducts electricity without any resistance. A closed wire loop,...