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 c

Question

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 ( ≫ 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 2 π μ 0 ​ ​ r 3 m ​ , where μ 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 r 4 k m 1 ​ m 2 ​ ​ , 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 distance r from the centre of the loop by the given process is proportional to:

TestHub · Accepted Answer

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,

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

Physics - Magnetism Question with Solution | TestHub

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