The work done in bringing the dipole from infinity to a distance from the center of the loop by the given process is proportional to [2021]

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 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 $r$ $(r ≫ a)$ 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 $\frac{μ_{0} m}{2 π r^{3}},$ 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 $\frac{k m_{1} m_{2}}{r^{4}},$ where $k$ is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles.

Q. 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 [2021]

Smart Achievers · Accepted Answer

(3)

$Magnetic flux due to dipole through ring$

$a$ $\Rightarrow i = \frac{μ_{0} m π a^{2}}{2 π r^{3} L}$

or $i \propto \frac{m}{r^{3}}$

$Magnetic moment, m' = i A = π a^{2} i = \frac{μ_{0} m π^{2} a^{4}}{2 π r^{3} L}$

$F = \frac{k m m'}{r^{4}} = \frac{k m^{2} π^{2} a^{4}}{2 π r^{7} L}$

$Therefore work done in bringing the dipole,$

$W = \int F d r \propto \int \frac{m^{2} d r}{r^{7}}$ or, $W \propto \frac{m^{2}}{r^{6}}$

Solution

1 $\frac{m}{r^{5}}$

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

3 $\frac{m^{2}}{r^{6}}$

4 $\frac{m^{2}}{r^{7}}$

Want Us to Email you About Special Offers & Updates?

Solution

1 mr5

2 m2r5

3 m2r6

4 m2r7

Ans. (3) Magnetic flux due to dipole through ring ϕ=Li=μ0m2πr3×πa2⇒i=μ0mπa22πr3L or i∝mr3 Magnetic moment, m'=iA=πa2i=μ0mπ2a42πr3L F=kmm'r4=km2π2a42πr7L Therefore work done in bringing the dipole, W=∫F dr∝∫m2drr7 or, W∝m2r6

Want Us to Email you About Special Offers & Updates?

1 $\frac{m}{r^{5}}$

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

3 $\frac{m^{2}}{r^{6}}$

4 $\frac{m^{2}}{r^{7}}$