Physics · Electromagnetic Induction

JEE Advanced 2021 — Paper 2 — Question 32

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 mm is brought along the axis of this loop from infinity to a point at distance r(≫r(\gg 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 mm, at a point on its axis at distance rr, is μ02πmr3\frac{\mu_{0}}{2 \pi} \frac{m}{r^{3}},

where μ0\mu_{0} is the permeability of free space. The magnitude of the force between two magnetic dipoles with moments, m1m_{1} and m2m_{2}, separated by a distance rr on the common axis, with their north poles facing each other, is km1m2r4\frac{k m_{1} m_{2}}{r^{4}}, where kk 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 rr from the center of the loop by the given process is proportional to

  1. Option A:

    m/r5\text{m}/{{\text{r}}^{5}}

  2. Option B:

    m2/r5{{m}^{2}}/{{r}^{5}}

  3. Option C:

    m2/r6{{\text{m}}^{2}}/{{\text{r}}^{6}}

    Correct
  4. Option D:

    m2/r7{{\text{m}}^{2}}/{{\text{r}}^{7}}

Answer: C

Step-by-step solution

dW=F.dx=−Km(i1πa2)r4(dr)\text{dW}=\text{F}.\text{dx}=\frac{-\text{Km}\left( {{\text{i}}_{1}}\pi {{\text{a}}^{2}} \right)}{{{\text{r}}^{4}}}\left( \text{dr} \right) i1=maπr3{{\text{i}}_{1}}=\frac{\text{ma}}{\pi {{\text{r}}^{3}}} W=∫∞rkm(maπr3×πa2)drr3=km2a3∫drr7\text{W}=\int _{\infty }^{\text{r}}\frac{\text{km}\left( \frac{\text{ma}}{\pi {{\text{r}}^{3}}}\times \pi {{\text{a}}^{2}} \right)\text{dr}}{{{\text{r}}^{3}}}=\text{k}{{\text{m}}^{2}}{{\text{a}}^{3}}\int \frac{\text{dr}}{{{\text{r}}^{7}}} Wαm2r6\text{W}\alpha \frac{{{\text{m}}^{2}}}{{{\text{r}}^{6}}}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2021
Paper
Paper 2
Subject
Physics
Chapter
Electromagnetic Induction
Topic
Magnetic Flux, Faraday's Law and Lenz's Law