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JEE Advanced 2021Paper 2
Exam
JEE Advanced
Year
2021
Paper
Paper 2
Subject
Physics
Chapter
Electromagnetic Induction

JEE Advanced 2021 · Paper 2 · Physics · Electromagnetic Induction

Paragraph A special metal S conducts electricity without any...

Paragraph
A special metal SS conducts electricity without any resistance. A closed wire loop, made of SS, 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 aa, with its center 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(≫a)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 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.
Paragraph A special metal S conducts electricity without any resistance. A closed wire loop, made of S , does not allow
QuestionWhen the dipole mm is placed at a distance rr from the centre of the loop (as shown in the figure), the current induced in the loop will be proportional to:

Options

  1. Amr3\dfrac{m}{{r}^{3}}
  2. Bm2r2\dfrac{{m}^{2}}{{r}^{2}}
  3. Cmr2\dfrac{m}{{r}^{2}}
  4. Dm2r\dfrac{{m}^{2}}{r}

Correct answer

A. mr3\dfrac{m}{{r}^{3}}

Step-by-step solution

Paragraph A special metal S conducts electricity without any resistance. A closed wire loop, made of S , does not allow

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

(ϕtotal)i=(ϕtotal)f{\left({ϕ}_{total}\right)}_{i} = {\left({ϕ}_{total}\right)}_{f}

As area is same, therefore,

μ0m2πr3=μ0I2a\dfrac{{\mu}_{0}m}{2{\pi r}^{3}} = \dfrac{{\mu}_{0}I}{2a}

⇒I=amπr3⇒ I = \dfrac{am}{{\pi r}^{3}}

⇒I∝mr3⇒ I ∝ \dfrac{m}{{r}^{3}}

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