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 \mu _0 \over 2\pi m \over r^3 , where \mu 0 is the permeability of free space. The magnitude of the force between two magnetic dipoles with moments, m1 and m2, separated by a distance r on the common axis, with their north poles facing each other, is km_1m_2 \over r^4 , where k 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 r from the center of the loop by the given process is proportional to

Magnetism PYQ — JEE Advanced 2021 · Shift 2 · Q52 with Solution

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Correct answer: C

Magnetic flux through the superconducting loop

Since the loop is made of a superconducting material, it does not allow any change in magnetic flux through itself.

So, when the magnetic dipole is brought near the loop, an induced current appears in the loop such that the net flux through the loop remains zero.

For $r \gg a$ , the magnetic field due to the dipole is nearly uniform across the loop.

Given axial field of the dipole at the loop center:

B_{\text{dipole}}=\frac{\mu_0}{2\pi}\frac{m}{r^3}

Hence flux through the loop due to the dipole is

\Phi_{\text{ext}}=B_{\text{dipole}}\cdot \pi a^2 =\frac{\mu_0}{2\pi}\frac{m}{r^3}\cdot \pi a^2 =\frac{\mu_0 a^2 m}{2r^3}

To keep total flux unchanged, the induced current in the loop creates an opposing magnetic field. Therefore the magnetic field at the center due to the induced loop must satisfy

B_{\text{loop}}\,\pi a^2 \propto \Phi_{\text{ext}}

B_{\text{loop}} \propto \frac{m}{r^3}

Magnetic moment of the induced current loop

For a circular current loop of radius $a$ , magnetic field at its center is proportional to its magnetic moment divided by $a^3$ :

B_{\text{loop}} \propto \frac{m_{\text{ind}}}{a^3}

Thus,

m_{\text{ind}} \propto a^3 B_{\text{loop}} \propto a^3\cdot \frac{m}{r^3}

Since $a$ is constant,

m_{\text{ind}} \propto \frac{m}{r^3}

Force between the dipole and induced dipole

Given that the force between two axial dipoles is

F=\frac{k m_1 m_2}{r^4}

Here,

$m_1 = m$
$m_2 = m_{\text{ind}} \propto \dfrac{m}{r^3}$

So,

F \propto \frac{m\left(\dfrac{m}{r^3}\right)}{r^4} =\frac{m^2}{r^7}

Work done in bringing the dipole from infinity to distance $r$

Work done against this repulsive force is

W=\int_{\infty}^{r} F\,dr

Since

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

we get

W \propto \int_{\infty}^{r} \frac{m^2}{x^7}dx \propto \frac{m^2}{r^6}

because

\int \frac{dx}{x^7} = -\frac{1}{6x^6}

Thus,

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

Option check

A: $\dfrac{m}{r^5}$ ❌
B: $\dfrac{m^2}{r^5}$ ❌
C: $\dfrac{m^2}{r^6}$ ✅
D: $\dfrac{m^2}{r^7}$ ❌

Therefore, the correct option is C.

More from Magnetism

View written solutionFree

Correct answer: C

Magnetic flux through the superconducting loop

Since the loop is made of a superconducting material, it does not allow any change in magnetic flux through itself.

So, when the magnetic dipole is brought near the loop, an induced current appears in the loop such that the net flux through the loop remains zero.

For $r \gg a$ , the magnetic field due to the dipole is nearly uniform across the loop.

Given axial field of the dipole at the loop center:

B_{\text{dipole}}=\frac{\mu_0}{2\pi}\frac{m}{r^3}

Hence flux through the loop due to the dipole is

\Phi_{\text{ext}}=B_{\text{dipole}}\cdot \pi a^2 =\frac{\mu_0}{2\pi}\frac{m}{r^3}\cdot \pi a^2 =\frac{\mu_0 a^2 m}{2r^3}

To keep total flux unchanged, the induced current in the loop creates an opposing magnetic field. Therefore the magnetic field at the center due to the induced loop must satisfy

B_{\text{loop}}\,\pi a^2 \propto \Phi_{\text{ext}}

B_{\text{loop}} \propto \frac{m}{r^3}

Magnetic moment of the induced current loop

For a circular current loop of radius $a$ , magnetic field at its center is proportional to its magnetic moment divided by $a^3$ :

B_{\text{loop}} \propto \frac{m_{\text{ind}}}{a^3}

Thus,

m_{\text{ind}} \propto a^3 B_{\text{loop}} \propto a^3\cdot \frac{m}{r^3}

Since $a$ is constant,

m_{\text{ind}} \propto \frac{m}{r^3}

Force between the dipole and induced dipole

Given that the force between two axial dipoles is

F=\frac{k m_1 m_2}{r^4}

Here,

$m_1 = m$
$m_2 = m_{\text{ind}} \propto \dfrac{m}{r^3}$

So,

F \propto \frac{m\left(\dfrac{m}{r^3}\right)}{r^4} =\frac{m^2}{r^7}

Work done in bringing the dipole from infinity to distance $r$

Work done against this repulsive force is

W=\int_{\infty}^{r} F\,dr

Since

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

we get

W \propto \int_{\infty}^{r} \frac{m^2}{x^7}dx \propto \frac{m^2}{r^6}

because

\int \frac{dx}{x^7} = -\frac{1}{6x^6}

Thus,

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

Option check

A: $\dfrac{m}{r^5}$ ❌
B: $\dfrac{m^2}{r^5}$ ❌
C: $\dfrac{m^2}{r^6}$ ✅
D: $\dfrac{m^2}{r^7}$ ❌

Therefore, the correct option is C.

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