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.

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Quantrex Academy · Accepted Answer

Correct answer: C. m ^ 2 r ^ 6

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.

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

As area is same, therefore,

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

$⇒ I = \dfrac{am}{{\pi r}^{3}}$

$⇒ I ∝ \dfrac{m}{{r}^{3}}$

The current carrying superconducting loop will also behave like a magnet, whose magnetic dipole moment is given by, ${m}_{1} = NIA = ( 1 ) \left(\dfrac{am}{{\pi r}^{3}}\right) \times \pi{a}^{2} = \dfrac{m{a}^{3}}{{r}^{3}}$

$⇒ {m}_{1} ∝ \dfrac{m}{{r}^{3}}$

The repulsive force felt on the magnet will be

$F = \dfrac{K{m}_{1}{m}_{2}}{{r}^{4}} = \dfrac{\left({K}^{'}\right)\left(\dfrac{m}{{r}^{3}}\right)\left(m\right)}{{r}^{4}} ∝ \dfrac{{m}^{2}}{{r}^{7}}$

${W}_{ext} = - ∫ Fdr = - K ' ∫ \dfrac{{m}^{2}}{{r}^{7}}dr$

Using $∫ {x}^{n}dx = \dfrac{{x}^{n + 1}}{n + 1}$

We get,

${W}_{ext} ∝ \dfrac{{m}^{2}}{{r}^{6}}$

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

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