An electricity and magnetism problem by Sudeep Salgia

The field due to the bar magnet can be written as $\displaystyle B = \frac{\mu _0 M}{2 \pi x^3}$ . Hence flux through the ring is given by $\displaystyle \phi = B.A. \cos 0 = \frac{\mu _0 M \pi a^2}{2 \pi x^3} = \frac{\mu _0 M a^2}{2 x^3}$ .

Thus using the Laws for Electromagnetic Induction, the induced current in the loop can be determined.

$\displaystyle i = - \frac{1}{R} . \frac{\text{d} \phi }{\text{d}t} \Rightarrow i = - \frac{1}{R} . \frac{\text{d}}{\text{d}t} \bigg( \frac{\mu _0 M a^2}{2 x^3} \bigg)$ $\displaystyle i = - \frac{\mu _0 M a^2}{2 R} \bigg( - \frac{3}{x^4}. \frac{\text{d} x}{\text{d}t} \bigg) \Rightarrow i= \frac{3 \mu _0 M a^2 v }{2R x^4}$

Using formula for the field due to a circular current carrying loop, we get,
$\displaystyle B_{\text{loop}} = \frac{\mu_0 i a^2}{2 (a^2 + x^2)^{3/2} } = \frac{\mu_0 a^2}{2 x^3} \bigg( \frac{3 \mu _0 M a^2 v }{2R x^4} \bigg) = \frac{3 \mu _0^2 M a^4 v }{4R x^7}$

For the force of interaction we will not use the mentioned formula because the magnetic moments are varying and not constant . The formula is actually only for constant moments $M_1$ and $M_2$ . Hence,
$\displaystyle F = - \frac{\text{d}}{\text{d}x} ( - \vec{M}.\vec{B_{\text{loop}}} )$
$\displaystyle \Rightarrow F = \frac{\text{d}}{\text{d}x} \bigg( \frac{3 \mu _0^2 M^2 a^4 v }{4R x^7} \bigg)$ $\displaystyle \Rightarrow |F| = \frac{21}{4}.\frac{\mu_0^2 M^2 a^4 v }{Rx^8}$ . Add the required values to get the answer.
NOTE: I have deliberately given the formula to bring out the importance of the question and the validity of the formula.