Question

An infinitesimally small bar magnet of dipole moment $M$ is pointing and moving with the speed $v$ is in the $X$-direction. A small closed circular conducting loop of radius $a$ is placed in the $Y-Z$ plane with its centre at $x = 0$, and its axis coinciding with the $X$-axis. Find the force opposing the motion of the magnet, if the resistance of the loop is $R$. Assume that the distance $x$ of the magnet from the centre of the loop is much greater than $a$.

Answer 1

$F_{\rm opp}=-\frac{9\mu_0^2a^4M^2v}{4R\,x^8}$

Answer 2

The force opposing the motion of the magnet can be found by following these steps:

1. Calculate the magnetic field $B_x$ due to the dipole at a distance $x$ from the loop's center:

   $$B_x=\frac{\mu_0}{4\pi}\frac{2M}{x^3}$$

2. Determine the magnetic flux $\Phi$ through the loop:

   $$\Phi = \pi a^2 B_x = \frac{\mu_0 a^2 M}{2x^3}$$

3. Compute the induced EMF $\mathcal{E}$ using Faraday's Law:

   $$\mathcal{E} = -\frac{d\Phi}{dt} = \frac{3\mu_0 a^2 Mv}{2x^4}$$

4. Calculate the induced current $I$ in the loop:

   $$I = \frac{\mathcal{E}}{R} = \frac{3\mu_0 a^2 Mv}{2Rx^4}$$

5. Find the power $P$ dissipated in the loop:

   $$P = I^2 R = \frac{9\mu_0^2 a^4 M^2 v^2}{4Rx^8}$$

6. Relate the power to the opposing force $F_{\rm opp}$:

   $$P = F_{\rm opp} v$$

7. Solve for the opposing force:

   $$F_{\rm opp} = \frac{P}{v} = \frac{9\mu_0^2 a^4 M^2 v}{4Rx^8}$$

Thus, the force opposing the motion is:

$$F_{\rm opp} = -\frac{9\mu_0^2 a^4 M^2 v}{4Rx^8}$$