An infinitesimally small bar magnet of dipole moment $\vec{M}$ is pointing and moving with the speed v in the x-direction. A small closed circular conducting loop of radius a and of negligible self-inductance lies 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.

Solution

Given,
A very small bar magnet of dipole moment

\overset{⇀}{M}

moving with speed v in the x-direction.

Field due to the bar magnet at distance x (near the loop)

B_{a} = \frac{μ_{0}}{4 π} . \frac{2 M}{x^{3}}

(axial line)

Flux linked with the loop,

ϕ

BA = {πa}^{2} . \frac{μ_{0}}{4 π} . \frac{2 M}{x^{3}}

Emf induced in the loop, e = - \frac{dϕ}{dt} = \frac{μ_{0}}{4 π} . \frac{6 π {Ma}^{2}}{x^{4}} . \frac{dx}{dt}

= \frac{μ_{0}}{4 π} . \frac{6 π {Ma}^{2}}{x^{4}} v .

Thus,
Induced current,

i = \frac{e}{R} = \frac{μ_{0}}{4 π} . \frac{6 π {Ma}^{2}}{{Rx}^{4}} v .

Let F be the force opposing the motion of the magnet.

Power due to the opposing force = Heat dissipated in the coil per second
This implies,

Fv = i^{2} R F = \frac{i^{2} R}{v} = {(\frac{μ_{0}}{4 π})}^{2} \times {(\frac{6 π {Ma}^{2}}{{Rx}^{4}})}^{2} x v^{2} \times \frac{R}{v}

F = \frac{9}{4} (\frac{μ_{0}^{2} M^{2} a^{4} v}{{Rx}^{8}}) .

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