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## Homework Statement

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 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.

## Homework Equations

## The Attempt at a Solution

Magnetic field due to the magnet of dipole moment ##\vec{M}## on its axis at a distance x (i.e. at the centre of the ring) = ##B = \dfrac{2μ_0M}{4πx^3} = \dfrac{μ_0M}{2πx^3}##

##φ_{ring} = B.A = \dfrac{μ_0M}{2πx^3}.πa^2 = \dfrac{μ_0Ma^2}{2x^3}##

##|ε| = \dfrac{dφ}{dt} = \dfrac{3μ_0Ma^2v}{2x^4}##

##i## (flowing in ring) = ## \dfrac{3μ_0Ma^2v}{2x^4R}##

After this, I'm confused.

My attempt:

## F = ilB = \dfrac{3μ_0Ma^2v}{2x^4R}.2a.\dfrac{μ_0M}{2πx^3}##

## F = \dfrac{3μ_0M^2a^3v}{2x^7Rπ}##

This answer is wrong. It is surely due to me having taken l = 2a. Usually, when we consider arbitrarily shaped conductors in a uniform magnetic field (it is uniform on the ring as x>>a), the force, i.e.

##\vec{F} = \int I \vec{dL} × \vec{B} = I \vec{L} × \vec{B}## where ##\vec{L}## is the length vector joining initial and final points of the conductor. Now, in case of a circular loop, this should've been zero. But that is clearly wrong as there is some force due to lenz's law. I believe it is wrong because the lorentz force equation can't be applied to currents induced due to the magnetic field. It can only be used when we have a current carrying conductor placed in an external magnetic field (I think). In any case, I'm not sure what to do. Please help.