Elementary problem but far too long to type out solution, i might do that after jee advanced

But the method is to consider a thin ring, find the flux through it, and then differentiate it, and then equate it to iR where i is the small amount of current that flows through it (the eddy current ) and R is the pretty big resistance of the thin ring

( $R = \frac {\rho 2 \pi r}{drdz}$

now u have the i that flows through it

now you find the force upon it (only the component of field in the plane of ring will contribute to force, once found , integrate throught from (r1 till r2) and then from (- to + infinity) to get total force,

now use newtons third law to claim the reverse, and then ofcourse equate to mg

The only trouble in the analsys is how to find the magnetic field in a simple fashion, because represting B in terms or 'r' and $\theta$ is useless here

so rather use the magnetic potential which is

$\dfrac { \mu M cos( \theta ) }{4 \pi r^2} = \dfrac { \mu M x}{4 \pi (x^2+y^2)^{3/2}}$

and partial differentiate with respect to x and y to find the components as needed

i dont suppose theres a better way? but how did you do it @Ronak Agarwal

And yes i saw that video, seen it before, amazing one, there is more however, checkout Messeiner effect

Ya.A past model of an APHO question.The approach makes use of the following. 1)First find the magnetic field due to a dipole(we don't want the exact form but we need the component parallel to the dipole axis) 2)Once we get the field we are ready to calculate the flux through a thin ring of width $dr$ and height $dy$ .The magnet being at a distance y from its centre.3)Find its resistance and noe from faradays law the current can be found out.Once we have the current we may find the force on the magnet.4)For calculating the force we know the magnetic field produced by a current carrying loop.So the force is given by $(f=MdB/dy)$ .Once we have the force on this small part we perform a double integration this looks of the sort of $d(dF)$ = $(9/8)$ $(M^2U^2v/πp)\int \int r^3y^2drdy/(r^2+y^2)^5$ .where the r varies from $r1 to r2$ and y from -infinite to +infinite.Once we have got the force equate with the weight to get the velocity v.The integral involving r can be easily found out but the one using $y$ needs to be computed carefully. So the final and came to be $(2^(10)/15)(mgp/M^2U^2)1/(r1^-3-r2^-3)=v$ Note that $p$ denoted the magnetic moment and $U$ denotes the permeability of vacuum.

While the consideration of flux and force has certainly helped as an understandable direct approach towards solving the problem, nevertheless, this question has got me thinking of using an alternative approach of employing the idea of energy conservation. In this aspect, I found a suitable research analysis that provides such a viewpoint which I think it is really elegant, considering the minimal integrals we would need to work through compared to our considerations of force and flux concepts. So, do take a look.

The Energy Approach