An infinitesimal bar magnet of dipole moment `M` is pointing and moving with speed `v` in the `x`-direction. A 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 `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 .

Explanation

A bar magnet can be thought of a current carrying coil of dipole moment M. The magnetic field of a current carrying coil at a distance x on its axis (x `gt gt` radius of the coil ) can be given as
`B=(mu_(0)(2M))/(4piX^(3))= (mu_(0)M)/(2piX^(3))`
`implies` The flux linked with the given loop
`phi` = B . (area of the loop )
`implies phi = B(pia^(2))= (mu_(0)Ma^(2))/(2x^(3))`
emf induced in the loop
`E=-(dphi)/(dt) = -(d)/(dt) [(mu_(0)Ma^(2))/(2x^(3))]`
`implies E = (3mu_(0)Ma^(2))/(2x^(4))(dx)/(dt) = (3mu_(0)Ma^(2))/(2x^(4))v`
The induced current in the loop =i = `(E)/(R) = (3mu_(0)Ma^(2)v)/(2Rx^(4))`
The induced dipole moment of the loop = M =I (`pia^(2))`
`implies M. =(3mu_(0)Ma^(2)v)/(2Rx^(4))(pia^(2))`
The force of interaction between two dipoles of dipole moment pointing along the line joining their centres can be written as
`F=(3mu_(0)MM)/(2pix^(4))` where x= separation between them.
Putting the values M we obtain
`F = (0mu_(0)^(2)M^(2)a^(4))/(4x^(8)R)` v