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

Magnetic feild intensity due to a bar magnet at a distance x can be written as follows:
`B=(mu_(0))/(4pi)(2M)/(x^(2))`
Magnetic flux linked with the loop can be written as follows:
`phi=BA=(mu_(0))/(4pi)(2M)/(x^(2))pia^(2)`
emf indueed in the loop can be written as follows:
`epsilon=-(dphi)/(dt)=(mu_(0))/(4pi)(6pia^(2)M)/(x^(4))(dx)/(dt)`
`implies epsilon=(mu_(0))/(4pi)(6pia^(2)M)/(x^(4)) v`
Induced current can be written as follows:
`i=(epsilon)/(R)= (mu_(0))/(4pi)(6pia^(2)M)/(Rx^(2)) v`
Let F be the force opposing the motion of magnet. Using energy conservation we can write the following:
Power invested due to opposing force = Rate of heat dissipation in the coil
`Fv=i^(2)R`
`Fv=[(mu_(0))/(4pi)(6pia^(2)M)/(Rx^(4)) v]^(2)R`
`Fv=[(mu_(0))/(2)(3a^(2)M)/(Rx^(4)) v]^(2)R`
`F=(9)/(4)[(mu_(0)^(2)M^(2)a^(4)v)/(Rx^(8))]`