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 .

Answer

A, B, D

Explanation

Given that `x gt gt a`
magnetic field at the centre of the coil due to the bar magnet is `B=(mu_(0))/(4pi)(2M)/x^(3)=(mu_(0))/(2pi)M/x^(3)`
Due to this, magnetic flux linked with the coil will be- `phi=BS=mu_(0)/(2pi)M/x^(3)(pia^(2))=(mu_(0)Ma^(2))/(2x^(3))`
`therefore` induced `emf` in the coil, due to motion of the magnet is `e=(-dphi)/(dt)=-((mu_(0)Ma^(2))/2)d/(dt)(l/x^(3))`
`=(mu_(0)Ma^(2))/2(3/(X^(4)))(dx)/(dt)=3/2(mu_(0)Ma^(2))/(Rx^(4))V ((dx)/(dt)=V)`
Therefore, induced current in the coil is `i=e/R=3/2(mu_(0)Ma^(2)V)/(Rx^(4))`
Potential energy of `vecM` in `vecB` will be `U=-M'B cos 180^(@)`
`U=M'B`
`=3/2(mu_(0)piMa^(4) V)/(Rx^(4))((mu_(0))/(2pi),M/x^(3))`
`U=3/4 (mu_(0)^(2)M^(2)a^(4) V)/R1/x^(7)`
`therefore F=-(dU)/(dx)=21/4(mu_(0)^(2)M^(2)a^(4) V)/(Rx^(8))`
Positive sign of `F` implies that there will be a repulsion between the magnet and the coil. Note that here we cannot apply `F=(mu_(0))/(4pi)(6MM')/x^(4)` (directly)...(i)
because here `M` is a funtion of `x` however equation (1) can be applied where `M` and `M'` both are constants.