A bar magnetic with poles `25cm` apart and pole strength `14*4A.m` rests with its centre on a frictionless pivot. It is held in equilibrium at `60^@` to a uniform magnetic field of induction `0*25T` by applying a force F at right angles to its axis, `10cm` from the pivot. Calculate the value of F. What will happen if the force is removed?

Explanation

The situation is shown in figure. In equilibrium the torque on M due to B is balanced by torque due to F, i.e., i.e., `vecMxxvecB=vecrxxvecF`

`MBsintheta=Frsin90^(@)orF=((mxx2l)Bsintheta)/r`
( as `M=mxx2l`) , So substituting the given data,
`F=(14.4xx(25xx10^(-2))xx0.25(sqrt3//2))/(10xx10^(-2))=7.8N`