A bar magnet with poles ` 25 cm` apart and of strength `14.4 amp - m` rests with centre on a frictionless pivot. It is held in equilibrium aat an angle of ` 60^@` with respect to a uniform magnetic field of induction `0.25 Wb//m^(2)` , by applying a force `F` at right angles to its axis at a point `12 cm` from pivot. Calculate `F`. What will happen if the force `F` 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., vec(M)xxvec(B)=vec(r )xxvec(F)`

`MB sin theta=Fr sin90^(@) or F=((mxx2l)B sin theta)/(r )`
`("as" M=mxx2l)`, So substituting the given data,
`F=(14.4xx(25xx10^(-2))xx0.25(sqrt3//2))/(10xx10^(-2))=7.8 N`
If the force `vec(F)` is removed, the torque `vec(M)xxvec(B)` will become unbalanced and under its action the magnet will execute oscillatory motion about the direction of `B` on its pivot `O` which will not be simple harmonic as `sin theta cancel(=) theta`