A magnetic needle having magnetic moment `10 A m^2` and length 2.0 cm is clamped at its centre in such a way that it can rotate in the vertical east-west plane. A horizontal force towards east is applied at the north pole to keep the needle fixed at an angle of `30^0` with the vertical. Find the magnitude of the applied force. The vertical component of the earth's magnetic field is `40 muT`.

Explanation

The situation is shown in figure. As the needle is in equilibrium, the torque of all the forces abou the centre should be zero. As the needle can rotate in the vertical east-west plane, the horizontal component of earth's magnetic field is ineffective. This gives, `mB_v l sin 30^@ + mB_v l sin 30^@ = Fl cos 30^@`
or, `F = 2 mB_v tan 30^@`
`= 2 M/2l B_v tan 30^@`
`= ((10 A-m^2) (40 xx 10^(-6) T))`/((1.0 xx 10^(-2) m) sqrt(3))`
`=2.3 xx 10^(-2) N`.