A magnetic needle of length 10cm, suspended at its middle point through a thread, stays at a an angle of `45^@` with the horizontal. The horizontal component of the earth's magnetic field is `18 muT`. (a) Find the vertical component of this field. (b) If the pole strength of the needle is `1.6 A-m`, what vertical force should be applied to an end so as to keep it in horizontal position?

Explanation

Without the applied force, the needle will stay in the direction of the resultant magnetic field of the earth. Thus, the dip `delta` at the plane is `45^@`. From figure,
`tan 45^@ = B_v/B_H`
or, `B_v = B_H = 18 muT`.
(b) When the force `F` is applied (figure), the needle stays in horizontal position. Taking torque about the centre of the magnet,
`2mB_v xx l = F xx l`
or, `F = 2mB_v`
`= 2 xx (1.6 A m) xx (18 xx 10^(-6) T)`
= 5.8 xx 10^(-5) N`.