A magnetic needle is hung by an untwisted wire, so that it can rotate freely in the magnetic meridian. In order to keep it in the horizontal position, a weight of 0.1 g is kept or one end of the needle. If the magnetic pole strength of this needle is 10 Am, find the value of the vertical component of the earth's magnetic field. `(g= 9.8 ms^(-2) )`

Explanation

Figure (a) shows the position of magnetic needle in the magnetic meridian without any weight.
In figure (b), a mass mis kept on the S-pole on the needle.

The vector sum of torques due to all forces must be zero for the equilibrium of the needle in horizontal direction.
`therefore -pB_(v) (l) + m.g(l) =0`
[ The torque producing rotations in clockwise direction is taken as negative ]
`m = 0.1 g = 10^(-4) kg, p=10 "Am"`
`therefore 2pB_(v) = mg`
`therefore B_(v) = (mg)/( 2p)`
`= (10^(-4) xx 9.8) /( 2 xx 10)`
`therefore B_(v) = 4.9 xx 10^(-5)` T