A bar magnet `30cm` long is placed in magnetic meridian with its north pole pointing south. The neutral point is observed at a distance of `30cm` from its centre. Calculate the pole strength of the magnet. Given horizontal component of earth's field is `0*34G`.

Answer

`8*61Am`

Explanation

Here, `2l=30cm, l=15cm=0*15m`
`d=30cm=0*3m, m=?`
`H=0*34G=0*34xx10^-4T`
When magnet is placed with its north pole pointing south, neutral points are obtained on the axial line, and `B_1=H`
`(mu_0)/(4pi)(2Md)/((d^2-l^2)^2)=H`
`M=(H(d^2-l^2)^2)/(2d(mu_0//4pi))=(0*34xx10^-4(0*3^2-0*15^2)^2)/(2xx0*3xx10^-7)`
`=2*58Am^2`
Also, `m=(M)/(2l)=(2*58)/(0*3)=8*61Am`