A bar magnet of length `10cm` is placed in the magnetic meridian with its north pole pointing towards the geographical north. A neutral point is obtained at a distance of `12cm` from the centre of the magnet. Find the magnetic moment of the magnet, when `H=0*34 gauss`.

Answer

`0*747JT^-1`

Explanation

Here, `2l=10cm`, `l=5cm=5xx10^-2m`,
`d=12cm, H=0*34G=0*34xx10^-4T`
As N-pole of magnet is towards geographic north, therefore neutral points lie on equatorial line of the magnet.
`:. H=(mu_0)/(4pi)(M)/((d^2+l^2)^(3//2))`
`M=(4pi)/(mu_0)(d^2+l^2)^(3//2)H`
`=10^-7[12^2+5^2]^(3//2)(10^-4)^(3//2)xx0*34xx10^-4`
`=10^-7xx13xx13xx13xx10^-6xx0*34xx10^-4`
`M=0*747JT^-1`