A neutral point is found on the axis of a bar magnet at a distance of `10cm` from one end. If the length of the magnet is `10cm` and `H=0*3gauss`, find the magnetic moment of the magnet.

Answer

`0*506Am^2`

Explanation

Here, `2l=10cm`, `H=0*3G`, `M=?`
`r=10+l=10+5=15cm=15xx10^-2m`
As neutral point is on the axis of bar magnet, therefore,
`H=(mu_0)/(4pi)(2M)/(r^3)=10^-7(2xxM)/((15xx10^-2)^3)`
`M=((15xx10^-2)^3H)/(2xx10^-7)=(3375xx10^-6xx0*3xx10^-4)/(2xx10^-7)`
`M=506*25xx10^-3=0*506Am^2`