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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.

Text Solution

Verified by Experts

Here, `2l=10cm`,
`d=10cm+10/2cm=15cm=15xx10^-2m`
`H=0*3gauss=0*3xx10^-4T, M=?`
As neutral point is on axial line of the magnet
`:. H=(mu_0)/(4pi)(2Md)/((d^2-l^2)^2)`
`M=(4pi)/(mu_0)((d^2-l^2)^2H)/(2d)`
`=(10^7[(15xx10^-2)^2-(5xx10^-2)^2]^2xx0*3xx10^-4)/(2xx15xx10^-2)`
`=0*4Am^2`
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