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A short bar magnet has a magnetic moment...

A short bar magnet has a magnetic moment of `0*48JT^-1`. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of `10cm` from the centre of the magnet on (i) the axis (ii) the equatorial line (normal bisector) of the magnet.

Text Solution

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Here, `M=0.48JT^-1`, `B=?`, `d=10cm=0.1m`
(i) On the axis of the magnet, `B=(mu_0)/(4pi)(2M)/(d^3)=10^-7xx(2xx0*48)/((0*1)^3)=0*96xx10^-4T` along S-N direction.
(ii) On the equatorial line of the magnet
`B=(mu_0)/(4pi)xx(M)/(d^3)=10^-7xx(0.48)/((0.1)^3)=0.48xx10^-4T`, along N-S direction.
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