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A bar magnet 30cm long is placed in magn...

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`.

Text Solution

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The correct Answer is:
`8.61`
Here ,2l = 30 cm ,l = 15 cm = 0.15 m
d = 30 cm = 0.3 m , m =?
H = 0.34 G = `0.34 xx 10^(-4) T `
When magnet is placed with its north pole pointing wouth , 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))=(0.34 xx 10^(-4)(0.3^(2)-0.15^(2))^(2))/(2xx0.3 xx 10^(-7))`
`= 2.58 "Am"^(2)`
Also , `m = M/(2l) = (2.58)/(0.3) = 8.61 `Am .
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