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A very small magnet is placed in the magnetic meridian with its S-pole pointing north. The null point is obtained 20 cm away from the centre of the magnet. What is the magnetic moment of the magnet if earth's field is 0.3xx 10^(-4) T? |
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Answer» Solution :As magnetic dipole is placed with its S-pole pointing north and at neutral POINT the field of dipole `(B_M)` is cancelled by earth.s field `B_H` , the neutral point will be on the axis of the dipoleas shown inFig. And as for an axial point , `B_M = (mu_0)/(4pi) (2M)/(R^(3))` , so the neutral point , ` (mu_0)/(4pi)(2M)/(r^(3)) = B_H` i.e.` M=(B_H xx r^3 )/(2 xx 10^(-7))[ as (mu_0)/(4pi) = 10^(-7)]`  So `, M = (0.3 xx 10^(-4) xx (0.2)^3)/(2) xx 10^(7) = 1.2 A-m^(2)` |
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