A short bar magnet is placed with its north pole pointing north. The neutral point is `10 cm` away from the centre of the magnet. If `H=0.4G`, calculate the magnetic moment of the magnet.
Text Solution
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When north pole of the towards magnetic north, pull point is obtained on perpendicular bisector of the magnet. Simultaneously, magnetic field due to the bar magnet should be equal to the horizontal components of earth's mangetic field H. `" " H=(mu_(0))/(4pi).(M)/(r^(3)) or M=(Hr^(3))/(mu_(0)//4pi)` Substituting the values, we have `" " M=((0.4xx20^(-4))(10xx10^(-2))^(3))/(10^(_7))=0.4A-m^(2)`
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