A bar magnet takes `pi/10` second to complete one oscillation in an oscillation magnetometer. The moment of inetria of the magnet about the axis of rotation is 1.2 xx 10^(-4) kg m^2` and the earth's horizontal magnetic field is `30 muT`. Find the magentic moment of the magnet.

Explanation

According to oscillation magnetometer
`T = 2 pi sqrt((1)/(MB_(H)))`
`rArr (pi)/(10) = 2 pisqrt((12 xx 10^(-4))/(M xx 30 xx 10^(-6)))`
`((1)/(20))^(2) = (12 xx 10^(-4))/(M xx 30 xx 10^(-6))`
` M = (12 xx 10^(-4) xx 400)/(30 xx 10^(-8))`
` = 16 xx 10^(2) A - m^(2) `
` = 1600 A - m^(2)`