A bar magnet of mass 100 g, length 7.0 cm, width 1.0 cm and height 0.50 cm takes `pi/2` seconds to complete an oscillation in an oscillation magnetometer placed in a horizontal magnetic field of `25 muT`. (a) Find the magnetic moment of the magnet. (b) If the magnet is put in the magnetometer with its `0.50cm` edge horizontal, what would be the time period?

Explanation

The moment of inertia of the magnet about the axis of rotation is
`I=(m')/(12)(L^(2)+b^(2))`
`=(100xx10^(-3))/(12)[(7xx10^(-2))^(2)+(1xx10^(-2))^(2)]kg-m^(2)`
`=(25)/(6)xx10^(-5)kg-m^(2)`
We have, `T=2pisqrt((I)/(MB))`
or `M=(4pi^(2)I)/(BT^(2))=(4pi^(2)xx25xx10^(-5)kg//m^(2))/(6xx(25xx10^(-6)T)xx(pi^(2))/(4)s^(2))`
`=27A-m^(2)`