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

In this case the moment of inertia becomes
`I'=(m')/(12)(L^(2)+b^(2))" where "b'=0.5cm`
The time period would be
`T'=sqrt((I')/(MB))...(ii)`
Dividing by equation (i),
`(T')/(T)=sqrt((I')/(I))=(sqrt((m')/(12)(L^(2)+b'^(2))))/(sqrt((m')/(12)(L^(2)+b^(2))))`
`=(sqrt((7cm)^(2)+(0.5cm)^(2)))/(sqrt((7cm)^(2)+(1.0cm)^(2)))=0.992`
or, `T'=(0.992xxpi)/(2)s=0.496pis`.