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

(a) The moment of inertia of the magnet about the axis of rotation is
`l=m/12(L^(2)-b^(2))`
`=(100xx10^(-3))/12[(7xx10^(-2))^2+(1xx10^(-2))^2]kg-m=25/6xx10 kg m`
We have, `T=2pisqrt(1/(MB))`
or `M=(4pi^(2)l)/(BT^(2))=(4pi^(2)xx25xx10^(-5)kg//m^(2))/(6xx(25xx10^(-6T)xxpi^(2)/4s^(2)))=27 A-m^(2)`
(b) In this case the moment of inertia becomes
` I=m/12(L^(2)+b^(2))` where `b=0.5 cm`
The time period would be
`T=sqrt(I/(MB))`...(ii)
Dividing the equation (i),
`(T')/T=sqrt((I')/I)=sqrt(m/12(L^(2)+b^(2)))/sqrt(m/12(L^(2)+b^(2)))=sqrt((7cm)^(2)+(0.5 cm)^(2))/sqrt((7cm)^(2)+(1.0 cm)^(2))=0.992`
or `T'=(0.992xxpi)/2s=0.496pi s`.