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
`I = m'/12 (L^2 + b^2)`
`= (100 xx 10^(-3))/(12) [(7 xx 10^(-2))^2 + (1 xx 10^(-2)^2] kg m^2`
`= 25/6 xx 10^(-5) kg m^2`.
We have,
`T= 2pi sqrt(I/MB)` … (i)
or, `M = (rpi^2 I)/(BT^2) = (4pi^2 xx 25 xx 10^(-5) kg m^2)/(6 xx (25 xx 10^(-6) T) xx (pi^2)/4 s^2))`
T` = sqrt((I')/MB)` ...(ii)
Dividing by equation (i), `T'/T' = sqrt(I')/I`
` = ((m'/12) (L^2 + b'^2))/((m'/12) (L^2 + b'^2)) = sqrt((7cm)^2 + (0.5cm)^2)/((7cm)^2 + (1.0 cm)^2)`
`= 0.992`
or, `T' = (0.992 xx pi)/2 s = 0.496pi s`.