Question

A bar magnet takes π/10 second the complete one oscillation in an oscillation magnetometer. The moment of inertia of the magnet about the axis of rotation is 1.2 × 10⁻⁴ kg m² and the earth's horizontal magnetic field is 30 μT. Find the magnetic moment of the magnet.

Answer 1

Given :
Time taken by the bar magnet to complete one oscillation, `T = pi/10 s`

Moment of inertia of the magnet about the axis of rotation, I = `1.2 xx 10^-4  "kgm"^2`

Horizontal component of Earth's magnetic field, B_H = 30 μT

Time period of oscillating magnetometer (T) is given by 

`T = 2pisqrt(I/(MB_H)`

Here ,

M = Magnetic moment of the magnet
On substituting the respective values, we get

`pi/10 = 2pisqrt((1.2 xx 10^-4)/(M xx 30 xx 10^-6)`

⇒ `(1/20)^2 = (1.2 xx 10^-4)/(M xx 30 xx 10^-6)`

⇒ `M = (1.2 xx 10^-4 xx 400)/(30 xx 10^-6)`

⇒ `M = 16 xx 10^2   "A-m"^2`

⇒ `M = 1600  "A-m"^2`