A magnetic needle pivoted through it centre of mass and free to rotate in a plane containing a uniform magnetic field of `100 G` is displaced slightly from its stable equilibrium . The frequency of its angular oscillations of small amplitudes is measured to be `1.5 s^(-1)` . If the moment of inertia of the needle about its axis of rotation is `0.75 xx 10^(-5) kg m^(2)` , determine the magnetic moment of the needle.

Answer

`0.07`

Explanation

We are given that B = 100 `G = 10^(-2)T`
`v = 1.5 s^(-1) , I = 0.75 xx 10^(-5) " kg "m^(2)`
We know that the time period of oscillation of the magnetic needle ,i.e.,
`T = 2pi sqrt(I/(mB))` , where m is the magnetic moment .
Clearly , `v = 1/T = 1/(2pi) sqrt((mB)/I)or v^(2)=1/(4pi^(2))(mB)/I`
or `m =(4pi^(2)v^(2)I)/B = (4(3.14)^(2)(1.5)^(2)(0.75xx10^(-5)))/(10^(-2))`
`= 0.067 approx 0.07` J/T