Question

A short bar magnet placed with its axis at `30^(@)` with an external field of 800 G experiences a torque of 0.016 Nm.
What is the magnetic moment of the magnet ?

Answer 1

From Eq., `tau = m B sin theta, theta =30^(@)`, hence `sin theta =1//2`.
Thus, `0.016 = m xx (800 xx 10^(-4)T) xx (1//2)`
`m = 160 xx 2//800 = 0.40 A m^(2)`