Question

The time period of angular oscillation of a bar magnet in a horizontal magnetic field is 3.14 s. If the horizontal component of the Earth’s magnetic field at the place is 40 µT and the moment of inertia of the magnet is 10^-4 kg∙m², calculate the magnetic moment of the magnet.

Answer 1

Data : T = 3.14 s, B_h = 40 µT = 4 × 10^-5 T

I = 10^-4 kg∙m²

T = 2\(\pi\) \(\sqrt{\cfrac{I}{MB}}\) 

The magnetic momentum of the magnet is