Question

A bar magnet of moment of inertia of 500 gcm² oscillates with a time period of 3.142 seconds in a horizontal plane. What is its magnetic moment if the horizontal component of earth’s magnetic field is 4 × 10⁻⁵T?

Answer 1

Data: T = 3.14 s, B_h = 4 × 10⁻⁵ T, I = 500 gcm² = 5 × 10⁻⁵ kgm², π = 3.14

T = `2π sqrt(I/(MB))`

The magnetic moment of the magnet is

M = `(4pi^2 I)/(BT^2)`

= `((4)(3.14)^2 (5 xx 10^-5  kgm^2))/((4 xx 10^-5  T)(3.14  s^2))`

= 5 A m²