A small rectangular bar magnet is suspended horizontally in a uniform magnetic field of 50 T , as shown in the figure. The moment of inertia of the magnet about its axis of suspension is 2 10⁻⁵ ^2 kg m ^2 . If the magnet takes 40 s to complete 10 small angular oscillations, what is the magnetic moment of the magnet?

Step-by-step solution

The time period of one oscillation is T = 40 10 = 4 s . The formula for the time period of a magnetic dipole oscillating in a uniform magnetic field is: T = 2 I MB Squaring both sides and rearranging for the magnetic moment M gives: M = 4 ^2 I T^2 B Substituting the given values I = 2 10⁻⁵ ^2 kg m ^2 , B = 50 10⁻⁶ T , and T = 4 s : M = 4 ^2 ( 2 10⁻⁵ ^2 ) 4^2 50 10⁻⁶ M = 8 10⁻⁵ 16 50 10⁻⁶ M = 8 10⁻⁵ 800 10⁻⁶ M = 10⁻⁵ 10⁻⁴ = 0.1 A m ^2 Answer: 0.1 A m ^2