In an oscillation magnetometer, a bar magnet requires /10 second to complete one oscillation. The moment of inertia of the magnet about its axis of rotation is 1.2 10⁻⁴ kg m ^2 , and the earth's horizontal magnetic field is 30 T . Determine the magnetic moment of the magnet.

Step-by-step solution

The time period of oscillation of a bar magnet in a magnetic field is given by: T = 2 I MB_H Squaring both sides, we get: T^2 = 4 ^2 I MB_H Rearranging for the magnetic moment M : M = 4 ^2 I T^2 B_H Substituting the given values T = 10 s , I = 1.2 10⁻⁴ kg m ^2 , and B_H = 30 10⁻⁶ T : M = 4 ^2 1.2 10⁻⁴ ( 10 )^2 30 10⁻⁶ M = 4 ^2 1.2 10⁻⁴ ^2 100 30 10⁻⁶ M = 4 1.2 10⁻⁴ 100 30 10⁻⁶ M = 4.8 10⁻² 30 10⁻⁶ M = 1600 A m ^2