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

Given:

• Moment of inertia: I = 500 g cm² = 500 × 10⁻⁷
• Time period: T = 3.142 s
• Horizontal component of Earth's magnetic field: B = 4 × 10⁻⁵

Calculation:

M = `(4pi^2 I)/(T^2B)`

M = `(4 xx (500 xx 10^-7))/(4 xx 10^-5)`

M = `(2000 xx 10^-7)/(4 xx 10^-5)`

M = `2000/4 xx 10^-2`

M = 500 × 10⁻²

M = 5 A m²

This is the required magnetic moment of the bar magnet.