Two bar magnets oscillate in a horizontal plane in earth's magnetic field with time periods of 3 s and 4 s respectively. If their moments of inertia are in the ratio of 3: 2, then the ratio of their magnetic moments will be:

Solution

Physics sir · 4 steps

Write the period formula, flip it to get M, and notice M goes as I divided by T squared.

1. Period of oscillation. A bar magnet oscillating in a uniform field B has a period that depends on its moment of inertia I and magnetic moment M.

   T = 2π√(I/MB) ⇒ M = (4π² I)/(B T²)

2. Take the ratio. Both magnets swing in the same earth's field, so B and 4π² cancel. That leaves M₁/M₂ = I₁/I₂ × (T₂/T₁)².
3. Put in numbers. M₁/M₂ = 3/2 × (4/3)² = 3/2 × 16/9 = 48/18 = 8/3.
4. So. The magnetic moments are in the ratio 8 : 3, option (b).

The idea to keep. For magnets oscillating in the same field, M ∝ I/T². The period sits squared in the denominator, so a slower magnet has a smaller moment for the same inertia.

Answer from the NTA final answer key. Solution by PrepWiser.