A short bar magnet having magnetic moment 4 Am², placed in a vibrating magneto meter, vibrates with a time period of 8 seconds. Another short bar magnet having a magnetic moment 8 Am² vibrates with a time period of 6 seconds. If the moment of inertia of the second magnet is 9 x 10^-2kg m² the moment of inertia of the first magnet is (assume that both magnets are kept in the same uniform magnetic induction field)

Explanation

Relate the moment of inertia (I) to the time period (T) and magnetic moment (μ) for a magnet in a vibrating magnetometer:The equation is:T² = (I * π²) / (μ * B)where:T is the time period (seconds)I is the moment of inertia (kg m²)μ is the magnetic moment (Am²)B is the magnetic field strength (T)Analyze the given information:The magnetic moment of the first magnet (μ₁) is 4 Am².The time period of the first magnet (T₁) is 8 seconds.The magnetic moment of the second magnet (μ₂) is 8 Am².The time period of the second magnet (T₂) is 6 seconds.The moment of inertia of the second magnet (I₂) is 9 x 10^-2 kg m².We are asked to find the moment of inertia of the first magnet (I₁).Solve for I₁:We can rewrite the equation to isolate I₁:I₁ = (T₁² * μ₁ * B) / π²Since both magnets are in the same uniform magnetic field (B), we can eliminate B from the equation. Additionally, the ratio of B cancels out when we divide the equation for I₂ by the equation for I₁:I₁ / I₂ = (T₁² * μ₁) / (T₂² * μ₂)Plug in the known values and solve:I₁ / (9 x 10^-2 kg m²) = (8 s)² * (4 Am²) / (6 s)² * (8 Am²)I₁ ≈ 8 x 10^-2 kg m²Therefore, the moment of inertia of the first magnet is 8 x 10^-2 kg m². This matches with answer choice b