In a uniform magnetic field, the magnetic needle has a magnetic moment 9.85 × 10⁻² A/m² and moment of inertia 5 × 10⁻⁶ kg m². If it performs 10 complete oscillations in 5 seconds then the magnitude of the magnetic field is _______________ mT. [Take π² as 9.85]

Solution

Physics sir · 4 steps

Get the period from the oscillation count, then turn T = 2π√(I/MB) around for B.

1. Find the period. 10 complete oscillations take 5 s, so one oscillation takes T = 5/10 = 0.5 s.
2. Use the oscillation formula. For small oscillations the restoring torque is MBθ, so T = 2π√(I/MB). Square both sides and solve for B.

   B = (4π² I)/(M T²)

3. Put in numbers. The magnetic moment is 9.85 × 10⁻² A m² and π² = 9.85, so the 9.85 cancels. That leaves B = (4 × 5 × 10⁻⁶)/(10⁻² × 0.25) = (2 × 10⁻⁵)/(2.5 × 10⁻³) = 8 × 10⁻³ T.
4. So. B = 8 × 10⁻³ T, which is 8 mT.

The common slip: 0.5. Taking the period as 10/5 = 2 s (that is the frequency) gives B = 0.5 mT instead of 8 mT.

The idea to keep. A dipole in a uniform field behaves like a pendulum with T = 2π√(I/MB). Get the period as time divided by number of oscillations, never the other way round.

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