Question

A bar magnet is oscillating in the earth’s magnetic field with a period \( T \). What happens to its period and motion, if its mass is quadrupled?

• A. Motion remains simple harmonic with time period \(4T\)
• B. Motion remains simple harmonic with time period nearly constant.
• C. Motion remains simple harmonic and time period becomes \( \frac{T}{2} \)
• D. Motion remains simple harmonic with time period \(2T\)

Hint:

For oscillations of a magnetic dipole in a magnetic field, \( T \propto \sqrt{I} \). If the mass increases but shape stays constant, moment of inertia scales proportionally.

Answer 1

The Correct Option is D

The time period of oscillation of a magnetic dipole (bar magnet) in a uniform magnetic field is given by: \[ T = 2\pi \sqrt{\frac{I}{MB}} \] Where:
- \( I \) is the moment of inertia of the magnet,
- \( M \) is the magnetic moment,
- \( B \) is the magnetic field strength. For a uniform bar magnet: \[ I \propto m \quad \text{(as the shape remains unchanged)} \] If the mass is quadrupled, \( m \rightarrow 4m \Rightarrow I \rightarrow 4I \) Hence, the new time period \( T' \) becomes: \[ T' = 2\pi \sqrt{\frac{4I}{MB}} = 2 \cdot T \] So, time period becomes twice, and motion remains simple harmonic.