Question

A magnet suspended in a uniform magnetic field is heated so as to reduce its magnetic moment by 19%. By doing this, the time period of the magnet approximately

• A. Increases by 11%
• B. Decreases by 19%
• C. Increases by 19%
• D. Decreases by 4%

Hint:

When magnetic moment decreases, the time period of oscillation increases as \( T \propto \frac{1}{\sqrt{M}} \).

Answer 1

The Correct Option is A

The time period of a magnet in a magnetic field is given by: \[ T = 2\pi \sqrt{\frac{I}{MB}} \] When \( M \) decreases by 19%, let \( M' = 0.81 M \): \[ T' = 2\pi \sqrt{\frac{I}{0.81 MB}} \] \[ T' = \frac{T}{\sqrt{0.81}} \] \[ T' \approx 1.11 T \] Thus, the time period increases by 11%.