Question

One bar magnet is in simple harmonic motion with time period \( T \) in an earth’s magnetic field. If its mass is increased by 9 times, the time period becomes:

• A. \( 3T \)
• B. \( 9T \)
• C. \( 4T \)
• D. \( \sqrt{3} T \)

Hint:

For a bar magnet oscillating in a magnetic field, the time period is proportional to the square root of the moment of inertia: \[ T \propto \sqrt{I}. \] If mass increases by \( n^2 \) times, the time period increases by \( n \) times.

Answer 1

The Correct Option is A

Step 1: Understanding the Formula The time period of oscillation of a 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.
Step 2: Effect of Increasing Mass Since \( I = m k^2 \), if mass increases by 9 times, then: \[ I' = 9I. \] \[ T' = 2\pi \sqrt{\frac{9I}{MB}}. \] \[ T' = 3T. \] Thus, the new time period is: \[ \boxed{3T}. \]