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: \[ \boxed{3T}. \]

Answer 2

The time period \( T \) of a simple harmonic motion of a bar magnet in the Earth's magnetic field is given by the equation:

\[ T = 2\pi \sqrt{\frac{I}{mB}} \]

where \( I \) is the moment of inertia of the magnet, \( m \) is the magnetic moment, and \( B \) is the magnetic field strength.

Assuming the bar magnet is a uniform rod, we have:

\[ I = \frac{1}{12} ML^2 \]

where \( M \) is the mass of the magnet and \( L \) is its length.

If the mass \( M \) is increased by 9 times, the new moment of inertia \( I' \) becomes:

\[ I' = \frac{1}{12} (9M)L^2 = 9 \left(\frac{1}{12} ML^2 \right) = 9I \]

The new time period \( T' \) is:

\[ T' = 2\pi \sqrt{\frac{I'}{mB}} = 2\pi \sqrt{\frac{9I}{mB}} = 2\pi \cdot 3 \cdot \sqrt{\frac{I}{mB}} = 3 \cdot 2\pi \sqrt{\frac{I}{mB}} = 3T \]

Thus, the time period becomes \( 3T \) when the mass of the bar magnet is increased by 9 times.

The correct option is \( 3T \).