Question

A bar magnet of moment of inertia \(9 \times 10^{-5} \text{ kg m}^2\) placed in a vibration magnetometer and oscillating in a uniform magnetic field \(16\pi^2 \times 10^{-5} \text{ T}\) makes 20 oscillations in 15 s. The magnetic moment of the bar magnet is

• A. \(3 \text{ Am}^2\)
• B. \(2 \text{ Am}^2\)
• C. \(5 \text{ Am}^2\)
• D. \(6 \text{ Am}^2\)
• E. \(4 \text{ Am}^2\)

Hint:

Always simplify common terms like \(\pi^2\) and powers of 10 before performing long multiplication. It prevents errors and saves significant time in entrance exams.

Answer 1

The Correct Option is A

Concept: A vibration magnetometer works on the principle of a magnetic dipole oscillating in a uniform external magnetic field. The motion is Simple Harmonic Motion (SHM) provided the angular displacement is small.
• Time Period (\(T\)): The time taken for one complete oscillation. It is related to the moment of inertia (\(I\)), magnetic moment (\(M\)), and magnetic field (\(B\)) by the formula: \[ T = 2\pi \sqrt{\frac{I}{MB}} \]
• Rearranging for \(M\): To find the magnetic moment, we square both sides and rearrange: \[ T^2 = 4\pi^2 \frac{I}{MB} \implies M = \frac{4\pi^2 I}{T^2 B} \]

Step 1: Calculate the Time Period (\(T\)).
The magnet makes 20 oscillations in 15 seconds. \[ T = \frac{\text{Total Time}}{\text{Number of oscillations}} = \frac{15}{20} = 0.75 \text{ s} \] Alternatively, in fractional form for easier calculation: \( T = \frac{3}{4} \text{ s} \).

Step 2: Substitute values into the rearranged formula.
Given: - \( I = 9 \times 10^{-5} \text{ kg m}^2 \) - \( B = 16\pi^2 \times 10^{-5} \text{ T} \) - \( T = 3/4 \text{ s} \) (so \( T^2 = 9/16 \)) \[ M = \frac{4\pi^2 (9 \times 10^{-5})}{\left( \frac{9}{16} \right) (16\pi^2 \times 10^{-5})} \]

Step 3: Simplify and solve for \(M\).
Canceling the terms \(\pi^2\), \(10^{-5}\), and \(9\): \[ M = \frac{4 \times 1}{\left( \frac{1}{16} \right) \times 16} \] \[ M = \frac{4}{1} = 4 \text{ Am}^2 \] *(Note: Re-calculating based on standard options; let's verify numerical simplification.)* If \( M = \frac{4\pi^2 \times 9 \times 10^{-5}}{\frac{9}{16} \times 16\pi^2 \times 10^{-5}} = \frac{36\pi^2 \times 10^{-5}}{9\pi^2 \times 10^{-5}} = 4 \text{ Am}^2 \). Option (E) is the correct match.