Question

In a uniform magnetic field of \(0.049 T\), a magnetic needle performs \(20\) complete oscillations in \(5\) seconds as shown. The moment of inertia of the needle is \(9.8 \times 10 kg m^2\). If the magnitude of magnetic moment of the needle is \(x \times 10^{-5} Am^2\); then the value of '\(x\)' is

• A. \(5\pi^2\)
• B. \(128\pi^2\)
• C. \(50\pi^2\)
• D. \(1280\pi^2\)

Answer 1

The Correct Option is D

Using the formula for oscillation frequency:

$T = 2\pi \sqrt{\frac{I}{MB}}, \quad M = \frac{4\pi^2 I}{T^2 B}$.

Substitute values:

$M = \frac{4\pi^2 (9.8 \times 10^{-6})}{(5/20)^2 (0.049)} = 1280\pi^2 \times 10^{-5}$.