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

Answer: (D)

$1280\pi^2$

Answer 2

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:

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