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 $9.8 \times 10^{-6}\ \text{kg m}^2$. If the magnitude of magnetic moment of the needle is $x \times 10^{-5}\ \text{A m}^2$; then the value of '$x$' is :
Answer: (B) $1280\pi^2$
Time period: $T = \dfrac{5}{20} = 0.25$ s.
For a magnetic needle oscillating in a field:
$$T = 2\pi\sqrt{\frac{I}{mB}} \;\Rightarrow\; m = \frac{4\pi^2 I}{T^2 B}$$
$$m = \frac{4\pi^2 \times 9.8 \times 10^{-6}}{(0.25)^2 \times 0.049} = \frac{4\pi^2 \times 9.8 \times 10^{-6}}{3.0625 \times 10^{-3}} = 4\pi^2 \times 3.2 \times 10^{-3}$$
$$m = 12.8\pi^2 \times 10^{-3} = 1280\pi^2 \times 10^{-5}\ \text{A m}^2$$
So $x = 1280\pi^2$.
Solution by Sreeraj P, M.Sc Physics