In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds. The moment of inertia of the needle is 9.8 imes 10^{-6}\ \mathrm{kg\,m^2}. If the magnitude

Question

In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds. The moment of inertia of the needle is 9.8 	imes 10^{-6}\ \mathrm{kg\,m^2}. If the magnitude of the magnetic moment of the needle is x 	imes 10^{-5}\ \mathrm{A\,m^2}, then the value of x is:

Vishal Gupta · Accepted Answer

C. Oscillation period T = \frac{5}{20} = 0.25 s. For a magnetic needle, T = 2\pi\sqrt{\frac{I}{MB}}, so M = \frac{4\pi^2 I}{T^2 B} = \frac{4\pi^2 	imes 9.8	imes10^{-6}}{0.0625 	imes 0.049} = 1280\pi^2 	imes 10^{-5} A m². Hence x = 1280\pi^2.

Magnetism and Matter

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