The magnetic moment and moment of inertia of a magnetic needle as shown are, respectively, $1.0 im

Question

The magnetic moment and moment of inertia of a magnetic needle as shown are, respectively, $1.0 	imes 10^{-2} \mathrm{Am}^2$ and $\frac{10^{-6}}{\pi^2} \mathrm{~kg} \mathrm{~m}^2$. If it completes 10 oscillations in 10s, the magnitude of the magnetic field is

Professor PC Thomas Classes · Answer

(c) : Magnetic moment, $M=1.0 	imes 10^{-2} \mathrm{Am}^2$ Moment of inertia, $I=\frac{10^{-6}}{\pi^2} \mathrm{Kg} \mathrm{m}^2$ Time period for one oscillation $=\frac{10}{10}=1 \mathrm{~s}$ We know that, $T=2 \pi \sqrt{\frac{I}{M B}}$ $$ \begin{aligned} & 	herefore \quad T^2=\frac{4 \pi^2 I}{M B} \ & B=\frac{4 \pi^2 I}{M T^2}=\frac{4 	imes(3.14)^2}{1 	imes 10^{-2}} 	imes \frac{\left(10^{-6}ight)}{(3.14)^2}=4 	imes 10^{-4}=0.4 \mathrm{mT} \end{aligned} $$