In a uniform magnetic field of 0.049 T , a magnetic needle performs 20 complete oscillations in 5 se

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 $9.8 	imes 10^{-6} \mathrm{~kg} \mathrm{~m}^2$. If the magnitude of magnetic moment of the needle is $x 	imes 10^{-5} \mathrm{Am}^2$; then the value of ' $x$ ' is

Professor PC Thomas Classes · Answer

(d) : Time period of the needle, $T=\frac{5}{20}=\frac{1}{4} \mathrm{~s}$ Given, $I=9.8 	imes 10^{-6} \mathrm{~kg} \mathrm{~m}^2, M=x 	imes 10^{-5} \mathrm{Am}^2$ and $B=0.049 \mathrm{~T}$ Since, $T=2 \pi \sqrt{\frac{I}{M B}}$ 
$\Rightarrow \frac{1}{4}=2 \pi \sqrt{\frac{9.8 	imes 10^{-6}}{x 	imes 10^{-5} 	imes 0.049}}$
$\Rightarrow \frac{1}{8 \pi}=\sqrt{\frac{9.8 	imes 10^{-6}}{x 	imes 10^{-5} 	imes 0.049}}$
$\Rightarrow \frac{1}{64 \pi^2}=\frac{9.8 	imes 10^{-6}}{x 	imes 10^{-5} 	imes 0.049}$
$\Rightarrow x=64 \pi^2 	imes \frac{9.8 	imes 10^{-1}}{0.049}=1280 \pi^2$