In a uniform magnetic field, the magnetic needle has a magnetic moment $9.85 imes 10^{-2} \mathrm{

Question

In a uniform magnetic field, the magnetic needle has a magnetic moment $9.85 	imes 10^{-2} \mathrm{~A} / \mathrm{m}^{2}$ and moment of inertia $5 	imes 10^{-6} \mathrm{~kg} \mathrm{~m}^{2}$. If it performs 10 complete oscillations in 5 seconds then the magnitude of the magnetic field is $\qquad$ mT. [Take $\pi^{2}$ as 9.85]\

Professor PC Thomas Classes · Answer

(8) : Magnetic moment,\ $M=9.85 	imes 10^{-2} \mathrm{~A} / \mathrm{m}^{2}$\ Moment of inertia, $I=5 	imes 10^{-6} \mathrm{~kg} \mathrm{~m}^{2}$\ Time period, $T=\frac{5}{10}=0.5 \mathrm{sec}$\ Let the magnetic field is $B$.\ Let the magnetic field is $B$.\ The formula of time period is $T=2 \pi \sqrt{\frac{I}{M B}}$\ $\Rightarrow 0.5=2 \pi \sqrt{\frac{5 	imes 10^{-6}}{9.85 	imes 10^{-2} 	imes B}}$\ $	herefore B=80 	imes 10^{-4} \mathrm{~T}=8 \mathrm{mT}$\