In a uniform magnetic field, the magnetic needle has a magnetic moment $9.85\times {10}^{-2}A/m^2$ and moment of inertia  $5\times {10}^{-6}kg{m}^2$. If it performs 10 complete oscillations in 5 seconds then the magnitude of the magnetic field is____________ mT.

Given: 

Magnetic moment,  $M=9.85\times {10}^{-2}A/m^2,$

 moment of inertia, $I=5\times {10}^{-6}kg{m}^2$

10 oscillations in 5 seconds

$\therefore$Time period $=\frac{5}{10}=0.5s$

$T=2\pi \sqrt{\frac{I}{MB}}\Rightarrow {\left(\frac{T}{2\pi }\right)}^2=\frac{I}{MB}\Rightarrow B=\frac{I\times 4{\pi }^2}{M\times T^2}=\frac{5\times {10}^{-6}\times 4\times 9.85\times {10}^{-2}}{9.85\times {10}^{-2}\times {\left(0.5\right)}^2}$

$\therefore B=80\times {10}^{-4}=8mT$