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The magnetic needle has magnetic moment  \(6.7\times 10^{-2}\;Am^2\)  and moment of inertia  \(\left(I\right)=7.5\times 10^{-6}\;\mathit{kg}m^2.\)  It performs \(10\) complete oscillations in  \(6.70\;\mathit{s.}\)  What is the magnitude of the magnetic field?

Answer

\(0.0098\;T\)
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Solution

Given:
Magnetic moment  \(\left(M\right)=6.7\times 10^{-2}\;Am^2\)
Moment of inertia \(\left(I\right)=7.5\times 10^{-6}\;\text{kg }m^2\)
We need to calculate first, the time period of oscillation of the needle:
 \(T=\frac{6.70}{10}=0.67\;s\) 
Now, the magnetic needle performs Simple harmonic motion  \(\text{(S.H.M)}\)  when placed in a magnetic field, and the time period of oscillation is given by:
\({\therefore}\;T=2\pi \sqrt{\frac I{\mathit{MB}}}\)
Squaring both sides, we get
\(T^2=4\pi^2 {\frac I{\mathit{MB}}}\) 
\(\Rightarrow B=\frac{4\pi ^2I}{MT^2}\)
\(=\frac{4\times \left(3.14\right)^2\times 7.5\times 10^{-6}}{6.7\times 10^{-2}\times (0.67)^2}\) 
\(=0.0098\;T\)  
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