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## QuestionPhysicsClass 12

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?

$$0.0098\;T$$
4.6
4.6

## 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
$$T^2=4\pi^2 {\frac I{\mathit{MB}}}$$
$$\Rightarrow B=\frac{4\pi ^2I}{MT^2}$$
$$=\frac{4\times \left(3.14\right)^2\times 7.5\times 10^{-6}}{6.7\times 10^{-2}\times (0.67)^2}$$
$$=0.0098\;T$$