91Ó°ÊÓ

$\mathrm{f}=0.312\mathrm{Hz}$

$\mathrm{B}=18.0×{10}^{-6}\mathrm{T}$

$\mathrm{μ}=0.680×{10}^{-3}\mathrm{J}/\mathrm{T}$

Here, we need to use the equation of torque relating with magnetic moment and magnetic field. We can compare this equation with the equation of restoring torque to find the restoring constant kappa. Then using the equation of period, we can find the rotational inertia.

Formulae:

Torque acting on magnetic needle due to external magnetic field$\mathrm{Ï„}=\mathrm{μ\mu þ²õ¾\pm ²Ô}\left(\mathrm{θ}\right)$

Restoring torque:$\text{Ï„}=−\text{°ìθ}$

Time period$\mathrm{T}=2\mathrm{Ï€}\sqrt{\frac{\mathrm{I}}{\mathrm{k}}}$

Equate both torque equations to find the relation for$\mathrm{k}$:

$\text{°ìθ}=\text{μµþ²õ¾±²Ô}\left(\text{θ}\right)$

As$\sin \left(\text{θ}\right)$is very small, put$\sin \left(\mathrm{θ}\right)=\mathrm{θ}$

Hence,

$\mathrm{k}=\mathrm{μ\mu þ}$

Now put this value in the equation of time period.

$\mathrm{T}=2\mathrm{Ï€}\sqrt{\frac{\mathrm{I}}{\mathrm{μ\mu þ}}}$

Rearranging the equation for:

$\text{I}={\left(\frac{\text{T}}{2\text{Ï€}}\right)}^2\left(\text{μµþ}\right)$

We have

$\begin{array}{c}\mathrm{T}=\frac{1}{\mathrm{f}}\\ =\frac{1}{0.312}\\ =3.205\mathrm{s}\end{array}$

Therefore,

$\begin{array}{c}\text{I}={\left(\frac{3.205}{2\mathrm{π}}\right)}^2(0.680×{10}^{−3}×18×{10}^{−6})\\ =0.318×{10}^{−8}\mathrm{kg}.{\mathrm{m}}^2\end{array}$

The needle’s rotational inertia about its (vertical) axis of rotation is $\mathrm{I}=0.318×{10}^{−8}\mathrm{kg}.{\mathrm{m}}^2$