Question

You place a magnetic compass on a horizontal surface, allow the needle to settle, and then give the compass a gentle wiggle to cause the needle to oscillate about its equilibrium position. The oscillation frequency is $\mathbf{0}\mathbf{.}\mathbf{312}\mathbf{}\mathbf{Hz}$. Earth’s magnetic field at the location of the compass has a horizontal component of$\mathbf{18}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\mu T}$. The needle has a magnetic moment of$\mathbf{0}\mathbf{.}\mathbf{680}\mathbf{}\mathbf{mJ}\mathbf{/}\mathbf{T}$. What is the needle’s rotational inertia about its (vertical) axis of rotation?

Short answer

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

Step-by-step solution

Listing the given quantities

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

$\mathrm{B}=18.0\times {10}^{-6}\mathrm{T}$

$\mathrm{\mu }=0.680\times {10}^{-3}\mathrm{J}/\mathrm{T}$

Understanding the concepts of torque related with magnetic moment

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$\mathrm{\tau }=\mathrm{\mu Bsin}\left(\mathrm{\theta }\right)$

Restoring torque:$\text{τ}=-\text{kθ}$

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

Step 3:Calculations of the needle’s rotational inertia about its (vertical) axis of rotation

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

$\text{kθ}=\text{μBsin}\left(\text{θ}\right)$

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

Hence,

$\mathrm{k}=\mathrm{\mu B}$

Now put this value in the equation of time period.

$\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{I}}{\mathrm{\mu B}}}$

Rearranging the equation for:

$\text{I}={\left(\frac{\text{T}}{2\text{π}}\right)}^2\left(\text{μB}\right)$

We have

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

Therefore,

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

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