Q57P - Fundamentals Of Physics

Question

A magnetic compass has its needle, of mass $\mathbf{0}\mathbf{.}\mathbf{050}\mathbf{ }\mathbf{kg}$ and length $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{cm}$, aligned with the horizontal component of Earth’s magnetic field at a place where that component has the value ${\mathbf{B}}_{\mathbf{h}}\mathbf{=}\mathbf{16}\mathbf{ }\mathbf{\mu T}$. After the compass is given a momentary gentle shake, the needle oscillates with angular frequency $\mathbf{\omega }\mathbf{=}\mathbf{45}\mathbf{ }\mathbf{rad}\mathbf{\backslash }\mathbf{sec}$. Assuming that the needle is a uniform thin rod mounted at its center, find the magnitude of its magnetic dipole moment.

Step-by-Step Solution

Answer

The magnitude of magnetic dipole moment is $8.4\times {10}^2 \mathrm{J}/\mathrm{T}$

Step 1: Listing the given quantities

$\mathrm{\omega }=45 \mathrm{rad}\backslash \sec$

${\mathrm{B}}_{\mathrm{h}}=16 \mathrm{\mu T}$

$\mathrm{L}=0.04 \mathrm{m}$

$\mathrm{m}=0.050 \mathrm{kg}$

Step 2: Understanding the concepts of magnetic dipole moment

Here we have to use the formula for torque. Then using equations of moment of inertia and angular acceleration, we can simplify the equation of torque to equation of oscillation. Then comparing it with the general equation of torque, we get the equation of angular velocity. We can rearrange that equation for magnetic dipole moment, and using the given values, we can solve it.

Formula:

$\mathrm{\tau }=\mathrm{\mu }\times {\mathrm{B}}_{\mathrm{h}}$

 $\mathrm{\tau }=\mathrm{I\alpha }$

Step 3: Calculations of the magnitude of magnetic dipole moment

Torque due to magnetic field is given by following formula:

$\begin{array}{c}\mathrm{\tau }=-\mathrm{\mu }\times {\mathrm{B}}_{\mathrm{h}}\\ =-{\mathrm{\mu B}}_{\mathrm{h}}\mathrm{sin\theta }\end{array}$

And we know that

$\mathrm{\tau }=\mathrm{I\alpha }$

$-{\mathrm{\mu B}}_{\mathrm{h}}\mathrm{sin\theta }=\mathrm{I\alpha }$

We know that 

$\mathrm{\alpha }=\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dt}}^2}$

$\mathrm{I}=\frac{{\mathrm{mL}}^2}{12}$

$-{\mathrm{\mu B}}_{\mathrm{h}}\mathrm{sin\theta }=\frac{{\mathrm{mL}}^2}{12}\times \frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dt}}^2}$

$\frac{{\mathrm{mL}}^2}{12}\times \frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dt}}^2}+{\mathrm{\mu B}}_{\mathrm{h}}\mathrm{sin\theta }=0$

$\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dt}}^2}+\frac{12{\mathrm{\mu B}}_{\mathrm{h}}\mathrm{sin\theta }}{{\mathrm{mL}}^2}=0$

$\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dt}}^2}+\frac{12{\mathrm{\mu B}}_{\mathrm{h}}}{{\mathrm{mL}}^2}\mathrm{\theta }=0$

Comparing this equation with the equation of oscillation, $\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dt}}^2}+{\mathrm{\omega }}^2\mathrm{\theta }=0$

We get

${\mathrm{\omega }}^2=\frac{12{\mathrm{\mu B}}_{\mathrm{h}}}{{\mathrm{mL}}^2}$

$\begin{array}{c}\mathrm{\mu }=\frac{{\mathrm{mL}}^2{\mathrm{\omega }}^2}{12{\mathrm{B}}_{\mathrm{h}}}\\ =\frac{0.050\times {0.04}^2\times {45}^2}{12\times 16\times {10}^{-6}}\\ =8.4\times {10}^2 \mathrm{J}/\mathrm{T}\end{array}$

 

The magnitude of magnetic dipole moment is $8.4\times {10}^2 \mathrm{J}/\mathrm{T}$ .