Question

A magnetic rod with length $\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{cm}$, radius $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{mm}$, and (uniform) magnetization$\mathbf{2.7}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}$A/mcan turn about its center like a compass needle. It is placed in a uniform magnetic field$\overrightarrow{\mathbf{B}}$ of magnitude $\mathbf{}\mathbf{35}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mT}$, such that the directions of its dipole moment and make an angle of $\mathbf{68}\mathbf{.}\mathbf{0}\mathbf{°}$. (a) What is the magnitude of the torque on the rod due to $\overrightarrow{\mathbf{B}}$? (b) What is the change in the orientation energy of the rod if the angle changes to $\mathbf{34}\mathbf{.}\mathbf{0}\mathbf{°}$?

Short answer

1. The magnitude of the torque on the rod due to$\overline{\mathrm{B}}$is$\mathrm{\tau }=1.49\times {10}^{-4}\mathrm{N}.\mathrm{m}$
2. The change in the orientation energy of the rod is$\mathrm{\Delta }\text{U}=-72.9\times {10}^{-6}\text{J}$

Step-by-step solution

Listing the given quantities

$\mathrm{L}=0.06\mathrm{m}$

$\mathrm{R}=3.00\times {10}^{-3}\mathrm{m}$

$\mathrm{B}=0.035\mathrm{T}$

The magnetization$\mathrm{M}=2.70\times {10}^3\mathrm{A}/\mathrm{m}$

Understanding the concepts of magnetic dipole moment

Here, we need to use the equation of torque due to magnetic field related with magnetic dipole moment and magnetic field and the equation of change in the orientation energy.

Formulae:

Torque due to magnetic field$\mathrm{\tau }=\mathrm{\mu Bsin}\left(\mathrm{\theta }\right)$

The change in the energy orientation:$\mathrm{\Delta U}=-\mathrm{\mu B}\times (\cos \left({\mathrm{\theta }}_{\mathrm{f}}\right)-\cos \left({\mathrm{\theta }}_{\mathrm{i}}\right))$

(a) Calculations of the magnitude of torque on the rod

Volume of rod$\left(\mathrm{V}\right)$is

$\begin{array}{c}\mathrm{V}={\mathrm{\pi r}}^2\mathrm{L}\\ =\mathrm{\pi }\times {(3.00\times {10}^{-3})}^2\times 0.06\\ =1.69\times {10}^{-6}{\mathrm{m}}^3\end{array}$

Dipole moment $\left(\text{μ}\right)$:

$\begin{array}{c}\mathrm{\mu }=\mathrm{MV}\\ =(2.70\times {10}^3)\times (1.69\times {10}^{-6})\\ =4.5\times {10}^{-3}{\text{A.m}}^2\end{array}$

Torque $\left(\text{τ}\right)$:

$\begin{array}{c}\mathrm{\tau }=\mathrm{\mu Bsin}\left(\mathrm{\theta }\right)\\ =(4.5\times {10}^{-3})\times \left(0.035\right)\times \left(\sin \left({68}^0\right)\right)\\ =1.49\times {10}^{-4}\mathrm{N}.\mathrm{m}\end{array}$

The magnitude of the torque on the rod due to $\overline{\mathrm{B}}$is$\mathrm{\tau }=1.49\times {10}^{-4}\mathrm{N}.\mathrm{m}$

(b) Calculations of the change in the orientation energy of the rod

$\begin{array}{c}\mathrm{\Delta U}=-\mathrm{\mu B}\times (\cos \left({\mathrm{\theta }}_{\mathrm{f}}\right)-\cos \left({\mathrm{\theta }}_{\mathrm{i}}\right))\\ =-4.5\times {10}^{-3}\times 0.035\times (\cos \left(68\right)-\cos \left(34\right))\\ =-72.9\times {10}^{-6}\text{J}\end{array}$

The change in the orientation energy of the rod is$\mathrm{\Delta }\text{U}=-72.9\times {10}^{-6}\text{J}$