91影视

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

$\mathrm{R}=3.00脳{10}^{-3}\mathrm{m}$

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

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

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$\mathrm{蟿}=\mathrm{渭叠蝉颈苍}\left(\mathrm{胃}\right)$

The change in the energy orientation:$\mathrm{螖鲍}=鈭�\mathrm{渭叠}脳(\cos \left({\mathrm{胃}}_{\mathrm{f}}\right)鈭�\cos \left({\mathrm{胃}}_{\mathrm{i}}\right))$

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

$\begin{array}{c}\mathrm{V}={\mathrm{蟺谤}}^2\mathrm{L}\\ =\mathrm{蟺}脳{(3.00脳{10}^{鈭�3})}^2脳0.06\\ =1.69脳{10}^{-6}{\mathrm{m}}^3\end{array}$

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

$\begin{array}{c}\mathrm{渭}=\mathrm{MV}\\ =(2.70脳{10}^3)脳(1.69脳{10}^{鈭�6})\\ =4.5脳{10}^{鈭�3}{\text{A.m}}^2\end{array}$

Torque $\left(\text{蟿}\right)$:

$\begin{array}{c}\mathrm{蟿}=\mathrm{渭叠蝉颈苍}\left(\mathrm{胃}\right)\\ =(4.5脳{10}^{鈭�3})脳\left(0.035\right)脳\left(\sin \left({68}^0\right)\right)\\ =1.49脳{10}^{鈭�4}\mathrm{N}.\mathrm{m}\end{array}$

The magnitude of the torque on the rod due to $\stackrel{炉}{\mathrm{B}}$is$\mathrm{蟿}=1.49脳{10}^{鈭�4}\mathrm{N}.\mathrm{m}$

$\begin{array}{c}\mathrm{螖鲍}=鈭�\mathrm{渭叠}脳(\cos \left({\mathrm{胃}}_{\mathrm{f}}\right)鈭�\cos \left({\mathrm{胃}}_{\mathrm{i}}\right))\\ =鈭�4.5脳{10}^{鈭�3}脳0.035脳(\cos \left(68\right)鈭�\cos \left(34\right))\\ =鈭�72.9脳{10}^{鈭�6}\text{J}\end{array}$

The change in the orientation energy of the rod is$\mathrm{螖}\text{U}=鈭�72.9脳{10}^{鈭�6}\text{J}$