91Ó°ÊÓ

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

$\mathrm{R}=3.00×{10}^{-3}\mathrm{m}$

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

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

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$\mathrm{Ï„}=\mathrm{μ\mu þ²õ¾\pm ²Ô}\left(\mathrm{θ}\right)$

The change in the energy orientation:$\mathrm{Δ\pm «}=−\mathrm{μ\mu þ}×(\cos \left({\mathrm{θ}}_{\mathrm{f}}\right)−\cos \left({\mathrm{θ}}_{\mathrm{i}}\right))$

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

$\begin{array}{c}\mathrm{V}={\mathrm{Ï€°ù}}^2\mathrm{L}\\ =\mathrm{Ï€}×{(3.00×{10}^{−3})}^2×0.06\\ =1.69×{10}^{-6}{\mathrm{m}}^3\end{array}$

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

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

Torque $\left(\text{Ï„}\right)$:

$\begin{array}{c}\mathrm{Ï„}=\mathrm{μ\mu þ²õ¾\pm ²Ô}\left(\mathrm{θ}\right)\\ =(4.5×{10}^{−3})×\left(0.035\right)×\left(\sin \left({68}^0\right)\right)\\ =1.49×{10}^{−4}\mathrm{N}.\mathrm{m}\end{array}$

The magnitude of the torque on the rod due to $\stackrel{¯}{\mathrm{B}}$is$\mathrm{τ}=1.49×{10}^{−4}\mathrm{N}.\mathrm{m}$

$\begin{array}{c}\mathrm{Δ\pm «}=−\mathrm{μ\mu þ}×(\cos \left({\mathrm{θ}}_{\mathrm{f}}\right)−\cos \left({\mathrm{θ}}_{\mathrm{i}}\right))\\ =−4.5×{10}^{−3}×0.035×(\cos \left(68\right)−\cos \left(34\right))\\ =−72.9×{10}^{−6}\text{J}\end{array}$

The change in the orientation energy of the rod is$\mathrm{Δ}\text{U}=−72.9×{10}^{−6}\text{J}$