Question

A uniform, constant magnetic field $\overrightarrow{B}$ directed at an angle of ${45}^o$ to the $x$-axis in the $x-y$ plane. $PQRS$ is a rigid, square wire frame carrying a steady current $I_0$ with its center at the origin $O$. At time $t=0$ the frame is at rest in the position shown in fig. The frame is of mass $M$ and length $L$.
If the torque $\overrightarrow{\tau }$ about $O$ acting on the frame due to the magnetic field is $\overrightarrow{\tau }=\frac{I_0{L}^{2}B}{\sqrt{X}}\times (-\hat{i}+\hat{j})$. Find $X$?

Answer 1

As magnetic field is in x-y plane and subtends an angle of ${45}^o$ with x-axis:
$B_x=Bcos{45}^o=B/\sqrt{2}$ and $B_y=Bsin{45}^o=B/\sqrt{2}$
So, in vector form $\overrightarrow{B}=\hat{i}\left(B/\sqrt{2}\right)+\hat{j}\left(B/\sqrt{2}\right)$
and $\overrightarrow{M}=I_{0}S\hat{k}=I_0{L}^2\hat{k}$
$\Rightarrow \overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}=I_0{L}^2\hat{k}\times (\frac{B}{\sqrt{2}}\hat{i}+\frac{B}{\sqrt{2}}\hat{j})$
$\Rightarrow \overrightarrow{\tau }=\frac{I_0{L}^{2}B}{\sqrt{2}}\times (-\hat{i}+\hat{j})$
i.e, torque has magnitude $I_0{L}^{2}B$ and is directed along the line QS from O to S.