Question

A loop carrying a current $i$, lying in the plane of the paper, is in the field of a long straight wire with current $i_0$ (inward) as shown in figure. If the torque acting on the loop is given by $\tau =\frac{{\mu }_0{ii}_0}{x\pi r}\left[\sin \theta \right](b-a)$. Find x.

Answer 1

Here the field is tangential at every point on the circular portions and hence the forces acting on these segments are zero.
Now consider two small elements of length $dr$ at a distance $r$ from the current element symmetrically as shown in figure.
The magnitude of the force experienced by each element is
$dF=iBdr$ ($\because id\overrightarrow{F}=id\overrightarrow{l}\times \overrightarrow{B}$)
where $B=\frac{{\mu }_0{i}_0}{2\pi r}$
$dF=\frac{{\mu }_0{ii}_0}{2\pi r}dr$
force on $fg$ will be in inward direction and on $eh$ will be in outward direction.
So, torque about x-axis:
$d\tau =(2r\sin \theta )\frac{{\mu }_0{ii}_0}{2\pi r}dr$
To obtain the total torque, we integrate the above expression within proper limit
$\tau =\frac{{\mu }_0{ii}_0}{\pi }\sin \theta {\int }_a^{b}dr=\frac{{\mu }_0{ii}_0}{2\pi r}\left[\sin \theta \right](b-a)$