Question

A loop with current $I$ is in field of a long straight wire with current $I_0$ The plane of the loop is perpendicular to the straight wire. Find torque acting on the loop.

• A. $\frac{2(b-a){\mu }_0{I}_{0}I\sin \theta }{\pi }$
  
• B. $\frac{(b-a){\mu }_0{I}_{0}I\sin \theta }{2\pi }$
  
• C. $\frac{(b-a){\mu }_0{I}_{0}I\sin \theta }{\pi }$
  
• D. $\frac{(b-a){\mu }_0{I}_{0}I\sin \theta }{4\pi }$
  

Answer 1

The correct option is C $\frac{(b-a){\mu }_0{I}_{0}I\sin \theta }{\pi }$


Magnetic force due to magnetic field $\overrightarrow{B}$ on a small current element, $I\overrightarrow{dl}$ is,

$\overrightarrow{d{F}_B}=I(\overrightarrow{dl}\times \overrightarrow{B})$

$\Rightarrow dF=IdlB\sin \theta$

Here, the field is tangential at every point on the circular portions, and hence the forces acting on these segments are zero.i.e.,

$\therefore \theta ={180}^{\circ }$ and $\theta =0^{\circ }$

$\Rightarrow F_{PQ}=0$ and $F_{RS}=0$

Now, consider two small elements of length $dr$ at a distance $r$ from the current element symmetrically as shown in the figure.


The magnitude of the force experienced by each element is,
$dF=IBdr$

As, $B=\frac{{\mu }_0{I}_0}{2\pi r}$

$\Rightarrow dF=\frac{{\mu }_0{I}_{0}Idr}{2\pi r}$

Force on $QR$ will be in inward direction and on $SP$ will be in outward direction.

So, torque about x-axis:

$d\tau =\left(2r\sin \alpha \right)\times \frac{{\mu }_0{I}_{0}Idr}{2\pi r}$

To obtain the total torque, we integrate the above expression within the proper limit,

$\tau =\int \tau =\frac{{\mu }_0{I}_{0}I\sin \theta }{\pi }{\int }_a^{b}dr=\frac{(b-a){\mu }_0{I}_{0}I\sin \theta }{\pi }$

Hence, option $\left(c\right)$ is the correct answer.