Question

A square loop of side $2a$ and carrying current $I$ is kept in $x-z$ plane with its centre at origin. A long wire carrying the same current $I$ is placed parallel to $z-$ axis and passing through point $(0,b,0),(b>>a)$. The magnitude of torque on the loop about $z-$ axis will be :

• A. $\frac{2{\mu }_0{I}^2{a}^2}{\pi b}$
• B. $\frac{2{\mu }_0{I}^2{a}^{2}b}{\pi (a^2+b^2)}$
• C. $\frac{{\mu }_0{I}^2{a}^{2}b}{2\pi (a^2+b^2)}$
• D. $\frac{{\mu }_0{I}^2{a}^2}{2\pi b}$

Answer 1

The correct option is A $\frac{2{\mu }_0{I}^2{a}^2}{\pi b}$


Force due to wire $AB$ on loop $abcd$ is given by-
$F=I(\overrightarrow{L}\times \overrightarrow{B})$

Wire $AB$ will develop repulsive force on segment $cd$ while attractive force on segment $ab$.

Force due to wire $AB$ on segment $cd$,

$F_{cd}=BI2a=\frac{{\mu }_{0}I}{2\pi r}I\times 2a$

$F_{cd}=\frac{{\mu }_0{I}^{2}a}{\pi \sqrt{b^2+a^2}}$

Similarly,

$F_{ab}=\frac{{\mu }_0{I}^{2}a}{\pi \sqrt{b^2+a^2}}$
So,
$F=F_{ab}=F_{cd}=\frac{{\mu }_0{I}^{2}a}{\pi \sqrt{b^2+a^2}}$

$F\cos \theta$ and $F\sin \theta$ are acting on wire $cd$ and $ab$ but torque will be caused by $F\cos \theta$.

Net torque about $z$-axis, $\tau =(F\cos \theta \times a)+(F\cos \theta \times a)=2F\cos \theta \times a$

$=\frac{{\mu }_0{I}^{2}a}{\pi \sqrt{b^2+a^2}}\times \frac{b}{\sqrt{b^2+a^2}}\times 2a$

$\Rightarrow \tau =\frac{2{\mu }_0{I}^2{a}^{2}b}{\pi (a^2+b^2)}$

If $b>>a$, then

$\tau =\frac{2{\mu }_0{I}^2{a}^2}{\pi b}$

Hence, option $\left(A\right)$ is right.