Question:
A square loop of side $2 a$ and carrying current $\mathrm{I}$ is kept is $x z$ plane with its centre at origin. A long wire carrying the same current $\mathrm{I}$ 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 :
Correct Option: , 2
Solution:
(2)
Force, $F=B I 2 a=\frac{\mu_{0} I}{2 \pi r} I \times 2 a$
Force, $F=\frac{\mu_{0} I^{2} a}{\pi \sqrt{b^{2}+a^{2}}}$
Torque, $\tau=F_{1} \times$ Perpendicular distance $=F \cos \theta \times 2 a$
$=\frac{\mu_{0} I^{2} a}{\pi \sqrt{b^{2}+a^{2}}} \times \frac{b}{\sqrt{b^{2}+a^{2}}} \times 2 a$
$\Rightarrow \tau=\frac{2 \mu_{0} I^{2} a^{2} b}{\pi\left(a^{2}+b^{2}\right)}$
If $b \gg a$ then $\tau=\frac{2 \mu_{0} I^{2} a^{2}}{\pi b}$