A square loop of side $2a$ and carrying current $I$ is kept in the $xz$ plane with its centre at the origin. A long wire carrying the same current $I$ is placed parallel to the $z$-axis and passes through the point $(0, b, 0)$, $(b \gg a)$. The magnitude of the torque on the loop about the $z$-axis will be:
Answer: (B) $\dfrac{2\mu_0I^2a^2b}{\pi(a^2+b^2)}$
Only the two sides of the loop parallel to the $z$-axis (at $x = \pm a$, length $2a$) give a torque about the $z$-axis. They carry opposite currents, so one is attracted towards the wire and the other repelled, each by
$$F = \frac{\mu_0I^2(2a)}{2\pi\sqrt{a^2+b^2}}$$
along the line joining it to the wire.
The component of this force along $y$ is $F\dfrac{b}{\sqrt{a^2+b^2}}$, with lever arm $a$ about the $z$-axis, and the two torques add:
$$\tau = 2\cdot a\cdot\frac{\mu_0I^2(2a)}{2\pi\sqrt{a^2+b^2}}\cdot\frac{b}{\sqrt{a^2+b^2}} = \frac{2\mu_0I^2a^2b}{\pi(a^2+b^2)}$$
(For $b \gg a$ this is close to $\dfrac{2\mu_0I^2a^2}{\pi b}$.)
Solution by Sreeraj P, M.Sc Physics