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

We have a square current-carrying loop of side $$2a$$ whose plane is the $$XZ$$-plane and whose centre is at the origin. Hence every point of the loop has coordinate $$y = 0$$. The loop carries current $$I$$. A very long straight conductor, also carrying the same current $$I$$, is parallel to the $$z$$-axis and passes through the fixed point $$(0,\, b,\, 0)$$ with $$b \gg a$$. Thus the straight wire is at a perpendicular distance $$b$$ from the centre of the square loop.

First we note the magnetic moment (also called magnetic dipole moment) of a current loop. For any closed loop the magnitude is given by the formula

$$m \;=\; I \, A,$$

where $$A$$ is the area enclosed by the loop and its direction is perpendicular to the plane of the loop according to the right-hand rule. Here the area of the square is

$$A \;=\; (2a)\times(2a) \;=\; 4a^{2}.$$

Hence the magnitude of the magnetic moment of the square loop is

$$m \;=\; I \,(4a^{2}) \;=\; 4 I a^{2}.$$

The loop lies in the $$XZ$$-plane, so its area vector (and therefore its magnetic-moment vector) points along the $$\pm y$$-axis. We shall simply denote its magnitude by $$m$$ and keep in mind that its direction is along $$\hat y$$ or $$-\hat y$$ depending on the sense of current; that sign will automatically be taken care of by the cross product.

Next we need the magnetic field produced by the long straight wire at a generic point of the loop. For an infinitely long wire the magnitude of the magnetic field at a perpendicular distance $$r$$ is given by the standard Biot-Savart result

$$B \;=\;\frac{\mu_{0} I}{2\pi r}.$$

The distance of the centre of the loop from the wire is exactly $$b$$, so the field at the centre due to the wire is

$$B_{0} \;=\;\frac{\mu_{0} I}{2\pi b}.$$

Because we are told that $$b \gg a$$, every point of the square loop is very nearly at the same distance $$b$$ from the wire; the fractional variation in distance is only of order $$a/b$$. Therefore, to an excellent approximation, we can treat the magnetic field over the whole loop as uniform and equal to $$B_{0}$$ in magnitude and direction. This simplification allows us to use the simple dipole-torque formula instead of integrating the force on each element.

The torque $$\vec\tau$$ acting on a current loop of magnetic moment $$\vec m$$ placed in a (uniform) magnetic field $$\vec B$$ is given by the vector relation

$$\vec\tau \;=\; \vec m \times \vec B.$$

The magnitude of the torque is therefore

$$\tau \;=\; m B \sin\theta,$$

where $$\theta$$ is the angle between $$\vec m$$ and $$\vec B$$. In our geometry the magnetic field of the straight wire is tangential to a circle centred on the wire; at the origin (centre of the square) that direction is along the $$+x$$-axis. Thus $$\vec B$$ is along $$\hat x$$ while $$\vec m$$ is along $$\pm\hat y$$, so the angle between them is exactly $$90^\circ$$ and $$\sin\theta = 1$$.

Putting everything together, the magnitude of the torque is

$$\tau \;=\; m B_{0} \;=\; \bigl(4 I a^{2}\bigr) \left(\frac{\mu_{0} I}{2\pi b}\right).$$

Multiplying the factors gives

$$\tau \;=\; \frac{4\,\mu_{0}\,I^{2}\,a^{2}}{2\pi b} \;=\; \frac{2\,\mu_{0}\,I^{2}\,a^{2}}{\pi b}.$$

This matches option C in the list provided.

Hence, the correct answer is Option 3.