Step-by-step Solution:

1. Determine the Magnetic Field (B) at the Loop's Center

   Let's set up a coordinate system. Let the circular loop and the two parallel wires lie in the xy-plane. Let the center of the loop be at the origin (0, 0). The two wires are parallel to the y-axis, located at x = -d (wire 1) and x = +d (wire 2).

   The currents in the wires are I and are in opposite directions. Let the current in wire 1 be in the +y direction and the current in wire 2 be in the -y direction.

   The magnetic field produced by a long straight wire at a distance r is given by $B = \frac{\mu_0 I}{2\pi r}$. The direction is determined by the right-hand rule.

   • Magnetic field from wire 1 at the origin: The distance is d. Using the right-hand rule (thumb in +y direction), the magnetic field at the origin points out of the page, i.e., in the +z direction. $B_1 = \frac{\mu_0 I}{2\pi d} \hat{k}$
   • Magnetic field from wire 2 at the origin: The distance is d. Using the right-hand rule (thumb in -y direction), the magnetic field at the origin also points out of the page, i.e., in the +z direction. $B_2 = \frac{\mu_0 I}{2\pi d} \hat{k}$
   • Net magnetic field: The two fields add up. $B_{net} = B_1 + B_2 = \frac{\mu_0 I}{2\pi d} \hat{k} + \frac{\mu_0 I}{2\pi d} \hat{k} = \frac{2\mu_0 I}{2\pi d} \hat{k} = \frac{\mu_0 I}{\pi d} \hat{k}$ Since d >> a, we can assume this magnetic field is uniform over the area of the loop. The magnitude of the magnetic field is $B = \frac{\mu_0 I}{\pi d}$.

2. Determine the Magnetic Dipole Moment (μ) of the Loop

   The magnetic dipole moment of a current loop is given by $\vec{\mu} = I \vec{A}$, where I is the current and $\vec{A}$ is the area vector.

   • The area of the circular loop of radius a is $A = \pi a^2$.
   • The current I in the loop is counter-clockwise. By the right-hand rule, the area vector (and thus the magnetic moment vector) points out of the page, in the +z direction.
   • The magnitude of the magnetic moment is $\mu = I A = I (\pi a^2) = \pi a^2 I$. In the initial position, $\vec{\mu}$ is parallel to $\vec{B}$.

3. Analyze the Rotation and the Angle for Torque Calculation

   The loop is rotated about its diameter parallel to the wires (i.e., about the y-axis) by an angle $\theta = 30^\circ$.

   • The magnetic field $\vec{B}$ remains in the +z direction.
   • The magnetic moment vector $\vec{\mu}$, which was initially along the +z axis, rotates with the loop. After a $30^\circ$ rotation, the angle between the new orientation of $\vec{\mu}$ and the fixed $\vec{B}$ vector is $\alpha = 30^\circ$.

4. Calculate the Torque (τ)

   The torque on a magnetic dipole in a uniform magnetic field is given by the formula $\vec{\tau} = \vec{\mu} \times \vec{B}$. The magnitude of the torque is $\tau = \mu B \sin(\alpha)$, where $\alpha$ is the angle between $\vec{\mu}$ and $\vec{B}$.

   Substituting the values we found:

   • $\mu = \pi a^2 I$
   • $B = \frac{\mu_0 I}{\pi d}$
   • $\alpha = 30^\circ$

   $\tau = (\pi a^2 I) \left( \frac{\mu_0 I}{\pi d} \right) \sin(30^\circ)$

   Simplify the expression: $\tau = \frac{\mu_0 I^2 a^2 \pi}{\pi d} \sin(30^\circ)$ $\tau = \frac{\mu_0 I^2 a^2}{d} \sin(30^\circ)$

   Since $\sin(30^\circ) = 1/2$: $\tau = \frac{\mu_0 I^2 a^2}{d} \left( \frac{1}{2} \right)$ $\tau = \frac{\mu_0 I^2 a^2}{2d}$

   This result matches option B.