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Step 1: Identify the Geometry and Given Quantities
A square loop of side length $2a$ is placed in the XZ-plane with its center at the origin. A long, straight wire carrying current $I$ runs parallel to the z-axis and passes through the point $(0, b, 0)$, with $b \gg a$. Both the wire and the loop carry the same current $I$.
Step 2: Approximate Magnetic Field at the Loop due to the Wire
For a long straight wire carrying current $I$, the magnetic field at a distance $r$ from the wire is
$$ B = \frac{\mu_0 \, I}{2\pi \, r}. $$
Since $b \gg a$, the loop is very close to the plane $y=0$ while the wire is at $y=b$, so each segment of the loop is approximately distance $b$ from the wire. Thus, we can treat the magnetic field across the loop as nearly uniform with magnitude
$$ B \approx \frac{\mu_0 \, I}{2 \pi \, b}. $$
Step 3: Determine the Magnetic Moment of the Loop
A square loop of side $2a$ has an area $A = (2a)\,(2a) = 4a^2$.
Because the loop carries current $I$, its magnetic dipole moment $\vec{M}$ is:
$$ M = I \times A = I \times 4a^2 = 4 I a^2. $$
The direction of this magnetic moment is perpendicular to the plane of the loop (here, along the $\pm y$-axis since the loop is in the XZ-plane).
Step 4: Calculate the Torque on the Loop
In a uniform magnetic field, the torque $\vec{\tau}$ on a current loop is given by:
$$ \vec{\tau} = \vec{M} \times \vec{B}, $$
and its magnitude is
$$ \tau = MB \,\sin \theta, $$
where $\theta$ is the angle between $\vec{M}$ and $\vec{B}$. In our setup, $\vec{M}$ is perpendicular to the XZ-plane (along $\pm y$), while the magnetic field $\vec{B}$ at the location of the loop is nearly in the $\pm x$-direction. Hence, $\theta = 90^\circ$, and $\sin \theta = 1.$
Step 5: Substitute Values to Obtain the Final Expression
Substituting $M = 4 I a^2$ and $B = \frac{\mu_0 \, I}{2 \pi b}$:
$$ \tau = \left(4 I a^2\right)\left(\frac{\mu_0 I}{2\pi b}\right) = \frac{4 \mu_0 I^2 a^2}{2\pi b} = \frac{2 \mu_0 I^2 a^2}{\pi b}. $$
This matches the given correct answer:
$$ \boxed{\frac{2\mu_0 I^2 a^2}{\pi b}}. $$
