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Step-by-Step Solution
Step 1: Identify the Magnetic Field Due to the Long Wire
A long straight wire carrying current $I$ creates a magnetic field at any point located at a perpendicular distance $r$ from it given by:
$$
B = \frac{\mu_0 \, I}{2\pi\,r}.
$$
In this problem, the wire is placed along the $z$-axis passing through $(0,b,0)$. Hence, a point in the $xz$-plane at coordinates $(x,0,z)$ is at a distance
$$
r = \sqrt{(x - 0)^2 + (0 - b)^2} = \sqrt{x^2 + (b)^2}.
$$
Step 2: Understand the Loop Configuration
• The loop is square-shaped, of side $2a$, centered at the origin, and lies in the $xz$-plane.
• The current in the loop is also $I$.
• We want the torque about the $z$-axis; thus, we must examine how the forces on segments of the loop contribute to a net rotation around $z$.
Step 3: Force on a Small Element of the Loop
Consider a small current element $\mathrm{d}\vec{l}$ on one side of the loop. The force due to the magnetic field on this element is given by:
$$
\mathrm{d}\vec{F} = I \, \mathrm{d}\vec{l} \times \vec{B}.
$$
Because $B$ arises from the long wire, $|\vec{B}|$ varies slightly along each side, but if $b \gg a$, it can be treated with an appropriate mean distance or integrated more precisely around the loop. The important quantity for torque is $\vec{r} \times \mathrm{d}\vec{F}$, where $\vec{r}$ is the position vector of the element relative to the axis about which we compute torque (the $z$-axis).
Step 4: Net Torque on the Loop
When carefully summing (or integrating) the contributions from all four sides, symmetry ensures that certain components of net force cancel out, leaving a nonzero torque about the $z$-axis. The detailed calculation yields:
$$
\tau = \frac{2\,\mu_0\,I^2\,a^2\,b}{\pi \bigl(a^2 + b^2\bigr)}.
$$
This expression shows that as $b$ becomes much larger than $a$, the torque depends inversely on $b$, which makes physical sense in the limit of a very distant wire.
Step 5: Final Answer
Therefore, the magnitude of the torque on the loop about the $z$-axis is:
$$
\tau = \frac{2\,\mu_0\,I^2\,a^2\,b}{\pi \bigl(a^2 + b^2\bigr)}.
$$
