Solution
(a) Force Experienced:
The magnetic field due to the infinite wire on the z-axis is circular in the xy-plane. Along the line of the second rod (parallel to y-axis at distance d), the field has components that vary symmetrically.
The force on an element $dy$ is $d\vec{F} = I d\vec{l} \times \vec{B}$. The z-component of the force is odd with respect to $y$, so it integrates to zero over the infinite length. The other components also vanish or cancel due to symmetry.
(b) Torque Experienced:
We consider a segment of length $2d$ centered at the point of closest approach. The torque is calculated about the center.
$$ d\tau = y dF_z = y \left( I \frac{\mu_0 I}{2\pi \sqrt{d^2+y^2}} \frac{y}{\sqrt{d^2+y^2}} dy \right) $$ $$ \tau = \frac{\mu_0 I^2}{2\pi} \int_{-d}^{d} \frac{y^2}{y^2+d^2} dy $$Solving the integral $\int \frac{y^2}{y^2+d^2} dy = y – d \arctan(y/d)$:
$$ \tau = \frac{\mu_0 I^2}{2\pi} \left[ y – d \arctan(y/d) \right]_{-d}^{d} $$ $$ \tau = \frac{\mu_0 I^2}{2\pi} \left[ (d – d\frac{\pi}{4}) – (-d + d\frac{\pi}{4}) \right] $$ $$ \tau = \frac{\mu_0 I^2}{2\pi} \left[ 2d – d\frac{\pi}{2} \right] = \frac{\mu_0 I^2 d}{4\pi} (4 – \pi) $$Note: The direction $\hat{j}$ corresponds to the coordinate definition in the specific problem context, representing the axis perpendicular to both the rod and the force couple.