A square coil ABCD of side ' L ' is carrying a current in clockwise direction. A straight conductor carrying current $\mathrm{I}_2$ (upward direction) is kept parallel to side $A B$ at a distance $\frac{L}{3}$ in the plane of $A B C D$. The net force on the coil ABCD is ( $\mu_0=$ magnetic permeability)

Magnetic force on a square coil near a straight current-carrying conductor

The straight conductor along the y-axis carrying current $I_2$ produces a magnetic field at perpendicular distance $x$ given by $B = \frac{\mu_0 I_2}{2\pi x}$, directed into the page ($-\hat{k}$) for $x > 0$ when $I_2$ flows upward.

For side AB, located at distance $L/3$ from the conductor with current $I_1$ flowing downward ($\vec{L}_{AB} = -L\hat{j}$), the magnetic field is $\vec{B}_{AB} = -\frac{3\mu_0 I_2}{2\pi L}\hat{k}$. The force is $\vec{F}_{AB} = I_1 (\vec{L}_{AB} \times \vec{B}_{AB}) = \frac{3\mu_0 I_1 I_2}{2\pi}\hat{i}$.

For side CD, at distance $4L/3$ with current flowing upward ($\vec{L}_{CD} = L\hat{j}$), the magnetic field is $\vec{B}_{CD} = -\frac{3\mu_0 I_2}{8\pi L}\hat{k}$. The force is $\vec{F}_{CD} = I_1 (\vec{L}_{CD} \times \vec{B}_{CD}) = -\frac{3\mu_0 I_1 I_2}{8\pi}\hat{i}$.

The horizontal sides BC and DA experience forces that cancel exactly. For side BC, $d\vec{F}_{BC} = \frac{\mu_0 I_1 I_2}{2\pi x}\hat{j}dx$, yielding $\vec{F}_{BC} = \frac{\mu_0 I_1 I_2}{2\pi}\ln 4\hat{j}$ after integration from $x = L/3$ to $4L/3$. For side DA, the integration from $x = 4L/3$ to $L/3$ gives $\vec{F}_{DA} = -\frac{\mu_0 I_1 I_2}{2\pi}\ln 4\hat{j}$, so $\vec{F}_{BC} + \vec{F}_{DA} = 0$.

The net force on the coil is therefore $\vec{F}_{net} = \vec{F}_{AB} + \vec{F}_{CD} = \frac{3\mu_0 I_1 I_2}{2\pi}\hat{i} - \frac{3\mu_0 I_1 I_2}{8\pi}\hat{i} = \frac{9\mu_0 I_1 I_2}{8\pi}\hat{i}$.

Final answer: $\boxed{\mathrm{D}}$