(D) The magnetic field $B_1$ produced by the long wire carrying current $I_1$ along the $x$-axis at a point $(0, 0, d)$ is given by the Biot-Savart Law. Using the right-hand rule,the direction of the magnetic field at $(0, 0, d)$ due to current $I_1$ along the $x$-axis is in the negative $y$-direction ($-j$ direction).
The magnetic force $dF$ on a current element $I_2 dl$ is given by $dF = I_2 (dl \times B_1)$.
Here,the current element $I_2 dl$ is directed along the $y$-axis (let's say $dl = dl \hat{j}$).
The magnetic field $B_1$ at $(0, 0, d)$ is directed along the negative $y$-axis $(B_1 = -B_1 \hat{j})$.
Since the current element $dl$ and the magnetic field $B_1$ are both parallel to the $y$-axis,the cross product $dl \times B_1$ is zero because $\hat{j} \times \hat{j} = 0$.
Therefore,the magnetic force $dF$ exerted on the segment of the second wire at $O_2$ is zero.