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(D) Let the current in each wire be $I$. The wires are at $(0, a, 0)$ and $(0, -a, 0)$ carrying current into the page ($-z$ direction).At a point $P(x

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(D) Let the current in each wire be $I$. The wires are at $(0, a, 0)$ and $(0, -a, 0)$ carrying current into the page ($-z$ direction).At a point $P(x, 0, 0)$ on the $x-$axis, the distance to each wire is $r = \sqrt{x^2 + a^2}$.The magnitude of the magnetic field due to each wire is $B = \frac{\mu_0 I}{2\pi r} = \frac{\mu_0 I}{2\pi \sqrt{x^2 + a^2}}$.Using the right-hand rule, the magnetic field vector $\vec{B}_1$ from the wire at $(0, a, 0)$ makes an angle $	heta$ with the $y-$axis, where $\cos 	heta = \frac{a}{r}$ and $\sin 	heta = \frac{x}{r}$.$\vec{B}_1 = \frac{\mu_0 I}{2\pi r} (\sin 	heta \hat{i} - \cos 	heta \hat{j}) = \frac{\mu_0 I}{2\pi r} (\frac{x}{r} \hat{i} - \frac{a}{r} \hat{j}) = \frac{\mu_0 I}{2\pi (x^2 + a^2)} (x \hat{i} - a \hat{j})$.Similarly, for the wire at $(0, -a, 0)$, $\vec{B}_2 = \frac{\mu_0 I}{2\pi (x^2 + a^2)} (x \hat{i} + a \hat{j})$.The net magnetic field is $\vec{B}_{net} = \vec{B}_1 + \vec{B}_2 = \frac{\mu_0 I}{2\pi (x^2 + a^2)} (2x \hat{i}) = \frac{\mu_0 I x}{\pi (x^2 + a^2)} \hat{i}$.Since the field is along the $x-$axis, the $y-$component of the field is zero. The $x-$component of the field is $B_x = \frac{\mu_0 I x}{\pi (x^2 + a^2)}$.At $x = 0$, $B_x = 0$. As $x 	o \infty$, $B_x 	o 0$. The function is odd, meaning it is positive for $x > 0$ and negative for $x < 0$, which matches the graph in option $D$.

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