A long straight wire along the z -axis carries a current ' I ' in...

The magnetic field due to a long straight wire carrying current $I$ is given by:

$\vec{B} = \dfrac{\mu_0 I}{2\pi d} (\hat{u} \times \hat{n})$

where $\hat{u}$ is the unit vector along the direction of current, $d$ is the perpendicular distance from the wire, and $\hat{n}$ is the unit vector along the position vector of the point with respect to the wire.

Since the current is in the negative z-direction, $\hat{u} = -\hat{k}$.

The position vector of the point $(x, y)$ on the $z = 0$ plane is $\vec{r} = x\hat{i} + y\hat{j}$.

The distance of the point from the wire is $d = |\vec{r}| = \sqrt{x^2 + y^2}$.

The unit vector $\hat{n} = \dfrac{\vec{r}}{|\vec{r}|} = \dfrac{x\hat{i} + y\hat{j}}{\sqrt{x^2 + y^2}}$.

Substituting these values:

$\vec{B} = \dfrac{\mu_0 I}{2\pi \sqrt{x^2 + y^2}} \left( -\hat{k} \times \dfrac{x\hat{i} + y\hat{j}}{\sqrt{x^2 + y^2}} \right)<br><br>\vec{B} = \dfrac{\mu_0 I}{2\pi (x^2 + y^2)} (-$ x(\hat{k} $\times \hat{i}) -$ y(\hat{k} $\times \hat{j}))<br><br>Using the standard cross products \hat{k} \times \hat{i} = \hat{j} and \hat{k} \times \hat{j} = -\hat{i}:<br><br>\vec{B} = \dfrac{\mu_0 I}{2\pi (x^2 + y^2)} (-$ x\hat{j} - y(-\hat{i}))$<br><br>\vec{B} = \dfrac{\mu_0 I (y\hat{i} - x\hat{j})}{2\pi (x² + y²)}$

Answer: $\dfrac{\mu_0 I(y\hat{i} - x\hat{j})}{2\pi(x² + y²)}$