A long straight wire along the Z-axis carries a current I in the negative z-direction. The magnetic field vector B at a point having coordinates $(x, y)$ in the z = 0 plane is

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$\frac{μ_{0} I (y \hat{i} - x \hat{j})}{2 π (x^{2} + y^{2})}$

$\frac{μ_{0} I (x \hat{j} - y \hat{i})}{2 π (x^{2} + y^{2})}$

$\frac{μ_{0} I (x \hat{i} - y \hat{j})}{2 π (x^{2} + y^{2})}$

$\frac{μ_{0} I (y \hat{i} + x \hat{j})}{2 π (x^{2} + y^{2})}$

answer is A.

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Detailed Solution

In the z = 0 plane, the situation is as follows
Question Image
Here, $P (x, y)$ is the point and $r = \sqrt{x^{2} + y^{2}}$
Magnetic field at P is perpendicular to OP, as it is shown in figure.
$B = B \sin θ \hat{i} - B \cos θ \hat{j}$
Here, $B = \frac{μ_{0}}{2 π} \frac{I}{r}$
$\sin θ = \frac{y}{r}, \cos θ = \frac{x}{r} B = \frac{μ_{0} I}{2 π r} (\frac{y}{r} \hat{i} - \frac{x}{r} \hat{j}) = \frac{μ_{0}}{2 π} \frac{I (y \hat{i} - x \hat{j})}{(x^{2} + y^{2})}$