First slide

Magnetic field due to a straight current carrying wire

Question

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

Moderate

A

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

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

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

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

Solution

Magnetic field at Pis $\vec{B}$ , perpendicular to OP in the direction shown in figure.
$\begin{array}{l} So, \vec{B} = B \sin θ \hat{i} - B \cos θ \hat{j} \\ Here B = \frac{μ_{0}}{2 π} \frac{I}{r} \\ \sin θ = \frac{y}{r} and \cos θ = \frac{x}{r} \\ ∴ \vec{B} = \frac{μ_{0}}{2 π} . \frac{I}{r^{2}} (y \hat{i} - x \hat{j}) = \frac{μ_{0} I (y \hat{i} - x \hat{j})}{2 π (x^{2} + y^{2})} \end{array}$
$(as r^{2} = x^{2} + y^{2})$

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