Question

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

• A. $\frac{{\mu }_{0}I(y\hat{i}-x\hat{j})}{2\pi (x^2+y^2)}$
• B. $\frac{{\mu }_{0}I(x\hat{i}-y\hat{j})}{2\pi (x^2+y^2)}$
• C. $\frac{{\mu }_{0}I(x\hat{i}+y\hat{j})}{2\pi (x^2+y^2)}$
• D. $\frac{{\mu }_{0}I(x\hat{i}-yx\hat{j})}{2\pi (x^2+y^2)}$

Answer 1

The correct option is A $\frac{{\mu }_{0}I(y\hat{i}-x\hat{j})}{2\pi (x^2+y^2)}$

The wire carries a current $I$ in the negative $z-direction$.
We have to consider the magnetic vector field $\overrightarrow{B}$ at $(x,y)$ in the $z=0$ plane.
Magnetic field $\overrightarrow{B}$ is perpendicular to $OP$.
$\therefore \overrightarrow{B}=B\sin \theta \hat{i}-B\cos \theta \hat{j}$
$\sin \theta =\frac{y}{r},\cos \theta =\frac{x}{r}B=\frac{{\mu }_{0}I}{2\pi r}$
$\therefore \overrightarrow{B}=\frac{{\mu }_{0}I}{2\pi r^2}(y\hat{i}-x\hat{j})$
or $\overrightarrow{B}=\frac{{\mu }_{0}I(y\hat{i}-x\hat{j})}{2\pi (x^2+y^2)}$.