Question

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

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

Answer 1

Answer: (A)

Magnetic field at $P$ is $B$, perpendicular to $OP$ in the direction
shown in figure.
So, $B=B\sin \theta \hat{i}-B\cos \theta \hat{j}$
Here, $B=\frac{{\mu }_{0}I}{2\pi r}$
$\sin \theta =\frac{y}{r}$ and $\cos \theta =\frac{x}{r}$
$\therefore B=\frac{{\mu }_{0}I}{2\pi }.\frac{1}{r^2}(y\hat{i}-x\hat{i})$
$=\frac{{\mu }_{0}I(y\hat{i}-x\hat{j})}{2\pi (x^2+y^2}$ (as $x^2+y^2$)