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

Answer 1

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

Magnetic feild at $P$ is $\overrightarrow{B}$ perpendicular to OP in the direction shown in the figure

So, $\overrightarrow{B}=B\sin \theta i-B\cos \theta j$

Here,

$B=\frac{{\mu }_0}{2\pi }.\frac{I}{r}\sin \theta =\frac{y}{r}\cos \theta =\frac{x}{r}$

$\overrightarrow{B}=\frac{{\mu }_{0}I}{2\pi }.\frac{1}{r^2}(y\hat{i}-x\hat{j})=\frac{{\mu }_{0}I(y\hat{i}-x\hat{j})}{2\pi (x^2+y^2)}$

As $r^2=x^2+y^2$

Hence the correct answer is $\frac{{\mu }_{0}I(y\hat{i}-x\hat{j})}{2\pi (x^2+y^2)}$