Question

A long straight wire along the z-axis carries a current I in the negative z direction. The magnetic vector field $\overline{B}$ 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. $\frac{{\mu }_{0}I(x\hat{j}-y\hat{i})}{4\pi (x^2+y^2)}$
• D. $\frac{{\mu }_{0}I(x\hat{i}-y\hat{j})}{4\pi (x^2+y^2)}$

Answer 1

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

An infinite wire carrying current in $(-z)$ axis

we have $\overrightarrow{l}=(-\hat{k})$ (current vector)

$\overrightarrow{l}=x\hat{i}+y\hat{j}\Rightarrow r^2=x^2+y^2$

using Biot-Savart law fo rinfinite wire

$\overrightarrow{B}=\frac{{\mu }_{0}I}{2\pi r}\left(\frac{\overrightarrow{l}\times \overrightarrow{r}}{r}\right)$ ($\frac{\overrightarrow{l}\times \overrightarrow{r}}{r}$ is the unit vector along $\overrightarrow{B}$)

$=\frac{{\mu }_{0}I}{2\pi r^2}[(-\hat{k})\times (x\hat{i}+y\hat{j})]\Rightarrow \overrightarrow{B}=\frac{{\mu }_{0}I}{2\pi (x^2+y^2)}(y\hat{i}-x\hat{j})$ (Ans)