Question

A long straight wire along the Z-axis carries a current ${}^{\prime }{I}^{\prime }$ in the negative Z-direction. The induced magnetic field $B$ at a point having coordinates $(x,y)$ is:

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

Answer 1

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

According to Biot Savart Law,

$\overrightarrow{B}\left(r\right)=\frac{{\mu }_o}{4\pi }{\int }_C\frac{I\overrightarrow{dl}\times \overrightarrow{r}}{r^3}$

$\overrightarrow{dl}=-dz\hat{k}$

$\overrightarrow{r}=x\hat{i}+y\hat{j}-z\hat{k}$

$r=(x^2+y^2+z^2{)}^{1/2}$

Let $d^2=x^2+y^2$

$\overrightarrow{B}=\frac{{\mu }_{o}I}{4\pi }{\int }_{z=-\infty }^{\infty }\frac{(-dz\hat{k})\times (x\hat{i}+y\hat{j}-z\hat{k})}{(d^2+z^2{)}^{3/2}}$

$\overrightarrow{B}=\frac{{\mu }_{o}I}{4\pi }{\int }_{z=-\infty }^{\infty }\frac{-xdz\hat{j}+ydz\hat{i}}{(d^2+z^2{)}^{3/2}}$

Let $z=d\tan \left(\theta \right)$

Then, $dz=d{\sec }^2\left(\theta \right)d\theta$

$\overrightarrow{B}=\frac{{\mu }_{o}I}{4\pi }{\int }_{\theta =-\pi /2}^{\pi /2}\frac{-xd\cos \left(\theta \right)d\theta \hat{j}+yd\cos \left(\theta \right)d\theta \hat{i}}{d^3}$

Solving the integration gives,

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