Question

A long solid cylindrical current carring conductor is placed along the $z-$ aixs, carrying current $I$ in the negative $z$ direction. The magnetic field $\overrightarrow{B}$ 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. $\frac{{\mu }_{0}I(x\hat{i}+y\hat{j})}{2\pi (x^2+y^2)}$
• C. $\frac{{\mu }_{0}I(x\hat{j}-y\hat{j})}{2\pi (x^2+y^2)}$
• D. $\frac{{\mu }_{0}I(x\hat{i}-y\hat{j})}{2\pi (x^2+y^2)}$

Answer 1

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

The plane representing $z=0$ is the $x-y$ plane. Let us represent the current carrying conductor in $x-y$ plane, carrying a current in $-z$ direction as shown in the figure.


The direction of magnetic field at point $\text{P}$ will be tangential to the circular field line in clockwise direction with radius $r$.

Here,

$\overrightarrow{B}=B\sin \theta \hat{i}-B\cos \theta \hat{j}......\left(i\right)$

The magnitude of magnetic field due to solid cylinder at an outside point is,

$B=\frac{{\mu }_{0}I}{2\pi r}$

Also, $\sin \theta =\frac{y}{r}$ and $\cos \theta =\frac{x}{r}$

Substituting these values in Eq. $\left(i\right)$

$\overrightarrow{B}=\frac{{\mu }_{0}I}{2\pi r}[\sin \theta \hat{i}-\cos \theta \hat{j}]$

$\Rightarrow \overrightarrow{B}=\frac{{\mu }_{0}I}{2\pi r}[\frac{y}{r}\hat{i}-\frac{x}{r}\hat{j}]$

or, $\overrightarrow{B}=\frac{{\mu }_{0}I}{2\pi r^2}[y\hat{i}-x\hat{j}]$

From geometry of figure,

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

Thus,
$\overrightarrow{B}=\frac{{\mu }_{0}I[y\hat{i}-x\hat{j}]}{2\pi (x^2+y^2)}$