Question

Obtain an expression for the magnetic induction at a point due to an infinitely long straight conductor carrying current.

Answer 1

Magnetic induction due to infinitely long straight conductor carrying current: XY, is an infinitely long straight conductor carrying a current $I$(Figure).

$P$ is a point at a distance $a$ from the conductor. $AB$ is a small element of length $dl$. $\theta$ is the angle between the current element $Idl$ and the line joining the element $dl$ and the point P. According to Biot-Savart law, the magnetic induction at the point P due to the current element $Idl$ is.

$dB=\frac{{\mu }_o}{4\pi }\frac{Idl\cdot \sin \theta }{r^2}$ ......$\left(1\right)$

AC is drawn perpendicular to BP from A.

$\angle OPA=\varphi$, $\angle APB=d\varphi$

In $\Delta$ABC, $\sin \theta =\frac{AC}{AB}=\frac{AC}{dl}$

$\therefore AC=dl\sin \theta$ ..........$\left(2\right)$

From $\Delta$APC, AC$=rd\varphi$ .........$\left(3\right)$

From equations $\left(2\right)$ and $\left(3\right)$,

$rd\varphi =dl\sin \theta$ ............$\left(4\right)$

Substituting equation $\left(4\right)$ in equation $\left(1\right)$

$dB=\frac{{\mu }_o}{4\pi }\frac{Ird\varphi }{r^2}=\frac{{\mu }_o}{4\pi }\frac{Id\varphi }{r}$ .............$\left(5\right)$

In $\Delta$OPA, $\cos \varphi =\frac{a}{r}$

$\therefore r=\frac{a}{cos\varphi }$ ..........$\left(6\right)$

Substituting equation $\left(6\right)$ in equation $\left(5\right)$

$dB=\frac{{\mu }_o}{4\pi }\frac{1}{a}\cos \varphi d\varphi$

The total magnetic induction at P due to the conductor XY is

$B=\stackrel{{\varphi }_2}{\underset{-{\varphi }_1}{\int }}dB=\stackrel{{\varphi }_2}{\underset{-{\varphi }_1}{\int }}\frac{{\mu }_{o}I}{4\pi a}\cos \varphi d\varphi$

$B=\frac{{\mu }_{o}I}{4\pi a}[\sin {\varphi }_1+\sin {\varphi }_2]$

For infinitely long conductor,

${\varphi }_1={\varphi }_2={90}^o$

$\therefore B=\frac{{\mu }_{o}I}{2\pi a}$

If the conductor is places in a medium of permeability ${\mu }_0$, $B=\frac{\mu I}{2\pi a}$.