Infinite Line of Current

An infinitely long Current carrying wire is considered to be a line with infinite length, which has a constantly moving current and uniform Current Density. The magnetic fields emanating from such a wire form concentric circles.

Magnetic Field §

Assume the wire carries a uniform current of $I$ (with uniform current density) along the Z-axis. If we observe the magnetic field at a point of distance $r$ away from the line, the magnetic field is given by:

Formula §

$$\vec{B}=\frac{{\mu }_0}{2\pi }\frac{I}{r}\hat{v}$$

• $\vec{B}$ = Magnetic Field (in $T$)
• $I$ = Current (in $A$)
• $r$ = Perpendicular distance from wire
• $\hat{v}$ = Unit vector which can be obtained via the right hand curl rule.

* Note that $\vec{B}=0$ if the point is directly on the line.

Derivation §

We can adapt the formula for Magnetic Field From A Current:

Note that here $r$ from the law is equal to $R$ in the diagram above.

We define the line to go along the Z-axis, so $\vec{l}$ can be given by $z\hat{z}$. Then, we know the cross product can be given by:

$$\hat{u}\times d\vec{l}=\hat{u}\times d\hat{z}z=dz\sin (\theta )\hat{v}$$

Where $\hat{v}$ is a unit vector on the X-Y plane, equal to $\hat{u}\times \hat{z}$

Then we have:

$$\vec{B}=\int \frac{{\mu }_0}{4\pi }\frac{I}{R^2}\sin (\theta )\hat{v}\text{d}z$$

$$\vec{B}=\frac{{\mu }_0}{4\pi }I\int \frac{\sin (\theta )}{R^2}\hat{v}\text{d}z$$

Using trigonometry:

$$\sin (\theta )=\frac{x}{R}$$

$$R=\sqrt{z^2+r^2}$$

$$\vec{B}=\frac{{\mu }_0}{4\pi }I{\int }_{-\infty }^{\infty }\frac{x}{(z^2+r^2{)}^{3/2}}\hat{v}\text{d}z$$

Using a U-Substitution, we can evaluate the integral:

$$\text{ Let }u=z^2+r^2\to \frac{du}{dr}=2r$$

$$\vec{B}=\frac{{\mu }_0}{4\pi }I\times \frac{1}{2}{\int }_{-\infty }^{\infty }{u}^{-3/2}du$$

Which evaluates to:

$$\vec{B}=\frac{{\mu }_0}{4\pi }I\times \frac{2}{R}\hat{v}$$

$$\vec{B}=\frac{{\mu }_{0}I}{2\pi R}\hat{v}$$