# 9.3: Long, Straight, Current-carrying Conductor

By way of example, let us use the expression \(\textbf{dA} = \frac{\mu I}{ 4 \pi r}\textbf{ds}\) , to calculate the magnetic vector potential in the vicinity of a long, straight, current-carrying conductor ("wire" for short!). We'll suppose that the wire lies along the \(z\)-axis, with the current flowing in the direction of positive \(z\). We'll work in cylindrical coordinates, and the symbols , **\(\hat{\rho},\,\hat{\phi},\,\hat{\textbf{z}}\)** will denote the unit orthogonal vectors. After we have calculated \(\textbf{A}\), we'll try and calculate its **curl** to give us the magnetic field \(\textbf{B}\). We already know, of course, that for a straight wire the field is \(\textbf{B}=\frac{\mu I}{2\pi \rho}\)**\(\hat{\phi}\)** , so this will serve as a check on our algebra.

Consider an element \(\hat{\textbf{z}}\,dz\) on the wire at a height \(z\) above the \(xy\)-plane. (The length of this element is \(dz\); the unit vector \(\hat{\textbf{z}}\) just indicates its direction.) Consider also a point P in the \(xy\)-plane at a distance \(\rho\) from the wire. The distance of P from the element \(dz\text{ is }\sqrt{\rho^2 +z^2}\) . The contribution to the magnetic vector potential is therefore

\[\textbf{dA}=\hat{\textbf{z}}\frac{\mu I}{4\pi}\cdot \frac{dz}{(\rho^2+z^2)^{1/2}}.\label{9.3.1}\]

The total magnetic vector potential is therefore

\[\textbf{A}=\hat{\textbf{z}}\frac{\mu I}{2\pi}\int_0^\infty \frac{dz}{(\rho^2+z^2)^{1/2}}.\label{9.3.2}\]

This integral is infinite, which at first may appear to be puzzling. Let us therefore first calculate the magnetic vector potential for a finite section of length \(2l\) of the wire. For this section, we have

\[\textbf{A}=\hat{\textbf{z}}\frac{\mu I}{2\pi}\cdot \int_0^l \frac{dz}{(\rho^2+z^2)^{1/2}}.\label{9.3.3}\]

To integrate this, let \(z = \rho \tan θ\), whence \(\textbf{A}=\hat{\textbf{z}}\frac{\mu I}{2\pi}\cdot \int_0^\alpha \sec \theta \, d\theta\) where \(l = \rho \tan \alpha\). From this we obtain \(\textbf{A}=\hat{\textbf{z}}\frac{\mu I}{2\pi}\cdot \ln (\sec \alpha +\tan \alpha )\), whence

\[\label{9.3.4}\textbf{A}=\hat{\textbf{z}}\frac{\mu I}{2\pi}\cdot \ln \left ( \frac{\sqrt{l^2+\rho^2}+l}{\rho}\right ) .\]

For \(l >> \rho\) this becomes

\[\label{9.3.5}\textbf{A}=\hat{\textbf{z}}\frac{\mu I}{2\pi}\cdot \ln \left ( \frac{2l}{\rho}\right ) =\hat{\textbf{z}}\frac{\mu I}{2\pi}(\ln 2l -\ln \rho ).\]

Thus we see that the magnetic vector potential in the vicinity of a straight wire is a vector field parallel to the wire. If the wire is of infinite length, the magnetic vector potential is infinite. For a finite length, the potential is given exactly by Equation \ref{9.3.4}, and, very close to a long wire, the potential is given approximately by Equation \ref{9.3.5}.

Now let us use Equation \ref{9.3.5} together with \(\textbf{B} = \textbf{curl A}\), to see if we can find the magnetic field \(\textbf{B}\). We'll have to use the expression for \(\textbf{curl A}\) in cylindrical coordinates, which is

\[\label{9.3.6}\textbf{curl A} = \left ( \frac{1}{\rho}\frac{∂A_\phi}{∂\phi}-\frac{∂A_\phi}{∂z}\right ) \hat{\boldsymbol{\rho}}+\left ( \frac{∂A_\rho}{∂z}-\frac{∂A_z}{∂\rho}\right ) \hat{\boldsymbol{\phi}}+\frac{1}{\rho}\left ( A_\phi +\rho \frac{∂A_\phi}{∂\rho}-\frac{∂A_\rho}{∂\phi }\right ) \hat{\textbf{z}}.\]

In our case, \(\textbf{A}\) has only a \(z\)-component, so this is much simplified:

\[\label{9.3.7}\textbf{curl A}=\frac{1}{\rho}\frac{∂A_z}{∂\phi}\hat{\boldsymbol{\rho}}-\frac{∂A_z}{∂\rho}\hat{\boldsymbol{\phi}}.\]

And since the \(z\)-component of \(\textbf{A}\) depends only on \(\rho\), the calculation becomes trivial, and we obtain, as expected

\[\label{9.3.8}\textbf{B}=\frac{\mu I}{2\pi \rho }\hat{\boldsymbol{\phi}}.\]

This is an approximate result for very close to a long wire – but it is exact for any distance for an infinite wire. This may strike you as a long palaver to derive Equation \ref{9.3.8} – but the object of the exercise was not to derive Equation \ref{9.3.8} (which is trivial from Ampère's theorem), but to derive the expression for \(\textbf{A}\). Calculating \(\textbf{B}\) subsequently was only to reassure us that our algebra was correct.