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# 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

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

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.