Loading [MathJax]/jax/output/HTML-CSS/jax.js
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Physics LibreTexts

9.3: Long, Straight, Current-carrying Conductor

( \newcommand{\kernel}{\mathrm{null}\,}\)

By way of example, let us use the expression dA=μI4πrds, 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 , ˆρ,ˆϕ,ˆz will denote the unit orthogonal vectors. After we have calculated A, we'll try and calculate its curl to give us the magnetic field B. We already know, of course, that for a straight wire the field is B=μI2πρˆϕ , so this will serve as a check on our algebra.

Consider an element ˆzdz on the wire at a height z above the xy-plane. (The length of this element is dz; the unit vector ˆz just indicates its direction.) Consider also a point P in the xy-plane at a distance ρ from the wire. The distance of P from the element dz is ρ2+z2. The contribution to the magnetic vector potential is therefore

dA=ˆzμI4πdz(ρ2+z2)1/2.

The total magnetic vector potential is therefore

A=ˆzμI2π0dz(ρ2+z2)1/2.

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

A=ˆzμI2πl0dz(ρ2+z2)1/2.

To integrate this, let z=ρtanθ, whence A=ˆzμI2πα0secθdθ where l=ρtanα. From this we obtain A=ˆzμI2πln(secα+tanα), whence

A=ˆzμI2πln(l2+ρ2+lρ).

For l>>ρ this becomes

A=ˆzμI2πln(2lρ)=ˆzμI2π(ln2llnρ).

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 ???, and, very close to a long wire, the potential is given approximately by Equation ???.

Now let us use Equation ??? together with B=curl A, to see if we can find the magnetic field B. We'll have to use the expression for curl A in cylindrical coordinates, which is

curl A=(1ρAzϕAϕz)ˆρ+(AρzAzρ)ˆϕ+1ρ(Aϕ+ρAϕρAρϕ)ˆz.

In our case, A has only a z-component, so this is much simplified:

curl A=1ρAzϕˆρAzρˆϕ.

And since the z-component of A depends only on ρ, the calculation becomes trivial, and we obtain, as expected

B=μI2πρˆϕ.

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 ??? – but the object of the exercise was not to derive Equation ??? (which is trivial from Ampère's theorem), but to derive the expression for A. Calculating B subsequently was only to reassure us that our algebra was correct.


This page titled 9.3: Long, Straight, Current-carrying Conductor is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jeremy Tatum via source content that was edited to the style and standards of the LibreTexts platform.

Support Center

How can we help?