# Magnetic potential derivation from B

π²³ ∞ ° → ~ µ ρ σ τ ω ∑ … √ ∫ ≤ ≥ ± ∃ · θ φ ψ Ω α β γ δ ∂ ∆ ∇ ε λ Λ Γ ô

## Homework Statement

Derive the vector potential A produced by an infinite straight wire of negligible thickness, located in free space and carrying a static electric current I.

## The Attempt at a Solution

I tried to start with the expression for Mag flux density B= µH
∇xH = jωD+J (ampere's law).
I substituted H=B/µ and couldn't proceed any further.

$$\vec{A}=\frac{\mu_0I}{4\pi}\int^\infty_{-\infty}\frac{\vec{dl}}{r}$$

and note the vector A is parallel to the wire since the vectors dl are all along the wire.

Last edited:
Cyosis
Homework Helper
I don't think that formula works if the current doesn't go to 0 in infinity. Try using Stokes instead.

$$\int_{\partial S} \vec{A} \cdot d\vec{s}= \iint_S (\nabla \times \vec{A}) \cdot d\vec{S}$$

The second integral should be very familiar.

I have to derive an expression for A. How can I start with A= something?

Cyosis
Homework Helper
I have to derive an expression for A. How can I start with A= something?

In the same way when you're asked to find the electric field of a charge distribution you start with E=something, something being the general expression for E. You then usually integrate and find E for that specific problem. Besides the hint I gave you in post 3 doesn't start with an expression A=something at all.

The vector potential can be expressed as follows:

B = $$\nabla$$$$\times$$A

Magnetic field curls around an infinitely long wire, so it's clear that the vector potential must be parallel to the wire current. Also, from symmetry the vector potential must be constant along the wire axis.

Using stokes' theorem, might be able to evaluate A as a function of the radius (or "cheat" mathematically to do so) by choosing a loop to be a square loop of a certain length parallel to the wire. Set one side of the rectangle at the distance where you want to evaluate the vector potential magnitude, and the other side (of length l) at an infinite distance from the wire. Then, the dotted integral around this loop will just be A times L, the length of the side of the rectangle.

It may be difficult to evaluate the flux because you could encounter infinities, but because the vector potential can differ by the gradient of an arbitrary function, you might be able to evaluate the loop integral with respect to a certain finite point rather than infinity.