Exercise 4.1.9: The vector potential of a long straight wire

About the further exercises of Section 4.1

Twenty further exercises in the spirit of the problems, the worked examples and the home exercises, graded from easy to harder within each group. Exercises 4.1.1–4.1.10 belong to the theory of Lecture 8 (the Lorentz force, Biot–Savart's law, flux, force and the vector potential) and Exercises 4.1.11–4.1.20 to the exercise session of Lecture 9 (Ampère's law, coils and rotating charges, measuring and shaping fields). They use only

F = q (E + v \times B), F = I \int d l \times B, B = \frac{μ _{0}}{4 π} \int \frac{I d l ^{'} \times R}{R ^{3}}, B = \nabla \times A, \nabla \cdot B = 0, \oint_{C} H \cdot d l = I_{enc}, Φ = \int_{S} B \cdot d S,

the two results of the section, $B = μ_{0} I / (2 π ρ)$ for a long straight wire and $B_{z} = μ_{0} I b^{2} / [2 (b^{2} + h^{2})^{3/2}]$ on the axis of a loop, and superposition.

Exercise 4.1.9

The vector potential of a long straight wire

cf. the vector potential

(a) Starting from $A = \frac{μ _{0}}{4 π} \int \frac{I d l ^{'}}{R}$ for a wire on the $z$ -axis, show that $A$ has only a $z$ -component and that for an infinitely long wire it diverges. Show that the difference $A_{z} (ρ) - A_{z} (ρ_{0})$ converges and equals $- \frac{μ _{0} I}{2 π} ln \frac{ρ}{ρ _{0}}$ . (b) Verify that $B = \nabla \times A$ reproduces Problem 4.1.1. (c) Check $\nabla \cdot A = 0$ and that $Δ A = - μ_{0} J$ holds outside the wire.

Try it yourself first.

Strategy

$A$ inherits the direction of the current, so it is along $a_{z}$ and its magnitude is the integral of $1/ R$ along the wire, which diverges logarithmically like the potential of an infinite line charge. Differences are finite, and the curl in cylindrical coordinates gives $B$ .

With $I d l^{'} = I d z^{'} a_{z}$ every element contributes along $a_{z}$ , so $A = A_{z} a_{z}$ with

A_{z} (ρ) = \frac{μ _{0} I}{4 π} \int_{- L}^{L} \frac{d z ^{'}}{ρ ^{2} + z ^{'2}} = \frac{μ _{0} I}{4 π} 2 ln \frac{L + L ^{2} + ρ ^{2}}{ρ}

for a wire of length $2 L$ (the integral is $2 arsinh (L / ρ)$ ). As $L \to \infty$ this grows like $ln (2 L / ρ)$ without bound: the potential of an infinite wire is infinite, exactly as the scalar potential of an infinite line charge is (Section 3.3). The difference between two points is finite:

A_{z} (ρ) - A_{z} (ρ_{0}) = \frac{μ _{0} I}{2 π} [ln \frac{L + L ^{2} + ρ ^{2}}{ρ} - ln \frac{L + L ^{2} + ρ _{0}^{2}}{ρ _{0}}] L \to \infty - \frac{μ _{0} I}{2 π} ln \frac{ρ}{ρ _{0}} .

Answer

A = - \frac{μ _{0} I}{2 π} ln \frac{ρ}{ρ _{0}} a_{z} [Wb/m], ρ_{0} arbitrary

In cylindrical coordinates (Section 2.3), for $A = A_{z} (ρ) a_{z}$ ,

\nabla \times A = \frac{1}{ρ} \frac{\partial A _{z}}{\partial ϕ} a_{ρ} - \frac{\partial A _{z}}{\partial ρ} a_{ϕ} = - \frac{d A _{z}}{d ρ} a_{ϕ} = \frac{μ _{0} I}{2 π ρ} a_{ϕ},

the field of Problem 4.1.1. The arbitrary $ρ_{0}$ disappears on differentiation: $A$ is defined up to a gradient (here a constant), $B$ is not.

$\nabla \cdot A = \partial A_{z} / \partial z = 0$ . Outside the wire ( $ρ > 0$ , $J = 0$ ) the Laplacian of $A_{z} (ρ)$ is

Δ A_{z} = \frac{1}{ρ} \frac{d}{d ρ} (ρ \frac{d A _{z}}{d ρ}) = \frac{1}{ρ} \frac{d}{d ρ} (- \frac{μ _{0} I}{2 π}) = 0 = - μ_{0} J_{z},

as required. On the axis itself the singularity carries the current: integrating $\nabla \times A = B$ round a small circle, $\oint A \cdot d l = 0$ but $\oint B \cdot d l = μ_{0} I$ , consistent with $Δ A_{z} = - μ_{0} I δ (x) δ (y)$ .

Check

Units: $μ_{0} I$ is (H/m)(A) $=$ Wb/m. $A_{z} = - \frac{μ _{0} I}{2 π} ln ρ$ is, up to the factor $μ_{0} ε_{0} I / ρ_{L}$ , the potential $V = - \frac{ρ _{L}}{2 π ε _{0}} ln ρ$ of the line charge: both solve the same two-dimensional Poisson equation. In two-dimensional problems each Cartesian component of $A$ is an electrostatic potential in disguise, and Exercise 4.1.7 used precisely this $A_{z}$ to get a flux without ever computing $B$ .