Question: Suppose you want to define a magnetic scalar potential U(Eq. 5.67)

in the vicinity of a current-carrying wire. First of all, you must stay away from the

wire itself (there $\mathbf{\nabla }\mathbf{}\mathbf{\times }\mathbf{}\mathit{B}\mathbf{}\mathbf{\ne }\mathbf{0}$); but that's not enough. Show, by applying Ampere's

law to a path that starts at a and circles the wire, returning to b (Fig. 5.47), that the

scalar potential cannot be single-valued (that is, $\mathit{U}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{\ne }\mathit{U}\mathbf{\left(}\mathit{b}\mathbf{\right)}$, even if they represent the same physical point). As an example, find the scalar potential for an infinite straight wire. (To avoid a multivalued potential, you must restrict yourself to simply connected regions that remain on one side or the other of every wire, never allowing you to go all the way around.)

Step by step solution

01

Given data

There is a straight wire carrying current I .

02

Magnetic potential and magnetic field of an infinite straight wire

The relation between magnetic field and magnetic potential is

$\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{-}\overrightarrow{\mathbf{\nabla }}\mathit{U}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Magnetic field of an infinite straight wire carrying current in cylindrical coordinates

$\overrightarrow{\mathbf{B}}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{I}}{\mathbf{2}\mathbf{\pi }\mathbf{s}}\hat{\mathbf{\varphi }}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(2\right)$

Here, is the permeability of free space.

03

Proof that vector potential in the vicinity of a wire carrying current is multivalued. Verification of this theorem for an infinite wire.

Apply Ampere's law on the loop shown in the figure

${\mu }_{0}I=\underset{a}{\overset{b}{\int }}\overrightarrow{B}·d\overrightarrow{l}$

Use equation (1) to get

$$\begin{gathered} {\mu }_{0}I=-\underset{a}{\overset{b}{\int }}\overrightarrow{\nabla }U·d\overrightarrow{l} \\ =U\left(a\right)-U\left(b\right) \end{gathered}$$

There is non-zero current flowing in the wire.

Thus,

$U\left(a\right)\ne U\left(b\right)$

Thus, the theorem is proved.

Assume that the magnetic potential of an infinite straight wire is

$U=-\frac{{\mu }_{0}I\varphi }{2\pi }$

Use equation (1)

$$\begin{gathered} \overrightarrow{B}=-\overrightarrow{\nabla }\left(-\frac{{\mu }_{0}I\varphi }{2\pi }\right) \\ =\frac{{\mu }_{0}I}{2\pi }\frac{1}{s}\frac{\partial }{\partial \varphi }\varphi \hat{\varphi } \\ =\frac{{\mu }_{0}I}{2\pi s}\hat{\varphi } \end{gathered}$$

This matches equation (2).

Thus the assumption was correct.

But for this magnetic potential,

$U\left(\varphi \right)\ne U\left(\varphi +2\pi \right)$

Thus, the theorem is verified.

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