91Ó°ÊÓ

There is a straight wire carrying current I .

The relation between magnetic field and magnetic potential is

$\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{=}\mathbf{-}\stackrel{\mathbf{→}}{\mathbf{∇}}\mathit{U}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

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

$\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}\mathbf{I}}{\mathbf{2}\mathbf{π}\mathbf{s}}\widehat{\mathbf{ϕ}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(2\right)$

Here, is the permeability of free space.

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

${\mathrm{μ}}_{0}I=\underset{a}{\overset{b}{∫}}\stackrel{→}{B}·d\stackrel{→}{l}$

Use equation (1) to get

$$\begin{gathered} {\mathrm{μ}}_{0}I=-\underset{a}{\overset{b}{∫}}\stackrel{→}{∇}U·d\stackrel{→}{l} \\ =U\left(a\right)-U\left(b\right) \end{gathered}$$

There is non-zero current flowing in the wire.

Thus,

$U\left(a\right)≠U\left(b\right)$

Thus, the theorem is proved.

Assume that the magnetic potential of an infinite straight wire is

$U=-\frac{{\mathrm{μ}}_{0}I\mathrm{ϕ}}{2\mathrm{π}}$

Use equation (1)

$$\begin{gathered} \stackrel{→}{B}=-\stackrel{→}{∇}\left(-\frac{{\mathrm{μ}}_{0}I\mathrm{ϕ}}{2\mathrm{π}}\right) \\ =\frac{{\mathrm{μ}}_{0}I}{2\mathrm{π}}\frac{1}{s}\frac{∂}{∂\mathrm{ϕ}}\mathrm{ϕ}\widehat{\mathrm{ϕ}} \\ =\frac{{\mathrm{μ}}_{0}I}{2\mathrm{π}s}\widehat{\mathrm{ϕ}} \end{gathered}$$

This matches equation (2).

Thus the assumption was correct.

But for this magnetic potential,

$U\left(\mathrm{ϕ}\right)≠U\left(\mathrm{ϕ}+2\mathrm{π}\right)$

Thus, the theorem is verified.