Another way to fill in the "missing link" in Fig. 5.48 is to look for a magnetostatic analog to Eq. 2.21. The obvious candidate would be

$\mathbf{A}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{r}}\mathbf{(}\mathbf{B}\mathbf{\times }\mathbf{dl}\mathbf{)}$

(a) Test this formula for the simplest possible case-uniform B (use the origin as your reference point). Is the result consistent with Prob. 5.25? You could cure this problem by throwing in a factor of localid="1657688349235" $\frac{\mathbf{1}}{\mathbf{2}}$, but the flaw in this equation runs deeper.

(b) Show that $\mathbf{\int }\mathbf{(}\mathbf{B}\mathbf{\times }\mathbf{dl}\mathbf{)}$is not independent of path, by calculating $\mathbf{\oint }\mathbf{(}\mathbf{B}\mathbf{\times }\mathbf{dl}\mathbf{)}$around the rectangular loop shown in Fig. 5.63.

Figure 5.63

As far as $\mathbf{l}$know${,}^{28}$the best one can do along these lines is the pair of equations

(i) localid="1657688931461" $\mathbf{v}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{r}\mathbf{\times }{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{1}}\mathbf{E}\mathbf{\left(}\mathit{\lambda }\mathbf{r}\mathbf{\right)}\mathbf{}\mathbf{d\lambda }$

(ii) $\mathbf{A}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{r}\mathbf{\times }{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{1}}\mathbf{\lambda B}\mathbf{\left(}\mathbf{\lambda r}\mathbf{\right)}\mathbf{d\lambda }$

[Equation (i) amounts to selecting a radial path for the integral in Eq. 2.21; equation (ii) constitutes a more "symmetrical" solution to Prob. 5.31.]

(c) Use (ii) to find the vector potential for uniform B.

(d) Use (ii) to find the vector potential of an infinite straight wire carrying a steady current. Does (ii) automatically satisfy $\mathbf{\nabla }\mathbf{‐}\mathbf{A}\mathbf{=}\mathbf{0}$[Answer:$\mathbf{(}{\mathbf{\mu }}_{\mathbf{o}}\mathbf{l}\mathbf{/}\mathbf{2}\mathbf{\pi s}\mathbf{)}\mathbf{(}\mathbf{z}\hat{\mathbf{s}}\mathbf{-}\mathbf{s}\hat{\mathbf{z}}\mathbf{)}$ ].

Consider the value of vector potential A is $\left(\mathrm{r}\right)={\int }_0^1(\mathrm{B}\times \mathrm{dl})$.

Consider the pair of equations are:

(i) $\mathrm{v}\left(\mathrm{r}\right)=-\mathrm{r}\times {\int }_0^1\mathrm{E}\left(\mathrm{\lambda r}\right)\mathrm{d\lambda }$

(ii) $\mathrm{A}\left(\mathrm{r}\right)=-\mathrm{r}\times {\int }_0^1\mathrm{\lambda B}\left(\mathrm{\lambda r}\right)\mathrm{d\lambda }$

Write the formula of vector potential.

$\mathbf{A}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{1}}\mathbf{(}\mathbf{B}\mathbf{\times }\mathit{d}\mathit{l}\mathbf{)}$ …… (1)

Here, $\mathbf{B}$is uniform magnetic field and is current through the wire.

Write the formula of magnetic field due to wire is,

$\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{l}}{\mathbf{2}\mathbf{\pi s}}\hat{\mathbf{\varphi }}$ …… (2)

Here, role="math" localid="1657691966896" $\mathbf{l}$is the current through the wire, $\mathbf{s}$is the length of wire, ${\mathit{\mu }}_{\mathbf{0}}$is the permeability of free space.

Write the formula of vector potential for uniform field.

$\mathbf{A}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{r}\mathbf{\times }{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{1}}\mathbf{\lambda B}\mathbf{\left(}\mathbf{\lambda r}\mathbf{\right)}\mathbf{d\lambda }$ …… (3)

Here, $\mathbf{r}$is distance, $\mathbf{B}$is magnetic field,$\mathbf{\lambda }$is vector function.

Write the formula ofvector potential of an infinite straight wire carrying a steady current.

$\mathbf{A}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{r}\mathbf{\times }{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{1}}\mathbf{\lambda B}\mathbf{\left(}\mathbf{\lambda r}\mathbf{\right)}\mathbf{d\lambda }$ …… (4)

Here, $\mathbf{r}$is distance, $\mathbf{B}$is magnetic field,$\mathbf{\lambda }$is vector function and $\mathbf{V}$is voltage.

Write the formula of divergence of vector potential$\mathbf{A}$ is,

$\mathbf{\nabla }\mathbf{‐}\mathbf{A}$ …… (5)

Here, $\mathbf{A}$is vector potential.

Determine thevector potential.

$\mathrm{A}\left(\mathrm{r}\right)=\mathrm{B}\times {\int }_0^{1}dl$

From the problem 5.25, the vector potential is,

$\mathrm{A}=-\frac{1}{2}(\mathrm{B}\times \mathrm{r})$

The vector potential is therefore incompatible with that of problem 5.25P. Negative signs are absent, indicating that the direction is also opposite.

Determine the magnitude field B due to wire is,

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{l}}{2\mathrm{\pi s}}\hat{\mathrm{\varphi }}$

To solve the integral.

$$\begin{gathered} \int \mathrm{B}\times \mathrm{dl}=\left(\frac{{\mu }_{0}l}{2\mathrm{\pi }a}\hat{s}-\frac{{\mu }_{0}l}{2\mathrm{\pi }b}\hat{s}\right)w \\ =\frac{{\mu }_{0}lw}{2\mathrm{\pi }}\left(\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}\right)\hat{s} \\ \ne 0 \end{gathered}$$

Thus $\int \mathrm{B}\times \mathrm{dl}\ne 0$is not independent of path.

Determine the vector potential for uniform field B can be calculated using the equation (ii).

Substitute $\lambda$for $\lambda$$\left(\lambda x\right)$into equation (3).

$$\begin{gathered} \mathrm{A}\left(\mathrm{r}\right)=-\mathrm{r}\times \mathrm{B}{\int }_0^1\mathrm{\lambda d\lambda } \\ =-\mathrm{r}\times \mathrm{B}{\left[\frac{{\mathrm{\lambda }}^2}{2}\right]}_0^1 \\ =-\mathrm{r}\times \mathrm{B}\left(\frac{1}{2}-0\right) \\ =-\frac{1}{2}(\mathrm{r}\times \mathrm{B}) \end{gathered}$$

Therefore, the value of vector potential for uniform field is $-\frac{1}{2}(\mathrm{r}\times \mathrm{B})$.

Use the expression of magnetic field, $\mathrm{B}=\frac{{\mu }_{0}l}{2\mathrm{\pi s}}\hat{\mathrm{\varphi }}$to find $\mathrm{B}\left(\mathrm{\lambda r}\right)$.

$\mathrm{B}=\left(\mathrm{\lambda r}\right)\frac{{\mu }_{0}l}{2\mathrm{\pi s}}\hat{\mathrm{\varphi }}$

The vector potential is,

role="math" localid="1657703468579" $$\begin{gathered} \mathrm{A}\left(\mathrm{r}\right)=-r\times {\int }_0^1\mathrm{\lambda R}\left(\mathrm{\lambda r}\right)d\mathrm{\lambda } \\ =\frac{-{\mu }_{01}l}{2\mathrm{\pi s}}(\mathrm{r}\times \hat{\mathrm{\varphi }}){\int }_0^1\lambda \frac{1}{\lambda }dx \\ =\frac{-{\mu }_{0}l}{2\mathrm{\pi s}}(\mathrm{r}\times \hat{\mathrm{\varphi }}){\left[x\right]}_0^1 \end{gathered}$$

Solve further as,

$$\begin{gathered} \mathrm{A}\left(\mathrm{r}\right)=-\frac{{\mu }_{0}l}{2\mathrm{\pi s}}(\mathrm{r}-\hat{\mathrm{\varphi }})\left[1-0\right] \\ =-\frac{{\mu }_{o}l}{2\mathrm{\pi s}}(\mathrm{r}-\hat{\mathrm{\varphi }}) \end{gathered}$$

The positive vector, r in cylindrical co-ordinates is,

$\mathrm{r}=s\hat{s}+z\hat{z}$

Userole="math" localid="1657704225332" $=s\hat{s}+z\hat{z}$, role="math" localid="1657704251282" $\hat{s}\times \hat{\varphi }=\hat{z}$and $\hat{z}\times \hat{\varphi }=-\hat{s}$in the vector potential equation.

$$\begin{gathered} \mathrm{A}\left(\mathrm{r}\right)=-\frac{{\mathrm{\mu }}_0\mathrm{l}}{2\mathrm{\pi s}}(\mathrm{r}\times \hat{\mathrm{\varphi }}) \\ \mathrm{A}=\frac{-{\mathrm{\mu }}_0\mathrm{l}}{2\mathrm{\pi s}}\left[\mathrm{s}\left[\hat{\mathrm{s}}\times \hat{\mathrm{\varphi }}\right]+\mathrm{z}\left[\hat{\mathrm{z}}+\hat{\mathrm{\varphi }}\right]\right] \\ =\frac{{\mu }_{0}l}{2\mathrm{\pi s}}(z\hat{s}-s\hat{z}) \end{gathered}$$

Use Ampere’s law to find the magnetic field .

$$\begin{gathered} B\left(2\mathrm{\pi s}\right)={\mathrm{\mu }}_0{\mathrm{l}}_{\mathrm{encl}} \\ B\left(2\mathrm{\pi s}\right)={\mathrm{\mu }}_0{\mathrm{J\pi s}}^2 \\ \mathrm{B}=\frac{{\mu }_0\mathrm{J}}{2}s\hat{\mathrm{\varphi }} \end{gathered}$$

Use $\mathrm{B}=\frac{{\mu }_0\mathrm{J}}{2}s\hat{\mathrm{\varphi }}$to in the vector potential.

Determine thevector potential for uniform field.

Substitute $\frac{{\mu }_0\mathrm{J}}{2}s\hat{\mathrm{\varphi }}$for into equation (4).

$$\begin{gathered} A=-r\times {\int }_0^1\lambda \left(\frac{{\mu }_{0}J}{2}\right)\lambda s\hat{\varphi }d\lambda \\ =\frac{-{\mu }_{0}J}{6}s(r\times \hat{\varphi }) \\ =\frac{{\mu }_{0}Js}{6}(z\hat{s}-s\hat{z}) \end{gathered}$$

Therefore, the value of vector potential of an infinite straight wire carrying a steady current is $\mathrm{A}=\frac{{\mu }_{0}Js}{6}(z\hat{s}-s\hat{z})$.

Determine the divergence of vector potential A is,

Substitute $\frac{{\mu }_{0}Js}{6}(z\hat{s}-s\hat{z})$for into equation (5).

$$\begin{gathered} \nabla .\mathrm{A}=\frac{{\mu }_{0}J}{6}\left[\frac{1}{s}\frac{\partial }{\partial s}\left(s^{2}z\right)+\frac{\partial }{\partial z}(-s^2)\right] \\ =\frac{-{\mu }_{0}J}{6}\left(\frac{1}{2}\left(2sz\right)\right) \\ =\frac{{\mu }_{0}Jz}{3} \\ \ne 0 \end{gathered}$$

Therefore, the value of divergence of vector potential is, $\nabla .\mathrm{A}\ne 0$.