Just as $\mathbf{V}\mathbf{.}\mathbf{B}\mathbf{=}\mathbf{0}$allows us to express B as the curl of a vector potential $\left(B=\nabla \times A\right)$, so $\mathbf{\nabla }\mathbf{.}\mathbf{A}\mathbf{=}\mathbf{0}$permits us to write A itself as the curl of a "higher" potential:$\mathbf{A}\mathbf{=}\mathbf{\nabla }\mathbf{\times }\mathbf{W}$. (And this hierarchy can be extended ad infinitum.)

(a) Find the general formula for W (as an integral over B), which holds when$\mathbf{B}\mathbf{\to }\mathbf{0}$ at $\mathbf{\infty }$.

(b) Determine for the case of a uniform magnetic field B. [Hint: see Prob. 5.25.]

(c) Find inside and outside an infinite solenoid. [Hint: see Ex. 5.12.]

Consider vector potential

Consider the "higher" potential:.

Write the formula ofgeneral formula for W.

$\mathbf{A}\mathbf{=}\mathbf{\nabla }\mathbf{\times }\mathbf{W}$ …… (1)

Here, $\mathbf{W}$is divergence.

Write the formula of W for the case of a uniform magnetic field B

$\mathbf{\nabla }\mathbf{.}\mathbf{W}\mathbf{=}\mathbf{\alpha }\left[\left(r.B\right)\left(\nabla \times r\right)-r\times \nabla \left(r.B\right)\right]\mathbf{+}\mathbf{\beta }\left[r^2\left(\nabla \times B\right)-B\times \nabla \left(r^2\right)\right]$ …… (2)

Here,B is constant, r is radius of spherical shell.

Write the formula of inside and outside an infinite solenoid.

$\mathbf{\oint }\mathbf{W}\mathbf{.}\mathbf{dl}\mathbf{=}\mathbf{\int }\mathbf{A}\mathbf{.}\mathbf{da}$ …… (3)

Here, W should point parallel to the axis andA is curl of higher potential.

The expression for magnetic field is,

$\mathrm{B}=\nabla \times \mathrm{A}$

Here,$\nabla .\mathrm{B}=0,\nabla \times \mathrm{B}={\mathrm{\mu }}_0\mathrm{J}$ , .

Determine the expression for vector potential is,

role="math" localid="1657534802785" $\mathrm{A}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\int \frac{\mathrm{J}}{\mathrm{r}}\mathrm{d\tau }$

Determine thegeneral formula for W.

Substitute role="math" localid="1657534865743" $\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\int \frac{\mathrm{J}}{\mathrm{r}}\mathrm{d\tau }$for A into equation (1).

$\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\int \frac{\mathrm{J}}{\mathrm{r}}\mathrm{d\tau }=\nabla \times \mathrm{W}$

role="math" localid="1657535168827" $\frac{1}{4\mathrm{\pi }}\int \frac{{\mathrm{\mu }}_0\mathrm{J}}{\mathrm{r}}\mathrm{d\tau }=\nabla \times \mathrm{W}$

$\frac{1}{4\pi }\int \frac{\nabla \times B}{r}d\tau =\nabla \times W$

$\nabla \times \left(\frac{1}{4\mathrm{\pi }}\int \frac{\mathrm{B}}{\mathrm{r}}\mathrm{d\tau }\right)=\nabla \times \mathrm{W}$

role="math" localid="1657535512760" $\mathrm{W}=\frac{1}{4\mathrm{\pi }}\int \frac{\mathrm{B}}{\mathrm{r}}\mathrm{d\tau }$

Thus, it is proved that,$\mathrm{W}=\frac{1}{4\mathrm{\pi }}\int \frac{\mathrm{B}}{\mathrm{r}}\mathrm{d\tau }$

Determine the divergence of the following expression:

$$\begin{gathered} W=\alpha r\left(r.B\right)+\beta r^{2}B \\ \nabla .W=\alpha \left[\left(r.B\right)\left(\nabla .r\right)+r.\nabla \left(r.B\right)\right]+\beta \left[r^2\left(\nabla .B\right)+B.\nabla \left(r^2\right)\right].\nabla r \\ =\frac{\partial \times }{\partial \times }+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z} \\ =3 \end{gathered}$$

Thus, B is constant and then all derivatives and $\nabla \times \mathrm{r}=0$

Determine

$$\begin{gathered} \nabla \left(r.B\right)=\left(B.\nabla \right)r \\ =\left(B_{\times }\frac{\partial }{\partial \times }+B_y\frac{\partial }{\partial y}+Bz\frac{\partial }{\partial z}\right)\left(x\stackrel{⏜}{x}+y\stackrel{⏜}{y}+z\stackrel{⏜}{z}\right) \\ =B_{\times }\stackrel{⏜}{x}+B_y\stackrel{⏜}{y}+B_z\stackrel{⏜}{Z} \\ =B \end{gathered}$$

Determine

$$\begin{gathered} \nabla \left(r^2\right)=\left(\stackrel{⏜}{x}\frac{\partial }{\partial x}+\stackrel{⏜}{y}\frac{\partial }{\partial y}\stackrel{⏜}{z}\frac{\partial }{z}\right)\left(x^2+y^2+z^2\right) \\ =2x\stackrel{⏜}{x}+2y\stackrel{⏜}{y}+2z\stackrel{⏜}{z} \\ =2r \end{gathered}$$

Determine the divergence of W as follow:

$$\begin{gathered} \nabla .W=\alpha \left[3\left(r.B\right)+\left(r.B\right)\right]+\beta \left[0+2\left(r.B\right)\right] \\ =2\left(r.B\right)\left(2\alpha +\beta \right) \end{gathered}$$

Determine the inside and outside an infinite solenoid.

Substitute 0 for $\left(r.B\right)\left(\nabla \times r\right)$, B for $\nabla \left(r.B\right)$, 0 for $r^2$$\left(\nabla \times B\right)$and $2\left(B\times r\right)$for $B\times \nabla \left(r^2\right)$into equation (2).

$$\begin{gathered} \nabla .W=\alpha \left[0-\left(r\times B\right)\right]+\beta \left[0-2\left(B\times r\right)\right] \\ =-\left(r\times B\right)\left(\alpha -2\beta \right) \\ =-\frac{1}{2}\left(r\times B\right) \end{gathered}$$

Here, $$\begin{gathered} \alpha -2\beta =\frac{1}{2} \\ \alpha -2\left(-2\alpha \right)=\frac{1}{2} \\ 5\alpha =\frac{1}{2} \\ \alpha =\frac{1}{10} \\ \beta =2\alpha =-\frac{1}{5} \end{gathered}$$

Thus, the value of for the case of a uniform magnetic field B is

$\mathrm{W}=\frac{1}{10}\left[\mathrm{r}\left(\mathrm{r}.\mathrm{B}\right)-2{\mathrm{r}}^2\mathrm{B}\right]$.

Determine the value of W as follows:

$\nabla \times \mathrm{W}=\mathrm{A}$

This,$\int \left(\nabla \times \mathrm{W}\right).\mathrm{da}=\int \mathrm{A}.\mathrm{da}$

Draw the circuit diagram of infinite solenoid as follows:

Figure 1

Determine the inside and outside an infinite solenoid.

Substitute$-Wl$ for$\oint W.dl$and $-Wl=\frac{-{\mu }_{0}nl{s}^2}{4}\stackrel{⏜}{z}$for into equation (3).

$$\begin{gathered} -Wl={\int }_{}^1\left(\frac{{\mu }_{0}nl}{2}\right)lsds \\ -Wl=\frac{{\mu }_{0}nl}{2}\frac{s^{2}l}{2} \end{gathered}$$

Thus, the value of inside solenoid is .

Determine the value of outside $\left(s>R\right)$solenoid as follows:

Substitute role="math" localid="1657540879964" $-Wlfor\oint W.dl,\frac{{\mu }_{0}nl{R}^{2}l}{4}Aand\left(\frac{{\mu }_{0}nl}{2}\right)\frac{R^2}{s}ld\overline{s}$for , for and for into equation (3).

role="math" localid="1657540985303" $-Wl,\frac{{\mu }_{0}nl{R}^{2}l}{4}Aand\left(\frac{{\mu }_{0}nl}{2}\right)\frac{R^2}{s}ld\overline{s}$

$$\begin{gathered} W\mathit{=}\frac{{\mathit{\mu }}_{\mathit{0}}\mathit{n}\mathit{l}{\mathit{R}}^{\mathit{2}}\mathit{l}}{\mathit{4}}\mathit{+}\frac{{\mathit{\mu }}_{\mathit{0}}\mathit{n}\mathit{l}{\mathit{R}}^{\mathit{2}}\mathit{l}}{\mathit{2}}Ins\mathit{/}R \\ W\mathit{=}\frac{{\mathit{\mu }}_{\mathit{0}}\mathit{n}\mathit{l}{\mathit{R}}^{\mathit{2}}\mathit{l}}{\mathit{4}}\left[1+2In\left(\frac{S}{R}\right)\right]\hat{\mathit{z}} \end{gathered}$$

Thus, the value of inside and outside an infinite solenoid is$\frac{{\mu }_{\mathit{0}}nl{R}^{\mathit{2}}l}{\mathit{4}}\mathit{[}\mathit{1}\mathit{+}\mathit{2}\mathit{I}\mathit{n}\left(\frac{S}{R}\right)\mathit{]}\hat{z}$ .