91Ó°ÊÓ

Consider vector potential

Consider the "higher" potential:.

Write the formula ofgeneral formula for W.

$\mathbf{A}\mathbf{=}\mathbf{∇}\mathbf{×}\mathbf{W}$ …… (1)

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

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

$\mathbf{∇}\mathbf{.}\mathbf{W}\mathbf{=}\mathbf{Î\pm }\left[\left(r.B\right)\left(∇×r\right)-r×∇\left(r.B\right)\right]\mathbf{+}\mathbf{β}\left[r^2\left(∇×B\right)-B×∇\left(r^2\right)\right]$ …… (2)

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

Write the formula of inside and outside an infinite solenoid.

$\mathbf{âˆ\circledR }\mathbf{W}\mathbf{.}\mathbf{dl}\mathbf{=}\mathbf{∫}\mathbf{A}\mathbf{.}\mathbf{da}$ …… (3)

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

The expression for magnetic field is,

$\mathrm{B}=∇×\mathrm{A}$

Here,$∇.\mathrm{B}=0,∇×\mathrm{B}={\mathrm{μ}}_0\mathrm{J}$ , .

Determine the expression for vector potential is,

role="math" localid="1657534802785" $\mathrm{A}=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}∫\frac{\mathrm{J}}{\mathrm{r}}\mathrm{»åÏ„}$

Determine thegeneral formula for W.

Substitute role="math" localid="1657534865743" $\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}∫\frac{\mathrm{J}}{\mathrm{r}}\mathrm{»åÏ„}$for A into equation (1).

$\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}∫\frac{\mathrm{J}}{\mathrm{r}}\mathrm{»åÏ„}=∇×\mathrm{W}$

role="math" localid="1657535168827" $\frac{1}{4\mathrm{Ï€}}∫\frac{{\mathrm{μ}}_0\mathrm{J}}{\mathrm{r}}\mathrm{»åÏ„}=∇×\mathrm{W}$

$\frac{1}{4\mathrm{π}}∫\frac{∇×B}{r}d\mathrm{τ}=∇×W$

$∇×\left(\frac{1}{4\mathrm{Ï€}}∫\frac{\mathrm{B}}{\mathrm{r}}\mathrm{»åÏ„}\right)=∇×\mathrm{W}$

role="math" localid="1657535512760" $\mathrm{W}=\frac{1}{4\mathrm{Ï€}}∫\frac{\mathrm{B}}{\mathrm{r}}\mathrm{»åÏ„}$

Thus, it is proved that,$\mathrm{W}=\frac{1}{4\mathrm{Ï€}}∫\frac{\mathrm{B}}{\mathrm{r}}\mathrm{»åÏ„}$

Determine the divergence of the following expression:

$$\begin{gathered} W=\mathrm{Î\pm }r\left(r.B\right)+\mathrm{β}{r}^{2}B \\ ∇.W=\mathrm{Î\pm }\left[\left(r.B\right)\left(∇.r\right)+r.∇\left(r.B\right)\right]+\mathrm{β}\left[r^2\left(∇.B\right)+B.∇\left(r^2\right)\right].∇r \\ =\frac{∂×}{∂×}+\frac{∂y}{∂y}+\frac{∂z}{∂z} \\ =3 \end{gathered}$$

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

Determine

$$\begin{gathered} ∇\left(r.B\right)=\left(B.∇\right)r \\ =\left(B_{×}\frac{∂}{∂×}+B_y\frac{∂}{∂y}+Bz\frac{∂}{∂z}\right)\left(x\stackrel{Áåœ}{x}+y\stackrel{Áåœ}{y}+z\stackrel{Áåœ}{z}\right) \\ =B_{×}\stackrel{Áåœ}{x}+B_y\stackrel{Áåœ}{y}+B_z\stackrel{Áåœ}{Z} \\ =B \end{gathered}$$

Determine

$$\begin{gathered} ∇\left(r^2\right)=\left(\stackrel{Áåœ}{x}\frac{∂}{∂x}+\stackrel{Áåœ}{y}\frac{∂}{∂y}\stackrel{Áåœ}{z}\frac{∂}{z}\right)\left(x^2+y^2+z^2\right) \\ =2x\stackrel{Áåœ}{x}+2y\stackrel{Áåœ}{y}+2z\stackrel{Áåœ}{z} \\ =2r \end{gathered}$$

Determine the divergence of W as follow:

$$\begin{gathered} ∇.W=\mathrm{Î\pm }\left[3\left(r.B\right)+\left(r.B\right)\right]+\mathrm{β}\left[0+2\left(r.B\right)\right] \\ =2\left(r.B\right)\left(2\mathrm{Î\pm }+\mathrm{β}\right) \end{gathered}$$

Determine the inside and outside an infinite solenoid.

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

$$\begin{gathered} ∇.W=\mathrm{Î\pm }\left[0-\left(r×B\right)\right]+\mathrm{β}\left[0-2\left(B×r\right)\right] \\ =-\left(r×B\right)\left(\mathrm{Î\pm }-2\mathrm{β}\right) \\ =-\frac{1}{2}\left(r×B\right) \end{gathered}$$

Here, $$\begin{gathered} \mathrm{Î\pm }-2\mathrm{β}=\frac{1}{2} \\ \mathrm{Î\pm }-2\left(-2\mathrm{Î\pm }\right)=\frac{1}{2} \\ 5\mathrm{Î\pm }=\frac{1}{2} \\ \mathrm{Î\pm }=\frac{1}{10} \\ \mathrm{β}=2\mathrm{Î\pm }=-\frac{1}{5} \end{gathered}$$

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:

$∇×\mathrm{W}=\mathrm{A}$

This,$∫\left(∇×\mathrm{W}\right).\mathrm{da}=∫\mathrm{A}.\mathrm{da}$

Draw the circuit diagram of infinite solenoid as follows:

Figure 1

Determine the inside and outside an infinite solenoid.

Substitute$-Wl$ for$âˆ\circledR W.dl$and $-Wl=\frac{-{\mathrm{μ}}_{0}nl{s}^2}{4}\stackrel{Áåœ}{z}$for into equation (3).

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

Thus, the value of inside solenoid is .

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

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

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

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

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