91影视

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

Consider the pair of equations are:

(i) $\mathrm{v}\left(\mathrm{r}\right)=-\mathrm{r}脳{鈭�}_0^1\mathrm{E}\left(\mathrm{位谤}\right)\mathrm{诲位}$

(ii) $\mathrm{A}\left(\mathrm{r}\right)=-\mathrm{r}脳{鈭�}_0^1\mathrm{位叠}\left(\mathrm{位谤}\right)\mathrm{诲位}$

Write the formula of vector potential.

$\mathbf{A}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}{\mathbf{鈭�}}_{\mathbf{0}}^{\mathbf{1}}\mathbf{(}\mathbf{B}\mathbf{脳}\mathit{d}\mathit{l}\mathbf{)}$ 鈥︹�� (1)

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

Write the formula of magnetic field due to wire is,

$\mathbf{B}\mathbf{=}\frac{{\mathbf{渭}}_{\mathbf{0}}\mathbf{l}}{\mathbf{2}\mathbf{蟺蝉}}\widehat{\mathbf{蠒}}$ 鈥︹�� (2)

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

Write the formula of vector potential for uniform field.

$\mathbf{A}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{r}\mathbf{脳}{\mathbf{鈭�}}_{\mathbf{0}}^{\mathbf{1}}\mathbf{位叠}\mathbf{\left(}\mathbf{位谤}\mathbf{\right)}\mathbf{诲位}$ 鈥︹�� (3)

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

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

$\mathbf{A}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{r}\mathbf{脳}{\mathbf{鈭�}}_{\mathbf{0}}^{\mathbf{1}}\mathbf{位叠}\mathbf{\left(}\mathbf{位谤}\mathbf{\right)}\mathbf{诲位}$ 鈥︹�� (4)

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

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

$\mathbf{鈭�}\mathbf{鈥�}\mathbf{A}$ 鈥︹�� (5)

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

Determine thevector potential.

$\mathrm{A}\left(\mathrm{r}\right)=\mathrm{B}脳{鈭�}_0^{1}dl$

From the problem 5.25, the vector potential is,

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

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,

$\mathrm{B}=\frac{{\mathrm{渭}}_0\mathrm{l}}{2\mathrm{蟺蝉}}\widehat{\mathrm{蠒}}$

To solve the integral.

$$\begin{gathered} 鈭�\mathrm{B}脳\mathrm{dl}=\left(\frac{{渭}_{0}l}{2\mathrm{蟺}a}\hat{s}-\frac{{渭}_{0}l}{2\mathrm{蟺}b}\hat{s}\right)w \\ =\frac{{渭}_{0}lw}{2\mathrm{蟺}}\left(\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}\right)\hat{s} \\ 鈮�0 \end{gathered}$$

Thus $鈭�\mathrm{B}脳\mathrm{dl}鈮�0$is not independent of path.

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

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

$$\begin{gathered} \mathrm{A}\left(\mathrm{r}\right)=-\mathrm{r}脳\mathrm{B}{鈭�}_0^1\mathrm{位诲位} \\ =-\mathrm{r}脳\mathrm{B}{\left[\frac{{\mathrm{位}}^2}{2}\right]}_0^1 \\ =-\mathrm{r}脳\mathrm{B}\left(\frac{1}{2}-0\right) \\ =-\frac{1}{2}(\mathrm{r}脳\mathrm{B}) \end{gathered}$$

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

Use the expression of magnetic field, $\mathrm{B}=\frac{{渭}_{0}l}{2\mathrm{蟺蝉}}\widehat{\mathrm{蠒}}$to find $\mathrm{B}\left(\mathrm{位谤}\right)$.

$\mathrm{B}=\left(\mathrm{位谤}\right)\frac{{渭}_{0}l}{2\mathrm{蟺蝉}}\widehat{\mathrm{蠒}}$

The vector potential is,

role="math" localid="1657703468579" $$\begin{gathered} \mathrm{A}\left(\mathrm{r}\right)=-r脳{鈭�}_0^1\mathrm{位搁}\left(\mathrm{位谤}\right)d\mathrm{位} \\ =\frac{-{渭}_{01}l}{2\mathrm{蟺蝉}}(\mathrm{r}脳\widehat{\mathrm{蠒}}){鈭�}_0^1位\frac{1}{位}dx \\ =\frac{-{渭}_{0}l}{2\mathrm{蟺蝉}}(\mathrm{r}脳\widehat{\mathrm{蠒}}){\left[x\right]}_0^1 \end{gathered}$$

Solve further as,

$$\begin{gathered} \mathrm{A}\left(\mathrm{r}\right)=-\frac{{渭}_{0}l}{2\mathrm{蟺蝉}}(\mathrm{r}-\widehat{\mathrm{蠒}})\left[1-0\right] \\ =-\frac{{渭}_{o}l}{2\mathrm{蟺蝉}}(\mathrm{r}-\widehat{\mathrm{蠒}}) \end{gathered}$$

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" $\hat{s}脳\widehat{蠒}=\hat{z}$and $\hat{z}脳\widehat{蠒}=-\hat{s}$in the vector potential equation.

$$\begin{gathered} \mathrm{A}\left(\mathrm{r}\right)=-\frac{{\mathrm{渭}}_0\mathrm{l}}{2\mathrm{蟺蝉}}(\mathrm{r}脳\widehat{\mathrm{蠒}}) \\ \mathrm{A}=\frac{-{\mathrm{渭}}_0\mathrm{l}}{2\mathrm{蟺蝉}}\left[\mathrm{s}\left[\hat{\mathrm{s}}脳\widehat{\mathrm{蠒}}\right]+\mathrm{z}\left[\hat{\mathrm{z}}+\widehat{\mathrm{蠒}}\right]\right] \\ =\frac{{渭}_{0}l}{2\mathrm{蟺蝉}}(z\hat{s}-s\hat{z}) \end{gathered}$$

Use Ampere鈥檚 law to find the magnetic field .

$$\begin{gathered} B\left(2\mathrm{蟺蝉}\right)={\mathrm{渭}}_0{\mathrm{l}}_{\mathrm{encl}} \\ B\left(2\mathrm{蟺蝉}\right)={\mathrm{渭}}_0{\mathrm{J蟺蝉}}^2 \\ \mathrm{B}=\frac{{渭}_0\mathrm{J}}{2}s\widehat{\mathrm{蠒}} \end{gathered}$$

Use $\mathrm{B}=\frac{{渭}_0\mathrm{J}}{2}s\widehat{\mathrm{蠒}}$to in the vector potential.

Determine thevector potential for uniform field.

Substitute $\frac{{渭}_0\mathrm{J}}{2}s\widehat{\mathrm{蠒}}$for into equation (4).

$$\begin{gathered} A=-r脳{鈭�}_0^1位\left(\frac{{渭}_{0}J}{2}\right)位s\widehat{蠒}d位 \\ =\frac{-{渭}_{0}J}{6}s(r脳\widehat{蠒}) \\ =\frac{{渭}_{0}Js}{6}(z\hat{s}-s\hat{z}) \end{gathered}$$

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

Determine the divergence of vector potential A is,

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

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

Therefore, the value of divergence of vector potential is, $鈭�.\mathrm{A}鈮�0$.