91影视

Vector potential is similar to scalar potential whose gradient gives the vector field.

If $蠀$is vector field, then the vector potential of vector field$\left(\mathrm{A}\right)$ . Write the expression for the vector field.

$\mathit{蠀}\mathbf{=}\mathbf{鈭�}\mathbf{脳}\mathit{A}$ 鈥︹�� (1)

It is also defined as curl of vector$A$ is numerically equal to the magnetic field.

Vector potential is given as,

$A=K\widehat{蠒}$

Write the expression for magnetic field.

$B=鈭�脳A$

Write the expression for the$鈭�脳A$in cylindrical coordinates.

$鈭�脳A=\left(\frac{1}{s}\frac{鈭�A_z}{鈭�蠒}鈭�\frac{鈭�A蠒}{鈭�z}\right)\widehat{s}+\left(\frac{鈭�A_s}{鈭�z}鈭�\frac{鈭�A_z}{鈭�s}\right)\widehat{蠒}+\frac{1}{s}\left[\frac{鈭�}{鈭�s}\left(A_{蠒}\right)鈭�\frac{鈭�A_s}{鈭�蠒}\right]\widehat{z}$

Substitute $A_s=0,\text{鈥�}{A}_{蠒}=K,\text{鈥�}{A}_z=0$

$\begin{array}{c}B=鈭�脳A\\ =\left(\frac{1}{s}\left(0\right)鈭�\frac{鈭�\left(K\right)}{鈭�z}\right)\widehat{s}+(0鈭�0)\widehat{蠒}+\frac{1}{s}\left[\frac{鈭�}{鈭�s}\left(sK\right)鈭�0\right]\widehat{z}\\ =0+0+\frac{K}{s}\widehat{z}\\ =\frac{K}{s}\widehat{z}\end{array}$

Write the expression for current density.

$J=\frac{1}{{渭}_0}(鈭�脳B)$

Write the expression for the $鈭�脳B$in cylindrical coordinates.

$鈭�脳B=\left(\frac{1}{s}\frac{鈭�B_z}{鈭�蠒}鈭�\frac{鈭�B蠒}{鈭�z}\right)\widehat{s}+\left(\frac{鈭�B_s}{鈭�z}鈭�\frac{鈭�B_z}{鈭�s}\right)\widehat{蠒}+\frac{1}{s}\left[\frac{鈭�}{鈭�s}\left(s_{蠒}\right)鈭�\frac{鈭�B_s}{鈭�蠒}\right]\widehat{z}$

Substitute $B_s=0,\text{鈥�}{B}_{蠒}=0,\text{鈥�}{B}_z=\frac{k}{s}$

$\begin{array}{c}鈭�脳B=\left(\frac{1}{s}\frac{鈭�}{鈭�蠒}\left(\frac{k}{s}\right)鈭�0\right)\widehat{s}+\left(0鈭�\frac{鈭�}{鈭�s}\left(\frac{k}{s}\right)\right)\widehat{蠒}+\frac{1}{s}\left[\frac{鈭�}{鈭�s}\left(0\right)鈭�0\right]\widehat{z}\\ =0+\frac{k}{s^2}\widehat{蠒}+0\\ =\frac{k}{s^2}\widehat{蠒}\end{array}$

Then,

$鈭�脳B=\frac{k}{s^2}\widehat{蠒}$

Then, the current density is,

$J=\frac{1}{{渭}_0}(鈭�脳B)$

Substitute$\frac{k}{s^2}\widehat{蠒}$for$鈭�脳B$in above equation.

$\begin{array}{c}J=\frac{1}{{渭}_0}\left(\frac{k}{s^2}\widehat{蠒}\right)\\ =\frac{k}{{渭}_0{s}^2}\widehat{蠒}\end{array}$

Therefore, the current density is $\frac{k}{{渭}_0{s}^2}\widehat{蠒}$.