91Ó°ÊÓ

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

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

$\mathit{Ï\dots }\mathbf{=}\mathbf{∇}\mathbf{×}\mathit{A}$ …… (1)

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

Vector potential is given as,

$A=K\widehat{\mathrm{Ï•}}$

Write the expression for magnetic field.

$B=∇×A$

Write the expression for the$∇×A$in cylindrical coordinates.

$∇×A=\left(\frac{1}{s}\frac{∂A_z}{∂\mathrm{ϕ}}−\frac{∂A\mathrm{ϕ}}{∂z}\right)\widehat{s}+\left(\frac{∂A_s}{∂z}−\frac{∂A_z}{∂s}\right)\widehat{\mathrm{ϕ}}+\frac{1}{s}\left[\frac{∂}{∂s}\left(A_{\mathrm{ϕ}}\right)−\frac{∂A_s}{∂\mathrm{ϕ}}\right]\widehat{z}$

Substitute $A_s=0,\text{ }{A}_{\mathrm{ϕ}}=K,\text{ }{A}_z=0$

$\begin{array}{c}B=∇×A\\ =\left(\frac{1}{s}\left(0\right)−\frac{∂\left(K\right)}{∂z}\right)\widehat{s}+(0−0)\widehat{\mathrm{ϕ}}+\frac{1}{s}\left[\frac{∂}{∂s}\left(sK\right)−0\right]\widehat{z}\\ =0+0+\frac{K}{s}\widehat{z}\\ =\frac{K}{s}\widehat{z}\end{array}$

Write the expression for current density.

$J=\frac{1}{{\mathrm{μ}}_0}(∇×B)$

Write the expression for the $∇×B$in cylindrical coordinates.

$∇×B=\left(\frac{1}{s}\frac{∂B_z}{∂\mathrm{ϕ}}−\frac{∂B\mathrm{ϕ}}{∂z}\right)\widehat{s}+\left(\frac{∂B_s}{∂z}−\frac{∂B_z}{∂s}\right)\widehat{\mathrm{ϕ}}+\frac{1}{s}\left[\frac{∂}{∂s}\left(s_{\mathrm{ϕ}}\right)−\frac{∂B_s}{∂\mathrm{ϕ}}\right]\widehat{z}$

Substitute $B_s=0,\text{ }{B}_{\mathrm{ϕ}}=0,\text{ }{B}_z=\frac{k}{s}$

$\begin{array}{c}∇×B=\left(\frac{1}{s}\frac{∂}{∂\mathrm{ϕ}}\left(\frac{k}{s}\right)−0\right)\widehat{s}+\left(0−\frac{∂}{∂s}\left(\frac{k}{s}\right)\right)\widehat{\mathrm{ϕ}}+\frac{1}{s}\left[\frac{∂}{∂s}\left(0\right)−0\right]\widehat{z}\\ =0+\frac{k}{s^2}\widehat{\mathrm{ϕ}}+0\\ =\frac{k}{s^2}\widehat{\mathrm{ϕ}}\end{array}$

Then,

$∇×B=\frac{k}{s^2}\widehat{\mathrm{ϕ}}$

Then, the current density is,

$J=\frac{1}{{\mathrm{μ}}_0}(∇×B)$

Substitute$\frac{k}{s^2}\widehat{\mathrm{ϕ}}$for$∇×B$in above equation.

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

Therefore, the current density is $\frac{k}{{\mathrm{μ}}_0{s}^2}\widehat{\mathrm{ϕ}}$.