Question

What current density would produce the vector potential, $\mathbf{A}\mathbf{=}\mathbf{k}\widehat{\mathbf{\varphi }}$(where $\mathbf{k}$is a constant), in cylindrical coordinates?

Short answer

The current density is $\frac{k}{{\mu }_0{s}^2}\widehat{\varphi }$.

Step-by-step solution

Define function

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

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

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

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

Determine magnetic field

Vector potential is given as,

$A=K\widehat{\varphi }$

Write the expression for magnetic field.

$B=\nabla \times A$

Write the expression for the$\nabla \times A$in cylindrical coordinates.

$\nabla \times A=\left(\frac{1}{s}\frac{\partial A_z}{\partial \varphi }-\frac{\partial A\varphi }{\partial z}\right)\widehat{s}+\left(\frac{\partial A_s}{\partial z}-\frac{\partial A_z}{\partial s}\right)\widehat{\varphi }+\frac{1}{s}\left[\frac{\partial }{\partial s}\left(A_{\varphi }\right)-\frac{\partial A_s}{\partial \varphi }\right]\widehat{z}$

Substitute $A_s=0,A_{\varphi }=K,A_z=0$

$\begin{array}{c}B=\nabla \times A\\ =\left(\frac{1}{s}\left(0\right)-\frac{\partial \left(K\right)}{\partial z}\right)\widehat{s}+(0-0)\widehat{\varphi }+\frac{1}{s}\left[\frac{\partial }{\partial s}\left(sK\right)-0\right]\widehat{z}\\ =0+0+\frac{K}{s}\widehat{z}\\ =\frac{K}{s}\widehat{z}\end{array}$

Write the expression for current density.

$J=\frac{1}{{\mu }_0}(\nabla \times B)$

Write the expression for the $\nabla \times B$in cylindrical coordinates.

$\nabla \times B=\left(\frac{1}{s}\frac{\partial B_z}{\partial \varphi }-\frac{\partial B\varphi }{\partial z}\right)\widehat{s}+\left(\frac{\partial B_s}{\partial z}-\frac{\partial B_z}{\partial s}\right)\widehat{\varphi }+\frac{1}{s}\left[\frac{\partial }{\partial s}\left(s_{\varphi }\right)-\frac{\partial B_s}{\partial \varphi }\right]\widehat{z}$

Substitute $B_s=0,B_{\varphi }=0,B_z=\frac{k}{s}$

$\begin{array}{c}\nabla \times B=\left(\frac{1}{s}\frac{\partial }{\partial \varphi }\left(\frac{k}{s}\right)-0\right)\widehat{s}+\left(0-\frac{\partial }{\partial s}\left(\frac{k}{s}\right)\right)\widehat{\varphi }+\frac{1}{s}\left[\frac{\partial }{\partial s}\left(0\right)-0\right]\widehat{z}\\ =0+\frac{k}{s^2}\widehat{\varphi }+0\\ =\frac{k}{s^2}\widehat{\varphi }\end{array}$

Then,

$\nabla \times B=\frac{k}{s^2}\widehat{\varphi }$

Then, the current density is,

$J=\frac{1}{{\mu }_0}(\nabla \times B)$

Substitute$\frac{k}{s^2}\widehat{\varphi }$for$\nabla \times B$in above equation.

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

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