0
$\begingroup$

Say I have an oscillating sinusoidal current flowing on the z-axis (say $I_0\sin{\omega t}$), in a straight line conductor. This current produces an alternating magnetic field in the azimuthal direction, which produces an changing electric field in the z direction, which creates an EM wave in the radial direction. How would one formally derive the expression for the induced magnetic and electric fields?

My initial response was that the magnetic field expression follows from the Ampere's Circuital law, i.e.

$$ B = \frac{\mu _o I_o \sin{\omega t}}{2\pi r} $$ in the azimuthal direction. However, this means ignoring the $\epsilon _o\frac{\partial E}{\partial t}$ part from the Maxwell's extension that is responsible for EM waves in the first place.

I tried to derive the wave equation from Maxwell's laws, but since it is not in free space and there is non-zero current density, I got stuck at

$$ \nabla ^2\overrightarrow{E} = \mu_o\epsilon_0 \frac{\partial ^2 \overrightarrow{E}}{\partial t^2} + \mu_0 \frac{\partial\overrightarrow{J}}{\partial t}$$

Any help?

$\endgroup$
1
  • $\begingroup$ Isn't this exactly done in every textbook or treatment of electrodynamics? $\endgroup$
    – ProfRob
    May 22 '20 at 8:03
0
$\begingroup$

$\mathbf{J}$ is non-zero on the wire but zero everywhere else. So you have your wave equation.

To compute the fields, you can get them from retarded potentials: $$\mathbf{A}(\mathbf{r},t)=\frac{\mu_0}{4\pi}\int \frac{\mathbf{J}(\mathbf{r'},t_r)}{|\mathbf{r}-\mathbf{r'}|}d\mathbf{r'}$$ $$V(\mathbf{r},t)=\frac{1}{4\pi\epsilon_0}\int \frac{\rho(\mathbf{r'},t_r)}{|\mathbf{r}-\mathbf{r'}|}d\mathbf{r'}$$ where $t_r\equiv t-|\mathbf{r}-\mathbf{r'}|/c$ is the retarded time. Finally, $$\mathbf{B}=\nabla\times\mathbf{A}$$ $$\mathbf{E}=-\nabla V-\frac{\partial \mathbf{A}}{\partial t}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.