Can the Last Maxwell's Equation Explain Polarization of a Wire's Insulator?

[BlackMelon]

Hi there!

Recently, I have been reading about polarization of a wire's insulator. First of all, I want to see a connection between the last Maxwell's Equation:
$$\nabla\times\\B\ =\mu_0\ J\ +\mu_0\ \epsilon_0\ \frac{\partial E}{\partial t}$$
and the polarization.

So I draw a simple cartoon below. Here, there is a wire carrying current J.
The inner cylinder is a conductor. The outer shell is an insulator.
Somehow the insulator get polarized (may be by external charges or whatever) such that + is on the red side - is on the black side.
The red and the black keeps swapping themselves back and forth, so does their electric field.

After I summed J and dE/dt and apply the Maxwell's last equation, I found the magnetic field B like a disc, tilting back and forth.

I would like to know if the scenario I made is correct?

If so, could you please suggest any scenario else to explain this equation?

Best
BlackMelon

 

[osilmag]

The B- Field around a current carrying wire is Ampere's Law and it should be a regular circle. The cross product means it is perpendicular.

The only way polarity flips is if the current is AC. Maxwell's version adds the electric field to it.

 

[BlackMelon]

The B- Field around a current carrying wire is Ampere's Law and it should be a regular circle. The cross product means it is perpendicular. View attachment 333794

The only way polarity flips is if the current is AC. Maxwell's version adds the electric field to it.

Curl (V1) = V2 means that V1 is perpendicular to V2, right?
From the Maxwell's Equation Curl(B) = u0J +u0e0(dE/dt)
means B is perpendicular to the summation of u0J and u0e0E (since dE/dt has either the same or reversed direction of E)

In my case, it is not the current that flips its polarity (B field is always counter clockwise) , but the electric field itself. And the electric field comes from the + charges and - charges inside the wire's insulator. Normally, they mixed together and create no total electric field. But this time, + grouped themselves together on one side, - did so on the other side. These two groups keep swapping themselves back and forth. The plane of B field tilts up and down. I just wonder if this case exist in real world situation?

 

[jasonRF]

Curl (V1) = V2 means that V1 is perpendicular to V2, right?

Wrong. Consider $\mathbf{V}_1 = y\mathbf{\hat{x}} + x\mathbf{\hat{y}} + y\mathbf{\hat{z}}$.

Hi there!

Recently, I have been reading about polarization of a wire's insulator. First of all, I want to see a connection between the last Maxwell's Equation:
$$\nabla\times\\B\ =\mu_0\ J\ +\mu_0\ \epsilon_0\ \frac{\partial E}{\partial t}$$
and the polarization.

What reading have you been doing? Because I haven't seen the polarization field in any of your equations. If you want to use that form of the equation, the polarization field must be in your current term so that $\mathbf{J} = \mathbf{J}_{f} + \frac{\partial \mathbf{P}}{\partial t}$. Here $\mathbf{J}_{f}$ is the free current that is only in the wire, and $\frac{\partial \mathbf{P}}{\partial t}$ is the polarization current that is only in the insulator. For a linear insulator of course $\mathbf{P} = \chi \epsilon_0 \mathbf{E}$ so that the relative permitivity is $\epsilon_r = 1+\chi$. I usually find it easier to work with $\mathbf{D} = \epsilon_0 \mathbf{E} + \mathbf{P} = \epsilon_r \epsilon_0 \mathbf{E}$. If you haven't seen this stuff before then you need to do more reading.

EDIT: if you want to understand the bound and free charge densities, then you will want to look at things like $-\mathbf{\nabla \cdot P} = \rho_{b}$ which is the volumetric bound charge density, $\mathbf{\hat{n} \cdot P} = \sigma_b$ is the bound surface charge density, $\mathbf{\hat{n} \cdot D} = \sigma_f$ is the free surface charge density, and $\mathbf{\nabla \cdot D} = \rho_{f}$ is the volumetric free charge density.

Note that your problem has azimuthal symmetry, but your sketches of what you think is going on with the polarization do not have that symmetry. That is a clue that you are not doing something right.

jason