1

Suppose that I have a perfect electrical conductor ($B=0$ inside conductor) in free space with a known magnetic field $\mathbf{B_s}$ outside of it, and no electric field. If I transform to a frame of reference moving non-relativistically with velocity $\mathbf{v_{s'}}$, I obtain an electric field $\mathbf{E_{s'}} = \mathbf{v_{s'}} \times \mathbf{B_s}$ outside the conductor. The boundary conditions imply that there is a surface charge density $\sigma$ in this frame on the conductor.

In general, \begin{equation} \sigma = \epsilon \mathbf{E}.\mathbf{\hat{n}}|_{\rm surface} \tag{1} \end{equation} where $\mathbf{\hat{n}}$ is the normal unit vector to the surface of the conductor.

My question(s) :

  1. Whether the electric field vector $\mathbf{E}$ in $Eq. (1)$ is just the vector $\mathbf{E_{s'}}$ or the composite (or total) electric field vector $\mathbf{E_{s'}} + \mathbf{E_{c'}}$, where $\mathbf{E_{c'}}$ is the secondary field due to the surface charges?

  2. In both cases, how to calculate the total electric field outside the conductor and the surface charge, since I do not know $\mathbf{E_{c'}}$ and $\sigma$ apriori?

OmG
  • 33
  • Related : Why don't stationary charge feel force from a current carrying wire?. – Frobenius Mar 11 '22 at 15:12
  • @Frobenius In the aforementioned related answer, there is a current in the stationary frame, which in the moving frame affects neutrality. However here, the conductor will still be charge neutral in moving frame. I just want to find the surface charge distribution and the external total electric field. Consider my conductor to be a sphere or a finite cylinder, if that supplements the explanation. But please help – OmG Mar 11 '22 at 18:03
  • You must mean (E=0 inside), not B. – Jerrold Franklin Mar 11 '22 at 19:22
  • @JerroldFranklin Both $E=0$ and $B=0$ inside the conductor. As user200143 correctly points out in point $2.$ of his answer below, a surface current will expel the magnetic field from inside the conductor. – OmG Mar 11 '22 at 19:30
  • What could be causing the surface current and keeping it going? I think the B=0 inside was just a misprint that now has a life. – Jerrold Franklin Mar 13 '22 at 12:07
  • @JerroldFranklin That would have been funny! But if you just utilize the EM interface (boundary) conditions at the conductor surface, it will become quite clear why there should be a surface current. See for example Griffiths' Introduction to Electrodynamics Section 3.6 (page 346) or see this link https://web.mit.edu/6.013_book/www/chapter8/8.4.html – OmG Mar 13 '22 at 18:15
  • MIT says, "in a time-varying magnetic field". Yes, that would change everything. – Jerrold Franklin Mar 14 '22 at 17:53
  • In a time varying B field "and no electric field" couldn't happen. . – Jerrold Franklin Mar 15 '22 at 19:34
  • Even with magnetostatic boundary conditions, we will get a steady surface current. The MIT page is referring to MQS fields, which is yet another approximation (you could look at that in detail). But do kindly refer to Griffiths or any other introductory text on EM first to clarify your doubt. – OmG Mar 15 '22 at 19:39

3 Answers3

0

Your problem is just the standard problem of a conducting sphere which is solved n every EM textbook for a uniform E field. If the original B field is not uniform, a Legendre polynomial, spherical harmonic expansion is necessary. In any event the total field is the one that counts.

Jerrold Franklin
  • 1,748
-1

  1. The surface charge density is from Maxwell's equations, $\vec \nabla \cdot \vec E = \rho/\epsilon_0$, so the electric field discontinuity is the total field, which is your $E'_s$.

  2. Since you have the fields in the rest frame, you Lorentz transform to your new frame in the standard way. The discontinuity in the transformed electric field is coming from the discontinuity in the original $\vec B$ field. That is, you say $\vec B = 0 $ in the conductor. Therefore the normal component of $\vec B$ is zero outside, but the tangential component is not. Since it is zero inside, the tangential component is discontinuous and there is a surface current which you calculate in the usual way from the curl of $\vec B$ Maxwell equation. If you Lorentz transform this surface current you will get the surface charge density.

user200143
  • 2,470
  • Thank you. I understood the 2nd point. In the 1st point, do you mean that there will be no extra field outside because of the surface charges, and that the total electric field outside would just be $\mathbf{E_{s'}}$? For example, my $\mathbf{E_{s'}}$ does not satisfy the tangential interface condition, i.e. $\hat{n} \times \mathbf{E_{s'}}$ is not zero. – OmG Mar 11 '22 at 19:01
  • The total field is NOT your E′s. – Jerrold Franklin Mar 11 '22 at 19:34
-1

Since $\vec{B}_\text{above} \neq 0$ above the surface and $\vec{B}_\text{below} = 0$ below, there must be a 2-D current density $\vec{K} = \frac{1}{\mu_0} \hat{n} \times \vec{B}_\text{above}$ on the surface. Note also that we must have $\hat{n} \cdot \vec{B}_\text{above} = 0$ (unless Cabrera was right all along.)

We can write $\vec{K}$ as a 3-D current density via throwing in a delta function: $$ \vec{J} = \vec{K} \delta( \vec{n} \cdot \vec{r}) $$ These will form the components of a four-current density $J^\mu = (0, \vec{J})$. By applying the Lorentz transformations to this four-vector, we can find the components of the four-current in another frame. In general, the $t$-component of the transformed four-current ${J'}^\mu$ will be non-zero, and it will correspond to the charge density $\rho$ in this frame; integrating the result over the normal to the surface (eliminating the delta function) will then yield the surface charge.

Michael Seifert
  • 48,772