1
$\begingroup$

My question:

If I have the Ampére-Maxwell law

$$\oint_\gamma \mathbf{B}\cdot d\mathbf{l}=\mu_0\left(I_{\text{enc.}}+\epsilon_0\frac{d\Phi_S(\mathbf{E})}{dt}\right) \tag 1$$ where $I_{\text{enc.}}$ is the current due to a difference in potential $\Delta V$ and the $$\epsilon_0\frac{d\Phi_S(\mathbf{E})}{dt} \tag 2$$ is the displacement current $I_S$, why when we are in the absence of sources have I only the quantity $(2)$?

If I have not a battery $\Delta V=0$ it's obvious that $I_{\text{enc.}}=0$. But the $(2)\neq 0$ why have I always a very small current near to zero that always circulates in a circuit or for other reasons?


Translation of the text into Italian language of a Physics textbook: in the particular case where there are neither charges nor currents in space, Maxwell's equations are simplified significantly.

enter image description here

$\endgroup$
3
  • 1
    $\begingroup$ The enclosed current is a "source" in this equation, so if we're talking about the "absence of sources" then by definition the current must be zero. On the other hand I'd also consider the displacement current to be a source in this equation, so I'm not clear why that term isn't also taken as zero. $\endgroup$
    – The Photon
    Commented Jun 8, 2020 at 21:38
  • $\begingroup$ @ThePhoton Hi, into my book of high school there is not an explanation. The last equation of Maxwell is written without sources as: $\oint_\gamma \mathbf{B}\cdot d\mathbf{l}=\mu_0\epsilon_0\frac{d\Phi_S(\mathbf{E})}{dt}$. $\endgroup$
    – Sebastiano
    Commented Jun 8, 2020 at 21:40
  • $\begingroup$ @ThePhoton I have edited my question. $\endgroup$
    – Sebastiano
    Commented Jun 8, 2020 at 21:51

2 Answers 2

1
$\begingroup$

The displacement current only exists when in your region of interest there exists an electric flux that varies over time. And if that is the case, then there’s a corresponding magnetic field. If there is no such flux, then $I_s=0$.

These fields can exist in the absence of charges in the form of electromagnetic waves. This is because Maxwell’s equations allow for EM waves to exist in free space.

$\endgroup$
2
  • $\begingroup$ For example when can I know "that there exists an electric flux that varies over time"? $\endgroup$
    – Sebastiano
    Commented Jun 8, 2020 at 21:42
  • $\begingroup$ EM waves will always be there. In the absence of these waves, in the context of circuits, you’ll have a time dependent flux if you have charge buildup somewhere, or if you have a capacitor or inductor. $\endgroup$ Commented Jun 8, 2020 at 21:51
1
$\begingroup$

Suppose you choose an area $A_c$ passing between the plates of a capacitor, and do the integration around the closed boundary of that area.

There is no current flowing through $A_c$, so no sources in the region of the integral, but there is a magnetic field around the boundary. Exactly as for an area $A_w$ with the same boundary, but passing a little before the capacitor, where there is a current $I$ through it.

$\endgroup$
2
  • $\begingroup$ Thank you very much for your reply. Can you, please, put a figure to understand better your response? Why is there a magnetic field around the boundary? $\endgroup$
    – Sebastiano
    Commented Jun 9, 2020 at 19:44
  • 1
    $\begingroup$ According to the Ampére-Maxwell Law, it is not required for the surface to be a plane. For the same boundary it is possible to have a surface cutting the wire, and another passing between the plates of the capacitor. The resulting magnetic field must be the same. $\endgroup$ Commented Jun 9, 2020 at 22:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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