Suppose I have a long solenoid with $n$ turns per unit length. The electric current through the solenoid is changing and I am interested in finding the magnetic field $(B)$ inside. So, we choose an Amperian path as in this figure and equate $\oint B.dl= \mu_0\times nli$ $\Rightarrow B = \mu_0\times ni$ and then differentiating wrt. time and using $\oint E_{induced}.dl= -\frac{d\phi}{dt}$ we get the E field at a distance $r$ from the centre of the loop equals $\dfrac{\mu_0\times nr}{2}\times \dfrac{di}{dt}$. And in the very first step of choosing Amperian path, we have an extra term $\mu_0\epsilon_0 \dfrac{d\phi_E}{dt}$ which depends upon the second time derivative of current. So, why is this term ignored in the derivation. enter image description here enter image description here


1 Answer 1


The extra term you are considering is called displacement current and it can be considered when an eletric field E is not costant in time. Ampere law can be applied to describe the system only when the current is costant in time (so yes its natural having an extra term, now it comes the explanation).

That therm comes from the fact that in the vacumm $i$ needs to be seen as $i_{tot} = i + \epsilon_o \frac{d \phi_E}{dt}$ (for more in depht view you can search "displacement current" on wikipedia), which acconts for the current not being costant in time in the ampere's law (to be precise all the discussion is made around the 4th maxwell equation, which is basically the non-integral version of ampere's law).

So the "complete" ampere's law should be like

\begin{equation} \int B u_n d\Sigma = \mu_oi_{tot} = \mu_oi + \mu_o\epsilon_o \frac{d \phi_E}{dt} \end{equation}

Most of the time the displacement current can be considered negligible (and $\epsilon_o$ is very small in comparison to $i$) and you case this term can be ignored.

  • $\begingroup$ Ok, so if I manage to use maxwell's equation in its full generality (in this case) (say using Jefimenko's equations) then the results obtained will be close and hence displacement terms ignored? $\endgroup$
    – user289443
    Commented Feb 21, 2021 at 12:05
  • $\begingroup$ yes the final result shoud be the same. But you can just not consider the displacement currenti from the beggining being 10^8 times smaller than $\mu_oi$ (you don't need to "demostrate" an approsimation like that, it's simply common sense) $\endgroup$ Commented Feb 21, 2021 at 14:58

Your Answer

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