# Amperian loop and magnetic field

Suppose I have a long solenoid with $$n$$ turns per unit length. The 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.

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).
$$$$\int B u_n d\Sigma = \mu_oi_{tot} = \mu_oi + \mu_o\epsilon_o \frac{d \phi_E}{dt}$$$$
Most of the time the displacement current can be considered negligible (and $$\epsilon_o$$ is very small in comparison to $$i$$) and you case this term can be ignored.
• 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) Feb 21, 2021 at 14:58