0
$\begingroup$

Consider a closed loop wire with current flowing through it:

enter image description here

Consider along the wire a battery $V$, a resistance $R$ and a capacitor $C$.

We know the integral form of Faraday's law:

$$\oint\limits_{\partial S} \vec E\cdot d\vec l = -\frac {d\Phi_B}{dt}=-\frac d {dt} \int\limits_S \vec B\cdot d\vec A$$

Because of the capacitor, the current through the wire will exponentially decay, therefore the magnetic field $B$ will also decay, therefore the right side of Faraday's law won't be 0.

However, in his lecture https://www.youtube.com/watch?v=6w3SzI_s5Sg&list=PLyQSN7X0ro2314mKyUiOILaOC2hk6Pc3j&index=26 at 5.15 Walter Lewin says that $$\oint\vec E\cdot d\vec l=0$$

What am I doing wrong?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
0
$\begingroup$

At 1:26 Lewin assumes that it is an $RC$ circuit, in other words he ignores any and all magnetic effects. Since the subject is "lumped" circuit theory, when speaking of magnetic effects he would have had to include in it some inductor, but he confined the problem to an $R$ and a $C$ and that is why he could write $\oint \mathbf E \cdot d{\ell}=0$. The magnetic moment of a closed loop of area $A$ is $\approx IA$ that usually carries quite small magnetic energy when compared to normal size capacitors and their electric energy. As you increase the local number of kinks, loops thereby making an inductor, the situation changes.

Improve this answer
$\endgroup$
2
  • $\begingroup$ So there is a self inductance, but it is so low that we ignore it, is that right? $\endgroup$
    – gluon
    Commented Jul 19, 2023 at 8:00
  • $\begingroup$ Yes, that is one way of putting it, it being the whole circuit. $\endgroup$
    – hyportnex
    Commented Jul 19, 2023 at 13:05

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.