0
$\begingroup$

We can clearly read in the first section of this chapter,

An inductance is made by winding many turns of wire in the form of a coil and bringing the two ends out to terminals at some distance from the coil, as shown in $\textrm{Fig.} 22–1$. We want to assume that the magnetic field produced by currents in the coil does not spread out strongly all over space and interact with other parts of the circuit. This is usually arranged by winding the coil in a doughnut-shaped form, or by confining the magnetic field by winding the coil on a suitable iron core, or by placing the coil in some suitable metal box, as indicated schematically in $\textrm{Fig.} 22–1$. In any case, we assume that there is a negligible magnetic field in the external region near the terminals a and b. We are also going to assume that we can neglect any electrical resistance in the wire of the coil. Finally, we will assume that we can neglect the amount of electrical charge that appears on the surface of a wire in building up the electric fields.

enter image description here

With all these approximations we have what we call an “ideal” inductance. (We will come back later and discuss what happens in a real inductance.) For an ideal inductance we say that the voltage across the terminals is equal to $L(dI/dt)$. Let’s see why that is so. When there is a current through the inductance, a magnetic field proportional to the current is built up inside the coil. If the current changes with time, the magnetic field also changes. In general, the curl of E is equal to $−∂\textbf{B}/∂t$; or, put differently, the line integral of $\textbf{E}$ all the way around any closed path is equal to the negative of the rate of change of the flux of $\textbf{B}$ through the loop. Now suppose we consider the following path: Begin at terminal $a$ and go along the coil (staying always inside the wire) to terminal $b$; then return from terminal b to terminal a through the air in the space outside the inductance. The line integral of E around this closed path can be written as the sum of two parts: $$\oint\textbf{E}\cdot d\textbf{s}= \underset{\substack{\text{via}\\\text{coil}}}{\int_a^b} \textbf{E}\cdot d\textbf{s}+ \underset{\text{outside}}{\int_b^a} \textbf{E}\cdot d\textbf{s}. $$

This the bit I'm having a hard time with, he says,

As we have seen before, there can be no electric fields inside a perfect conductor. (The smallest fields would produce infinite currents.) Therefore the integral from $a$ to $b$ via the coil is zero. The whole contribution to the line integral of $\textbf{E}$ comes from the path outside the inductance from terminal $b$ to terminal $a$.The whole contribution to the line integral of E comes from the path outside the inductance from terminal b to terminal a. Since we have assumed that there are no magnetic fields in the space outside of the “box,” this part of the integral is independent of the path chosen and we can define the potentials of the two terminals.

The bold italics in the first and last paragraphs are mine. So we assumed that, one, there is no charge build up on the surface of the coil, and, two, there is no magnetic field outside said coil. Where does the $\textbf{E}$ outside the coil come from then?

The question here is different from this question, even though the same text is referenced.

$\endgroup$
2
  • 1
    $\begingroup$ There is no magnetic field outside the box. There is one outside the coil inside the box. $\endgroup$
    – Dale
    May 15, 2020 at 21:58
  • 1
    $\begingroup$ B is changing , it is not constant $\endgroup$
    – user65081
    May 16, 2020 at 16:30

2 Answers 2

0
$\begingroup$

I also read that in the Feynman's lectures and didn't understand. If there is a constant current in a circuit as it is the case of a superconductor, it is true that the electric field inside the conductor must be zero. Otherwise the charges would be accelerated.

But the example shows a different case. The presence of $I$ results in a $\nabla \times B$. But as we are talking about a changing $I$ that means a changing $B$. A changing $B$ results in a local $\nabla \times E$.

That $E$ (in the coil wires) opposes to the changing of $I$, and while whatever makes $I$ changes keeps doing it, $E$ will be available at the terminals.

The mechanical equivalent would be a mass being accelerated (a changing current). The mass (coil) reacts with a force ($E$) over the agent.

No electric field in a conductor is equivalent to no net forces on a body in uniform velocity.

$\endgroup$
0
$\begingroup$

Electric field outside the inductor is sum of two components, conservative electric field $\mathbf E_C$, and induced electric field $\mathbf E_i$. The induced electric field is assumed negligible far enough from the inductor, where the integration path is assumed. So only the conservative component $\mathbf E_C$ is important there.

This conservative field is due to all electric charges on surfaces of conductors in the system, including the inductor. Thus it is not true, if we want to understand inductor, that we should "assume that we can neglect the amount of electrical charge that appears on the surface of a wire in building up the electric field". On the contrary, electric charges on surface of the inductor, its terminals and the rest of the circuit are important, as they generate the conservative electric field present.

$\endgroup$

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.