2
$\begingroup$

My textbook is using Faraday's law to explain the self inductance that happens in a solenoid with changing current.

According to Wikipedia, Faraday's Law is:

The induced electromotive force in any closed circuit is equal to the negative of the time rate of change of the magnetic flux through the circuit.

The loops in a solenoid aren't closed - it's loops are "almost closed", but the beginning and end don't touch each other.

So how can Faraday's Law be used?

$\endgroup$
1
  • 1
    $\begingroup$ The article also has this immediately after your quoted text: This version of Faraday's law strictly holds only when the closed circuit is a loop of infinitely thin wire,[13] and is invalid in other circumstances as discussed below. A different version, the Maxwell–Faraday equation (discussed below), is valid in all circumstances. $\endgroup$ Mar 5, 2014 at 19:24

3 Answers 3

1
$\begingroup$

Faraday's law is given as follows

$$\oint \vec{E}\cdot\vec{dl} = -\frac{\mathrm{d} }{\mathrm{d} t}\iint_{ }^{ }\vec{B}\cdot\vec{dA}$$

Look at the R.H.S of the equation. It is the changing magnetic flux( w.r.t time ) that is responsible for the induced $E.M.F$. Although the coil of the solenoid does not consist of perfect conducting loops, the magnetic flux associated with the conducting loops depends on the $surface$ $area$ $enclosed$ by the coil of the solenoid.

enter image description here

You clearly see that the loop I have drawn encloses some area. The magnetic flux is associated with this area.

enter image description here

Here, I have drawn a loop somewhere in the middle of the solenoid. Although the loops are not perfect, they enclose some area. We are concerned about the area enclosed while calculating the magnetic flux.

In solenoids you'll find that the coil consists of loops that are tightly wound. This configuration can be approximately treated as loops arranged close together.

$\endgroup$
0
$\begingroup$

Inside the solenoid there's no (component of the) magnetic field that isn't perpendicular to the loops, so treating a solenoid as if it has closed loops is valid approximation.

Perhaps you can understand it better if you approach the solenoid as a single wire and see what the emf does from that point of view.

$\endgroup$
0
$\begingroup$

I suspect this is an approximation that works well.

Each piece of the coils is almost exactly aligned with the induced electric field $\vec{E}$ at that position. (To be more accurate, some are exactly aligned, but most aren't. Try imagining a coil wrapped around a cylinder.) In this way, the EMF $\oint \vec{E}\cdot d\vec{l}$ for the closed loop is nearly identical for a single "loop" of wire that is open. This effect is minimized if the coils are tightly wound.

$\endgroup$

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.