-1
$\begingroup$

I don't get it. Is the area linked different? Or is it something less?

This is the kind of closed spiral I'm thinking of:

enter image description here

$\endgroup$
  • $\begingroup$ This question is unclear. I would recommend a drawing for showing the configuration. I am not even sure what a closed spiral is. How can a spiral be closed? $\endgroup$ – Dale Nov 15 '18 at 12:01
  • 1
    $\begingroup$ @Dale I think the diagram from OP's post here would describe the (strange) concept of closed spirals. Ideally, that could be edited into the question; I'll do that. $\endgroup$ – user191954 Nov 15 '18 at 13:53
1
$\begingroup$

You're probably familiar with Faraday's law of induction:$$\varepsilon=-N\frac{\mathrm{d}\Phi}{\mathrm{d}t}$$ $N$ is the number of turns in the coil. One way I like to think of Faraday's law is by visualizing a coil as $N$ distinct single-loops, each of which follows $\varepsilon_0=-\frac{\mathrm{d}\Phi}{\mathrm{d}t}$ (since we're taking $N=1$). Then, these independent potential sources are taken in series, and the final potential, $\varepsilon$, is $N\times\varepsilon_0$, because you add the individual voltages for a set of sources in series.

Let's say that the vector area of the closed loop is $S$, and the number of turns in that is $N$. For the spiral, we need to assume that there are $N$ loops too (i.e. the coil passes a certain 'starting' point $N$ times), and you mentioned that the outer radius will be the same as the closed loop's.

We can think of the spiral as a series of individual loops, whose radii decrease continuously. This isn't strictly exact, since a spiral isn't the same as a set of concentric circles, but it's a good approximation. Clearly, the outer coil's vector area will be $S$, since the radius is the same as the closed loop's, but each inner circle will have a smaller area, so the corresponding EMFs will be smaller, since for constant areas, $$\frac{\mathrm{d}\Phi}{\mathrm{d}t}=S'\cos\theta\frac{\mathrm{d}B}{\mathrm{d}t}$$ And that's why their sum will be less than the emf for the conventional closed loop, which is $N\times S\cos\theta\frac{\mathrm{d}B}{\mathrm{d}t}$.

It's nice that you fixed your closed question here; this one looks like it fits with the homework policy.

$\endgroup$
  • $\begingroup$ Wait, do you mean that I should assume a regular circular loop as the juxtaposition of several loops (all of the same radius), while the spiral is to be thought of as (again) a combination of loops, but with decreasing radii? $\endgroup$ – Aabesh Ghosh Nov 15 '18 at 16:58
  • $\begingroup$ That might account for the lesser flux noticed. $\endgroup$ – Aabesh Ghosh Nov 15 '18 at 16:58
  • $\begingroup$ @AabeshGhosh That's exactly what I was trying to say. $\endgroup$ – user191954 Nov 16 '18 at 0:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.