# Why is the emf induced in a closed spiral in a time-changing magnetic field less than that induced in a closed loop of the same radius?

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:

• 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? – Dale Nov 15 '18 at 12:01
• @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. – user191954 Nov 15 '18 at 13:53

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 $$N\times S\cos\theta\frac{\mathrm{d}B}{\mathrm{d}t}$$.