1. Jul 21, 2009

### ibc

Hello
Say there's a uniform magnetic field changing in time (B=Asin(t) or something), and a ring (pic 1).
so obviously by faraday's law there would be current running through the ring caused by the induced emf.
Now, say we connect 2 wires from the center of the ring up to the top and to the left (pic 2), now I want to know what current would be flowing through these 2 new wires (or generally what would happen with these circuit, but what can I do with it? (all wires have some resistivity, or just put little resistors on each section)
I can still use faraday's law to get the same induced emf through the whole ring, and calculate it's new resistance therefore find the current, by this way, I would get some current I which splits into I1 and I2 (on the ring and into the new wires). but how would I find I1 and I2? I'd want to say:
I1R1=V1=V2=I2R2
but do we know V1=V2? since an emf caused by a changing magnetic field is not a conservative field.
Also, how can I even be sure that there's a current I through the ring which splits into I1 and I2, maybe the current runs in cycles inside this section enclosed by the wires and the ring? after all I can use faraday's law only for this section, and get some different (proportional to the area) emf, which should cause current flowing through this section.

So the question is, what do I do in such case?

thanks

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2. Jul 22, 2009

### gabbagabbahey

You now essentially have 3 loops (do you see them?), and you can use Faraday's law to calculate the emf around each of them. This, in turn, will allow you to find the current in each wire.

3. Jul 22, 2009

### ibc

first of all, if I think of them as loops, faraday's law gives inconclusive answers, I must calculate the integral E*dl at each point, and that would be equal the the change of flux.
so naming the currents I, I1, I2 as in the first post, I can calculate that integral on each loop and I have I=I1+I2
so if I do it for the whole ring, I get that I=I1, I2=0, because the integral must be the same as without the two wires, and this solutions also solves the case for every other loops (the integral equals the flux change in that loop), so I can say weepee nothing changes, same solution as without the new wires?

4. Jul 22, 2009

### gabbagabbahey

Really? Why do you say that?

The emf around the whole circle (you call it the "integral", but since you are not actually performing any integration, just calculating the change in flux, which equals the emf, I see no reason to call it an integral) must equal the voltage drop around the entire circle and that emf is the same with or without the wires, but that doesn't mean the current is the same does it? In fact, there is no longer a current around the entire loop, but rather one current $I$ through one quarter of the loop and another (presumably different) current $i_1$ through the other 3/4 of the loop. If you say that the resistance of the entire circle is $R$, then it seems reasonable to assume that the resistance of the upper-left 1/4 of the loop is $\frac{R}{4}$ and the resistance of the remaining portion of the circle is $\frac{3R}{4}$....so, applying Kirchoff's voltage law to the circle you have:

$$\text{emf}_{circle}+\frac{IR}{4}+\frac{3i_1R}{4}=0$$

Even after you use Faraday's law to calculate $\text{emf}_{circle}$, in order to find $I$ and $i_1$, you need at least one more equation, and that's why you need to use Faraday's law to calculate the $\text{emf}$ through one of the smaller loops and then analyze that loop in the same manner (using Kirchoff's laws).

5. Jul 22, 2009

### ibc

the voltage drop is what I was calling the integral, as mentioned it's the integral E*dl

I don't see how you can say that there's the same current in all parts of these quarter of a ring.
anyway, kirchoff's voltage law does not apply here since the field we get out of a changing magnetic flux is NOT a conservative field.

6. Jul 22, 2009

### gabbagabbahey

I was referring to just the curved outer 1/4 circle....do you argue that the current could be different at different points along that section of wire?

No, I think you need to review Kirchoff's circuit laws; they can be derived directly from Maxwell's equations and apply to non-conservative electric fields just as well as conservative ones.

Just because,

$$\oint_{\text{circuit}} \textbf{E}\cdot\text{d}\textbf{l}=0$$

does not mean that the field is necessarily conservative--- it just means that at each point along the path of the circuit you can define a unique potential (relative to some chosen ground). For the field to be conservative, you would have to be able to say

$$\oint_{\mathcal{C}} \textbf{E}\cdot\text{d}\textbf{l}=0$$

for all possible closed paths $\mathcal{C}$, not just the path corresponding to a loop in a circuit.

7. Jul 24, 2009

### ibc

as far as I know, the kirchoff's law which regards the voltage is simply a result of the electric field being a conservative one. I don't see any reason for this to be true when speaking of a non-conservative field.
I think the problem here is an example, I didn't understand your solution, but when I solved it (by applying faraday's law on each one of the 3 possible loops) I got that the current in the two new wires is zero, and around the ring is just as it was without the wires.
which means that in this circuit the voltage between two points depends on the path (if you go through the new wires you just get a zero, and if you go through the quarter of a circle you get some voltage drop, i.e the integral of E*dl is not zero for this loop (the quarter of a circle + new wires)