# Signal Induction in a Wire due to Alternating Current

I wanted to make sure I understand induction well enough.

Assume we have two wires running parallel to each other. Wire A has a signal of $f(t)$, wire B has a signal of $\hat{f}(t)$.

Let's connect a signal generator to wire A, therefore putting $$f(t) = A \cdot sin(2\pi f_{c} t)$$ where $A$ is the amplitude of the wave, and $f_{c}$ is its frequency.

This will induce changing current $\hat{f}(t)$ in wire B. My question is: How will $\hat{f}(t)$ look like?

My guess is that it will have the form $$\hat{f}(t) = \hat{A} \cdot sin(2\pi f_{c} t + \phi)$$ where $\hat{A} \leq A$ and is proportional to $A$ and $\phi$ is an additional phase factor due to the fact that radio waves travel at finite speed.

Is that a correct guess? Does it only happen if the wires are infinitely long? I haven't derived this properly and I don't really need a detailed derivation (although it wouldn't hurt). I just want to know if there is a good way to describe the received $\hat{f}(t)$.

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It's unclear which of your signals are voltages and which are currents. Maybe use $v_A(t)$ or $i_A(t)$ instead of $f(t)$ to make it clear? – The Photon May 30 '14 at 16:36
I guess it should be voltage – Martin Marinov May 30 '14 at 17:28
But in your fourth paragraph you say $\hat{f}(t)$ is a current. Can you see why I'm confused? – The Photon May 30 '14 at 17:51
Yes, I mean, I am doing signal processing and I think radio receivers measures voltages. What would be the difference if that was current? – Martin Marinov May 30 '14 at 18:47

TL;DR They shift but only if you have non perfect system, phase difference is compensated in a perfect system.

First how to get no shift. Imagine it is an ideal transformer. You apply the induction law once and get the $B(t)$.

$$U(t)=\int_\ell E(t)\mathrm{d}\ell=-\frac{\mathrm{d}B(t)}{\mathrm{d}t}$$

Now the $B(t)$ is shifted in relation to $U(t)$ because of the derivation, but when we do the calculation of $\hat U$ we shift it back:

$$\hat U(t)=-\frac{\mathrm{d}B(t)}{\mathrm{d}t}$$

So two shifts in opposite directions give zero phase difference. Perfect. (You should now think: but the second conductor influences the first...in some way, so I think this may be wrong! That is what I was thinking, then I got over it. Explanation: you would use the same formula over and over again (ping-pong) always getting net zero shift.)

If you really want to have a shift, you can look at it as a non-ideal transformer(if you like add capacitances). $$\hat U(t)=\hat L\frac{\mathrm{d}\hat i(t)}{\mathrm{d}t}+ M\frac{\mathrm{d} i(t)}{\mathrm{d}t}+R\hat i$$ $$U(t)=L\frac{\mathrm{d} i(t)}{\mathrm{d}t}+ M\frac{\mathrm{d}\hat i(t)}{\mathrm{d}t}+Ri$$ For a constant frequency $\omega$ you can write: $$\hat U(t)=j\omega \hat L \hat i+ j\omega M i+R\hat i$$ $$U(t)=j\omega L i+j\omega M\hat i+Ri$$ If you know how to solve simple circuits these equations shouldn't be a problem for you.

The question remains, what are the values of $\hat L, M$ and $L$?

The derivation of these formulas is something I can't remember, but you'll find it for sure somewhere. The resistances can be calculated easier. What would happen if the resistances were the same?

If you add the capacitances you'll have a characteristic impedance rating. Even more fun stuff to think about!

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So in the perfect case where $\phi = 0$, do I expect to see the same amplitude in wire B, namely will $\hat{A} = A$ ? Does the distance between the wire change anything? – Martin Marinov May 30 '14 at 17:27
Perfect coupling means that absolutely everything from the first wire is coupled to the second wire. Think of a transformer as a 4 pole element. When it is perfect you simply connect the terminals as if there was absolutely no resistance/inductance whatever. So yes the amplitudes will be the same as they will practically be connected to each other on each end($-\infty$ and $+\infty$). It's a really ridiculous oversimplified case. – Kurtovic May 30 '14 at 17:34
You can for a detailed analysis use power line models. These incorporate segmentation into 4-pole $\Pi$ elements. That way the mutual elements(inductance, capacitance) will be in every $\Pi$ element, where every of these elements represents a defined length. The other thing you could consider is wave propagation equations(mentioned at the end). That would probably be an overkill, but it would surely be interesting. – Kurtovic May 30 '14 at 17:44