# 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 $\hat{f}(t)$.

• 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? May 30, 2014 at 16:36
• I guess it should be voltage May 30, 2014 at 17:28
• But in your fourth paragraph you say $\hat{f}(t)$ is a current. Can you see why I'm confused? May 30, 2014 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? May 30, 2014 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!

• 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? May 30, 2014 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. May 30, 2014 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. May 30, 2014 at 17:44