# Induced current in a loop due to varying flux

Suppose you have a loop of wire beside a infinite long wire carrying a current I. The current in the infinite wire is given by: $$I(t) = I_0(1-e^{(-\alpha t)})$$ Where $\alpha$ is a constant and $I_0$ is the current obtained after a long time. I know the flux through the loop is given by $\Phi(t) = KI(t)$, and that the loop has the risistance $R$.

So I guess the induced current in the loop must be given by:

$$I_{induced} = V_{induced}\frac{1}{R}= - \frac{d}{dt}\Phi \frac{1}{R} = - \frac{I_0K \alpha}{R}e^{-\alpha t}$$

Now comes the real problem what if the loop also has an inductance, $L$. Will the induced current then be given by.

$I = I_1 + I_2$ With $I_1$ equal to the $I_{induced}$ above, and

$$I_2 = -\frac{R}{L}\frac{dI_2}{dt}$$

solving the differential equation giving me: $I_2 = e^{-\frac{L}{R}t}$

Making $I = - \frac{I_0K \alpha}{R}e^{-\alpha t} + e^{-\frac{L}{R}t}$

Or do I need need to do something completely different?

You do have a problem, in the sense that even if you had no wire, a loop with self inductance could have lots of different currents running through it. But let's see what we get.

We need names to be clear, so let $I_w$ be the current in the wire and let $I_l$ be the current in the loop. Let the flux due to $I_w$ be $KI_w$ and let the flux due to $I_l$ be $LI_l$ (both of which require approximations before being valid just for your information).

Then $I_l=-\frac{1}{R}\left(K\frac{d I_w}{d t}+L\frac{d I_l}{dt}\right)$ possibly. Check the signs, both overall and relative.

Now it is easy to tell if your answer is correct, see if it solves the above differential equation. OK so what about the problem if there were no wire. One solution is to note that at $t=0$ there was no current in the wire so we could choose to have no current in the loop at the same time.

If we do that then it looks like we have enough information even when there is a wire. To check the signs make sure both the induced currents $-\frac{K}{R}\frac{d I_w}{d t}$ and $-\frac{L}{R}\frac{d I_l}{dt}$ are currents that oppose the increase in the magnetic flux.

• Perhabs it is not clear, but there is no current in the loop beside what is being induced from change in the infinite wires magnetic field. (can't really tell if you already took that into account?). But What did you conclude? Is my answer wrong..? And more importantly was my approach wrong? – Nillo Jun 11 '15 at 19:15
• @Nillo It wasn't clear but I suggested that the current be zero at t=0 to reproduce the "no wire implies no loop current." But if you look at your answer and set $I_0=0$ do you get that? Check my work type questions are not allowed here, conceptual questions that haven't already been asked are on topic here. But you can take you equation and put it in my differential equation to check your solution though yours must be wrong if setting $I_0=0$ doesn't give you zero current, right? I tried to explain the correct way to think about the problem. – Timaeus Jun 11 '15 at 19:25
• Alright so my solution is completely wrong... As far as I can tell (Doesn't fit into the differential equation). So I guess I should find the solution to $I_l$ by solving the differential equation you just wrote? How do you get that equation btw? (or is it a long story?) – Nillo Jun 11 '15 at 19:39
• I got it to work now, thank you so much for your answer! – Nillo Jun 11 '15 at 20:12
• @Nillo That's nice, because I added more details to my answer but they didn't take and got lost. – Timaeus Jun 11 '15 at 22:21