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Inductance Hell
by Tyler Clarke on 2025-4-5
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Inductance Hell

By Tyler Clarke in Physics on 2025-4-5

I intended to get a lot done today. I needed to write the multivariable 16.4 review, build my company webpage, solve the collatz conjecture and prove that the real part of every nontrivial zero of the reimann zeta function is `frac 1 2`. And get my group to actually do some work on the project due in like three days.

I did not do those things.

This week, physics was hellish. Specifically, one of the problems in the physics homework was hellish. I generally read the textbook regularly, and attend lectures and pay attention and allat jazz, so I can usually finish the homework effortlessly. This week, however, we've just 'bout finished Maxwell's equations, and inductance is... bad. I've missed family obligations, killed orphans, and had slightly less time than usual to play 0 a.d. (in order of awfulness), purely for this one question. The really annoying thing is that it seems conceptually simple: we're given a bunch of parameters about a solenoid and a resistive ring around it, and we need to find the induced current in the ring at a given time after a voltage source is applied to the solenoid. Easy, right? The law of inductance should just about solve it.

I started this nightmare with high hopes. The resistance in the ring is known, and we know that `V = frac {d phi} {dt}`, so by Ohm's law `I_{"ring"} = frac {frac {d phi} {dt}} {R_{"ring"}}`. Because the b-field outside the solenoid is essentially negligible, flux in this case is just going to be `A_{"sol"} * B_{"sol"}`, and `B_{"sol"}` is easy to find- it's just `mu_0 * frac N l * I_{"sol"}`. Taking the derivative with respect to `t` is also quite simple: `frac {d phi} {dt} = A_{"sol"} * mu_0 * frac N l * frac {d I_{"sol"}} {dt}`. Not too pleasant, but not too hard either.

It was about this point that Sally Ride came flaming down.

See, the problem wants to know the induced current at some time very close to 0 but not equal to 0. Because current increases by an exponential function, this means we're dealing with some nasty, nasty exponents. It took me embarrassingly long just to find the damn function to use: `I_"sol"(t) = I_"max" * (1 - e^{-t frac {R_"sol"} L})`. Ohshitohshitohshit abort abort abort! What even is that! Where the hell did `L` come from! It turns out that the awful, awful `frac {R_"sol"} L` thingy is so pervasive it has a name: the time constant. If you're familiar at all with exponential functions, this is obviously a damper on decay rate; the question is, how the hell do we actually find it? What is `L`?

For whatever insane reason, `L` stands for "inductance". Inductance is the tendency of a given circuit to, well, induct: a measure of how the magnetic field produced by a changing current inside the wire resists the current in the wire. This property is the actual reason current ramps up slowly; inductance only resists change in current. I really, really, really don't want to work this out the hard way; fortunately, there's a ready-made equation for specifically a solenoid's inductance: `L = frac {mu_0 N^2 A} {l}`, where `N` is the number of coils, `A` is the area of the solenoid's circular cross-section, and `l` is the length of the solenoid. I don't want to calculate that (or even think about it, to be honest), but calling that `L` is just asking to confuse inductance with length, so I'm going to call it something sensible: `L_"induc"`.

Back to current. We need to find `frac {d I_{"sol"}} {dt}`, and for once this is straightforward. Expanded out, our formula for current is `I_"sol"(t) = I_"max" - I_"max"e^{-t frac {R_"sol"} {L_"induc"}}` (`I_"max"` is just the current we'll achieve after a very long time: ohm's law tells us that it's `frac V {R_"sol"}`). The derivative of this is easy enough to take: `frac {d I_{"sol"}} {dt} = -frac V R_"sol" e^{-t frac {R_"sol"} {L_"induc"}} * frac {-R_"sol"} {L_"induc"}`. We can cancel and simplify a bit to get `frac {d I_{"sol"}} {dt} = frac V {L_"induc"} e^{-t frac {R_"sol"} {L_"induc"}}`.

Now that we know the derivative of I with respect to time, we can smush it into our previous equation for magnetic flux, to get `frac {d phi} {dt} = A_{"sol"} * mu_0 * frac N l * frac V {L_"induc"} e^{-t frac {R_"sol"} {L_"induc"}}`. Is it too late to change majors?

But wait- it gets worse! This isn't actually the final equation we need. What we need is `I_{"ring"} = frac {frac {d phi} {dt}} {R_{"ring"}}`. Substituting in `frac {d phi} {dt}` gives us `I_{"ring"} = frac {A_{"sol"} * mu_0 * frac N l * frac V {L_"induc"} e^{-t frac {R_"sol"} {L_"induc"}}} {R_{"ring"}}`. Time to start filling in the blanks. We already know that `L_"induc" = frac {mu_0 N^2 A} {l}`, and the solenoid is known to have 1840 turns, a length of 0.1525 meters, and an area of `pi * 0.0139^2`. I'll spare you the calculation: the result is `L_"induc" = 0.01693382781`. I don't like that number, Sam-I-Am, but at least it's a number and not a disgusting derivative. Substituting in some more (voltage: 33.63 volts, time: 2.14 microseconds, `R_"sol"`: 153.7ohms, `R_"ring"`: 1637.1 ohms) gives us `I_"ring" = 0.00001094959`. You know what I absolutely hate? That that answer is right. It has no business being right. That's a disgusting answer.

I tried exactly those same steps with exactly those same equations a total of 6 times, having rederived them almost entirely every time. I had Gemma2 (locally) calculate an answer, and when it failed, resorted to ChatGPT. I read discussion boards, and wiki pages, and LibreTexts, and even those shitty Hyper Physics thingies from GSU. The only thing that worked was getting out my least favorite editor (surely you've moved to Lapce/Zed/Builder from the awful that is Code by now, right? If only any of them had web editing worth a damn...), grinding out some templates, and writing down my thought process here in long and sarcastic form. Specifically, I rederived from scratch equations that I've been trying and failing to find and apply for the last four days.

I'm increasingly glad I didn't try to refund that domain name.

There's still time in the day to do some things that I very much need to do, but I'll try to get Thomas 16.4 out tonight. 16.3 was awesome and 16.4 is looking to be a continuation of the awesome. Until then, cheers, and happy Saturday!