Wind a wire into a long tight helix, push a current through it, and the field inside comes out remarkably uniform — that is a solenoid, and it is the working heart of every relay, contactor and solenoid valve you will ever meet. Its field is $B = \dfrac{\mu_0 N I}{L}$, read aloud B equals mu-nought N I over L.

The letters. $B$ is the field inside the coil in tesla. $N$ is the total number of turns — a plain count, no units. $I$ is the current in amperes. $L$ is the length of the coil in metres, end to end along its axis. And $\mu_0$ is the permeability of free space, $4\pi\times 10^{-7}\ \mathrm{T\cdot m/A}$ — a constant, like $g$, that comes free with the universe. Turns, current and length are given here; $B$ is what you solve for.

Here is the distinction the whole lesson turns on. Textbooks often write the same law as $B = \mu_0 n I$, and that $n$ is not $N$. Little $n$ is $N/L$, the turns per metre, measured in $\mathrm{m^{-1}}$; big $N$ is the bare count. Confuse them and your answer is wrong by a whole factor of the coil's length — which, for a 20 cm coil, means five times too big. Think of it physically: take a finished coil and stretch it out to twice the length without adding a single turn, and the field inside halves. Nothing about $N$ changed. Everything about $n$ did.

Units guide, they do not confess. Push them through: $\mathrm{T\cdot m/A}$ times amps, divided by metres, leaves tesla — the rearrangement survives. If the units refuse to land on tesla the algebra is wrong, no appeal. But units that do work out never prove you right; the check runs one way only.