A field on its own induces nothing. What generates voltage is a field CHANGING through a loop — and to talk about that you first need a word for how much field a loop catches. That word is flux: $\Phi = B A \cos\theta$, read aloud phi equals B A cosine theta, where $\Phi$ is the Greek capital phi.

$\Phi$ is the magnetic flux in webers (Wb); one weber is a tesla-square metre, and benchwork runs in milliwebers. $B$ is the field in tesla, $A$ is the loop's area in square metres, and $\theta$ is the tilt angle — and read this one carefully, because it is measured from the loop's normal, the line sticking straight out of its face, NOT from the loop's surface. So a loop facing the field square-on is $\theta = 0$, $\cos\theta = 1$, and it catches every last line. Turn it edge-on at $90^\circ$ and it catches none. Most problems quietly assume face-on and never mention the angle; when a question does mention it, it is mentioning it for a reason.

Now the law that pays out. Faraday's law: $\varepsilon = N \dfrac{\Delta\Phi}{\Delta t}$ — epsilon equals N delta-phi over delta-t, with $\varepsilon$ the Greek letter epsilon, the induced EMF in volts. $N$ is the number of turns in the coil, a plain count. $\Delta\Phi$ — delta-phi — is the CHANGE in flux in webers, not the flux that was there; $\Delta t$ is the time the change took, in seconds. The delta always means “how much this shifted”.

Two nuggets. $\Delta t$ sits underneath, so speed is everything: the same flux change over a tenth of the time gives ten times the volts, which is exactly how an ignition coil turns 12 V into a spark. And this is a two-step problem by nature — flux first, its rate second — so write the flux down on its own line before you go anywhere near a voltage.