When solving for mutual induction, how do you know what surface to take the flux over?
You are allowed to choose any surface* as long as its boundary is the loop around which you are computing the emf. In general you choose a flat surface if that's possible, so as to make calculations easier, but you could choose whatever surface you like as long as it has the right boundary.
* okay, not quite any surface -- you have to choose an orientable surface, but you don't need to worry about that unless you are deliberately being perverse.
So you ALWAYS make the boundary the loop where you're computing the emf?
Can you think, by any chance, of an example of this? (besides coaxial cables which i made a post about a few days/hours ago)
Yes, the loop around which you are computing the emf must always be the boundary of the surface.
If you are confused about this, I would suggest that you watch Walter Lewin talk about it (start at 5:00) if you haven't already.
As for an example, basically any Faraday's Law problem will do the trick.
Here's an example: A circuit consisting of a circular loop of wire (radius 1 cm) and an LED is placed inside a solenoid, with the loop of wire concentric with the coils of the solenoid. The solenoid is turned on and the magnetic field inside smoothly increases to 0.1 T over 0.001 seconds. The LED has a 1 ohm resistance and will explode if a current greater than 50 mA flows through it. Does the LED survive?
Answer: We want to calculate the EMF of the circuit, so the circuit has to be the boundary of my surface. I choose the flat surface, which is just a circle, and in this case I am in luck -- the magnetic field is perpendicular to that surface. So the magnetic flux is BA. The rate of change of magnetic flux is 0.1 T * pi(0.01 m)2/0.001 s = 0.01*pi Tm2/s. Therefore, the emf is 0.031 V, which produces a current of 31 mA. That is not enough to explode the LED.
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