EMF Induced in a solenoid When is the EMF induced in a solenoid higher, when making or breaking the circuit?
Isn't it the same magnitude since change in flux is the same in both cases? 
 A: The emf of an inductor (solenoid) is directly proportional to the change in current per time. When you connect a solenoid to a voltage source, the back-emf is initially exactly equal to the voltage applied. So, if you apply a perfect voltage source to a perfect inductor, the current rises linearly forever. In the real world, after a while the resistance of the circuit (solenoid and voltage source) comes into effect, and the current rise slows until it finally stops where the current times the resistance equals the applied voltage.
If you suddenly disconnect a perfect solenoid that has current flowing through it, the change in current per time becomes infinite, and you get infinite voltage across the now-open terminals. While this would be entertainingly cataclysmic, it isn't real: as I said here, when the voltage gets high enough it will jump across the slowly increasing gap, and even once the gap gets large enough to stop the spark there's capacitance across the gap that will slow down the dI/dt and reduce the voltage to terrestrial values.
Final answer: unless you have a humongous voltage source that you're planning on hooking across your solenoid, you can probably get more back EMF when you break the circuit than when you make it.
