Why is induced current through an inductor more when switch is put off than when the switch is put on? This problem I saw somewhere got me thinking. I thought very hard about this but couldn't get to any conclusion. (here opened and closed are verbs, i mean when the current is flowing in the circuit and we cut the key off, its opened, and when there is no current in the circuit and we switch it on, its closed.)
 A: Imagine a circuit consisting of a battery, a wire, a switch, and an inductor, all in series. For "resistor" you could simply sum the internal resistance of battery and wire - it doesn't really matter (I just don't like "unrealistic" circuits for simple explanations).
When you close the switch, current will attempt to flow. The most current that could flow (if the inductor were a perfect conductor) is $I_0=\frac{V}{R}$, but the inductor will try to resist the change in current and therefore generate a reverse e.m.f. that is initially no greater than V, the voltage of the battery (because when it reaches that value, there is no force left to drive current).
When the switch is opened, the current through the inductor attempts to go to zero "in an instant". Unfortunately, just generating a back e.m.f. of V will not be sufficient to stop the current change - the circuit is broken, and with an infinite resistance in the loop you need an infinite back e.m.f. to keep the current flowing.
In reality there will be a little bit of stray capacitance in any inductor (if only the turn-to-turn capacitance); that acts to create a "short circuit" for the current, so the change in current through the inductor when the switch is opened is not infinite, and a finite voltage spike ensues.
But either way, the back e.m.f. is indeed greater when the switch is opened than when it's closed, because the circuit impedance is greater.
