The time constant $\tau$ of a circuit which contains inductance $L$ and resistance $R$ is $\tau = \frac LR$.
This parameter tells you how rapidly a current can change in the circuit.
In the circuit you describe the inductance has a fixed value and so the rate at which current changes is determined by the resistance in the circuit.
The other important thing about the circuit is that Faraday's law relates the induced emf in the circuit $\mathcal E$ to the rate of change of current in the circuit $\mathcal E= (-)L\dfrac {dI}{st}$.
So if the rate of change of current in the circuit is very high then the circuit tries very hard to reduce that rate of change of current by inducing a large emf which produces a large induced current to counter the changing current producing it.
When you open the switch the circuit resistance $R$ suddenly becomes very high as the air between the contacts of the switch is a dreadful conductor of electricity.
So the current in the circuit has to fall very rapidly.
This in turn induces a large enough emf in the circuit to make the air between the contacts of the switch a conductor (the arc) thus reducing the resistance of the circuit which in turn means that the current will now be reduced at a lower rate.
If you have fluorescent lights which have an inductor in the circuit you might notice that when the lights are switched off a nearby radio suffers from interference.
The interference is the result of arc formed when the switch is opened.
Indeed you may sometimes actually be able to hear the arc.
