# Induced EMF in an inductor

In the above figure, the circuit is closed for a very long time and is then opened. It is stated that "the emf across the inductor just after opening will be very high so as to oppose the change of current". I am aware of the mathematical expression of an inductor during charging and discharging, is there any similar expression to justify this behaviour?

• Are you aware of this? $$v(t)=L\frac{di(t)}{dt}$$ – Bob D May 7 at 10:29
• @BobD . Yes I am, so you mean to say that since rate of change of current is high, due to the equation, the voltage across the inductor will also be high, Thanks. – Vaishakh Sreekanth Menon May 7 at 11:05
• Exactly. An attempt to change current in zero time theoretically results in an infinitely high voltage. In reality it doesn’t because opening an air gap switch results in an arc ionizing the air allowing the current to continue. The breakdown of air is evidence that thousands of volts were generated. Hope this helps. – Bob D May 7 at 11:49