The current in self inductance L = 40 mH is to be increased uniformly from 1A to 11A in 4 ms. The e.m.f. induced in the inductor during the process is

Concept:

Self-inductance:

• Self Inductance is the property of a coil by which an emf is induced in it by changing the associated magnetic flux
• The change in magnetic flux is depicted by the change in current in the coil.
• The polarity of emf induced is opposite to the applied potential difference.
• The unit of self-inductance is Henery.

\( e=L \frac{dI}{dt} \)

e is emf induced, L is self-inductance, \(\frac{di}{dt}\) is the rate of change in current with time.

Calculation:

Given 

Self-inductance L = 40 mH

Change in current: 11 A - 1A = 10 A

Time interval = 4 ms = 4 × 10 ^-3 sec

The rate change of current:

\(\frac{di}{dt} = \frac{10}{4 \times 10^{-3}}\)

emf induced 

\(\implies e =(40\times 10^{-3}) \frac{10}{4 \times 10^{-3}}\)

⇒ e = 100 Volt

So, the emf induced in the coil is 100 Volt