The current in the coil of a large electromagnet falls from 6 A to 2 A in 10 ms. The induced emf across the coil is 100 V. Find the self-inductance of the coil.

• A. 1.25 H
• B. 0.25 H
• C. 0.5 H
• D. 0.75 H

Answer: (B)

Equation of self inductance:
V = Ls*(di/dt)
Ls = V/(di/dt)
= 100/((6 - 2)/(0.01))
= 0.250 H
= 250 mH