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Question
When current flowing through a solenoid decreases from 5A to 0 in 20 milliseconds, an emf of 500V is induced in it.
- What is this phenomenon called?
- Calculate coefficient of self-inductance of the solenoid.
Solution
1. The phenomenon is called self-induction.
2. ΔI = (5 – 0) A = 5 A, Δt = 20 × 10-3 sec.
`e = L (ΔI)/(Δt)`
`L = e × (Δt)/(ΔI)`
= `(500 xx 20 xx 10^(-3))/5 "H"`
= 2H
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