Question

A current of 1.5 amperes in a 800 turns coil produces 1.5×10/-5 Weber magnetized flux in each turn. Find out the Self-induction of coil.​

Answer 1

Given:

Current (I) = 1.5 A

Number of turns (N) = 800

Magnetic flux (Ф) 1.5 × $10^{-5}$ weber

To find:

Self-induction (L)

Solution:

• N×Ф = L×I

      ⇒800 × 1.5 × $10^{-5}$ = L × 1.5

• L = 8 × $10^{-3}$ H