Question

A current of 2.5 amperes in a 1000 turns coil produces
0.5x10-5 weber magnetized flux in each turn. Find out the self-
inductance of coil.​

Answer 1

Answer:

Find out self-inductance of a coil.

Explanation:

Self inductance-It defined as efficiency/tendency of the coil in the inductance resist or oppose the changes in current by itself. It is represented as "L".

L = NΦ/I

Given:

I=2.5 A

N =1000

Φ=0.5×$10^{-5}$ weber.

Calculation:

L = 1000×$\frac{0.5}{2.5}$×$10^{-5}$

L=200×$10^{-5}$

L = 2×$10^{-3}$

L = 2 mH

Result:

Hence self inductance is equal to 2 mH