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An iron cored coil with 1000 turns generates a magnetic flux of 500 μwb in the core while carrying an electric current of 50 A. What will be the self-inductance of the coil?
1. 0.01 H
2. 0.1 H
3. 0.1 mH
4. 0.01 mH

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Correct Answer - Option 1 : 0.01 H

Concept:

Self-inductance

 \(L = \frac{{{{\rm{μ }}_0}{{\rm{μ }}_{\rm{r}}}{\rm{A}}{{\rm{N}}^2}}}{l}\)

Here, A = Area

N = Number of turns

l = Length of coil

magnetic flux \(ϕ = \frac{{{{\rm{μ }}_0}{{\rm{μ }}_{\rm{r}}}{\rm{A}}{{\rm{N}}{I}}}}{l}\)

Calculation:

N= 1000, ϕ = 500 μwb, I = 50 A

Rl = reluctance

ϕ = (1000 × 50 ) / Rl

\( \frac{{1000× 50}}{500× 10^{-6}} = R_l\;\)

R= 108

\(L= \frac{N^2}{R_l}=\frac{{1000\times 1000}}{{10^8}}\)

L = 0.01 H

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