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
\( \frac{{1000× 50}}{500× 10^{-6}} = R_l\;\)
Rl = 108
\(L= \frac{N^2}{R_l}=\frac{{1000\times 1000}}{{10^8}}\)
L = 0.01 H
