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

R_{l = }reluctance

ϕ = (1000 × 50 ) / Rl

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

R_{l }= 10^{8}

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

**L = 0.01 H**