A long solenoid has 1000 turns. When a current of 4 A flows through...

In the case of self-inductance of the solenoid, magnetic flux $\left(ϕ\right)$ is directly proportional to the current $\left(i\right)$ flowing in the coil. The flux through each turn is equal to the flux linkage.

The relation is given as $Nϕ = Li$, here $N$ is the number of turns and $L$ is inductance.

So putting the given values, we get

$1000 \times 4 \times {10}^{- 3} = L \times 4$

Thus, the self inductance is $1H$.