Self Inductance

As presented in NCERT Class 12, Chapter 6 (Electromagnetic Induction), the self-inductance of a long solenoid is $L={\mu }_0{n}^{2}Al={\mu }_0\frac{N^2}{l}A$, where the inductance measures the flux linkage produced per unit current due to the solenoid's own field. Here $N=800$, $l=0.40\text{m}$, and $A=5\times 10^{-4}{\text{m}}^2$. Substituting: $L=(4\pi \times 10^{-7})\times \frac{800^2}{0.40}\times 5\times 10^{-4}$. Computing $\frac{800^2}{0.40}=\frac{640000}{0.40}=1.6\times 10^6$, then $L=(1.2566\times 10^{-6})\times (1.6\times 10^6)\times (5\times 10^{-4})\approx 1.0\times 10^{-3}\text{H}=1.0\text{mH}$. The option 0.5 mH halves the result by dropping a factor. The option 2.0 mH doubles it, perhaps by mis-squaring the turns. The option 4.0 mH overcounts area or turns. A dimensional check confirms ${\mu }_0$ (H/m) times $N^{2}A/l$ (m) yields henries, and about 1 mH is typical for such a solenoid. Notice that inductance depends only on geometry and the core material, never on the current flowing, so the same solenoid always has the same self-inductance whether it carries 1 A or 10 A. The quadratic dependence on turns is especially important: doubling the winding count quadruples the inductance, because both the flux produced per turn and the number of turns linking that flux increase together. Inserting a ferromagnetic core in place of air would multiply this value by the relative permeability, dramatically boosting the inductance for the same dimensions.