Magnetic Field Inside A Solenoid

Inside a long solenoid the field is nearly uniform and parallel to the axis, given by $B={\mu }_{0}nI$, where $n$ is the number of turns per unit length. First compute $n=\frac{N}{L}=\frac{1000}{0.5}=2000{\text{turns m}}^{-1}$. Substituting into the formula, $B=(4\pi \times 10^{-7})(2000)(3)=7.54\times 10^{-3}\text{T}$. The option $3.77\times 10^{-3}\text{T}$ halves the result, as if only one end of the solenoid were considered. The option $1.20\times 10^{-3}\text{T}$ uses the total turns without dividing by length. The option $2.40\times 10^{-2}\text{T}$ overcounts by treating $N$ rather than $n$. This expression follows directly from applying Ampere's circuital law to an ideal solenoid, as derived in NCERT. A plausibility check confirms the field of a few millitesla, consistent with a tightly wound solenoid carrying a modest current and far larger than a single straight wire would produce. A subtle but important feature is that the interior field of an ideal long solenoid is independent of the cross-sectional radius and is uniform across the bore, which is why solenoids are the preferred way to create a known, controllable field for experiments. Near the open ends the field weakens to roughly half its central value, so the formula strictly applies to points well inside, away from the fringing field at the mouths of the coil.