An ideal air-cored solenoid has a self-inductance of $50 \mathrm{mH}$ and comprises $N$ closely woun

Question

An ideal air-cored solenoid has a self-inductance of $50 \mathrm{mH}$ and comprises $N$ closely wound turns. If the number of turns is doubled while meticulously keeping its length and cross-sectional area constant, its new self-inductance will be

Professor PC Thomas Classes · Answer

(b) : The self-inductance of a solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$. Since the core material ($\mu_0$), cross-sectional area ($A$), and length ($l$) remain constant, $L$ is directly proportional to the square of the number of turns ($L \propto N^2$). If the number of turns is doubled ($2N$), the self-inductance increases by a factor of $2^2 = 4$. Thus, the new self-inductance is $4 	imes 50 \mathrm{mH} = 200 \mathrm{mH}$.