An air-cored solenoid has a self-inductance of 2 mH. Its number of turns is then halved, but its length and cross-section stay the same. Its core is then filled with a material of relative permeability 400. The new self-inductance of the solenoid is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(200\ \mathrm{mH}\)
A long solenoid with a core of relative permeability \( \mu_r \) has \( L=\mu_r\mu_0n^{2}Al \). Here \( n \) is the number of turns per unit length. Half the turns over the same length make \( n \) half. The length and the area stay the same.
For the air core, \( \mu_r=1 \) and \( \mu_0n^{2}Al=2\ \mathrm{mH} \).
\( \begin{aligned} L^{\prime} &= \mu_r\mu_0\left(\frac{n}{2}\right)^{2}Al \\ &= \frac{400}{4}\times\mu_0n^{2}Al \\ &= 100\times2\ \mathrm{mH} \\ &= 200\ \mathrm{mH} \end{aligned} \)
NCERT (2026-27):
Self-inductance – “If we fill the inside of the solenoid with a material of relative permeability μr (for example soft iron, which has a high value of relative permeability), then,”
\( L=\mu_r\mu_0n^{2}Al \)\( \qquad (6.16) \)