An air-cored solenoid is 40 cm long and has an area of cross-section of 5 cm2. Its self-inductance is to be \(0.5\pi\ \mathrm{mH}\). The total number of turns it needs is (take \(\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}\))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1000
For a long solenoid \( L=\mu_0n^{2}Al \), where \( n \) is the number of turns per unit length. With \( N \) turns in all, \( n=N/l \).
\( \begin{aligned} L &= \mu_0\left(\frac{N}{l}\right)^{2}Al \\ &= \frac{\mu_0N^{2}A}{l} \end{aligned} \)
\( l=40\ \mathrm{cm}=0.4\ \mathrm{m} \)
\( A=5\ \mathrm{cm^{2}}=5\times10^{-4}\ \mathrm{m^{2}} \)
\( L=0.5\pi\times10^{-3}\ \mathrm{H} \)
\( \begin{aligned} N^{2} &= \frac{Ll}{\mu_0A} \\ &= \frac{0.5\pi\times10^{-3}\times0.4}{4\pi\times10^{-7}\times5\times10^{-4}} \\ &= 1\times10^{6} \\ N &= 1000 \end{aligned} \)
NCERT (2026-27):
Self-inductance – “… The total flux linked with the solenoid is”
\( N\Phi_B=(nl)(\mu_0nI)(A)=\mu_0n^{2}AlI \)
“where nl is the total number of turns. Thus, the self-inductance is,”
\( \displaystyle L=\frac{N\Phi_B}{I}= \)\( \mu_0n^{2}Al \)\( \qquad (6.15) \)