The radius of cross-section of a long solenoid is doubled, keeping its length and number of turns per unit length the same. Its new self-inductance compared with the old one is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → four times
For a long solenoid of cross-sectional area \( A \), length \( l \) and \( n \) turns per unit length, the self-inductance is \( L=\mu_0 n^{2}Al \). With \( n \) and \( l \) fixed, \( L \) is proportional to the area, and the area of a circular cross-section is \( \pi r^{2} \).
Doubling the radius gives:
\( \begin{aligned} \frac{L'}{L} &= \frac{\pi(2r)^{2}}{\pi r^{2}} \\ &= 4 \end{aligned} \)
Therefore, the new self-inductance is four times the old one.
NCERT (2026-27):
Self-inductance – “… Let us calculate the self-inductance of a long solenoid of cross-sectional area A and length l, having n turns per unit length. … the self-inductance is,
\( \displaystyle L=\frac{N\Phi_B}{I}=\mu_0 n^{2}Al \)\( \qquad (6.15) \)”