A 10 cm long air core solenoid of cross-sectional area \( 1.0\,\mathrm{cm^2} \) contains 300 turns. Its inductance is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 36\pi \times 10^{-6}\,\mathrm{H} \)
The self-inductance of a long solenoid with \( N \) turns, length \( l \) and area of cross-section \( A \), with air inside, is
\( \displaystyle L = \frac{\mu_0 N^{2} A}{l} \)
\( N = 300 \)
\( A = 1.0\,\mathrm{cm^2} = 1.0\times10^{-4}\,\mathrm{m^2} \)
\( l = 10\,\mathrm{cm} = 0.1\,\mathrm{m} \)
\( \displaystyle L = \frac{4\pi\times10^{-7} \times (300)^{2} \times 1.0\times10^{-4}}{0.1} \)
\( \displaystyle L = \frac{4\pi\times10^{-7} \times 9}{0.1} = 36\pi\times10^{-6}\,\mathrm{H} \)
The inductance grows as the square of the number of turns, \( N^{2} \), because the field and the number of turns linked by it both grow with \( N \).