A 50 cm long air core solenoid has a 2 cm radius and 500 turns. The self-inductance of the solenoid would be (given \( \mu_0 = 4\pi \times 10^{-7}\,\mathrm{T\,m\,A^{-1}} \) and \( \pi^2 = 10 \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 8.0 \times 10^{-4}\,\mathrm{H} \)
The self-inductance of a long air-core solenoid with \( N \) turns, length \( l \) and cross-sectional area \( A \) is
\( \displaystyle L = \frac{\mu_0 N^{2} A}{l} \)
\( N = 500 \)
\( l = 50\,\mathrm{cm} = 0.5\,\mathrm{m} \)
\( r = 2\,\mathrm{cm} = 0.02\,\mathrm{m} \)
The cross-sectional area is that of a circle of radius 2 cm:
\( \displaystyle A = \pi r^{2} = \pi \times (0.02)^{2} = 4\pi\times10^{-4}\,\mathrm{m^2} \)
\( \displaystyle L = \frac{4\pi\times10^{-7} \times (500)^{2} \times 4\pi\times10^{-4}}{0.5} \)
\( \displaystyle L = \frac{16\pi^{2} \times 2.5\times10^{5} \times 10^{-11}}{0.5} = 8\pi^{2}\times10^{-5}\,\mathrm{H} \)
Using \( \pi^{2} = 10 \):
\( \displaystyle L = 80\times10^{-5}\,\mathrm{H} = 8.0\times10^{-4}\,\mathrm{H} \)