What is the self inductance of a solenoid of length 50 cm, if area of cross section is \( 10\,\mathrm{cm^2} \) and total number of turns is 600?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 9.04\times10^{-4}\,\mathrm{H} \)
The self inductance of a long solenoid with \( N \) turns, length \( l \) and area of cross-section \( A \) depends on the square of the number of turns:
\( \displaystyle L = \frac{\mu_0 N^{2} A}{l} \)
\( l = 50\,\mathrm{cm} = 0.5\,\mathrm{m} \)
\( A = 10\,\mathrm{cm^2} = 10\times10^{-4}\,\mathrm{m^2} = 10^{-3}\,\mathrm{m^2} \)
\( N = 600, \ N^{2} = 3.6\times10^{5} \)
\( \displaystyle L = \frac{4\pi\times10^{-7} \times 3.6\times10^{5} \times 10^{-3}}{0.5} \)
\( \displaystyle L = \frac{4.52\times10^{-4}}{0.5} = 9.04\times10^{-4}\,\mathrm{H} \)