The self-inductance of a solenoid of length 0.5 m, area of cross-sect…

The correct answer is Option (1) → 3.7 mH

For a long solenoid the field inside is \( B = \mu_0 nI \) with \( n = N/l \), and the flux linked with all N turns gives the self-inductance of a solenoid.

    \( \displaystyle L = \frac{\mu_0 N^{2} A}{l} \)

    \( N = 700 \)
    \( l = 0.5\,\mathrm{m} \)
    \( A = 30\,\mathrm{cm^{2}} = 30\times10^{-4}\,\mathrm{m^{2}} = 3.0\times10^{-3}\,\mathrm{m^{2}} \)

    \( \displaystyle L = \frac{(4\pi\times10^{-7}\,\mathrm{T\,m\,A^{-1}})(700)^{2}(3.0\times10^{-3}\,\mathrm{m^{2}})}{0.5\,\mathrm{m}} \)

    \( \displaystyle L = \frac{4\pi\times10^{-7}\times4.9\times10^{5}\times3.0\times10^{-3}}{0.5}\,\mathrm{H} \)

    \( \displaystyle L \approx 3.7\times10^{-3}\,\mathrm{H} \)

So \( L \approx 3.7\,\mathrm{mH} \).