A solenoid of length 1.5 m contains 1400 turns of cross-sectional area \( 30 \mathrm{~cm}^{2} \) each carries 8 A current which decrease to 2 A at 0.2 sec and the coefficient of the permeability is \( 4 \pi \times 10^{-7} \mathrm{~Wb} / \mathrm{Amp} . \mathrm{m} \) and \( \pi=22 / 7 \), so: \begin{tabular}{|l|c|c|} \hline & The self-induction coefficient & \begin{tabular}{c} The induced current if the resistance of \\ the solenoid equals \( \mathbf{2} \boldsymbol{\Omega} \) \end{tabular} \\ \hline (A) & \( 7.919 \times 10^{-3} \mathrm{H} \) & 0.07392 A \\ \hline (B) & \( 7.919 \times 10^{-3} \mathrm{H} \) & 0.152 A \\ \hline (C) & \( 4.928 \times 10^{-3} \mathrm{H} \) & 0.07392 A \\ \hline (D) & \( 4.928 \times 10^{-3} \mathrm{H} \) & 0.152 A \\ \hline \end{tabular}
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m} \) - \( N = 1400 \) (number of turns) - \( A = 30 \, \text{cm}^2 = 30 \times 10^{-4} \, \text{m}^2 \) - \( l = 1.5 \, \text{m} \) Substitute the values: \[ L = \frac{4\pi \times 10^{-7} \times 1400^2 \times 30 \times 10^{-4}}{1.5} \] Show more…
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