A 200 turn coil of self inductance 20 mH carries a current of 4 mA. The magnetic flux linked with each turn of the coil would be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 4\times10^{-7}\,\mathrm{Wb} \)
The self-inductance relates the total flux linkage of the coil to its current: \( N\Phi = L I \), where \( \Phi \) is the flux through each turn.
\( L = 20\,\mathrm{mH} = 20\times10^{-3}\,\mathrm{H} \)
\( I = 4\,\mathrm{mA} = 4\times10^{-3}\,\mathrm{A} \)
\( N = 200 \)
\( \displaystyle N\Phi = L I = 20\times10^{-3} \times 4\times10^{-3} = 8\times10^{-5}\,\mathrm{Wb} \)
\( \displaystyle \Phi = \frac{8\times10^{-5}}{200} = 4\times10^{-7}\,\mathrm{Wb} \)