A magnetic flux of \( 12\,\mu\mathrm{Wb} \) is linked with a coil when a current of 0.2 A passes through it. What is the self-inductance of the coil?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 6\times10^{-5}\,\mathrm{H} \)
The magnetic flux linked with a coil is proportional to the current through it, \( \Phi=LI \), and the constant \( L \) is the self-inductance of the coil.
\( \Phi=12\,\mu\mathrm{Wb}=12\times10^{-6}\,\mathrm{Wb} \)
\( I=0.2\,\mathrm{A} \)
\( \begin{aligned} L &= \frac{\Phi}{I} \\ &= \frac{12\times10^{-6}\,\mathrm{Wb}}{0.2\,\mathrm{A}} \\ &= 6\times10^{-5}\,\mathrm{H} \end{aligned} \)