Physics - EMI/AC Question with Solution | TestHub

Here, $N=400,L=10\mathrm{mH}=10\times 10^{-3}\mathrm{H}$ $I=2\mathrm{mA}=2\times 10^{-3}\mathrm{A}$ Total magnetic flux linked with the coil, $$\begin{array}{rl}\varphi =NLI& =400\times \left(10\times 10^{-3}\right)\times 2\times 10^{-3}\\ & =8\times 10^{-3}\mathrm{Wb}\end{array}$$ Magnetic flux through the cross-section of the coil = Magnetic flux linked with each turn $$=\frac{\varphi }{N}=\frac{8\times 10^{-3}}{400}=2\times 10^{-5}\mathrm{Wb}$$