A coil of N = 100 turns carries a current I = | vedclass

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(D) The total magnetic flux linkage in a coil is given by the formula $N\phi = LI$,where $N$ is the number of turns,$\phi$ is the magnetic flux per tu

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(D) The total magnetic flux linkage in a coil is given by the formula $N\phi = LI$,where $N$ is the number of turns,$\phi$ is the magnetic flux per turn,$L$ is the self-inductance,and $I$ is the current.Given values are $N = 100$,$I = 5 \, A$,and $\phi = 10^{-5} \, Wb$.Substituting these values into the formula: $100 	imes 10^{-5} = L 	imes 5$.$10^{-3} = 5L$.$L = \frac{10^{-3}}{5} = 0.2 	imes 10^{-3} \, H$.Since $1 \, H = 1000 \, mH$,we have $L = 0.2 \, mH$.

$A$ coil of $N = 100$ turns carries a current $I = 5 \, A$ and creates a magnetic flux $\phi = 10^{-5} \, Wb$ per turn. The value of its inductance $L$ will be ...... $mH$.

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