A coil of 100 turns carries a current of 5, m | vedclass

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(D) The total magnetic flux linkage $(\Phi_T)$ in a coil is given by the product of the number of turns $(N)$ and the magnetic flux through a single t

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None of these

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(D) The total magnetic flux linkage $(\Phi_T)$ in a coil is given by the product of the number of turns $(N)$ and the magnetic flux through a single turn $(\phi)$: $\Phi_T = N\phi$.Given: $N = 100$, $\phi = 10^{-5} \, Wb$, and current $i = 5 \, mA = 5 	imes 10^{-3} \, A$.The formula for self-inductance $(L)$ is $\Phi_T = Li$.Substituting the values: $N\phi = Li$.$100 	imes 10^{-5} = L 	imes (5 	imes 10^{-3})$.$10^{-3} = L 	imes 5 	imes 10^{-3}$.$L = \frac{10^{-3}}{5 	imes 10^{-3}} = \frac{1}{5} = 0.2 \, H$.Converting to $mH$: $0.2 \, H = 0.2 	imes 1000 \, mH = 200 \, mH$.Since $200 \, mH$ is not among the options, the correct choice is $D$.

$A$ coil of $100$ turns carries a current of $5\, mA$ and creates a magnetic flux of $10^{-5} \, Wb$. The inductance is.....$mH$.

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