The current passing through a coil of 120 tur | vedclass

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(D) Given: Number of turns $N = 120$,Inductance $L = 40 	ext{ mH} = 40 	imes 10^{-3} 	ext{ H}$,Current $I = 30 	ext{ mA} = 30 	imes 10^{-3} 	ext

Vedclass · Accepted Answer

$10 	imes 10^{-6} 	ext{ Wb}$

Vedclass · Answer

(D) Given: Number of turns $N = 120$,Inductance $L = 40 	ext{ mH} = 40 	imes 10^{-3} 	ext{ H}$,Current $I = 30 	ext{ mA} = 30 	imes 10^{-3} 	ext{ A}$.We know that the total flux linkage is given by $N\phi = LI$.Therefore,the magnetic flux $\phi$ linked with the coil is given by $\phi = \frac{LI}{N}$.Substituting the values: $\phi = \frac{40 	imes 10^{-3} 	imes 30 	imes 10^{-3}}{120}$.$\phi = \frac{1200 	imes 10^{-6}}{120} = 10 	imes 10^{-6} 	ext{ Wb}$.

The current passing through a coil of $120$ turns and inductance $40 \text{ mH}$ is $30 \text{ mA}$. The magnetic flux linked with the coil is:

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