A long solenoid has 1000 turns. When a curren | vedclass

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(C) Given: Number of turns $N = 1000$,Current $I = 4\, A$,Magnetic flux per turn $\phi_{0} = 4 	imes 10^{-3}\, Wb$.The total magnetic flux $\phi$ lin

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(C) Given: Number of turns $N = 1000$,Current $I = 4\, A$,Magnetic flux per turn $\phi_{0} = 4 	imes 10^{-3}\, Wb$.The total magnetic flux $\phi$ linked with the solenoid is given by $\phi = N \phi_{0}$.Substituting the values,$\phi = 1000 	imes 4 	imes 10^{-3}\, Wb = 4\, Wb$.We know that the relation between flux and self-inductance $L$ is $\phi = L I$.Therefore,the self-inductance $L = \frac{\phi}{I} = \frac{4\, Wb}{4\, A} = 1\, H$.

$A$ long solenoid has $1000$ turns. When a current of $4\, A$ flows through it,the magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3}\, Wb$. The self-inductance of the solenoid is......$H$

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