A long solenoid has 500 turns. When a current | vedclass

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(A) The self-inductance $L$ of a solenoid is defined by the relation $\Phi_{total} = L \cdot I$,where $\Phi_{total}$ is the total magnetic flux linked

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$1$

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(A) The self-inductance $L$ of a solenoid is defined by the relation $\Phi_{total} = L \cdot I$,where $\Phi_{total}$ is the total magnetic flux linked with the solenoid.The total magnetic flux linked with the solenoid is given by $\Phi_{total} = N \cdot \phi$,where $N$ is the number of turns and $\phi$ is the magnetic flux linked with each turn.Given:Number of turns $N = 500$Current $I = 2 \, A$Flux per turn $\phi = 4 	imes 10^{-3} \, Wb$Total flux $\Phi_{total} = 500 	imes 4 	imes 10^{-3} \, Wb = 2 \, Wb$.Using the formula $L = \frac{\Phi_{total}}{I}$:$L = \frac{2 \, Wb}{2 \, A} = 1 \, H$.Therefore,the self-inductance of the solenoid is $1 \, H$.

$A$ long solenoid has $500$ turns. When a current of $2 \, A$ is passed through it,the resulting 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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