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

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(A) The total magnetic flux linkage $\Phi$ is given by the product of the number of turns $N$ and the flux linked with each turn $\phi$.$\Phi = N \phi

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$1$

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(A) The total magnetic flux linkage $\Phi$ is given by the product of the number of turns $N$ and the flux linked with each turn $\phi$.$\Phi = N \phi$Given $N = 500$ and $\phi = 4 	imes 10^{-3}$ Wb,the total flux is:$\Phi = 500 	imes 4 	imes 10^{-3} = 2$ Wb.The self-inductance $L$ is defined by the relation $\Phi = LI$,where $I$ is the current.$L = \frac{\Phi}{I} = \frac{2}{2} = 1$ $H$.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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