A long solenoid has 1500 turns. When a curren | vedclass

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(A) The flux linked with each turn of the solenoid is given as $\phi = 2.8 	imes 10^{-3} \, Wb$. The total magnetic flux $(\Phi_{net})$ linked with t

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$1.2$

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(A) The flux linked with each turn of the solenoid is given as $\phi = 2.8 	imes 10^{-3} \, Wb$. The total magnetic flux $(\Phi_{net})$ linked with the solenoid having $N = 1500$ turns is calculated as: $\Phi_{net} = N 	imes \phi = 1500 	imes 2.8 	imes 10^{-3} = 4.2 \, Wb$. The relationship between total flux, self-inductance $(L)$, and current $(I)$ is given by $\Phi_{net} = L 	imes I$. Therefore, the self-inductance $L$ is: $L = \frac{\Phi_{net}}{I} = \frac{4.2}{3.5} = 1.2 \, H$.

$A$ long solenoid has $1500$ turns. When a current of $3.5 \, A$ flows through it, the magnetic flux linked with each turn of the solenoid is $2.8 \times 10^{-3} \, Wb$. The self-inductance of the solenoid is (in $ \, H$)

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