An air-cored coil has a self-inductance of 0. | vedclass

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(C) The formula for the self-inductance of a solenoid is $L = \frac{\mu_0 N^2 A}{l}$.When an iron core is introduced,the permeability becomes $\mu = \

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$1 \ H$

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(C) The formula for the self-inductance of a solenoid is $L = \frac{\mu_0 N^2 A}{l}$.When an iron core is introduced,the permeability becomes $\mu = \mu_0 \mu_r$. The new inductance $L'$ is given by $L' = \frac{\mu_0 \mu_r (N')^2 A}{l}$.Given $\mu_r = 1000$ and $N' = \frac{N}{10}$.Substituting these values: $L' = \frac{\mu_0 	imes 1000 	imes (N/10)^2 A}{l} = \frac{\mu_0 	imes 1000 	imes N^2 A}{100 	imes l} = 10 	imes \left( \frac{\mu_0 N^2 A}{l} ight)$.Since the initial inductance $L = 0.1 \ H$,we have $L' = 10 	imes L = 10 	imes 0.1 \ H = 1 \ H$.

An air-cored coil has a self-inductance of $0.1 \ H$. $A$ soft iron core of relative permeability $1000$ is introduced,and the number of turns is reduced to $\left(\frac{1}{10}\right)^{\text{th}}$. The value of the new self-inductance is:

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