(B) The self-inductance $L$ of a solenoid is given by $L = \frac{\mu N^2 A}{l}$,where $\mu = \mu_0 \mu_r$.Initially,$L_1 = \frac{\mu_0 N^2 A}{l} = 0.1

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(B) The self-inductance $L$ of a solenoid is given by $L = \frac{\mu N^2 A}{l}$,where $\mu = \mu_0 \mu_r$.Initially,$L_1 = \frac{\mu_0 N^2 A}{l} = 0.18\, mH$.When a core of relative permeability $\mu_r = 900$ is inserted,the new self-inductance $L_2$ becomes $L_2 = \frac{\mu_0 \mu_r N^2 A}{l} = \mu_r L_1$.Substituting the values,we get $L_2 = 900 \times 0.18\, mH = 162\, mH$.

Class 12 Physics Electromagnetic Induction Self Induction

Easy

The coefficient of self-inductance of a solenoid is $0.18\, mH$. If a core of soft iron of relative permeability $900$ is inserted,then the coefficient of self-inductance will become nearly.....$mH$.

A
$5.4$
B
$162$
C
$0.006$
D
$0.0002$

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Electromagnetic Induction

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Self Induction

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The coefficient of self-inductance of a solenoid is $0.18\, mH$. If a core of soft iron of relative permeability $900$ is inserted,then the coefficient of self-inductance will become nearly.....$mH$.

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