Magnetic flux of 10 mu Wb is linked with a co | vedclass

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(B) The magnetic flux $\phi$ linked with a coil is given by the relation $\phi = L \cdot I$,where $L$ is the self-inductance and $I$ is the current fl

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$5$

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(B) The magnetic flux $\phi$ linked with a coil is given by the relation $\phi = L \cdot I$,where $L$ is the self-inductance and $I$ is the current flowing through the coil.Given:$\phi = 10 \ \mu Wb = 10 	imes 10^{-6} \ Wb$$I = 2 \ mA = 2 	imes 10^{-3} \ A$Rearranging the formula for self-inductance $L$:$L = \frac{\phi}{I}$Substituting the values:$L = \frac{10 	imes 10^{-6}}{2 	imes 10^{-3}}$$L = 5 	imes 10^{-3} \ H$$L = 5 \ mH$Therefore,the self-inductance of the coil is $5 \ mH$.

Magnetic flux of $10 \mu Wb$ is linked with a coil,when a current of $2 \ mA$ flows through it. What is the self-inductance of the coil (in $mH$)?

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