A magnetic flux of 5 mu Wb is linked with a c | vedclass

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(A) The self-inductance $L$ of a coil is defined by the relation $\phi = L I$,where $\phi$ is the magnetic flux and $I$ is the current flowing through

Vedclass · Accepted Answer

$5 	imes 10^{-3} \ H$

Vedclass · Answer

(A) The self-inductance $L$ of a coil is defined by the relation $\phi = L I$,where $\phi$ is the magnetic flux and $I$ is the current flowing through the coil.Rearranging for $L$,we get $L = \frac{\phi}{I}$.Given values are:$\phi = 5 \ \mu Wb = 5 	imes 10^{-6} \ Wb$$I = 1 \ mA = 1 	imes 10^{-3} \ A$Substituting these values into the formula:$L = \frac{5 	imes 10^{-6} \ Wb}{1 	imes 10^{-3} \ A} = 5 	imes 10^{-3} \ H$.Therefore,the self-inductance of the coil is $5 	imes 10^{-3} \ H$.

$A$ magnetic flux of $5 \mu Wb$ is linked with a coil when a current of $1 \ mA$ flows through it. Calculate the self inductance of the coil.

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