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(C) The mutual inductance $M$ of two concentric coils where $r \ll R$ is given by the formula: $M = \frac{\mu_0 \pi N_1 N_2 r^2}{2 R}$.Here,$N_1$ and

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$4$

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(C) The mutual inductance $M$ of two concentric coils where $r \ll R$ is given by the formula: $M = \frac{\mu_0 \pi N_1 N_2 r^2}{2 R}$.Here,$N_1$ and $N_2$ are the number of turns in the inner and outer coils respectively,$r$ is the radius of the inner coil,and $R$ is the radius of the outer coil.From the formula,we observe that $M \propto r^2$.Using the concept of relative error,we have $\frac{\Delta M}{M} = 2 \frac{\Delta r}{r}$.Given that the percentage change in radius is $\frac{\Delta r}{r} 	imes 100 \% = 2 \%$.Therefore,the percentage change in mutual inductance is $\frac{\Delta M}{M} 	imes 100 \% = 2 	imes (2 \%) = 4 \%$.

Two concentric circular coils,one of small radius $r$ and the other of large radius $R$,are placed coaxially with their centers coinciding. If the radius $r$ is changed by $2 \%$,then the change in mutual inductance of the arrangement is (assume $r \ll R$). (in $\%$)

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