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(C) The magnetic field at the center of a circular coil carrying current $I$ is given by $B = \frac{\mu_0 NI}{2R}$.The total magnetic flux $\phi$ link

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$\frac{\mu_0 N^2 \pi R}{2}$

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(C) The magnetic field at the center of a circular coil carrying current $I$ is given by $B = \frac{\mu_0 NI}{2R}$.The total magnetic flux $\phi$ linked with the coil is the product of the number of turns $N$, the magnetic field $B$, and the area of the coil $A = \pi R^2$.$\phi = N \cdot B \cdot A = N \cdot \left( \frac{\mu_0 NI}{2R} \right) \cdot \pi R^2 = \frac{\mu_0 N^2 \pi R I}{2}$.The coefficient of self-induction $L$ is defined as $L = \frac{\phi}{I}$.Substituting the value of $\phi$, we get $L = \frac{\mu_0 N^2 \pi R}{2}$.

$A$ circular coil of radius $R$ has $N$ turns of a wire. The coefficient of self-induction of the coil will be $(\mu_0 = \text{permeability of free space})$

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