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(B) The self-inductance $L$ of a circular coil is given by $L = \frac{N \Phi_B}{I}$.For a circular coil,the magnetic field at the center is $B = \frac

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$75$

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(B) The self-inductance $L$ of a circular coil is given by $L = \frac{N \Phi_B}{I}$.For a circular coil,the magnetic field at the center is $B = \frac{\mu_0 N I}{2r}$.The magnetic flux through the coil is $\Phi_B = B \cdot A = \left( \frac{\mu_0 N I}{2r} ight) (\pi r^2) = \frac{\mu_0 N I \pi r}{2}$.Thus,$L = \frac{N}{I} \left( \frac{\mu_0 N I \pi r}{2} ight) = \frac{\mu_0 \pi r}{2} N^2$.This shows that $L \propto N^2$.Therefore,$\frac{L_2}{L_1} = \left( \frac{N_2}{N_1} ight)^2$.Given $L_1 = 108 \ mH$,$N_1 = 600$,and $N_2 = 500$:$L_2 = L_1 \left( \frac{N_2}{N_1} ight)^2 = 108 	imes \left( \frac{500}{600} ight)^2 = 108 	imes \left( \frac{5}{6} ight)^2 = 108 	imes \frac{25}{36} = 3 	imes 25 = 75 \ mH$.

$A$ coil of wire of radius $r$ has $600$ turns and self-inductance of $108 \ mH$. The self-inductance of a coil with the same radius and $500$ turns is (in $mH$)

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