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(B) The self-inductance $L$ of a coil is proportional to the square of the number of turns $N$,given by the formula $L \propto N^2$,assuming the radiu

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$75$

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(B) The self-inductance $L$ of a coil is proportional to the square of the number of turns $N$,given by the formula $L \propto N^2$,assuming the radius and length remain constant.Therefore,the ratio of self-inductances for two similar coils is given by $\frac{L_B}{L_A} = \left( \frac{N_B}{N_A} ight)^2$.Given: $N_A = 600$,$L_A = 108 \, mH$,and $N_B = 500$.Substituting the values into the formula:$L_B = L_A 	imes \left( \frac{N_B}{N_A} ight)^2$$L_B = 108 	imes \left( \frac{500}{600} ight)^2$$L_B = 108 	imes \left( \frac{5}{6} ight)^2$$L_B = 108 	imes \frac{25}{36}$$L_B = 3 	imes 25 = 75 \, mH$.Thus,the self-inductance of the second coil is $75 \, mH$.

$A$ coil of wire of a certain radius has $600$ turns and a self-inductance of $108 \, mH$. The self-inductance of a $2^{nd}$ similar coil of $500$ turns will be.........$mH$.

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