A coil of wire of radius r has 600 turns and | vedclass

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(B) The self-inductance $L$ of a solenoid or a coil is given by the formula $L = \frac{\mu_0 N^2 A}{l}$,where $N$ is the number of turns,$A$ is the ar

Vedclass · Accepted Answer

$75$

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(B) The self-inductance $L$ of a solenoid or a coil is given by the formula $L = \frac{\mu_0 N^2 A}{l}$,where $N$ is the number of turns,$A$ is the area,and $l$ is the length. For a coil of a fixed radius and length,$L \propto N^2$.Given $L_1 = 108 \ mH$ for $N_1 = 600$ turns.We need to find $L_2$ for $N_2 = 500$ turns.Using the ratio: $\frac{L_2}{L_1} = \left( \frac{N_2}{N_1} ight)^2$.$\frac{L_2}{108} = \left( \frac{500}{600} ight)^2 = \left( \frac{5}{6} ight)^2 = \frac{25}{36}$.$L_2 = 108 	imes \frac{25}{36} = 3 	imes 25 = 75 \ mH$.

$A$ coil of wire of radius $r$ has $600$ turns and a 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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