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

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(B) Given: $L_1 = 108 \ mH$,$N_1 = 600$ turns,$N_2 = 500$ turns,and $L_2 = ?$For a circular coil,the self-inductance $L$ is proportional to the square

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(B) Given: $L_1 = 108 \ mH$,$N_1 = 600$ turns,$N_2 = 500$ turns,and $L_2 = ?$For a circular coil,the self-inductance $L$ is proportional to the square of the number of turns $N^2$,given by $L = \frac{\mu_0 \pi N^2 r}{2}$.Since the radius $r$ is the same for both coils,we have the relation:$\frac{L_1}{L_2} = \left(\frac{N_1}{N_2}ight)^2$Substituting the given values:$\frac{108}{L_2} = \left(\frac{600}{500}ight)^2$$\frac{108}{L_2} = \left(\frac{6}{5}ight)^2 = \frac{36}{25}$$L_2 = \frac{108 	imes 25}{36}$$L_2 = 3 	imes 25 = 75 \ mH$Thus,the self-inductance of the coil is $75 \ mH$.

$A$ 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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