Two solenoids A and B of equal number of turn | vedclass

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

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$1 : 3$

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(B) The self-inductance $L$ of a solenoid is given by the formula $L = \frac{\mu_0 N^2 A}{l}$, where $N$ is the number of turns, $A$ is the cross-sectional area, and $l$ is the length of the solenoid.Since $A = \pi r^2$, the formula becomes $L = \frac{\mu_0 N^2 \pi r^2}{l}$.Given that the number of turns $N$ is equal for both solenoids, we have $L \propto \frac{r^2}{l}$.Let $l_A, r_A$ be the length and radius of solenoid $A$, and $l_B, r_B$ be the length and radius of solenoid $B$.We are given $\frac{l_A}{l_B} = \frac{1}{3}$ and $\frac{r_A}{r_B} = \frac{1}{3}$.Therefore, the ratio of self-inductances is $\frac{L_A}{L_B} = \frac{r_A^2}{l_A} 	imes \frac{l_B}{r_B^2} = \left( \frac{r_A}{r_B} ight)^2 	imes \left( \frac{l_B}{l_A} ight)$.Substituting the given ratios: $\frac{L_A}{L_B} = \left( \frac{1}{3} ight)^2 	imes \left( \frac{3}{1} ight) = \frac{1}{9} 	imes 3 = \frac{1}{3}$.Thus, the ratio of the self-inductance of solenoid $A$ to that of $B$ is $1 : 3$.

Two solenoids $A$ and $B$ of equal number of turns have their lengths and radii in the same ratio $1 : 3$. The ratio of the self-inductance of solenoid $A$ to that of $B$ will be

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