Two solenoids of equal number of turns having their lengths and radii in the same ratio 1 : 3. The ratio of their self inductances $L_1 : L_2$ will be:

1 : 3

The correct answer is Option (2) → 1 : 3

Given,

$L_1$ and $L_2$, two solenoid with equal number of turns (N).

$\frac{R_1,\,Radii\,of\,L_1}{R_2,\,Radii\,of\,L_2}=\frac{1}{3}$

Now,

$A=πr^2$

$∴\frac{A_1}{A_2}=\frac{π(1)^2}{π(3)^2}=\frac{1}{9}$

self inductance, $L=μ_0-μ_r\frac{N^2A}{l}⇒\frac{A}{l}$

Now,

$\frac{L_1}{L_2}=\frac{\frac{A_1}{l_1}}{\frac{A_2}{l_2}}=\frac{A_1.l_2}{A_2.l_1}$

$\frac{L_1}{L_2}=\frac{1.3}{9.1}$

$=\frac{1}{3}$