The ratio of the number of turns per unit length of two solenoids $A$ and $B$ is $1: 3$ and the lengths of…

The ratio of the number of turns per unit length of two solenoids $A$ and $B$ is $1: 3$ and the lengths of $A$ and Bare in the ratio $1: 2$. If the two solenoids have same cross sectional area, the ratio of the self inductances of the solenoids A and B is
  1. $1: 12$
  2. $1: 6$
  3. $1: 18$
  4. $1: 9$

Solution

For solenoids A and B , $\mathrm{n}_1: \mathrm{n}_2=1: 3, \mathrm{l}_1: \mathrm{l}_2=1: 2, \mathrm{~A}_1=\mathrm{A}_2$
Self inductance, $L=\mu_0 n^2 A 1 \Rightarrow L \mu n^2 1$ $\therefore \frac{\mathrm{L}_1}{\mathrm{~L}_2}=\left(\frac{\mathrm{n}_1}{\mathrm{n}_2}\right)^2\left(\frac{\mathrm{l}_1}{\mathrm{I}_2}\right)=\left(\frac{1}{3}\right)^2\left(\frac{1}{2}\right)=1: 18$

Asked in: AP EAMCET 2024 (20 May Shift 2)

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