First slide

Inductance

Question

Two coils are made of copper wires of same length. In the first coil the number of turns is 3 n and radius is r. In the second coil number of turns is n and the radius is 3r. The ratio of self-inductances of the coils will be

Easy

A

3 : 1
B

1 : 3
C

9 : 1
D

1 : 9

Solution

The self-inductance of the coil is
$L = \frac{μ_{0} {πN}^{2} r}{2} or L \propto N^{2} r ∴ \frac{L_{1}}{L_{2}} = {(\frac{N_{1}}{N_{2}})}^{2} (\frac{r_{1}}{r_{2}}) = {(\frac{3 n}{n})}^{2} (\frac{r}{3 r}) = 3 : 1$

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