First slide

Inductance

Question

When the number of turns in a coil is doubled without any change in the length of the coil, its self-inductance

Easy

A

becomes four times
B

is doubled
C

is halved
D

is unchanged

Solution

$L \propto N^{2}, i.e., \frac{L_{1}}{L_{2}} = {(\frac{N_{1}}{N_{2}})}^{2} \Rightarrow L_{2} = L_{1} {(\frac{N_{2}}{N_{1}})}^{2} = 4 L_{1}$

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