If $$N$$ is the number of turns in a coil, the value of self-inductance varies as
A.
$${N^0}$$
B.
$$N$$
C.
$${N^2}$$
D.
$${N^{ - 2}}$$
Answer :
$${N^2}$$
Solution :
Magnetic flux, $$\phi = BA$$
and magnetic field due to circular coil is $$B = \frac{{{\mu _0}Ni}}{{2R}}$$
As self-inductance, $$L = \frac{{N\phi }}{i}$$
$$\eqalign{
& \therefore L = \frac{N}{i}\left( {BA} \right) = \frac{N}{i}\left( {\frac{{{\mu _0}Ni}}{{2R}}} \right)A \cr
& = \frac{{{\mu _0}{N^2}A}}{{2R}} \cr
& \therefore L \propto {N^2} \cr} $$
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