If ' $N$ ' is the number of turns in a circular coil, the value of its self-inductance varies as
If ' $N$ ' is the number of turns in a circular coil, the value of its self-inductance varies as
$\mathrm{N}^1$
$\mathrm{N}^3$
$\mathrm{N}^2$
$\mathrm{N}^0$
Solution
The correct option is (C)
Concept: The flux associated with the loop is $\phi=\mathrm{BA}$ and the coefficient of self-inductance $\mathrm{L}$ is defined as $\mathrm{L}=\frac{\phi}{\mathrm{I}}$, where the I is the current flowing through the coil.
The magnetic magnetic field $B$ at the center of a coil carrying current $\mathrm{i}$, with radius $\mathrm{r}$ is given by,
$B=\frac{\mu_0 I}{2 r}$
So for a coil with $\mathrm{n}$ no. of turns: $\mathrm{B}_{\mathrm{N}}=\mathrm{N}\left(\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}\right)$
The self flux associated with one coil is : $\phi 1=\mathrm{B}_{\mathrm{N}} \mathrm{A}$
Now, the self flux associated with coil of $\mathrm{N}$ turns is: $\phi \mathrm{N}=\mathrm{N} \phi_1$
Therefore,
$\phi N=N\left\{N\left(\frac{\mu_0 I}{2 r}\right) A\right\}_1$
Now using the definition of self-inductance: $L=\frac{\theta}{I}$, we get, $\mathrm{L} \propto \mathrm{N}^2$
So, self-inductance is proportional to square of number of turns in the coil.