# Self-induction in a circular coil

There's a formula for self inductance: $$L=\frac{n\Phi}{i}$$ where n is the number of loops.

But the book also says self inductance is directly proportional to $$n^2$$

I totally agree with the second sentence. But the formula seems to show $$L$$ is directly proprtional to $$n$$ and not $$n^2$$.

I am confused. Can you help?

Notice, the magnetic magnetic field $$B$$ at the center of a coil carrying current $$i$$, with radius $$r$$ & having $$n$$ no. of turns $$B=\frac{\mu_0}{2}\frac{ni}{r}$$ hence, magnetic flux $$\phi$$ linked to the coil is given as $$\Phi=BA=\frac{\mu_0}{2}\frac{ni}{r}\pi r^2=\frac{\mu_0 \pi nir}{2}$$ Now, setting the value of $$\phi$$, we get $$L=\frac{n\Phi}{i}=\frac{n\frac{\mu_0 \pi nir}{2}}{i}=\frac{\mu_0 \pi n^2r}{2}$$ $$L\propto n^2$$ It is obvious that keeping other parameters constant, the self inductance $$\color{red}{L}$$ of a coil is directly proportional to $$\color{red}{n^2}$$
Self inductance $$L$$ : $$L=\frac{n\phi}{i}$$ Magnetic flux, is the product of Magnetic field and the area of cross section intercepted by magnetic field lines $$\phi=BA$$
Magnetic field is directly proportional to the number of turns in the inductor $$B\propto n$$
$$\phi\propto n$$ $$L\propto n\cdot n$$ $$\therefore L\propto n^2$$ I hope it helps you