Animated Solution for Physics - Magnetic Effects of Current: A coil having $N$ turns is wound tightly in the form of a spiral with inner and outer radii $a$ and $b$, respectively. Find the magnetic field at centre, when a current $I$ passes through coil

It starts at an inner radius $a$ and ends at an outer radius $b$, with a total of $N$ turns carrying a current $I$. We need to find the magnetic field exactly at its center.

Since the radius is continuously changing from $a$ to $b$, we cannot use the standard formula for a single circular loop directly. Instead, we must take a small elemental ring of thickness $dr$ at a distance $r$ from the center. This allows us to treat the continuously varying radius as a series of infinitesimally thin circular loops.

The total $N$ turns are spread over a radial width of $(b-a)$. So, the number of turns per unit width is $\frac{N}{b-a}$. Therefore, the number of turns $dN$ in our element of thickness $dr$ will be this ratio multiplied by $dr$:

Now, what is the magnetic field $dB$ produced by this elemental ring at the center? We know the formula for a circular coil: it is ${\mu }_0$ times the current times the number of turns, divided by twice the radius. Substituting our $dN$, we get the expression for $dB$:

This means we integrate $dB$ from the inner radius $a$ to the outer radius $b$. We can pull all the constant terms outside the integral, leaving us with the integral of $\frac{1}{r}$ with respect to $r$:

The integral of $\frac{1}{r}$ is simply the natural logarithm of $r$. Applying the upper limit $b$ and lower limit $a$, we get $\ln b$ minus $\ln a$, which simplifies to $\ln \left(\frac{b}{a}\right)$:

And there we have it! The total magnetic field at the center of the spiral coil.