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. When a current $I$ passes through the coil, the magnetic field at the centre is

Imagine you are looking at a tightly wound spiral coil. It's not just a single loop, nor is it a standard solenoid. It's a flat spiral, like a coiled watch spring.

The coil starts at an inner radius $a$ and winds its way outwards until it reaches an outer radius $b$. Across this entire radial width, there are a total of $N$ turns of wire, and a steady current $I$ flows through them.

Our mission is to find the exact magnetic field produced at the very center of this spiral.

Because the radius of each turn is constantly changing, the magnetic field contribution from each turn is also different. A turn closer to the center (near radius $a$) will produce a stronger magnetic field than a turn further away (near radius $b$). This continuous variation is a classic signal that we need to use the power of calculus!

To tackle this, we can't look at the whole spiral at once. We need to break it down into infinitesimally small, manageable pieces. Let's consider a tiny elemental ring of the spiral at a distance $r$ from the center, having an infinitesimally small thickness $dr$.

For a standard circular coil of radius $r$ carrying a current $I$ with $dN$ turns, the magnetic field $dB$ at its center is given by the well-known formula:

This is our master equation. But to use it, we need to figure out exactly how many turns, $dN$, are packed into our tiny elemental ring of thickness $dr$.

Think of the turns like a population spread across a city. The total "population" is $N$ turns, and they are spread evenly across a "distance" equal to the radial width of the spiral, which is $(b-a)$.

Therefore, the number of turns per unit radial length (the turn density) is simply:

Now, if we want to find the number of turns $dN$ in our tiny thickness $dr$, we just multiply the turn density by that thickness:

Now we have everything we need. Let's substitute our expression for $dN$ back into the master equation for the magnetic field $dB$:

Rearranging this to group the constants together, we get:

This equation tells us the tiny magnetic field $dB$ produced by just one elemental ring. To find the total magnetic field $B$ from the entire spiral, we must add up (integrate) the contributions from all such rings.

Since the spiral starts at the inner radius $a$ and ends at the outer radius $b$, these will be our limits of integration. Let's set up the integral:

The term $\frac{{\mu }_{0}NI}{2(b-a)}$ is completely constant, so we can pull it outside the integral:

I know integration can sometimes look intimidating, but this is one of the most beautiful and simple integrals in calculus! The integral of $\frac{1}{r}$ with respect to $r$ is simply the natural logarithm, $\ln (r)$.

Finally, applying the upper and lower limits, we get:

Using the property of logarithms that $\ln x-\ln y=\ln (x/y)$, we arrive at our elegant final answer:

This result perfectly captures how the magnetic field depends on the total turns, the current, and the logarithmic ratio of the outer to inner radii. It's a brilliant demonstration of how calculus helps us solve complex physical geometries!