Animated Solution for Physics - Magnetic Effects of Current: A long insulated copper wire is closely wound as a spiral of $N$ turns. The spiral has inner radius $a$ and outer radius $b$. The spiral lies in the $X-Y$ plane and a steady current $I$ flows through the wire. The Z-component of the magnetic field at the centre of the spiral is

Imagine a tightly wound spiral coil lying flat in the $X-Y$ plane. It has an inner radius $a$ and an outer radius $b$, with a total of $N$ turns carrying a steady current $I$. Our goal is to find the exact magnetic field produced at the very center of this spiral.

At first glance, you might be tempted to use the standard formula for the magnetic field of a circular loop, $B=\frac{{\mu }_{0}I}{2R}$. However, there is a catch here! The radius of our spiral is not constant; it continuously increases from $a$ to $b$. Because the radius is a variable, we cannot rely on a single algebraic formula. We must invoke the power of calculus.

To solve this, we break the complex spiral down into infinitesimally thin, perfect circular rings. Let's consider one such elemental ring at a distance $r$ from the center, having a minuscule radial thickness $dr$.

First, we need to determine how many turns of the wire are packed into this tiny thickness $dr$. The total $N$ turns are uniformly distributed over the entire radial width of the spiral, which is $(b-a)$. Therefore, the number of turns per unit radial length (the turn density) is $\frac{N}{b-a}$.

Multiplying this density by our elemental thickness $dr$ gives us the number of turns in our specific ring:

Since each individual turn carries a steady current $I$, the total effective current $dI$ circulating within our elemental ring is simply the current per turn multiplied by the number of turns:

Now, we can treat this elemental ring as a standard circular current loop. According to the Biot-Savart law, the magnetic field $dB$ produced at the center by this specific ring is:

Substituting our expression for $dI$ into this equation, we get the raw setup for our integral:

To find the total magnetic field $B$ generated by the entire spiral, we must sum up the contributions from all such elemental rings. We do this by integrating our expression for $dB$ from the innermost radius $r=a$ to the outermost radius $r=b$:

The terms ${\mu }_0$, $N$, $I$, and $2(b-a)$ are all constants, so they can be pulled safely outside the integral:

The integral of $\frac{1}{r}$ with respect to $r$ is the natural logarithm, $\ln (r)$. Evaluating this from $a$ to $b$ yields:

Using the logarithmic identity $\ln x-\ln y=\ln \left(\frac{x}{y}\right)$, we arrive at our final, elegant result:

By the right-hand grip rule, since the current flows in the $X-Y$ plane, the resulting magnetic field at the center points perfectly perpendicular to the plane, directly along the Z-axis.