Animated Solution for Physics - Magnetic Effects of Current: Magnitude of magnetic field (in SI unit) at the centre of a hexagonal shaped coil of side 10 cm, 50 turns and carrying current $I$ ampere in units of $\frac{{\mu }_{0}I}{\pi }$ is

A Journey into Biot-Savart Law

Imagine you are standing at the exact center of a massive hexagonal arena. All around you, electric current is surging through the boundary walls. How intensely do you feel the magnetic field at the center? This is the essence of our problem, a classic application of the Biot-Savart Law wrapped in elegant geometry.

We are given a hexagonal coil with a side length of $a=10\text{ cm}$ and $N=50$ turns. The goal is to find the magnetic field at the center. Instead of tackling the entire hexagon at once, we break it down into its fundamental building blocks: six identical finite straight wires.

By the principle of superposition, the total magnetic field at the center is simply the sum of the magnetic fields produced by each of these six sides, multiplied by the total number of turns.

For a finite straight wire carrying current $I$, the magnetic field at a perpendicular distance $r$ is given by the master formula:

In a regular hexagon, if we draw lines from the center to the ends of one side, we form an equilateral triangle. The perpendicular from the center bisects the $60^{\circ }$ angle, giving us symmetric angles of ${\theta }_1={\theta }_2=30^{\circ }$.

Before we plug numbers into our formula, we need the perpendicular distance $r$. Looking at the right-angled triangle formed by the center, the midpoint of the side, and a vertex, we can use basic trigonometry:

Substituting $a=10\text{ cm}$, we get:

To keep our units strictly SI, we convert this to meters: $r=5\sqrt{3}\times 10^{-2}\text{ m}$.

Now, let's substitute $r$ and our angles into the magnetic field formula for a single side:

Since $\sin 30^{\circ }=\frac{1}{2}$, the sum inside the bracket is simply $1$.

This $B_1$ is the field from just one side of one turn. To find the total field $B_{net}$, we must account for all 6 sides of the hexagon and all 50 turns of the coil. This means we multiply our single-side result by $6\times 50=300$.

Let's carefully simplify the expression. The $4$ in the denominator multiplies with $5$ to give $20$. The $10^{-2}$ in the denominator moves to the numerator as $100$:

Rationalizing the denominator by multiplying the numerator and denominator by $\sqrt{3}$, we arrive at our elegant final answer:

The question asks for the coefficient of $\frac{{\mu }_{0}I}{\pi }$, which is exactly $500\sqrt{3}$. By systematically breaking down the geometry and applying the Biot-Savart law, what seemed like a complex multi-turn polygon problem unraveled into a beautiful, straightforward calculation.