Question

Current $I$ is moving in a regular hexagon-shaped loop of side length $a$. What is the magnetic field at the center of the hexagon?

Answer 1

Step 1: Given

$\text{side}=a$

$\theta =30°$

Step 2: To Find

The magnetic field is at the center of the hexagon.

Step 3: Calculate the magnetic field

In the triangle AMO, the angle can be written as,

$$\begin{gathered} \tan \theta =\frac{{\displaystyle \frac{a}{2}}}{r} \\ \tan 30°=\frac{a}{2r} \\ r=\frac{\sqrt{3}a}{2} \end{gathered}$$

The magnetic field due to the side AB is given as,

$$\begin{gathered} B_{AB}=\frac{{\mu }_0}{4\pi }\times \frac{I}{r}\left(\sin \theta +\sin \theta \right) \\ =\frac{{\mu }_0}{4\pi }\times \frac{I}{{\displaystyle \frac{\sqrt{3}a}{2}}}\times 2\sin 30° \\ =\frac{{\mu }_{0}I}{2\pi \sqrt{3}a} \end{gathered}$$

The total magnetic field at the center is given as,

$$\begin{gathered} B_0=6B_{AB} \\ =6\times \frac{{\mu }_{0}I}{2\pi \sqrt{3}a} \\ =\frac{{\mu }_{0}I\sqrt{3}}{\pi a} \end{gathered}$$