Question

An equilateral triangular loop of wire on each side ℓ carries a current i. The magnetic field produced at the circumcentre of the loop is:

Answer 1

Step 1:

Given:

Each side of the triangular loop $=\mathcal{l}$

current that is carried by the loop $=i$

Step 2:

Finding the magnetic field produced at the circumcentre (O) of the loop:

If $\mathcal{l}$ is the side of the triangle, the distance of circumcenter from each of the sides of the triangle is, $r=\frac{\mathcal{l}\sqrt{3}}{6}$

The magnetic induction due to each of the sides of the triangle carrying a current $i$ is given as,

$$\begin{gathered} \frac{{\mu }_0}{4\pi }\frac{i}{r}(\sin 60°+\sin 60°)=\frac{{\mu }_0}{4\pi }\frac{i}{r}\frac{\sqrt{3}}{1} \\ =\frac{{\mu }_0}{4\pi }\frac{6i}{\mathcal{l}} \end{gathered}$$

Since the corresponding direction of the magnetic field at each side of the triangle is the same.

So the total magnetic induction at point O will be = $3\times \left(\frac{{\mu }_0}{4\pi }\frac{6i}{\mathcal{l}}\right)$

$$\begin{gathered} =\frac{{\mu }_0}{4\pi }\frac{18i}{\mathcal{l}} \\ =\frac{{\mu }_0}{2\pi }\frac{9i}{\mathcal{l}} \end{gathered}$$

The magnetic field produced at the circumcentre of the loop is $\frac{{\mu }_0}{2\pi }\frac{9i}{\mathcal{l}}$.