Question

A current of $1amp$ is flowing in the sides of an equilateral triangle of side $4.5\times {10}^{-2}m$. Find the magnetic field at the centroid of triangle.

Answer 1

Magnetic field at the centroid of triangle is given by:

$B=\frac{3{\mu }_{0}I}{4\pi a}(\sin {\theta }_1-\sin {\theta }_2)$

For an equilateral triangle, ${\theta }_1={\theta }_2={60}^0$

From the figure,

$a=\frac{1}{2}\cot 60$

$a=\frac{4.5\times {10}^{-2}}{2\sqrt{3}}$

So,

$B=\frac{3\times {10}^{-7}\times 2\sqrt{3}\times 2\sqrt{3}}{4.5\times {10}^{-2}\times 2}$

$B=4\times {10}^{-5}T$