Question

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

• A. $1.15\times {10}^{-5}T$
• B. $40T$
• C. $0.4\times {10}^{-3}T$
• D. $4\times {10}^{-2}T$

Answer 1

The correct option is A $1.15\times {10}^{-5}T$

Given: I=1 A, side =4.5$\times {10}^{-2}$

Solution: As we know the formula of magnetic field at the centroid of triangle from one side :

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

Then for 3 sides :$B=3\times \frac{{\mu }_{0}I}{4\pi a}\times (sin{\theta }_1+sin{\theta }_2)$

So, the angle is 60${}^0$ and to find a i.e. the distance between centroid and center of one side is given by:

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

So,$B=\frac{3\times {10}^{-7}\times 1\times 2\sqrt{3}}{4.5\times {10}^{-2}}\times [\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}]$ (sin 60${}^0$)=$\frac{\sqrt{3}}{2}$

$B=4.5\times {10}^{-5}T$