Question

A current $1A$ is flowing in the sides of equilateral triangle of side $4.5\times {!0}^{-2}m$. The magnetic induction at centroid of the triangle is

• A. $4\times {10}^{-5}T$
• B. $40T$
• C. $0.4\times {10}^{5}T$
• D. $4\times {!0}^{-2}T$

Answer 1

The correct option is A $4\times {10}^{-5}T$

Magnetic field at the centroid is given by,

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

Here, ${\theta }_1={\theta }_2=60$

Length $=4.5\times {10}^{-2}m$

$a=\frac{1}{2}cot60$

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

Therefore,

$B=\frac{3\times {10}^{-7}\times 1\times 2\sqrt{3}}{4.5\times {10}^{-2}}\frac{2\sqrt{3}}{2}=4\times {10}^{-5}T$

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