Question

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

• A. $2\times {10}^{-5}T$
• B. $4\times {10}^{-5}T$
• C. $8\times {10}^{-5}T$
• D. $1.2\times {10}^{-4}T$

Answer 1

Answer: (B)

$B_1=\frac{{\mu }_{0}I}{4\pi a}\left(\sin {\varphi }_1+\sin {\varphi }_2\right)$
$\tan 30^{\circ }=\frac{a}{\left[\frac{l}{2}\right]}$
$\Rightarrow a=\frac{1}{2\sqrt{3}}$
$\Rightarrow B_1=\frac{{\mu }_{0}I}{4\pi \left(\frac{l}{2\sqrt{3}}\right)}\left(\sin 60^{\circ }+\sin 60^{\circ }\right)$
$\Rightarrow B_1=\frac{6{\mu }_{0}I}{4\pi l}$
$B_{\text{total }}=3B_1$
$\Rightarrow B_{\text{total }}=4\times 10^{-5}T$