Question

A current of $1 \,A$ 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)

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

Answer 2

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