Question

An equilateral triangle of side length $l$ is formed from a piece of wire of uniform resistance. The current I is fed as shown in the figure. Then the magnitude of the magnetic field at its centre $O$ is :

• A. $\frac{\sqrt{3}{\mu }_{0}I}{2\pi l}$
• B. $\frac{3\sqrt{3}{\mu }_{0}I}{2\pi l}$
• C. $\frac{{\mu }_{0}I}{2\pi l}$
• D. Zero

Answer 1

Answer: (D)

$B_{AB}=\frac{{\mu }_0}{4\pi r}\left(\frac{I}{3}\right)\left[\sin 60^{\circ }+\sin 60^{\circ }\right]\otimes$
$B_{BC}=\frac{{\mu }_0}{4\pi r}\left(\frac{I}{3}\right)\left[\sin 60^0+\sin 60^{\circ }\right]\otimes$
$B_{AC}=\frac{{\mu }_0}{4\pi r}\left(\frac{2I}{3}\right)\left[\sin 60^0+\sin 60^{\circ }\right]\odot$
$\therefore So{B}_{net}$ at $O=$ zero