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)

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