Question

An equilateral triangular loop of wire of side $l$ carries a current $I$. The magnetic field produced at the circumcentre of the loop is

• A. $\frac{{\mu }_0}{4\pi }\frac{3\sqrt{3}I}{l}$
• B. $\frac{{\mu }_0}{4\pi }\frac{9I}{l}$
• C. $\frac{{\mu }_0}{4\pi }\frac{18I}{l}$
• D. $\frac{{\mu }_0}{4\pi }\frac{6I}{l}$

Answer 1

Answer: (C)

If $l$ is the side of the triangle, the distance of the circumcentre from each of the sides of the triangle is
$r=\frac{1}{3}l\sin 60^{\circ }=\frac{l\sqrt{3}}{6}$
The magnetic induction at the circumcentre $O$ due to each of the side of the triangle carrying a current $I$ is
$B=\frac{{\mu }_0}{4\pi }\frac{I}{r}\left(\sin 60^{\circ }+\sin 60^{\circ }\right)=\frac{{\mu }_0}{4\pi }\frac{6I}{l}$
Since the direction of magnetic field in each case is the same. Hence, the total magnetic induction at the circumcentre $O$ is
$B_O=3B=3\times \frac{{\mu }_0}{4\pi }\frac{6I}{l}=\frac{{\mu }_0}{4\pi }\frac{18I}{l}$