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Q.
An equilateral triangular loop of wire of side $l$ carries a current $I$. The magnetic field produced at the circumcentre of the loop is
Moving Charges and Magnetism
Solution:
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{6 I}{l} $
Since the direction of magnetic field in each case is the same. Hence, the total magnetic induction at the circumcentre $O$ is
$B_{O}=3 B=3 \times \frac{\mu_{0}}{4 \pi} \frac{6 I}{l}=\frac{\mu_{0}}{4 \pi} \frac{18 I}{l}$
