Q.
An equilateral triangle of side $l$ is formed from a piece of wire of uniform resistance. The current $i$ is fed as shown in the figure. The magnitude of the magnetic field as its centre $O\ $ is
A) $\frac{\sqrt{3}\mu_{0}i}{2\pi l}$
B) $\frac{3\sqrt{2}\mu_{0}i}{2\pi l}\ $
C) $\frac{\mu_{0}i}{2\pi l}$
D) zero
Explanation
(d) Magnetic field at $O\ due\ to\ PR$
$B_{1} = \frac{\mu_{0}}{4\pi}\ \frac{2i/3}{r}$ $\lbrack\sin{30{^\circ} + \sin{30{^\circ}\rbrack\ }}$
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$\ = \frac{\mu_{0}}{4\pi}\frac{2i}{3r}$
It is directed outside the paper. Magnetic field at $O,$ due to $PQR$,
$B_{2}\ \ = 2 \times \frac{\mu_{0}}{4\pi}\ \frac{(i/3)}{\mu}\ \lbrack\sin{30{^\circ} + \sin{30{^\circ}\rbrack}}$
$\ \ $= $\frac{\mu_{0}}{4\pi}\frac{2i}{3r}$
It is directed inside the paper.
$\therefore\ \ \ \ $Resultant magnetic field at $O,$
$B = B_{1} - B_{2} =$0
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.