An equilateral triangle of side 'a' carries a | vedclass

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(B) The magnetic field at point $P$ is produced by the three sides of the triangle. $1$. The two sides connected to point $P$ produce zero magnetic fi

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$\frac{{{\mu _0}i}}{{2\sqrt 3 \pi a}} \odot $

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(B) The magnetic field at point $P$ is produced by the three sides of the triangle. $1$. The two sides connected to point $P$ produce zero magnetic field at $P$ because the point $P$ lies on the axis of these wires. $2$. The third side (base) is at a perpendicular distance $d = \frac{\sqrt{3}a}{2}$ from point $P$. $3$. The magnetic field due to a finite wire at a distance $d$ is given by $B = \frac{\mu_0 i}{4\pi d} (\sin 	heta_1 + \sin 	heta_2)$. $4$. For the base,the angles subtended at $P$ are $	heta_1 = 30^\circ$ and $	heta_2 = 30^\circ$. $5$. Substituting the values: $B = \frac{\mu_0 i}{4\pi (\frac{\sqrt{3}a}{2})} (\sin 30^\circ + \sin 30^\circ) = \frac{\mu_0 i}{2\sqrt{3}\pi a} (\frac{1}{2} + \frac{1}{2}) = \frac{\mu_0 i}{2\sqrt{3}\pi a}$. $6$. Using the right-hand rule,the direction of the magnetic field at $P$ is outward $(\odot)$.

An equilateral triangle of side '$a$' carries a current '$i$'. What is the magnetic field at point '$P$' (a vertex of the triangle)?

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