The current passing through a conducting loop in the form of equilateral triangle of side $$4\sqrt{3}$$ cm is 2A. The magnetic field at its centroid is $$\alpha\times10^{-5}T.$$ The value of $$\alpha$$ is______.
(Given :$$\mu_{o}=4\pi\times 10^{-7} SI$$ units)

B at centroid is sum of B from each straight wire (side of triangle) of triangle. As direction of B from all three wires is same.

Calculate B from a single side, 3B is the required answer

B = $$\frac{\mu_0 I}{4 \pi r} (\sin \theta_1 + \sin \theta_2) $$

Substitute the values $$\theta_1$$=$$\theta_2$$=60, you obtain

B=$$\sqrt{3}$$

$$B_{net}$$=$$3\sqrt{3}$$

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The current passing through a conducting loop in the form of equilateral triangle of side $$4\sqrt{3}$$ cm is 2A. The magnetic field at its centroid is $$\alpha\times10^{-5}T.$$ The value of $$\alpha$$ is______.(Given :$$\mu_{o}=4\pi\times 10^{-7} SI$$ units)