Magnetic Effects & Magnetism

JEE Main level, single correct, medium

21 of 276

Question 21 of 276

The current passing through a conducting loop in the form of equilateral triangle of side 43 cm4 \sqrt{3} \mathrm{~cm} is 2 A. The magnetic field at its centroid is α×10−5 T\alpha \times 10^{-5} \mathrm{~T}. The value of α\alpha is ____\_\_\_\_.
(Given : μo=4π×10−7\mu_{\mathrm{o}}=4 \pi \times 10^{-7} SI units)

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Answer: (C) 333 \sqrt{3}

For an equilateral triangle of side aa, the distance of the centroid from any side is r=a23r = \frac{a}{2\sqrt{3}}.
Given side a=43a = 4\sqrt{3} cm =43×10−2= 4\sqrt{3} \times 10^{-2} m.
So, r=43×10−223=2×10−2r = \frac{4\sqrt{3} \times 10^{-2}}{2\sqrt{3}} = 2 \times 10^{-2} m.
The magnetic field at the centroid due to one side of the triangle is B1=μ0I4πr(sin⁡θ1+sin⁡θ2)B_1 = \frac{\mu_0 I}{4\pi r} (\sin \theta_1 + \sin \theta_2).
For an equilateral triangle, θ1=θ2=60∘\theta_1 = \theta_2 = 60^\circ.
B1=10−7×22×10−2(sin⁡60∘+sin⁡60∘)=10−5×(32+32)=3×10−5B_1 = \frac{10^{-7} \times 2}{2 \times 10^{-2}} (\sin 60^\circ + \sin 60^\circ) = 10^{-5} \times (\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}) = \sqrt{3} \times 10^{-5} T.
The total magnetic field at the centroid due to all three sides is B=3B1=33×10−5B = 3B_1 = 3\sqrt{3} \times 10^{-5} T.
Comparing with B=α×10−5B = \alpha \times 10^{-5} T, we get α=33\alpha = 3\sqrt{3}.

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