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Magnetics question

2023 · 30 Jan · Shift 2 · Q58
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  5. /2023 · 30 Jan · Shift 2 · Q58

Magnetics question

2023 · 30 Jan · Shift 2 · Q58

JEE MainPhysicsMagneticsMCQ+4 / −1
As shown in the figure, a current of 2 A2 \mathrm{~A}2 A flowing in an equilateral triangle of side 43 cm4 \sqrt{3} \mathrm{~cm}43​ cm. The magnetic field at the centroid O\mathrm{O}O of the triangle is JEE Main 2023 (Online) 30th January Evening Shift Physics - Magnetic Effect of Current Question 78 English (Neglect the effect of earth's magnetic field)
  1. A
    43×10−4 T4 \sqrt{3} \times 10^{-4} \mathrm{~T}43​×10−4 T
  2. B
    43×10−5 T4 \sqrt{3} \times 10^{-5} \mathrm{~T}43​×10−5 T
  3. C
    33×10−5 T3 \sqrt{3} \times 10^{-5} \mathrm{~T}33​×10−5 T
  4. D
    3×10−4 T\sqrt{3} \times 10^{-4} \mathrm{~T}3​×10−4 T
View written solutionFree

Correct answer: C

  1. Given data
  • Current in the triangular loop: I=2 AI = 2\,\text{A}I=2A
  • Side of equilateral triangle: a=43 cm=43×10−2 ma = 4\sqrt{3}\,\text{cm} = 4\sqrt{3}\times 10^{-2}\,\text{m}a=43​cm=43​×10−2m
  • We need magnetic field at the centroid OOO.
  1. Distance of centroid from each side

For an equilateral triangle of side aaa, altitude is h=32ah = \frac{\sqrt{3}}{2}ah=23​​a

The centroid divides altitude in the ratio 2:12:12:1, so perpendicular distance from centroid to each side is r=h3=13⋅32a=a36r = \frac{h}{3} = \frac{1}{3}\cdot \frac{\sqrt{3}}{2}a = \frac{a\sqrt{3}}{6}r=3h​=31​⋅23​​a=6a3​​

Substitute a=43×10−2a = 4\sqrt{3}\times 10^{-2}a=43​×10−2 m: r=43×10−2⋅36=12×10−26=2×10−2 mr = \frac{4\sqrt{3}\times 10^{-2}\cdot \sqrt{3}}{6} = \frac{12\times 10^{-2}}{6} = 2\times 10^{-2}\,\text{m}r=643​×10−2⋅3​​=612×10−2​=2×10−2m

So, r=0.02 mr = 0.02\,\text{m}r=0.02m

  1. Magnetic field due to one side of the triangle

For a finite straight wire, magnetic field at perpendicular distance rrr is B=μ0I4πr(sin⁡θ1+sin⁡θ2)B = \frac{\mu_0 I}{4\pi r}(\sin\theta_1 + \sin\theta_2)B=4πrμ0​I​(sinθ1​+sinθ2​)

At the centroid, for one side of an equilateral triangle, the point lies on the perpendicular bisector of the side, so θ1=θ2=θ\theta_1 = \theta_2 = \thetaθ1​=θ2​=θ

Now half the side is a2=23×10−2 m\frac{a}{2} = 2\sqrt{3}\times 10^{-2}\,\text{m}2a​=23​×10−2m

Thus tan⁡θ=a/2r=23×10−22×10−2=3\tan\theta = \frac{a/2}{r} = \frac{2\sqrt{3}\times 10^{-2}}{2\times 10^{-2}} = \sqrt{3}tanθ=ra/2​=2×10−223​×10−2​=3​

So, θ=60∘\theta = 60^\circθ=60∘

Hence field due to one side: B1=μ0I4πr(sin⁡60∘+sin⁡60∘)B_1 = \frac{\mu_0 I}{4\pi r}(\sin 60^\circ + \sin 60^\circ)B1​=4πrμ0​I​(sin60∘+sin60∘) B1=μ0I4πr(2⋅32)B_1 = \frac{\mu_0 I}{4\pi r}(2\cdot \frac{\sqrt{3}}{2})B1​=4πrμ0​I​(2⋅23​​) B1=μ0I34πrB_1 = \frac{\mu_0 I\sqrt{3}}{4\pi r}B1​=4πrμ0​I3​​

Using μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7}μ0​=4π×10−7: B1=4π×10−7⋅2⋅34π⋅0.02B_1 = \frac{4\pi \times 10^{-7} \cdot 2 \cdot \sqrt{3}}{4\pi \cdot 0.02}B1​=4π⋅0.024π×10−7⋅2⋅3​​ B1=23×10−70.02B_1 = \frac{2\sqrt{3}\times 10^{-7}}{0.02}B1​=0.0223​×10−7​ B1=1003×10−7B_1 = 100\sqrt{3}\times 10^{-7}B1​=1003​×10−7 B1=3×10−5 TB_1 = \sqrt{3}\times 10^{-5}\,\text{T}B1​=3​×10−5T

  1. Net magnetic field due to all three sides

At the centroid, magnetic field due to each side is perpendicular to the plane and in the same direction (by right-hand rule), so they add directly: B=3B1B = 3B_1B=3B1​ B=3×3×10−5B = 3\times \sqrt{3}\times 10^{-5}B=3×3​×10−5 B=33×10−5 TB = 3\sqrt{3}\times 10^{-5}\,\text{T}B=33​×10−5T

  1. Compare with options

The value obtained is 33×10−5 T\boxed{3\sqrt{3}\times 10^{-5}\,\text{T}}33​×10−5T​

This matches Option C.

  1. Verification with stored correct answer

Stored correct answer: C

Our derived answer: C

So the answer agrees with the stored correct answer.

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