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Magnetics question

2021 · 31 Aug · Shift 2 · Q44
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  5. /2021 · 31 Aug · Shift 2 · Q44

Magnetics question

2021 · 31 Aug · Shift 2 · Q44

JEE MainPhysicsMagneticsMCQ+4 / −1
A current of 1.5 A is flowing through a triangle, of side 9 cm each. The magnetic field at the centroid of the triangle is : (Assume that the current is flowing in the clockwise direction.)
  1. A
    3 ×\times× 10 −-− 7 T, outside the plane of triangle
  2. B
    23×2\sqrt 3 \times23​× 10 −-− 7 T, outside the plane of triangle
  3. C
    23×2\sqrt 3 \times23​× 10 −-− 5 T, inside the plane of triangle
  4. D
    3 ×\times× 10 −-− 5 T, inside the plane of triangle
View written solutionFree

Correct answer: D

  1. Given data
  • Current, I=1.5 AI = 1.5\,\text{A}I=1.5A
  • Side of equilateral triangle, a=9 cm=0.09 ma = 9\,\text{cm} = 0.09\,\text{m}a=9cm=0.09m
  • We need magnetic field at the centroid.

For an equilateral triangle, the centroid, incentre and circumcentre coincide. So the perpendicular distance from centroid to each side is the inradius:

r=a36r = \frac{a\sqrt{3}}{6}r=6a3​​

Thus,

r=0.0936=0.0153 mr = \frac{0.09\sqrt{3}}{6} = 0.015\sqrt{3}\,\text{m}r=60.093​​=0.0153​m


  1. Magnetic field due to one side of the triangle

For a finite straight wire, magnetic field at a point 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 of an equilateral triangle, for each side the point lies symmetrically, so

θ1=θ2=θ\theta_1 = \theta_2 = \thetaθ1​=θ2​=θ

From geometry,

  • Half side =a/2= a/2=a/2
  • Perpendicular distance =r=a3/6= r = a\sqrt{3}/6=r=a3​/6

So,

tan⁡θ=a/2r=a/2a3/6=3\tan\theta = \frac{a/2}{r} = \frac{a/2}{a\sqrt{3}/6} = \sqrt{3}tanθ=ra/2​=a3​/6a/2​=3​

Hence,

θ=60∘\theta = 60^\circθ=60∘

Therefore field due to one side:

B1=μ0I4πr(sin⁡60∘+sin⁡60∘)B_1 = \frac{\mu_0 I}{4\pi r}(\sin60^\circ + \sin60^\circ)B1​=4πrμ0​I​(sin60∘+sin60∘)

B1=μ0I4πr(2⋅32)=μ0I34πrB_1 = \frac{\mu_0 I}{4\pi r}(2\cdot \frac{\sqrt{3}}{2}) = \frac{\mu_0 I\sqrt{3}}{4\pi r}B1​=4πrμ0​I​(2⋅23​​)=4πrμ0​I3​​

Now substitute r=a3/6r = a\sqrt{3}/6r=a3​/6:

B1=μ0I34π(a3/6)=6μ0I4πa=3μ0I2πaB_1 = \frac{\mu_0 I\sqrt{3}}{4\pi \left(a\sqrt{3}/6\right)} = \frac{6\mu_0 I}{4\pi a} = \frac{3\mu_0 I}{2\pi a}B1​=4π(a3​/6)μ0​I3​​=4πa6μ0​I​=2πa3μ0​I​


  1. Total magnetic field due to all three sides

All three sides produce magnetic field at the centroid in the same direction, so

B=3B1=3⋅3μ0I2πa=9μ0I2πaB = 3B_1 = 3\cdot \frac{3\mu_0 I}{2\pi a} = \frac{9\mu_0 I}{2\pi a}B=3B1​=3⋅2πa3μ0​I​=2πa9μ0​I​

Using μ0=4π×10−7 T m/A\mu_0 = 4\pi \times 10^{-7}\,\text{T m/A}μ0​=4π×10−7T m/A,

B=9(4π×10−7)(1.5)2π(0.09)B = \frac{9(4\pi \times 10^{-7})(1.5)}{2\pi(0.09)}B=2π(0.09)9(4π×10−7)(1.5)​

Cancel π\piπ:

B=9⋅4⋅1.5×10−72⋅0.09B = \frac{9\cdot 4 \cdot 1.5 \times 10^{-7}}{2\cdot 0.09}B=2⋅0.099⋅4⋅1.5×10−7​

B=54×10−70.18B = \frac{54\times 10^{-7}}{0.18}B=0.1854×10−7​

B=300×10−7=3×10−5 TB = 300 \times 10^{-7} = 3\times 10^{-5}\,\text{T}B=300×10−7=3×10−5T


  1. Direction of magnetic field

The current is flowing clockwise in the plane of the triangle. By the right-hand rule, magnetic field at the centroid is into the plane.

So the direction is inside the plane of the triangle.


  1. Matching with options

We found:

B=3×10−5 TB = 3\times 10^{-5}\,\text{T}B=3×10−5T

Direction: inside the plane

This matches:

  • Option D: 3×10−5 T3 \times 10^{-5}\,\text{T}3×10−5T, inside the plane of triangle

  1. Comparison with stored correct answer

Stored correct answer: D

My derived answer: D

They agree.

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