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Magnetics question

2018 · 15 Apr · Shift 2 · Q48
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Magnetics question

2018 · 15 Apr · Shift 2 · Q48

JEE MainPhysicsMagneticsMCQ+4 / −1
A current of 1 A is flowing on the sides of an equilateral triangle of side 4.5 ×\times× 10-2 m. The magnetic field at the center of the triangle will be :
  1. A
    2 ×\times× 10-5 Wb/m2
  2. B
    Zero
  3. C
    8 ×\times× 10-5 Wb/m2
  4. D
    4 ×\times× 10-5 Wb/m2
View written solutionFree

Correct answer: D

  1. Given data
  • Current in each side: I=1 AI = 1\,\text{A}I=1A
  • Side of equilateral triangle: a=4.5×10−2 ma = 4.5 \times 10^{-2}\,\text{m}a=4.5×10−2m

We need the magnetic field at the center of the equilateral triangle due to currents in all three sides.

  1. Magnetic field due to one finite straight side

For a finite straight conductor, 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​)

Here, for the center of an equilateral triangle, the perpendicular from the center to any side bisects that side. So,

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

Hence,

B1=μ0I2πrsin⁡θB_1 = \frac{\mu_0 I}{2\pi r}\sin\thetaB1​=2πrμ0​I​sinθ
  1. Geometry of the equilateral triangle

The distance from the center to any side (inradius) is

r=a36r = \frac{a\sqrt{3}}{6}r=6a3​​

Substituting a=4.5×10−2a = 4.5 \times 10^{-2}a=4.5×10−2 m,

r=4.5×10−236r = \frac{4.5 \times 10^{-2}\sqrt{3}}{6}r=64.5×10−23​​

Also, half the side is

a2=2.25×10−2 m\frac{a}{2} = 2.25 \times 10^{-2}\,\text{m}2a​=2.25×10−2m

From the right triangle,

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

Using r=a36r = \frac{a\sqrt{3}}{6}r=6a3​​,

tan⁡θ=a/2a3/6=33=3\tan\theta = \frac{a/2}{a\sqrt{3}/6} = \frac{3}{\sqrt{3}} = \sqrt{3}tanθ=a3​/6a/2​=3​3​=3​

So,

θ=60∘\theta = 60^\circθ=60∘

Thus,

sin⁡θ=sin⁡60∘=32\sin\theta = \sin 60^\circ = \frac{\sqrt{3}}{2}sinθ=sin60∘=23​​
  1. Field due to one side
B1=μ0I2πr⋅32B_1 = \frac{\mu_0 I}{2\pi r}\cdot \frac{\sqrt{3}}{2}B1​=2πrμ0​I​⋅23​​

Using r=a36r = \frac{a\sqrt{3}}{6}r=6a3​​,

B1=μ0I2π(a36)⋅32B_1 = \frac{\mu_0 I}{2\pi \left(\frac{a\sqrt{3}}{6}\right)}\cdot \frac{\sqrt{3}}{2}B1​=2π(6a3​​)μ0​I​⋅23​​ B1=μ0I2π⋅6a3⋅32B_1 = \frac{\mu_0 I}{2\pi}\cdot \frac{6}{a\sqrt{3}} \cdot \frac{\sqrt{3}}{2}B1​=2πμ0​I​⋅a3​6​⋅23​​ B1=3μ0I2πaB_1 = \frac{3\mu_0 I}{2\pi a}B1​=2πa3μ0​I​

Now substitute μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7}μ0​=4π×10−7, I=1I=1I=1, a=4.5×10−2a=4.5\times10^{-2}a=4.5×10−2:

B1=3(4π×10−7)2π(4.5×10−2)B_1 = \frac{3(4\pi \times 10^{-7})}{2\pi(4.5\times10^{-2})}B1​=2π(4.5×10−2)3(4π×10−7)​ B1=12π×10−79π×10−2=129×10−5B_1 = \frac{12\pi \times 10^{-7}}{9\pi \times 10^{-2}} = \frac{12}{9}\times 10^{-5}B1​=9π×10−212π×10−7​=912​×10−5 B1=43×10−5 TB_1 = \frac{4}{3}\times 10^{-5}\,\text{T}B1​=34​×10−5T
  1. Total field due to all three sides

The magnetic field due to each side at the center is in the same direction (perpendicular to the plane), so they add:

B=3B1=3(43×10−5)B = 3B_1 = 3\left(\frac{4}{3}\times10^{-5}\right)B=3B1​=3(34​×10−5) B=4×10−5 TB = 4\times10^{-5}\,\text{T}B=4×10−5T

Since 1 T=1 Wb/m21\,\text{T} = 1\,\text{Wb/m}^21T=1Wb/m2,

B=4×10−5 Wb/m2B = 4\times10^{-5}\,\text{Wb/m}^2B=4×10−5Wb/m2
  1. Option check
  • A: 2×10−52\times10^{-5}2×10−5 ❌
  • B: Zero ❌
  • C: 8×10−58\times10^{-5}8×10−5 ❌
  • D: 4×10−54\times10^{-5}4×10−5 ✅
  1. Final answer

The magnetic field at the center of the triangle is

4×10−5 Wb/m2\boxed{4\times10^{-5}\,\text{Wb/m}^2}4×10−5Wb/m2​

So the correct option is D.

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