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Magnetics question

2020 · 3 Sep · Shift 1 · Q49
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Magnetics question

2020 · 3 Sep · Shift 1 · Q49

JEE MainPhysicsMagneticsMCQ+4 / −1
Magnitude of magnetic field (in SI units) at the centre of a hexagonal shape coil of side 10 cm, 50 turns and carrying current I (Ampere) in units of μ0Iπ{{{\mu _0}I} \over \pi }πμ0​I​ is :
  1. A
    250 3\sqrt 33​
  2. B
    5 3\sqrt 33​
  3. C
    500 3\sqrt 33​
  4. D
    50 3\sqrt 33​
View written solutionFree

Correct answer: C

  1. Magnetic field due to one straight side of a regular hexagon

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 centre of a regular hexagon, for each side the point lies on the perpendicular bisector, so

θ1=θ2=θRightarrowBone side=μ0I2πrsin⁡θ\theta_1=\theta_2=\theta Rightarrow B_{\text{one side}}=\frac{\mu_0 I}{2\pi r}\sin\thetaθ1​=θ2​=θRightarrowBone side​=2πrμ0​I​sinθ
  1. Geometry of the regular hexagon

Given side length

a=10 cm=0.1 ma=10\text{ cm}=0.1\text{ m}a=10 cm=0.1 m

For a regular hexagon:

  • Circumradius =a=a=a
  • Apothem (distance from centre to a side) is
r=acos⁡30∘=32ar=a\cos 30^\circ=\frac{\sqrt{3}}{2}ar=acos30∘=23​​a

Also, half of one side is

a2\frac{a}{2}2a​

From the right triangle formed,

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

Substitute r=32ar=\frac{\sqrt{3}}{2}ar=23​​a:

tan⁡θ=a/2(3/2)a=13\tan\theta=\frac{a/2}{(\sqrt{3}/2)a}=\frac{1}{\sqrt{3}}tanθ=(3​/2)aa/2​=3​1​

Hence,

θ=30∘\theta=30^\circθ=30∘

So,

sin⁡θ=12\sin\theta=\frac{1}{2}sinθ=21​
  1. Field due to one side
Bone side=μ0I2πr⋅12=μ0I4πrB_{\text{one side}}=\frac{\mu_0 I}{2\pi r}\cdot \frac12 =\frac{\mu_0 I}{4\pi r}Bone side​=2πrμ0​I​⋅21​=4πrμ0​I​

Now put r=32ar=\frac{\sqrt3}{2}ar=23​​a:

Bone side=μ0I4π(32a)=μ0I2π3 aB_{\text{one side}}=\frac{\mu_0 I}{4\pi \left(\frac{\sqrt3}{2}a\right)} =\frac{\mu_0 I}{2\pi \sqrt3\, a}Bone side​=4π(23​​a)μ0​I​=2π3​aμ0​I​
  1. Field due to all 6 sides of one turn
Bone turn=6×Bone side=6⋅μ0I2π3 a=3μ0Iπ3 a=3 μ0IπaB_{\text{one turn}}=6\times B_{\text{one side}} =6\cdot \frac{\mu_0 I}{2\pi \sqrt3\, a} =\frac{3\mu_0 I}{\pi \sqrt3\, a} =\frac{\sqrt3\,\mu_0 I}{\pi a}Bone turn​=6×Bone side​=6⋅2π3​aμ0​I​=π3​a3μ0​I​=πa3​μ0​I​
  1. For 50 turns
B=50×3 μ0IπaB=50\times \frac{\sqrt3\,\mu_0 I}{\pi a}B=50×πa3​μ0​I​

With a=0.1a=0.1a=0.1 m,

B=50×3 μ0Iπ×0.1=5003(μ0Iπ)B=50\times \frac{\sqrt3\,\mu_0 I}{\pi \times 0.1} =500\sqrt3\left(\frac{\mu_0 I}{\pi}\right)B=50×π×0.13​μ0​I​=5003​(πμ0​I​)
  1. Required value in units of μ0Iπ\dfrac{\mu_0 I}{\pi}πμ0​I​

Thus the magnitude is

5003500\sqrt35003​

So the correct option is C.

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