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Electromagnetic Induction question

2021 · 26 Feb · Shift 2 · Q49
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  5. /2021 · 26 Feb · Shift 2 · Q49

Electromagnetic Induction question

2021 · 26 Feb · Shift 2 · Q49

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
An aeroplane, with its wings spread 10 m, is flying at a speed of 180 km/h in a horizontal direction. The total intensity of earth's field at that part is 2.5 ×\times× 10 −-− 4 Wb/m2 and the angle of dip is 60 ∘^\circ∘. The emf induced between the tips of the plane wings will be ‾\underline{\hspace{2cm}}​.
  1. A
    88.37 mV
  2. B
    62.50 mV
  3. C
    54.125 mV
  4. D
    108.25 mV
View written solutionFree

Correct answer: D

  1. Given data
  • Wing span of aeroplane: l=10 ml = 10\,\text{m}l=10m
  • Speed of plane: v=180 km/h=180×10003600=50 m/sv = 180\,\text{km/h} = 180\times \frac{1000}{3600} = 50\,\text{m/s}v=180km/h=180×36001000​=50m/s
  • Earth’s magnetic field: B=2.5×10−4 TB = 2.5\times 10^{-4}\,\text{T}B=2.5×10−4T
  • Angle of dip: δ=60∘\delta = 60^\circδ=60∘
  1. Relevant component of magnetic field

The plane is moving horizontally, and the wings are spread horizontally. The induced emf across the wing tips is due to the vertical component of Earth’s magnetic field.

If the angle of dip is 60∘60^\circ60∘, then the vertical component is

Bv=Bsin⁡δ=2.5×10−4sin⁡60∘B_v = B\sin\delta = 2.5\times 10^{-4}\sin 60^\circBv​=Bsinδ=2.5×10−4sin60∘

Using sin⁡60∘=32≈0.866\sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866sin60∘=23​​≈0.866,

Bv=2.5×10−4×0.866=2.165×10−4 TB_v = 2.5\times 10^{-4} \times 0.866 = 2.165\times 10^{-4}\,\text{T}Bv​=2.5×10−4×0.866=2.165×10−4T
  1. Motional emf formula

The motional emf across the wing tips is

E=Bvlv\mathcal{E} = B_v l vE=Bv​lv

Substituting the values,

E=(2.165×10−4)(10)(50)\mathcal{E} = (2.165\times 10^{-4})(10)(50)E=(2.165×10−4)(10)(50) E=2.165×10−4×500\mathcal{E} = 2.165\times 10^{-4} \times 500E=2.165×10−4×500 E=1.0825×10−1 V\mathcal{E} = 1.0825\times 10^{-1}\,\text{V}E=1.0825×10−1V E=0.10825 V=108.25 mV\mathcal{E} = 0.10825\,\text{V} = 108.25\,\text{mV}E=0.10825V=108.25mV
  1. Option matching

Thus the induced emf is

108.25 mV\boxed{108.25\,\text{mV}}108.25mV​

So the correct option is D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.

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