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JEE Main 2016 (10 Apr, Shift 1) · Motional EMF

Motional EMF question from JEE Main 2016 (10 Apr, Shift 1)

Q51JEE Main 2016 · 10 APR · SHIFT 1Difficult

Motional EMF

A fighter plane of length 20 m, wing span (distance from tip of one wing to the tip of the other wing) of 15 m and height 5 m is flying towards east over Delhi. Its speed is 240 ms−1. The earth’s magnetic field over Delhi is 5 ×\times 10−5 T with the declination angle ~ 0° and dip of θ\theta such that sin θ\theta = 2/3. If the voltage developed is VB between the lower and upper side of the plane and VW between the tips of the wings then VB and VW are close to:

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Answer: (b)

  1. BV=Bsin⁡θ=5×10−5×23=3.33×10−5 TB_V = B\sin\theta = 5 \times 10^{-5} \times \tfrac{2}{3} = 3.33 \times 10^{-5}\text{ T}, BH=Bcos⁡θ=5×10−5×53=3.73×10−5 TB_H = B\cos\theta = 5 \times 10^{-5} \times \tfrac{\sqrt{5}}{3} = 3.73 \times 10^{-5}\text{ T}
  2. Wings (horizontal, along N-S) cut BVB_V: VW=BVlv=3.33×10−5×15×240≈0.12 V=120 mVV_W = B_V l v = 3.33 \times 10^{-5} \times 15 \times 240 \approx 0.12\text{ V} = 120\text{ mV}
  3. Height (vertical) cuts BHB_H: VB=BHhv=3.73×10−5×5×240≈45 mVV_B = B_H h v = 3.73 \times 10^{-5} \times 5 \times 240 \approx 45\text{ mV}
  4. Flying east with BVB_V downward, v⃗×B⃗\vec{v} \times \vec{B} points north, which is the pilot's left, so the left wing tip is at higher potential.
  5. So the answer is (b).
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