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Question:

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 m/s. The earth's magnetic field over Delhi is 5 × 10⁻⁵ T with the declination angle ≈ 0° and dip of θ such that sinθ = 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

VB=45mV;VW=120mVwith left side of pilot at higher voltage

VB=40mV;VW=135mVwith right side of pilot at high voltage

VB=40mV;VW=135mVwith left side of pilot at higher voltage

VB=45mV;VW=120mVwith right side of pilot at higher voltage

Solution:

VB = VBHl = 240 × 5 × 10⁻⁵ cos(θ) × 5 = 44.7mV
By right hand rule, the change moves to the left of pilot