JEE Main2016PhysicsMagnetic Properties of MatterActual
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 - 1 . The earth's magnetic field over Delhi is 5 × 10 - 5 T with the declination angle ~ 0 o and dip of θ such that sin ⁡ θ = 2 3 . If the voltage developed is V B between the lower and upper side of the plane and
Options
- AV B = 40 m V ; V W = 135 m V with left side of pilot at higher voltage
- BV B = 45 m V ; V W = 120 m V with right side of pilot at higher voltage
- CV B = 40 m V ; V W = 135 m V with right side of pilot at high voltage
- DV B = 45 m V ; V W = 120 m V with left side of pilot at higher voltage
Correct answer
D. V B = 45 m V ; V W = 120 m V with left side of pilot at higher voltage
Step-by-step solution
V B = B H 5 2 4 0 B H = B cos θ, .... B v = B s i n θ B H = 5 5 × 1 0 - 5 3 B v = 1 0 3 × 1 0 - 5 T V B = 5 5 3 × 1 0 - 5 × 5 × 2 4 0 V B = 44.6 mV = 4 5 mV V w = B v ℓ V   = 10 3 × 10 − 5 × 15 × 240 = 1 0 - 4 × 1 2 0 0 V w = 1 2 0 mV (left side at fighter voltage)