JEE Main 2016 — Magnetic Properties Of Matter Question with Solution
From: JEE Main 2016 (Online) 10th April Morning Slot
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 ms−1. The earth’s magnetic field over Delhi is 5 10−5 T with the declination angle ~ 0o and dip of such that sin = 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 :
Choose an option
Show full solutionCorrect option: B
Correct answer
BVB = 45 mV; VW = 120 mV with left side of pilot at higher voltage
Step-by-step explanation
VB = vhBcos
= 240 5 5 105
= 44.7 103 V
= 45 mV
Vw = vB sin
= 15 240 5 105
= 1200 104 V
= 120 mV
From right hand rule, the charge moves to the left side of the pilot.
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