JEE Main 2016PhysicsMagnetic Properties of MatterHardMCQ

JEE Main 2016 — Magnetic Properties of Matter Question with Solution

JEE Main 2016 (10 Apr Online)

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 5×10-5T with the declination angle ~0o and dip of θ such that sin⁡θ=23. 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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Show full solutionCorrect option: D
Correct answer
DVB=45 mV;VW=120 mV with left side of pilot at higher voltage

Step-by-step explanation



V B = B H 5 2 4 0

B H = B cos θ, ....

Bv=Bsinθ

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)

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About this question

This is a previous-year question from JEE Main 2016, covering the Magnetic Properties of Matter chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.