Question 103: Magnetism and Matter

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 $$ 	imes $$10−5 T with the declination angle ~ 0o and dip of $$	heta $$ such that sin $$	heta $$ = 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 :

Accepted Answer

V
B
= vhBcos$$	heta $$
= 240 $$ 	imes $$ 5 $$ 	imes $$ 5 $$ 	imes $$ 10
$$-$$5
$$ 	imes $$ $${{\sqrt 5 } \over 3}$$
= 44.7 $$ 	imes $$ 10
$$-$$3
V
= 45 mV
V
w
= $$l$$vB sin$$	heta $$
= 15 $$ 	imes $$ 240 $$ 	imes $$ 5 $$ 	imes $$ 10
$$-$$5
$$ 	imes $$ $${2 \over 3}$$
= 1200 $$ 	imes $$ 10
$$-$$4
V
= 120 mV
From right hand rule, the charge moves to the left side of the pilot.

Answer

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

Answer

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

Answer

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

Answer

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

Magnetism and Matter

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