Question

A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 × 10–4 T and the dip angle is 30°......

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Answer 1

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Speed of the jet plane, v = 1800 km/h = 500 m/s

Wing spanof jet plane, l = 25 m

Earth’s magnetic field strength, B = 5.0 × 10−4 T

Angle of dip = 30°

Vertical component of Earth’s magnetic field,

BV = B sin

= 5 × 10−4 sin 30°

= 2.5 × 10−4 T

Voltage difference between the ends of the wing can be calculated as:

e = (BV) × l × v

= 2.5 × 10−4 × 25 × 500

= 3.125 V

Hence, the voltage difference developed between the ends of the wings is

3.125 V.

Answer 2

Answer:

V

=

$= \: 1800 \: kmph$

$= 1800 \: \times \frac{5}{18} \: kmph$

$= 500 \: m$

L = distance between the ends of the wings

$= \: 25 \: m$

Earth’s magnetic field at the location has a magnitude of =

$5 \times {10}^{ - 4}$

Angle

$= {30}^{°}$

Motional emf = e =

$= B_{v}vl \: \: \:$

Where

$B_{v} \: = vertical \: component \: of \: the \: earth \: magnetic \: field$

Hence the voltage difference developed between the ends is

= 5 × 10^-4 × sin30° ×500 × 25

= 3.1 V