Voltage difference of the wing of an airplane in Earth's magnetic fiel

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SUMMARY

The discussion focuses on calculating the voltage difference across the wing of an airplane in Earth's magnetic field using the equation V = vBL sin θ. The parameters include a velocity of 223 m/s, a wing length of 35 m, and a magnetic field strength of 0.50 gauss, which should be converted to tesla for accuracy. Participants emphasize the need to use the correct units and equations relevant to electromagnetic induction, specifically addressing the relationship between magnetic fields and induced voltage.

PREREQUISITES
  • Understanding of electromagnetic induction principles
  • Familiarity with the equation V = vBL sin θ
  • Knowledge of unit conversions, specifically gauss to tesla
  • Basic proficiency in physics equations related to force and charge
NEXT STEPS
  • Study the conversion of gauss to tesla for accurate magnetic field measurements
  • Explore the derivation and applications of the equation V = vBL sin θ
  • Research the effects of Earth's magnetic field on aircraft performance
  • Learn about the principles of electromagnetic induction in practical applications
USEFUL FOR

Aerospace engineers, physics students, and anyone interested in the interaction between magnetic fields and moving conductive materials, particularly in aviation contexts.

aChordate
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Homework Statement



5.jpg


Homework Equations



B=F/|q0|vsinθ

V=IR

The Attempt at a Solution



V=500m/h = 8.05m/h = 223m/s
L=35m
B=0.50gauss

0.50g=F/|q0|*223m/s*sin90
 
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aChordate said:
B=F/|q0|vsinθ
Wouldn't V = vBL sin θ be more to the point?
V=500m/h = 8.05m/h = 223m/s
I guess you mean v=500mph = 805kph = 223m/s
 
aChordate said:

Homework Statement



View attachment 59730

Homework Equations



B=F/|q0|vsinθ

V=IR

The Attempt at a Solution



V=500m/h = 8.05m/h = 223m/s
L=35m
B=0.50gauss

0.50g=F/|q0|*223m/s*sin90

You need to change gauss to tesla.
Also, none of your proffered equations are directly relevant. Consider haruspex's suggestion.
 

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