Charge q Relativity: Force Calculation & Observation

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SUMMARY

The discussion focuses on the force acting on a charge q placed at a distance d from an infinite charged wire with a linear charge density λ. It establishes that the force on charge q is calculated using the electric field generated by the wire. Additionally, it addresses the transformation of electric (E) and magnetic (B) fields according to the Lorentz Transformation, specifically under the influence of an observer moving at speed u along the wire. Notably, it concludes that there is no relativistic force acting on charge q in this scenario.

PREREQUISITES
  • Understanding of electric fields generated by charged wires
  • Familiarity with Lorentz Transformation in special relativity
  • Knowledge of linear charge density (λ)
  • Basic concepts of force in electromagnetism
NEXT STEPS
  • Study the derivation of electric fields from infinite charged wires
  • Learn about the implications of Lorentz Transformation on electromagnetic fields
  • Explore the concept of relativistic forces in different reference frames
  • Investigate applications of charge interactions in relativistic physics
USEFUL FOR

This discussion is beneficial for physicists, electrical engineers, and students studying electromagnetism and special relativity, particularly those interested in the interactions of charged particles in varying reference frames.

chaotic
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A charge q is placed at a distance d from an innite charged wire carrying lamdba charge per
unit length.
(a)What is the force acting on the charge q? (b)Find the force on charge q with respect
to an observer moving moving at speed u along the wire.
 
Last edited:
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chaotic said:
A charge q is placed at a distance d from an in nite charged wire carrying  charge per
unit length.
(a)What is the force acting on the charge q? (b)Find the force on charge q with respect
to an observer moving moving at speed u along the wire.

E and B fields transform according to the Lorentz Transformation:

[tex]\mathbf{E}'=\gamma (\mathbf{E}+\mathbf{v}\times \mathbf{B})+(1-\gamma)\frac{\mathbf{E}\cdot \mathbf{v}}{v^2}\mathbf{v}[/tex]

[tex]\mathbf{B}'=\gamma (\mathbf{B}-\frac{1}{c^2}\mathbf{v}\times \mathbf{E})+(1-\gamma)\frac{\mathbf{B}\cdot \mathbf{v}}{v^2}\mathbf{v}[/tex]
 
In a,there is no relativistic force!
 
Last edited:

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