Does the Magnetic Field Affect the Dielectric Constant in the Lorentz Model?

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SUMMARY

The discussion clarifies that in the Lorentz model for calculating the dielectric constant, the magnetic field is typically disregarded due to its relative weakness compared to the electric field. Specifically, the Lorentz force, represented as qE, is significantly stronger than the magnetic force, qvB, at room temperature. However, in high-energy particle fields, the magnetic field does play a role. Additionally, while the magnetic field's temporal change can be comparable to the electric field, its effects are negligible in visible optics due to the larger wavelength of the electric field compared to atomic sizes.

PREREQUISITES
  • Understanding of the Lorentz model in electromagnetism
  • Knowledge of the Lorentz force equation (qE and qvB)
  • Familiarity with electromagnetic wave properties
  • Basic principles of optical activity
NEXT STEPS
  • Research the role of magnetic fields in high-energy particle physics
  • Study the effects of temporal changes in magnetic fields on dielectric properties
  • Explore the relationship between electric field wavelengths and atomic dimensions
  • Investigate optical activity and its dependence on electromagnetic fields
USEFUL FOR

This discussion is beneficial for physicists, electrical engineers, and students studying electromagnetism, particularly those interested in the interactions between electric and magnetic fields in various contexts.

bahaar
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Hi
I read in text for calculation dielectric constant in Lorentz model, we only regard electric field that coupled with each oscillator. So the magnetic field of input wave has no effect in calculation of dielectric constant in this model?
 
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Hi bahaar, the reason we don't consider magnetic field is because it is too weak in EM wave situation. E/B=c which is speed of light. Lorentz force qE is much greater than qvB, because v is not high enough in room temperature. But in high energy particle field, magnetic field does matter.

-Zephyron
 
That's not the only reason. While B itself may be weak, the change of B with t \partial B/\partial t=\rot E may be comparable to the electric field itself. However in the case of visible optics, the wavelength over which the electric field changes is much larger than the size of atoms or molecules, so that this effect is usually negligible, too. However it explains e.g. phenomena like optical activity.
 
Thanks both of you.:smile:
 

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