I understand that the magnetic force exists as a consequence of length contraction between 2 charge carriers in a current carrying wire (wrt the frame of the moving charge); but I don't understand how to arrive at the equation $$F =qvB\sin (\theta).$$


1 Answer 1


Actually there no force name qvb ,it is lorentz force ,that is
In physics (specifically in electromagnetism) the Lorentz force (or electromagnetic force) is the combination of electric and magnetic force on a point charge due to electromagnetic fields. A particle of charge q moving with a velocity v in an electric field E and a magnetic field B experiences a force of


Maxwell's equations do not contain any information about the effect of fields on charges. One can imagine an alternate universe where electric and magnetic fields create no forces on any charges, yet Maxwell's equations still hold. (E and B would be unobservable and totally pointless to calculate in this universe, but you could still calculate them!) So you can't derive the Lorentz force law from Maxwell's equations alone. It is a separate law.


--> Some people count a broad version of "Faraday's law" as part of "Maxwell's equations". The broad version of Faraday's law is "EMF = derivative of flux" (as opposed to the narrow version "curl E = derivative of B"). EMF is defined as the energy gain of charges traveling through a circuit, so this law gives information about forces on charges, and I think you can derive the Lorentz force starting from here. (By comparison, "curl E = dB/dt" talks about electric and magnetic fields, but doesn't explicitly say how or whether those fields affect charges.)

--> Some people take the Lorentz force law to be essentially the definition of electric and magnetic fields, in which case it's part of the foundation on which Maxwell's equations are built.

--> If you assume the electric force part of the Lorentz force law (F=qE), AND you assume special relativity, you can derive the magnetic force part (F=qv x B) from Maxwell's equations, because an electric force in one frame is magnetic in other frames. The reverse is also true: If you assume the magnetic force formula and you assume special relativity, then you can derive the electric force formula.

--> If you assume the formulas for the energy and/or momentum of electromagnetic fields, then conservation of energy and/or momentum implies that the fields have to generate forces on charges, and presumably you can derive the exact Lorentz force law. To think about this i have article i read on website,the QM Lorentz force. Given $$H=\frac{1}{2m}\left(\mathbf{p}-\frac{e\mathbf{A}}{c}\right)^2+e\phi = \frac{\mathbf{\Pi}^2}{2m}+e\phi$$

the text claims

$$\frac{d\mathbf{\Pi}}{dt} = e\left[\mathbf{E}+\frac{1}{2c}\left(\frac{d\mathbf{x}}{dt}\times\mathbf{B}-\mathbf{B}\times\frac{d\mathbf{x}}{dt}\right)\right].$$\begin{align} \frac{d\mathbf{\Pi}}{dt} &= \frac{1}{i\hbar}[\mathbf{\Pi},H] \\ &= \frac{1}{i\hbar}\left[\mathbf{\Pi},\frac{\mathbf{\Pi}^2}{2m}+e\phi\right] \\ &= \frac{e}{i\hbar}\left[\mathbf{\Pi},\phi\right] \\ &= \frac{e}{i\hbar}\left[\mathbf{p}-\frac{e\mathbf{A}}{c},\phi\right] \\ &= \frac{e}{i\hbar}\left[\mathbf{p},\phi\right]- \frac{e}{i\hbar}\left[\frac{e\mathbf{A}}{c},\phi\right] \\ &= -e\nabla\phi- \frac{e^2}{i\hbar c}\left[\mathbf{A},\phi\right] \\ &= e\mathbf{E} - \frac{e^2}{i\hbar c}\left[\mathbf{A},\phi\right] \end{align}$$

  • $\begingroup$ If you are copying from an article, you need to mention it in "references", as it have copyright. $\endgroup$
    – user600016
    Sep 14, 2019 at 7:52
  • $\begingroup$ For formula i have written it $\endgroup$
    – Jack Rod
    Sep 14, 2019 at 8:46

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