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Why magnetic field never do work?
if you pull a loop of wire (to the right) with velocity v and charge is moving perpendicular to your motion (upright) with velocity u, so which velocity does the magnetic force refer to? the loop of wire is immersed in an uniform magnetic field (toward the page)Magnetic force is always perpendicular to velocity so there is no work. (cos 90°=0)
Are you familiar with Poyntings theorem?Why magnetic field never do work?
No, I haven't learned that theorem yet.Are you familiar with Poyntings theorem?
OK, so Poynting's theorem is the key theorem about energy and work in electromagnetism. I will briefly derive it here (using natural units so that I don't have to keep track of constants) for the "microscopic" Maxwell's equations. Often it is derived for the macroscopic equations, so you can easily look those up also.
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The second term is the energy density of the electromagnetic field, and the third term is the power density of the work done on matter. So the first term is interpreted as the flow of electromagnetic energy from one location to another. So basically what it says is that any change in the electromagnetic energy density must be associated with a corresponding flux of electromagnetic energy or work done on matter.
Regarding your specific question in the OP, note that the term for work done on matter is ##E \cdot J## so the magnetic field is not involved.
Work is a non-thermal transfer of energy. Poyntings theorem involves a non-thermal transfer of energy. Therefore it describes work.There is no concept of work involved
Poynting's theorem is an equation relating EM fields with charge and current densities. It does not "involve a transfer of energy" by itself.Work is a non-thermal transfer of energy. Poyntings theorem involves a non-thermal transfer of energy. Therefore it describes work.
Sure. similar statements can be made of any equation in physics.Adopting the terms occurring in this theorem as EM energy and work done on matter is an interpretation introduced on top of this theorem; it is not part of it.
Well, the particular one that I showed was the microscopic expression, so it works for those situations at a microscopic level.This interpretation assumes power given to matter is given by j⋅E\mathbf j\cdot \mathbf E. This assumption is valid in some cases like Ohmic dissipation in conductor but deficient in others, like when electro-chemical and other non-EM electromotive and ponderomotive phenomena are involved.
I don't doubt that.In macroscopic theory with regular sources, Poynting's theorem is always valid. Its work-energy interpretation is not always valid; there are electro-chemical, thermo-electric effects where work on matter cannot be handled properly with simple expressions from the Poynting theorem.
By the way, what is your favorite version of the macroscopic Poyntings?In macroscopic theory with regular sources, Poynting's theorem is always valid.
By the way, what is your favorite version of the macroscopic Poyntings?