Why magnetic force isn't doing any work I know it's a similar case to centripetal force but can someone give me best possible answer not giving the dot product analogy but some high level mathematical proof and a useful analogy
Why charge doesn't speed up or speed down because of magnetic force because if magnetic field is acting like a Potential field then charge velocity should change e.g in projectile motion when taken in potential field it's independent axes of motion become dependent on each other
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2 Answers
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I think the work is $\int (\int\rho(E+\frac{\partial r}{\partial t}\times B)\cdot dr)dV=\int( \int\rho E\cdot dr)dV$ since (if I understand correctly) dr points in the same direction as $\frac{\partial r}{\partial t}$ and the cross product makes the resulting vector normal to dr (and B too for that matter).
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Why charge doesn't speed up or speed down because of magnetic force because if magnetic field is acting like a potential field then charge velocity should change
The charge velocity does change but the charge does not speed up (or slow down). The component of the magnetic force parallel to the velocity is zero and thus, only the direction of the velocity changes, not the magnitude (speed).
Why magnetic force isn't doing any work I know it's a similar case to centripetal force but can someone give me best possible answer not giving the dot product analogy
If the charge does not speed up (or slow down), the kinetic energy of the charge does not change - KE depends on the magnitude of the velocity (squared) - and thus, no work is done on or by the charge by the magnetic force.
but some high level mathematical proof
Why???
answered Aug 27, 2016 at 2:22
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$\begingroup$ I require high mathematical proof because I think it's the place where symmetry of physics lies it would help me look at this concept a lot better $\endgroup$– MahinAug 27, 2016 at 2:25
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$\begingroup$ @user112820, it doesn't work that way here. $\endgroup$ Aug 27, 2016 at 3:28
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$\begingroup$ I have modified my question now can you give me a better answer $\endgroup$– MahinAug 27, 2016 at 5:00