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According to my text book:

The Lorentz Force F on a charged particle moving in an electromagnetic field is given by:

$$F= qE + q(v\times B)$$

then it states that "it is to be pointed out that only the electric force does work, while the no work is done by the magnetic force which is simply a deflecting force."

How can this by true won't the the magnetic force have a considerable magnitude, will it now impart kinetic energy to the charge? Any explanations will help a lot. Please keep the terminology simple, thanks.

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    $\begingroup$ Possible duplicate of Does a magnetic field do work on an intrinsic magnetic dipole? $\endgroup$ – Danu Aug 15 '16 at 12:25
  • $\begingroup$ Please try to search the site for similar questions first, before you ask a new one. $\endgroup$ – Danu Aug 15 '16 at 12:26
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    $\begingroup$ @Danu: I don't think this is a duplicate, the question is about why the Lorentz force doesn't do any work on a lone charge. Which is a question I don't think we have a duplicate of it, all I can find are those who suppose we already know that the Lorentz force does not work on a charge $\endgroup$ – ACuriousMind Aug 15 '16 at 12:28
  • $\begingroup$ @ACuriousMind It's not clear to me from the OP that this question is really about point charges. It seems very broad (and honestly speaking it sounds like it hasn't really been thought through). $\endgroup$ – Danu Aug 15 '16 at 12:36
  • $\begingroup$ @Danu thanks for linking that article but my question is why there is no work by the magnetic force, but in the link you provided the questioner already knows that bit, please read it again. $\endgroup$ – Batwayne Aug 15 '16 at 12:37
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Work is done at a rate of $\vec{F} \cdot \vec{v}$. So any force component perpendicular to the velocity at all times cannot do any work classically. The kinetic energy remains unchanged because for the kinetic energy, only the magnitude of the velocity and not its direction is important. All that magnetic force does is change the direction of the velocity while keeping the velocity magnitude constant and that does not need any work.

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  • $\begingroup$ I want to point out that work is actually $\vec F\cdot\vec d$. Alternatively, you can use $\int\vec F\cdot\vec v dt$. But what you have written is power, not work $\endgroup$ – Jim Aug 15 '16 at 12:54
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    $\begingroup$ @Jim "work is done at a rate of" and the work rate is power, is it not? $\endgroup$ – Sanya Aug 15 '16 at 12:58
  • $\begingroup$ Very sketchy wording. I'll give it to you, but you're a hair away from it not being acceptable. Way to be technically correct, the best kind of correct $\endgroup$ – Jim Aug 15 '16 at 13:00
  • $\begingroup$ @Jim I'm very sorry about my poor expression then - English is not my first language. How would you express that clearly? $\endgroup$ – Sanya Aug 15 '16 at 14:58
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    $\begingroup$ Since you are not a native speaker, I'll add this comment: I, for one, see nothing wrong with your phrasing. It doesn't look unusual to me. $\endgroup$ – garyp Aug 15 '16 at 17:40
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The fact that the magnetic forces do not work is not owing to the smallness or largeness of the magnitude of the magnetic field or magnetic force. But it is owing to the direction of the magnetic force relative to the direction of the velocity of the charged particle.

The magnetic field $\vec{B}$ produces a force $\vec{F}$ on a charged particle of charge $q$ and velocity $v$ according to the formula,

$\vec{F}=q(\vec{v}$x$\vec{B})$

As implied by this formula, this force will always be perpendicular to the velocity of the particle and thus the power imparted to the particle via this force, $\vec{F}.\vec{v}$ $=0$, always. Thus no amount of energy is transferred by the means of magnetic forces to the particle and thus the magnetic forces do not alter the speed of the particles.

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