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An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. Why is it so?

Qmechanic
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Niharika
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    This is definitely a duplicate. See: http://physics.stackexchange.com/q/73945/ – DanielSank Sep 28 '14 at 06:19
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    Not really helpful to mark as a duplicate to a question that has no accepted answer and is also marked as a duplicate. This is a better link - though the answer to this question is in the question itself - the magnetic part of the Lorentz force is always perpendicular to velocity and so ${\bf F} \cdot {\bf v}$ will always be zero. http://physics.stackexchange.com/q/16326/ – ProfRob Sep 28 '14 at 08:48
  • We actually do take into account the mutual magnetic forces. Co-incidentally, it is an example where newton's third law does not hold. – Gaurav Sep 28 '14 at 09:44
  • @RobJeffries Your justification that F.v=0 implies no magnetic force acts is an incorrect one. In that case, during uniform circular motion the force acting is zero simply because the force and velocity are mutually perpendicular.(?) – Gaurav Sep 28 '14 at 09:58
  • @PhysicistByLogic Get a grip and read what I wrote. It wasn't what you say. The force is always perpendicular to the velocity. No work is done. – ProfRob Sep 28 '14 at 16:12
  • As pointed out by Rob, no work is done. However, it does not mean that the force can be ignored altogether. It may still affect the particle's trajectory. (in the same way the gravitational force on a planet in circular orbit around a star does no work, but makes the planet follow a circular orbit). – Gaurav Sep 29 '14 at 03:51
  • @RobJeffries Pardon me for any misunderstanding, I didn't mean to offend. – Gaurav Sep 29 '14 at 04:32

1 Answers1

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The electric and magnetic forces are defined via the Lorentz force on a charged particle

$$\vec F = q(\vec E + \vec v \times \vec B)$$

The magnetic force comes from the second term which defines it to be perpendicular to the velocity and therefore displacement $\vec {dr}$; so it doesn't do any work.

John McAndrew
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