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Why do charged particles only produce magnetic fields while in motion?

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Assuming you mean the Lorentz force, Because that's the way that nature works. –  Dimensio1n0 Aug 23 '13 at 18:34
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@DImension10AbhimanyuPS I don't think OP is talking about the Lorentz force; he's asking why static charge distributions don't produce magnetic fields. Where does the OP say anything about forces? Also, I'm baffled by the multiple downvotes. Is this not a legitimate question? –  joshphysics Aug 23 '13 at 18:52
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@Dilaton Slightly reordered, the question asks "Why do charged particles only produce magnetic fields when in motion?" How is that unclear? I edited the post to make it more clear. –  joshphysics Aug 23 '13 at 19:02
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Whoever it is, please stop flagging this as low-quality. It's appeared 3 (or 4?) times in the queue alreadys. –  Dimensio1n0 Aug 23 '13 at 19:45
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duplicate of: physics.stackexchange.com/q/51346 –  Physiks lover Aug 25 '13 at 15:33
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marked as duplicate by Ben Crowell, Emilio Pisanty, Dan, Waffle's Crazy Peanut, Qmechanic Aug 28 '13 at 9:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1 Answer

Edit. This answer is not quite right, see my answer in the question duplicate

No magnetic field from a static charge - Is there a simple physical argument to show why?

This is an answer from the perspective of classical electrodynamics.

Basically, the answer is "because Maxwell's equations say so," and these are the fundamental, empirically tested equations of classical electrodynamics. To be precise we prove the following:

Claim. For any localized charge distribution $\rho$ for which the corresponding current $\mathbf J$ vanishes, the magnetic field is everywhere zero.

Proof. First, recall the continuity equation: \begin{align} \frac{\partial \rho}{\partial t} +\nabla\cdot\mathbf J = 0 \end{align} The hypothesis $\mathbf J = \mathbf 0$ therefore implies that the charge distribution is static; $\partial\rho/\partial t = 0$. By Gauss's Law \begin{align} \nabla\cdot \mathbf E = \frac{\rho}{\epsilon_0}, \end{align} the electric field of a localized charge distribution has the following integral expression: \begin{align} \mathbf E(t,\mathbf x) = \int_{\mathbb R^3}d^3x'\,\rho(t,\mathbf x')\frac{\mathbf x - \mathbf x'}{|\mathbf x - \mathbf x'|} \end{align} It follows that since $\rho$ is time-independent, then so is $\mathbf E$: \begin{align} \frac{\partial\mathbf E}{\partial t} = \mathbf 0 \end{align} That fact, the fact that the current vanishes, and Ampere's Law \begin{align} \nabla\times \mathbf B = \mu_0\mathbf J +\mu_0\epsilon_0 \frac{\partial \mathbf E}{\partial t} \end{align} combine to require the magnetic field to have zero curl; \begin{align} \nabla\times\mathbf B = 0 \end{align} But now recall that one of Maxwell's equations tells us that the divergence of the magnetic field is zero; \begin{align} \nabla\cdot \mathbf B = 0 \end{align} It follows from the Helmholtz Decomposition, that the magnetic field must vanish everywhere provided it falls off sufficiently rapidly at infinity, a reasonable property that should be true of physical charge distributions.

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Thank you very much joshphysics for your answer and support. –  APARAJITA Aug 24 '13 at 6:29
    
@APARAJITA Sure thing. I hope my edit of your question was appropriate. –  joshphysics Aug 24 '13 at 7:47
    
Yes it was, so is this the reason for charged particle to have magnetic field when in motion? –  APARAJITA Aug 24 '13 at 12:40
    
Josh it might be stupid for me to ask but don't you think there will be some kind of similarity between a moving charged particle and magnet? –  APARAJITA Aug 24 '13 at 12:46
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@APARAJITA Well the answer is yes; they are similar in the sense that they both produce magnetic fields. –  joshphysics Aug 24 '13 at 18:30
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