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There is a widely known formula for the magnetic field due to a moving charged particle. $$\frac{\mu_0}{4\pi} q \frac{\vec{v}\times\vec{r}}{r^3} $$

The usual derivation is as follows.

$$ dB = \frac{\mu_0}{4\pi} i \frac{\vec{dl}\times\vec{r}}{r^3}$$ (Biot Savart Law) And then

$$ i = \frac{dq}{dt}$$ so $$ i\vec{dl} = \frac{dq}{dt}\vec{dl} = dq\frac{\vec{dl}}{dt} = dq\vec{v} $$

Finally, $$ dB = \frac{\mu_0}{4\pi} dq \frac{\vec{v}\times\vec{r}}{r^3} $$

which on integration gives the above formula.

However, my teacher says that this formula is not correct since Biot Savart Law itself is applicable only for continuous flows, whereas a charged particle constitutes a discrete current. Is that true? If yes, is there any similar formula for the field due to a moving charged particle? Please show the derivation too in that case.

Edit: Griffiths himself writes at one point in his book that this equation is "simply wrong". In a footnote, he also writes that it is wrong in principle wheras it is true for non-relativistic speeds, and later on in his book, he goes on to prove that. (Example 10.4) What my confusion is that this "true for non-relativistic speeds" is also true for Coulomb's law. Why isn't that law also "simply wrong" then ?

Thanks.

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    $\begingroup$ For a moving charged particle, you can write the current in terms of a Dirac delta function, just like you can write the charge density of the particle as a Dirac delta function. This can be found, for instance, in the book Introduction to Electrodynamics by David Griffiths. Then, formally, the current is "continuous", and you can use those formulae. $\endgroup$ – leastaction Jun 15 '15 at 16:36
  • $\begingroup$ @leastaction Would you mind giving me some reference as to where Griffiths actually says that? (Like some chapter, or topic, ...). $\endgroup$ – Aritra Das Jun 16 '15 at 15:05
  • $\begingroup$ Section 1.5 (page 45) in Griffiths's book introduces the Dirac Delta function. See problem 1.46 for an orientation. $\endgroup$ – leastaction Jun 16 '15 at 20:31
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The problem I have with your (phrasing of your) teacher's statement is the concept "discrete current". What does that even mean?

When you look at the usual Biot-Savart law with continuous current, you consider an infinitesimally small line segment. As the size of the line segment becomes smaller, so does the amount of charge that you consider - until in the limit, you consider an amount of charge $dq$ that tends to zero as $dl$ tends to zero.

The only thing different when you have discrete particles is that the charge never tends to zero - it tends to a finite value. But that in no way invalidates the rest of the analysis.

As @leastaction@ said, if you consider the charge "lumpy" (a delta function) rather than continuous, the equations are virtually unchanged.

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  • $\begingroup$ Well, I am a bit confused. Griffiths himself writes at one point in his book that this equation is "simply wrong". In a footnote, he writes that it was wrong in principle wheras it is true for non-relativistic speeds, and later on in his book, he goes on to prove that. (Section 10.4) What my confusion is that this "true for non-relativistic speeds" is also true for Coulomb's law. Why isn't that law also "simply wrong" then ? $\endgroup$ – Aritra Das Jun 16 '15 at 15:00
  • $\begingroup$ @AritraDas - I don't know the answer to the question in your comment, sorry. $\endgroup$ – Floris Jun 16 '15 at 15:12
  • $\begingroup$ @Aritra, I think there is some confusion in your post. The equations you write are indeed valid in the non-relativistic regime. The example you refer to (Example 10.4) asks you to write down the fields of a point charge in a frame where it moves with a constant velocity. If you take the limit $c \rightarrow \infty$ in the result of this calculation (eqn. 10.68 of Griffiths), you will recover the "usual" Coulomb law. If you like, this is Coulomb's law for a charge moving at a constant velocity. $\endgroup$ – leastaction Jun 16 '15 at 20:34
  • $\begingroup$ @leastaction Well, then why does Griffith write that this is 'simply wrong' ? Does he actually mean that deriving the equation in that way (the way I showed in my original post) is wrong ? $\endgroup$ – Aritra Das Jun 17 '15 at 17:13
  • $\begingroup$ @AritraDas, did you read the footnote on the same page (or possibly the next page, depending on which edition you have) in which the author says "I say this loud and clear to emphasize the point of principle; actually, Eq. 5.43 is approximately right for nonrelativistic charges ($v << c$), under conditions where retardation can be neglected (see Ex. 10.4)." This is the point that was made in the above responses. Perhaps you are confusing the regimes of validity of these equations. $\endgroup$ – leastaction Jun 17 '15 at 17:21
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Perhaps this will help. Consider this from the point of view that Electrostatics is about the E and B fields being constant in time. The Coulombs expression for the E field is for static charge. Coulombs 1/r^2 E-field is accurate some of the time because static charge (in statistical bulk) is possible. A single moving electron (unlike the current in a long wire) definitely creates a changing B field at any position with respect to time, regardless of speed v, therefore it is not a "statics" situation for creating a magnetic field (for magnetism, statics requires a constant extended current such that the B field is a constant in time). This was my interpretation of Griffith's statement "simply wrong", in that for a moving point charge the expression will never be exact (though as you said, it is a good approx. for $v<<c$). I hope this adds value to the discussion.

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