# Is the Biot Savart Law applicable only for continuous currents?

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.

• 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. – leastaction Jun 15 '15 at 16:36
• @leastaction Would you mind giving me some reference as to where Griffiths actually says that? (Like some chapter, or topic, ...). – Aritra Das Jun 16 '15 at 15:05
• Section 1.5 (page 45) in Griffiths's book introduces the Dirac Delta function. See problem 1.46 for an orientation. – leastaction Jun 16 '15 at 20:31

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.

• 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 ? – Aritra Das Jun 16 '15 at 15:00
• @AritraDas - I don't know the answer to the question in your comment, sorry. – Floris Jun 16 '15 at 15:12
• @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. – leastaction Jun 16 '15 at 20:34
• @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 ? – Aritra Das Jun 17 '15 at 17:13
• As long as we are quoting books, Jackson's Classical Electrodynamics (often regarded as the authority on anything classical E&M), happily uses a point current source without comment to get a version of the Biot-Savart law for a single particle. Nonetheless, as I found the last time this question came up (physics.stackexchange.com/questions/166318/…), there are some people (like Aritra's teacher) who will get upset by this. – Rococo Sep 18 '15 at 5:21

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.