# Biot-savart law

1. Jan 15, 2007

### theName()

I am curious how you derive this law. It seems that you could derive coulomb  s law from gauss  s law, but are there are no similar analogy for Bio-Savart law.

2. Jan 15, 2007

### arunbg

Both Coulomb's law and Biot-Savart's laws were framed from experimental analysis of electomagnetic phenomena. They are fundamental and cannot be "derived" as such. However it can be said that coulomb's law is consistent with gauss's law, and biot-savart's with ampere's circuital law.

3. Jan 15, 2007

### theName()

if coulomb law be derived, from gauss law, then i don t see why the B.S law cannot be derive from a more general law? If the B.S law cannot be derived, then how do we know the equation at all? I think about this for the last two nights. How is it possible that there is a pie in the denominator...

Last edited: Jan 15, 2007
4. Jun 15, 2008

### Ataman

*Bump*

I am interested in this as well.

-Ataman

5. Jun 15, 2008

### Nick89

As said, the laws are formed by experimental analysis. The 4 pi r^2 is (probably, I don't know this for 100% certain) due to the geometry of the magnetic field.

Also, I believe Coulomb's law was not derived from Gauss' law. It can be derived, for a certain amount of problems involving the necessary symmetry, because they describe the same thing.

Last edited: Jun 15, 2008
6. Jun 15, 2008

### Ataman

But since Coulomb's law for point charges can be derived from Gauss' Law, you can use Gauss's Law for anything (I mean on a practical level). Which is why many books insist on Gauss's Law being more fundamental than Coulumb's Law.

Isn't this the case with the modified Ampere's Law as well? Couldn't you derive the magnetic field from a moving point charge or an infinitesimal bit of current, and get the Biot-Savart? Since Ampere's Law is considered more fundamental?

-Ataman

7. Jul 5, 2008

8. Jul 6, 2008

### Stuart Downs

Even though I am a EE with a MS degree I was teaching the second semester of E&M at a local junior college (Palomar Junior College, San Marcos, Ca.). That's when it was like I was in the matrix and saw things clearly one day --- when working with the Biot Sabvart law. I saw that I could obtain the Biot Savart law from elementary principles. I do not think now that this is an entirely experimental soultion --- it can't be --- see my post --- but obtainable from elementary physics --- confirmed by a UIUC Physicist. I'm sure this must have been done before --- perhaps not I do not know --- One thing for sure though it was very satifying --- THERE IS ONLY "ONE" FIELD !

All the best

Stu

9. Jul 20, 2010

### KishorB

Saw the posts here. Most are saying that Ampere's law is more fundamental than Biot Savart law and is applicable in practical problems too. One question to those: can you please tell me the way to find the magnetic field due to a circular current loop at its center as well as at a point on its axis using Ampere's law? please give me the detail.

If you can't, then why Ampere's law is fundamental even it is not applicable in practical situation?

10. Jul 20, 2010

### stevenb

I'm not sure how rigorous this is, and it probably isn't too rigorous, but one can psuedo-derive Biot Savart Law by analogy with Coulomb's Law. This is done by looking at the field of an infinite line of charge using superposition and Coulomb's Law. For a linear charge distribution, the static electric field drops off as 1/r with a line charge, rather than as 1/r^2 for a point charge. Similarly, Ampere's Law (static version) can be used to derive the field of an infinitely long straight wire with current, and again there is a 1/r dropoff in the magnetic field. Based on this, it is not hard to formulate Biot Savart for an infinitesimal current element, with a 1/r^2 dropoff in field. Other aspects of the law such as constant of proportionality and cross product rule are experimentally confirmed, and are necessary for the formulas to agree in all geometries.

Rather than thinking of Biot Savart as similar to Coulomb's Law, perhaps we should think of the field for a long current wire as the simpler case, and hence a better analogy. In practice, it's not (experimentally) easy to put current into a short wire segment without having the current in any connection wires interfere with a measurement. However, a long wire is easier. The local field measurements of a long wire won't be corrupted by the fields due to the connection wires that are very far away.

I expect that a mathematician can develop a formal proof that is less reliant on intuition and analogy. I too would be interested to see it, if someone knows of it. I think, if such a proof were straightforward, it would be in every electromagnetics book. I'm guessing the formal process is easier using a covariant formulation of Maxwell's Equations. As Stu said, there is really only one field entity, the electromagnetic field tensor, and this viewpoint opens up alternate ways of proving equivalence of various empirical (special case) laws. If I were going to try and prove this, I'd approach the problem from this point of view. However, I'm an engineer, not a mathematician, so I'll just wait until a smart guy shows up to give us the answer.

11. Jul 30, 2010

### stevenb

Well, that smart guy never did show up, so I decided to take a couple of hours and pretend to be one. I think I came up with an illustrative derivation of Biot Savart from Coulomb's Law. It doesn't seem to be fully rigorous because I used a low speed (non relativistic) limit at one point to simplify the derivation. I started off thinking to use a Lorentz transformation of the EM field tensor as a way to convert Coulomb's law to Biot Savart's law. This triggered some old memories where I saw people use the Lorentz length contraction to demonstrate how magnetic forces can be viewed as manifestations of electric forces. So, that's the basic idea, and it turns out to be relatively straightforward, much to my own surprise. What I show here is an outline of the derivation with mention of the key concepts.

Start by considering a long wire with with current $$I$$ and a test charge $$Q$$ moving parallel to it with velocity $$v$$. The idea is to consider the magnetic force on the test charge due to the current element that is perpendicular to the radial vector, or in other words, the current element closest to the test charge. This will result in a formula for Biot Savart which does not include the cross product (sine of the angle dependence), however, the cross product rule will be intuitively obvious from the geometry and can be inserted in later. I do this just to keep the derivation brief and more understandable.

First, note that the magnetic effects of currents depend on net charge flow, not on the number of charges, type of charge carrier, nor drift velocity, independently. Thus, we can simplify the derivation by assuming positive charge carriers with a suitable linear charge density $$q$$ chosen to match the assumption that the charges also move with velocity $$v$$ matched to the test charge. Charge neutrality will require that the charge density of moving positive charges are canceled by an equal, but stationary relative to the conductor, charge density of negative charges.

Now, note that there is no magnetic force on the test charge in a reference frame that moves with the test charge. That is F=QVxB is zero in that reference frame. This requires that the force in that reference frame is due to electric forces only. The electric forces will result, despite the charge neutrality, because of the Lorentz contraction from motion. Even at nonrelativistic speeds this is true. Effectively the average separation between positive charges will increase due to length contraction of the moving conductor (relative to the moving frame). Also the average separation of negative charges will decrease because they are also moving relative to the moving reference frame. Hence, we can write the following net effect of Coulombs law to calculate the electric force. This is the net effect from both positive and negative charges.

$${{F}\over{Q}}=E= {{q\; dl}\over{4\pi\epsilon_0 r^2 \sqrt{1- {{v^2}\over{c^2}}} }} - {{q\; dl \sqrt{1- {{v^2}\over{c^2}}} }\over{4\pi\epsilon_0 r^2}}$$

To simplify, use a first order expansion of the square root using the binomial expansion. This implies a non-relativistic limit for speeds much less than the speed of light.

$${{F}\over{Q}}=E\approx {{q\; dl (1+{{v^2}\over{2c^2}} ) }\over{4\pi\epsilon_0 r^2 }} - {{q\; dl (1-{{v^2}\over{2c^2}} ) }\over{4\pi\epsilon_0 r^2}}$$

This simplifies to

$${{F}\over{Q}}=E\approx {{q\;v^2 \;dl }\over{4\pi\epsilon_0 r^2 c^2 }} = {{I v \; dl}\over{4\pi\epsilon_0 r^2 c^2 }}$$

Noting that the speed of light can be related to permeability and permittivity and reinterpreting electric force as magnetic force QvB in the stationary frame, results in the following.

$$dB={{\mu_0 I \; dl}\over{4\pi r^2 }}$$

Now the cross product rule can be restored by noting that the length contraction only occurs in the direction of motion and not perpendicular to that. Hence for a parallel orientation, the electric field is equal from both positive and negative charges, so forces are zero. It's not hard to show that :

$$dB={{\mu_0 I \sin\theta \; dl}\over{4\pi r^2 }}$$

EDIT: thinking furthur on this, I think I have a mistake on the above. I think my calculation of electric force is off by a factor of two by not applying length contraction correctly. Then I need to equate this to magnetic force in a reference frame that has no electric force. I'll try this and update the derivation.

EDIT: OK, so here is my correction to the above.

In a reference frame moving with the test charge, the positive charges are at rest and the negative charges appear to move at velocity $$-v$$. Hence, the negative charges appear to be separated by a shorter distance due to length contraction. This results in net charge imbalance and stronger field due to negative charges, in a small current element. Applying coulombs law results in the following.

$${{dF}\over{Q}}=dE= {{-q\; dl}\over{4\pi\epsilon_0 r^2 \sqrt{1- {{v^2}\over{c^2}}} }} + {{q\; dl }\over{4\pi\epsilon_0 r^2}}$$

Using the non-relativistic limit to first order results in the following.

$${{dF}\over{Q}}=dE\approx {{-q\; dl (1+{{v^2}\over{2c^2}} ) }\over{4\pi\epsilon_0 r^2 }} + {{q\; dl }\over{4\pi\epsilon_0 r^2}}$$

and simplifying results in the following (which differs by a factor of 2 from before).

$${{dF}\over{Q}}=dE\approx {{-q\;v^2 \;dl }\over{8\pi\epsilon_0 r^2 c^2 }} = {{-I v \; dl}\over{8\pi\epsilon_0 r^2 c^2 }}$$

Now consider a frame of reference moving at half of the speed of the test particle. In this frame the charges in the wire are moving in opposite directions with equal speed $$v/2$$. Hence, the Lorentz contraction is the same for each and there is no net electric charge to generate electric field. This means that all force can be interpreted as due to magnetic field. Since the test charge is also moving as speed $$v/2$$ in this frame, the magnetic force is dF/Q=vB/2. Equating this to the above relation results in the following, if it is noting that the speed of light can be related to permeability and permittivity.

$$dB={{\mu_0 I \; dl}\over{4\pi r^2 }}$$

Considering non perpendicular current elements will result in a cross product relation since length contraction does not occur in a direction perpendicular to motion. The end result is the same as before.

$$dB={{\mu_0 I \sin\theta \; dl}\over{4\pi r^2 }}$$

Last edited: Jul 31, 2010
12. Nov 2, 2011

### m_gignac

It can be derived from Maxwell's equation for magetostatics. If you look at the divergence of the magnetic field, it is a divergenceless field, meaning there exists a vector potential such that the curl of the vector potential is that of the magnetic field. This can then be inserted into the curl of the magnetic field. I don't feel like typing it out completely, but this is a place to start. Once you get an express for the vector potential, just take the curl of it, and you get Biot Savart Law..

Cheers,

13. Jul 9, 2012

### Trifis

Biot-Savart law is not so general as Coulomb's law...

In relativity an observer can indicate a pure electric field but never a pure magnetic field. Thus deriving Biot-Savart within this scope wouldn't be possible without a leftover electric field!

Now how does this leftover electric field is compensated for a vast amount of particles (classical example: the straight current-carrying wire) is something I would personaly like to know too!

14. Oct 14, 2012

### gamebm

Here is a brief answer(like exercise with hints). The conclusion is these two laws are equivalent, and it is basically a (good) mathematics/physics exercise to show this.

(1) Let us forget about relativity, just concern static magnetic field, think about it, the problem is well define in this context already. Let us not talk about which one is more experimentally/theoretically fundamental, since I am a theorist and will show you below, they are equivalent, given one of them you get the other.

(2) Firstly, given Biot-Savart law, we can derive Ampere's law in its derivative form (then I guess you understand that by using Stokes's theorem we arrive to its integral form in no time). It is straightforward, calculate the curl of magnetic induction B on the l.h.s. of Biot-sarvat law. Now it is a good exercise of vector calculus, for simplicity, we can assume that we only have one circuit with constant current i (later you can convince yourself this assumption is not really necessary), take care of the fact that the derivative is acted on the spacial point where the B is observed (evaluated). Now you will face an expression like AX(BXC) which can be simplified using
AX(BXC)=B(A.C)-C(A.B)
Without further going into details, I just point out here one term will give you a delta function, while the other term vanishes. The term which contains a delta function will alrighty give you the r.h.s of differential form of Ampere's law. Consider it as an exercise if you are studying Jackson, if you are undergraduate, still it is a workable problem.

(3) Now the hard part, how to show Biot-Savart from Ampere's law. I read in a book which gives a detailed derivation, unfortunately the book is not in English, for who reads Portugues, the third volume of Moses's fisica basica contains such a derivation, which is at undergraduate level. Here I will outline how it works out. The (non-trivial) trick is to make use of scalar potential of magnetic field, by discuss the similarity (some rigorous arguments can be inserted when you think with care) between electric dipole and magnetic dipole, one can write down the expression of magnetic scalar potential for magnetic dipole. By summing infinitesimal elements (or integral in space), one arrives at an integral form of magnetic potential for any given spatial current distribution. Now to evaluate the magnetic induction, one only need to calculate the divergence of this potential, which can be carried out quite straightforwardly given the mathematical tool we are employed. I have no doubt that it can be derived by various methods, plz post some references if you know and pass by.

Last edited: Oct 14, 2012