# Unit definitions for E and B

1. Jul 13, 2009

### DeepSeeded

Why is the electric field defind as per meter while the magnetic field is defined as per square meter?

Does Ampere's law only contain the current density in the solution because of this fact? Couldn't your simplify the Maxwell equations if you changed the SI definitions?

2. Jul 13, 2009

### clem

Yes, if you use Gaussian units.

3. Jul 13, 2009

### DeepSeeded

There is still a current density in the solution to the curl of the magenetic field whicle the curl of the electric field does not have any dependence on current density.

4. Jul 13, 2009

### clem

Why is that a problem? E and B each have units q/r^2 in Gaussian.

5. Jul 14, 2009

### Born2bwire

The magnetic analog (dual) of E is H, not B. B would be associated with D.

6. Jul 14, 2009

### clem

F=q[E+(v/c)XB]. Where is H?

7. Jul 14, 2009

### Born2bwire

$$\mathbf{B} = \mu \mathbf{H}$$

We can see dimensionally that the E and B will not be similar due to the velocity and acting on the B field. Maxwell's equations are:

$$\nabla\cdot\mathbf{D} = \rho_e$$
$$\nabla\cdot\mathbf{B} = \rho_m$$
$$\nabla\times\mathbf{E} = \mathbf{M}-\frac{\partial \mathbf{B}}{\partial t}$$
$$\nabla\times\mathbf{H} = \mathbf{J}+\frac{\partial \mathbf{D}}{\partial t}$$

The divergence of D and B are their associated monopole charges. The cross products of E and H are associated with the time derivative of the B and D fields (and current sources) respectively. The symmetry of the equations allows us to use the dual to easily convert between magnetic and electric field equations by

$$\mathbf{E} \rightarrow \mathbf{H}$$, $$\mathbf{H} \rightarrow \mathbf{E}$$, $$\mathbf{B} \rightarrow -\mathbf{D}$$, $$\mathbf{D} \rightarrow -\mathbf{B}$$,

$$\mathbf{J} \rightarrow \mathbf{M}$$, $$\mathbf{M} \rightarrow \mathbf{J}$$, $$\rho_e \rightarrow -\rho_m$$, $$\rho_m \rightarrow -\rho_e$$

So we would expect that the H and E fields be dimensionally similar like the OP is asking about. Which is true, the E field is V/m, H field is A/m, B field is Wb/m2 and D is C/m2. Maxwell's equations has the symmetry that I think that the OP was looking for, it is just that he was looking for symmetry between the wrong field quantities.

8. Jul 14, 2009

### DeepSeeded

Ok those equations are looking a lot more symetric, but I have no idea what H, D, and M are.....

9. Jul 14, 2009

### Bob S

There are two quantities, E and H, which have the following units in the mks system:
E units are volts per meter
H units are amp-turns per meter.
So both of these are per meter.

We also have the permeability and permittivity of free space
u0 has units Henrys per meter
e0 has units Farads per meter.

Now B = u0 H (in vacuum), so it has units (amp-turns) x Henrys per meter squared.
B also has units weber per meter squared and Tesla in the MKS system.

There is an electric equivalent called the displacement D = e0E, so
D has units volt-Farads per meter squared.

10. Jul 14, 2009

### DeepSeeded

Sweet, so what is M then? Magnetic field density?

11. Jul 14, 2009

### Bob S

M , called the magnetization, has the same units as H, but is related to H by a dimensionless constant called the magnetic susceptability χ:

M = χ H

In general. χ depends on H so we have

B = u0(H + M) = u u0 H

where u is the relative permeability (dimensionless)

Last edited: Jul 14, 2009
12. Jul 15, 2009

### Born2bwire

That is a correct definition for M, but I was not using it for magnetization but instead as a ficticious magnetic current.

For the OP, it is a common tool in electrical engineering to allow for ficticious magnetic charges and currents. In reality, classical electrodynamics sets the magnetic charges and currents to 0. However, they are a useful tool in analyzing problems. The first reason is that the symmetry can allow us to easily find the magnetic field equations from an electric field equations (and vice-versa) thus saving us some conversion time. This is done by taking the "dual" of the equation as I described above. A second reason is due to the use of the equivalence principle which states that any electromagnetic field can be described by a set of currents. That is, if I have an imaginary closed surface that has a set of fields passing through, I can replicate these fieds by impressing a set of currents on the imaginary surface. Say I have a cavity with an opening that will allow fields in the cavity to propagate out, then I can perfectly describe the fields leaving the opening by considering the opening to have a set of currents across it. These currents will excite the equivalent fields that would normally leave the cavity. In this case, we sometimes make use of the ficticious magnetic currents. The use of this technique is that if I want to simulate the fields leaving the cavity, I can solve for the cavity's behavior by itself and find the equivalent currents for the fields passing trhrough the opening. Then, I can just use this sheet of currents to model the fields outside of the cavity in any situation I want as opposed to having to solve for the cavity first and then propagate the fields out of the cavity and so on.

13. Jul 15, 2009

### cmos

Some on this thread have suggested that the OP was in error by comparing E to B rather than E to H. I don't necessarily think this is the case. After all, E and B are the fundamental quantities and only E and B are necessary while analyzing the microscopic Maxwell equations (i.e. the equations in free space; D and H are only necessary when considering fields in media).

As clem has alluded to above, E and B happen to be dimensionally equivalent in the Gaussian system of units (both have dimensions of dyne1/2/cm). So if we used this system of units the first time we learned E&M, the OP would not have asked the original question.

It is in SI units that E has dimensions of V/m, B has dimensions of Wb/m2, H has dimensions of A/m, and D has dimensions of C/m2. It is in this system of units that Born2bwire writes the macroscopic Maxwell equations above and asserts that E should be compared not to B but to H. The thing to note is that, experimentally, H is much easier to measure and much easier to set up than B. So from an engineer's perspective, it makes H a much more useful quantity than B. Similarly, E is a much more useful quantity than D.

It therefore pleases me to imagine that those who setup the SI definitions of electromagnetism did so with an intent to make the symmetry between E and H (as well as between D and B) as Born2bwire pointed out.

14. Jul 15, 2009

### cabraham

E is in volts/meter, & H is in amp-turn/meter. D is in amp-seconds/meter^2, while B is in volt-seconds/turn*meter^2.

The commonality is that E*D & H*B are both joules/meter^2.

Did this help? It's all about energy. By the way, is H the counterpart to E, or is it B? There is no answer because, E & H have similar units, as does D & B. In unit terms E relates to H and D to B. But the forces acting on an electron are q(E + u X B). Hence E & B appear together. But if we examinw the fields at the boundaries of 2 media we get this. The normal components of D and B must be the same for each media (no electric polarization, otherwise Dn1-Dn2 = rho). The tangential components of E & H are equal for both media. Hence E behaves like H, and D like B.

So E & H are correlated 2 of 3 ways, just as D & B.

There is really no answer as to "is E the counterpart of H or B?"

Claude

15. Jul 15, 2009

### Bob S

Or is it joules/meter3?

16. Jul 15, 2009

### cabraham

It is indeeed joules/meter^3, energy per volume.

Claude

17. Jul 16, 2009

### DeepSeeded

My direct confusion is that according to the 3rd maxwell equation in SI units it looks as if though the Magnetic field generated due to the curl of the E field has no relation to the strength of the E field or anything besides its curl.

In the 4th equation the addition of J gives some relation to the current of the magnetic field and the resulting Electric field's power output.

18. Jul 16, 2009

### cmos

One problem is that you are reading the equations slightly wrong. When one writes 'curl E = - dB/dt', it is actually the (changing) magnetic field that generates the solenoidal electric field (you have it the other way around). One reason for writing the Maxwell equations with the vector operators on the left hand side of the equation and the rest on the right hand side is to separate the 'causes' from the 'effects'. Sadly, placing the vector operators on the left hand side makes you look at the effect before you see the cause on the right hand side.

I think one other point to mention is that you have to remember that the Maxwell equations are differential equations. It is true that the equations by themselves do not uniquely determine the field quantities (essentially what you said in your first sentence). You need to specify boundary conditions in addition to Maxwell's equations to uniquely determine the fields.

In Ampere's law, the current density plays a similar role to the charge density in Gauss' law. The current density is a source of solenoidal magnetic fields as the charge density is a source of diverging electric fields.

19. Jul 16, 2009

### DeepSeeded

Wait a minute! You mean to tell me that if you have an electric field going around in a circle no magnetic field will be generated? How do speakers work then?

The equation is symmetric, it works both ways, it must.

20. Jul 16, 2009

### cmos

In your example, it is the current in those wires that generates the magnetic field (à la Ampère's law). True, there is an electric field that drives that current but ultimately it is the current that induces the magnetic field; for if you had an electric field permeating wires of zero conductivity (and thus driving zero current), you will not have an induced magnetic field.

Think of it another way: a simple DC electromagnet, i.e. a solenoid wrapped around an iron core such that a DC/constant current pass through the solenoid. Somehow, the current through the solenoid magnetizes the iron core. (Think of a wire wrapped around a pencil and the north/south poles being at either end of the pencil).

If you use Faraday's law as you have tried, then the solenoidal electric field (curl E) must produce a time-varying magnetic field (dB/dt). But since 'curl E' is constant in this DC example, then dB/dt must also be constant. This implies that the value of the magnetic field will monotonically increase towards infinity.

If you use Ampère's law (as you should), then the constant current (J) implies a magnetic field that is constant in time but only varies with position in space (curl B).