2

Suppose we have a plane and a wire with current passing perpendicular to it. If we consider magnitude field of any point at a perpendicular distance $r$ from the wire, we have the following expression as field for points outside the wire:

$$ B = \frac{\kappa}{r} \hat{\theta}$$

Where $\kappa$ is some constant.

Now, let's suppose a similar configuration of a cylinder with uniform charge density and let's say we consider a point far away, then the field is given as:

$$ E= \frac{ \kappa'}{r} \hat{r}$$

Where $\kappa'$ is another constant.

In the derivation, the fact that both of these fields are symmetric are used in argument to find the magnitude, but my mind boggling doubt is, why is it that , on a fundamental level, that the the unit vectors are different?!

For instance, suppose we draw a line from origin to point on cylinder, then if we reflect the point and vector attached of a point not on line, then we get the correct answer for eletric field but with opposite sign for magnetic. Here is a pic for the magnetic field case:

enter image description here

In the above, I have reflected the vector on top most point by horizontal line. I get the bottom point and bottom vector but with wrong direction.

But had I done such a thing for electrostatic field, I'd have right vector for the reflected point. Very roughly speaking, why does electrostatic field have this symmetric which magnetic field doesn't?

tryst with freedom
  • 7,940
  • Current causes curl, charge causes divergence – jensen paull Aug 06 '22 at 07:30
  • In what you note about reflection, you are correctly finding that magnetic and electric field are different kinds of vectors, called axial and polar, respectively. A polar vector is the 'ordinary' kind. An axial vector has the same behaviour under rotations as a polar vector, and the opposite behaviour under reflections. – Andrew Steane Aug 18 '22 at 09:37

3 Answers3

2

The magnetic field is not really a vector field; it is a bivector field, meaning that at each point it has a magnitude and a pair of directions. However, since we are in three dimensions, we usually represent the magnetic field as a vector field pointing in a direction normal to the bivector.

In the case of a long, straight, stead-current-carrying wire, the true bivector form of the magnetic field points in the following two directions: the radial direction, and the direction of the current.

The electric field is truly a vector field. In the case of a long, straight, uniformly charged wire, the electric field points radially.

So in fact the electric and magnetic fields both point radially in these two analogous scenarios, but the magnetic field also has a second direction that it gets from the direction of the current. The fact that magnetic fields are sourced by currents (which have direction) and electric fields by charges (which do not) gives rise to magnetic fields having an additional direction.

The bidirectional nature of the magnetic field becomes obscured when we replace it by a normal vector. When the two directions are the radial direction and the longitudinal direction, the normal direction is the circumferential one that you usually associate with the magnetic field around a wire.

Brian Bi
  • 6,491
  • 27
  • 62
1

There are a few directions we can go for an answer. One is that the two problems have different symmetry: The electric case is spherical, while the magnetic case is cylindrical. But we may, as you might, think of the magnetic case as a $2D$ spherical problem, and ask why there is such a difference between the two cases.

Well, first there is a physical difference: the magnetic field has zero divergence, and therefore cannot have a radial component. If it had a radial component, then by $2D$ spherical symmetry it should be constant around some 1-sphere (the 1-sphere is just the circle), and give a nonzero net divergence. So instead of a combination of a radial and angular field, we only get the angular part in the magnetic case.

But the question still remains: Why does the $3D$ electric case have only a radial component, and vanishing angular part? One can use the fact that $\vec{E}$ has vanishing curl (in the stationary case), but there is another reason. Let's suppose that the electric field has some angular part $\vec{E_\Omega}$. Let's also look at the electric field on a 2-sphere of some radius. $\vec{E_\Omega}$ is a vector field defined on the 2-sphere - meaning that it is always tangent to the sphere (by definition). According to the Hairy Ball Theorem, it vanishes somewhere. By spherical symmetry, it should vanish everywhere on the sphere. Thus, the electric field is radial.

Note that we do not have the same problem in the $2D$ magnetic case, as the 1-sphere does have a constant vector field defined on it (meaning- tangent to it), which is nonvanishing- that is just $B \ \hat{\theta}$

Rd Basha
  • 2,122
  • 5
  • 17
  • Couldn't you use the same argument for magnetic field's angular component with this 3D sphere? – tryst with freedom Aug 09 '22 at 01:17
  • Only in a 3D spherically symmetric current setup... – Rd Basha Aug 09 '22 at 10:04
  • @Beautifullyirrational For such a tetup, the magnetic field is indeed zero (so there is no radial part and no angular part). see: https://physics.stackexchange.com/questions/225847/why-is-the-magnetic-field-of-a-spherically-symmetric-current-zero – Rd Basha Aug 09 '22 at 11:44
0

It's simply because the $E$ and $B$ fields don't satisfy the same equations: just look at the 4 Maxwell equations and see the 'asymmetry' in the roles of $E$ and $B$. If you start with different equations, then it's no surprise that you get different solutions.

peek-a-boo
  • 6,235
  • Hmm well yeah that's true. But rhe physical symmetry idea of each point on circle being indistinguishable from each other leads to same magnitude but different unit vector. I thought maybe something cause this which is deep – tryst with freedom Aug 06 '22 at 09:17