Is there an intrinsic difference between an electric and a magnetic field?

Question

Or is the only difference that a magnetic field is an oscillating electric field?

Answer 1

There are a couple of ways of answering this, but at the most obvious level, the force laws in the presence of an electric and magnetic field are very different. In the presence of an electric field, a charge of magnitude $q$ feels a force:

$${\vec F} = q {\vec E}$$

while the same charge feels a force:

$${\vec F} = q {\vec v} \times {\vec B}$$

in the presence of a magnetic field. ($\vec v$ is the velocity of the charge in the relevant reference frame). So, these force equations are very different, so electric and magentic forces manifest quite differently.

More fundamentally, Einstein was able to show that the electric and magnetic fields are not independent entities, but rather belong to an object called the electromagnetic tensor:

$${\bf F} = {\vec E}\cdot dt \wedge d{\vec x} + \epsilon_{abc}B^{a}dx^{b}dx^{c}$$

which is capable of explaining the weird appearance of the velocity in the magnetic force equation, and gives the electromagnetic force the same form in every reference frame, amongst other things.

Answer 2

Is there an intrinsic difference between an electric and a magnetic field?

Yes, the force (on an electric charge) due to a magnetic field can do no work while the force due to an electric field can.

As Jerry Schirmer points out, the magnetic force on a charge $q$ in a magnetic field $\vec B$ is

$$\vec F_m = q\vec v \times \vec B$$

The power is then

$$P_m = \vec F_m \cdot \vec v = q\, \vec v \cdot (\vec v \times \vec B) \equiv 0 $$

Answer 3

According to one of Maxwell's equations, $\vec \nabla \cdot \vec B=0$.

As others have suggested, in an $(n+1)$-dimensional spacetime, the electric field is a [polar] vector with dimensionality $n$, whereas the magnetic field is a bivector with dimensionality $n(n-1)/2$ [which, in three spatial dimensions, can be thought of as a pseudovector (with 3 components)].