Electrodynamics - I'm trapped in a stupid paradox - pleas help

[coquelicot]

Homework Statement

B = variable magnetic field, of constant direction (e.g B = b cos(wt) N, where N=const vect)

A = magnetic potential vector of B

R = vector from some origin O to a point M

It is assumed that no exterior electric potential is present.

Problem: to find the electric field E at M caused by the variable magnetic field.

Homework Equations

A=1/2 B x R (A depends on the origin O, but this is OK since the magnetic potential vector is not uniquely determined).

E = -dA/dt (Maxwell-Faraday, since there no exterior electric potential is present)

It follows immediately that E = 1/2 dB/dt x R, but then, the electric field depends on the position of the origin O, an absurdity since the electric field is uniquely defined.

What's wrong?

The Attempt at a Solution

 

[vanhees71]

Let's write this in a bit more readable way (which may enhance the chance to get an answer to questions). Given is a magnetic field of the form

$$\vec{B}=\vec{B}_0 \cos (\omega t),$$

where $\vec{B}_0=\text{const}.$ Asked is after the electric field, given that there is no other than the induced field. To answer this question you first need Faradays Law (written in Heaviside-Lorentz units with $c=1$),

$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}=\vec{B}_0 \omega \sin(\omega t).$$

This determines $\vec{E}$ only up to the gradient of a scalar field:

$$\vec{E}=-\frac{\omega}{2} \vec{x} \times \vec{B}_0 \sin(\omega t) -\vec{\nabla} \Phi.$$

and thus you also need Gauss's Law

$$\vec{\nabla} \cdot \vec{E}=\rho=0,$$

where I assumed that there are no charges present, which I took as the meaning of the statement that there is no "external electric potential". From this you get

$$\Delta \Phi=0.$$

Finally we need to fulfill also the Ampere-Maxwell Law. Assuming that there's no current

$$\vec{\nabla} \times \vec{B} = \partial_t \vec{E}-\vec{\nabla} \partial_t \Phi.$$

Obviously the left-hand side is 0, and thus

$$\vec{\nabla} \partial_t \Phi=\frac{\omega^2}{2} \vec{x} \times \vec{B}_0 \cos(\omega t).$$

This leads to a contradiction since the right-hand side is not curl free, but the left-hand side is. Thus there doesn't exist any full solution of all Maxwell equations under the assumption that $\rho=0$ and $\vec{j}=0$. You can find a solution when assuming $\vec{j} \neq 0$, but this leads to something $\vec{j} \propto \vec{x} \times \vec{B}_0$, and then of course there's no translation symmetry anymore, and no contradiction occurs. Of course with the given magnetic field, this current is not unique in the first place since you can add any time-dependent gradient field to the electric field (which should be harmonic if you insist on $\rho=0$).

In conclusion one must say the problem is not well posed and has no unique solution. Of course, already the magnetic field is not very realistic.

 

[coquelicot]

Thank you for your answer Vanhees71. I am not entirely satisfied: indeed, what you have proved is that there can't exists a (time) variable uniform magnetic field in the void space. This seems to me strange, even if you're probably right.