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Let me explain in details. Consider a region in space with no free charges and no free currents, so that the charge density is $\rho=0$ and the volumetric current is $\vec{J}=\vec{0} \text{ }$ in the entire space. Let $\vec{B}=\vec{B}(\vec{x},t) \text{ }$ be the magnetic field in the space. Suppose that $\vec{B} \text{ }$ is uniform in space, that is, suppose that $\vec{B} \text{ }$ is only a function of time: $\vec{B}(\vec{x},t)=\vec{B}(t) \text{ }$. Moreover, suppose that $\vec{B}(t)=at\hat{e}_z$ for some constant $a$. Notice that although $\vec{B} \text{ }$ is uniform, $\vec{B} \text{ }$ varies with time. Under these circumstances, what is the electric field generated by $\vec{B} \text{ }$ in the entire space?

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Something like betatron? – Vladimir Kalitvianski Oct 27 '11 at 6:38

Im sorry, but the answer cannot be determined from the given information. We know $ E $ is constant in time, that the curl of $E$ is constant in time and space, but that is not enough to determine $E$. I can write down an infinite number of solutions which satisfy those requirements. You need to specify some additional boundary conditions.

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Hm, you're right. There are no boundary conditions for this problem, so there is no reason for us to expect one solution. Besides, no mention was made as to how $\frac{\partial{\vec{B}}}{\partial{t}}$ behaves. Nevertheless, I have fixed this in the problem's statement. However, I am still intrigued. Could you please write me one possible solution (regardless of its boundary conditions)? I couldn't even find one possible solution for this problem. Notice that there are both axial and translational simmetry, which suggests that $\vec{E}$ is uniform, an absurd! – Physicist Student Oct 27 '11 at 3:37
@FalsoFalseta: for simple case, you need go no farther than the electric field inside solenoid with time-dependent current. – Jerry Schirmer Oct 27 '11 at 3:59
+1 and this is expanded further in… – Physiks lover Oct 27 '11 at 12:30

$\nabla\times B = \mu_0 J + \mu_0\epsilon_0 \frac{\partial E}{\partial t}$

Since the magnetic field is constant in position, $\nabla \times B=0$, and so, since $J=0$, you have $E$ doesn't change (and stays zero).

If the obvious implications of what I wrote above isn't clear to the reader, let me spell it out here. The OP asks what electric field is "generated". This means that I am not to assume that there's some previously existing electric field.

So at time $t=0$, we assume $\vec{E} = 0$ everywhere. But the above shows that $\vec{E}$ stays constant. Therefore, $\vec{E}$ remains zero, and no electric field is generated.

As a general comment that the reader will find useful for all sorts of coupled sets of partial differential equations describing waves, if one is interested in the evolution over time of a solution, one should pay attention to the partial derivatives with respect to time and solve for them. This is a different form of Maxwell's equations.

This is the sort of thing you have to do if you are going to write a computer program which approximates the evolution of a solution to Maxwell's equations. That is, solve for $\partial E/\partial t$ and $\partial B/\partial t$.

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What about the other Maxwell equation? Curl of E must be nonzero. – user1631 Oct 27 '11 at 0:38
That proves that $\vec{E}$ is constant in time. But what's the behavior of $\vec{E}$ in space? – Physicist Student Oct 27 '11 at 0:56
Hmmm. I thought what I wrote was pretty clear. I'll add some more comments. – Carl Brannen Oct 27 '11 at 21:01

The unique axisymmetric field is

$$ E_x(t) = - {ay/2} $$ $$ E_y(t) = {ax/2} $$

Which obviously satisfies Maxwell's equations. This would be the electric field generated in a circular solenoid centered at the origin. Note that the field has a center, it breaks the naive translation invariance.

You can add any electrostatic field $E'$ to the electric field above, and the electrostatic field to add is determined by the boundary conditions. Using a constant electrostatic field, you shift the center of rotational invariance from the origin to another point.

Using the general 2+1 quadratic electrostatic potential (the real and imaginary parts of the complex squaring function)

$$ \phi(x,y) = C xy + D (y^2-x^2)$$

You get a more general solution

$$ E_x = ({-a\over 2} + C) y - D x$$ $$ E_y = ({a\over 2} + C) x - D y $$

Setting D=0 and C= a/2, you get the field which is correct for an infinitely long elliptical solenoid along the z-axis, centered at the origin (two parallel plates of current in opposite directions). These cases are determined by symmetry, to specify the electrostatic field for a general configuration requires more information than is given, as user1631 said, in particular, the boundary conditions for E.

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That does it. Problem solved. Thanks! – Physicist Student Nov 30 '11 at 4:26
@PhysicistStudent: You don't need to say it, there's a check-box for that. – Ron Maimon Jul 21 '12 at 8:48

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