# Calculate electric field induced by a changing magnetic field

I want to plot the electric field (as a vector field plot) which is induced by a changing magnetic field for some simple cases.

Suppose for example that the magnetic field changes linearly (or quatratically) with t. Then you may calculate the curl $\nabla \times \vec{E}$ of $\vec{E}$ via:

$$\nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t}$$

as in this question. However this doesn't give you a unique solution for $\vec{E}$.

I guess that the solution will become unique if I add some boundary conditions. But I don't know how to do this in detail.

Suppose you have the experimental setup, that the magnetic field is (homogenous and) perpendicular to a given area $A$ in the $x_1$,$x_2$ plane (for example $A$ being a circular or rectangular area) and zero for every point $P$ which $x_1$ and $x_2$ coordinates are outside of $A$.

What is needed for the experimental setup to make the solution unique. How to calculate the solution in detail, what will it look like?

Edit: Since there are no charges in my example you have also the equation $\nabla\cdot \vec{E} = 0$. But I don't see how this helps.

Edit2

Here is what I tried so far:

If you suppose that the electric field is symmetric around the $x_3$ axis (but why can I assume this?) one could procceed as follows:

Let $\gamma$ be a circular path in the $x_1$-$x_2$-plane with radius $r$ and center $(0,0,0)$:

$$2\pi r |\vec{E}| = \int_{\gamma} \vec{E} d\vec{s} = -\frac{d\Phi}{dt} = -\pi r^2 \frac{d |\vec{B}(t)|}{d t}$$

which implies:

$$|\vec{E}| = - \frac{r}{2} \frac{d |\vec{B}(t)|}{d t}$$

But this gives only the absolute value of $\vec{E}$ and not the direction. Furthermore it depends linearly on $r$ which would means that it goes to $\infty$ if $r$ goes to $\infty$. But that seems to be very unintuitive to me.

Also there is no distinguished point, so I should perhaps better assume translation invariance...

Edit 3

Bonus question: How to solve it via differential equations (not using the integral form)?

• The easiest way is (probably) to first use symmetry arguments to induce the direction of the $E$-field (easiest if $A$ is circular) and then use the integral version of Faraday's law. – ScroogeMcDuck Jul 10 '14 at 8:02
• @ScroogeMcDuck How would you do this in detail? – Julia Jul 10 '14 at 14:31

You're almost there. For the symmetry argument: first notice that Faraday's law, $\oint\textbf{E} \cdot d\textbf{l}=-\frac{d\Phi}{dt}$, looks the same as Ampère's law from electrostatics: $\oint\textbf{B}\cdot d\textbf{l}=\mu_0 I$.

Now consider a current (or a homogeneous current density) pointing in the positive $x_3$-direction. What is the direction of the magnetic field such a current would produce? Indeed, using the right-hand rule (or any of your favorite symmetry arguments) it readily follows that the $\textbf{B}$-field encircles the current, i.e. it is symmetric around the $x_3$-axis and points anticlockwise. I trust that you are familiar with symmetries of magnetic fields produced by steady currents.

Now compare the form of Faraday's and Ampère's laws. Because the laws look exactly the same, it's easy to see that the electric field due to a flux decrease in the $x_3$-direction will have the same symmetry as a magnetic field due to a current (density) in the $x_3$ -direction. Hence here the $\bf{E}$-field will also be symmetric around the $x_3$-axis and will point in the azimuthal direction! (It'll point clockwise if the flux increases, but this will follow from the calculation.)

Therefore, we can do the calculation in just the way you did, yielding

$\textbf{E}=-\frac{r}{2}\frac{dB}{dt}\hat{\phi}$ as you noted. (Here $\textbf{B}(t)=B(t) \hat{x}_3$.)

Note that in your calculation you had already assumed that $\textbf{E}$ is in the $\hat{\phi}$ direction when you said that $2\pi r |\vec{E}| = \int_{\gamma} \vec{E} d\vec{s}$, since this assumes that $\bf{E}$ and $d\bf{s}$ are parallel.

The last thing to notice is that your calculation holds when your contour $\gamma$ is in the circle $A$ where the flux changes. Let $R$ be the radius of $A$. If $\gamma$ is outside $A$, then it encloses all of the flux, hence $\frac{d\Phi}{dt}=\pi R^2\frac{dB}{dt}$ so that for $r>R$ we get

$\textbf{E}=-\frac{R^2}{2r}\frac{dB}{dt}\hat{\phi}$,

which nicely vanishes as $r\rightarrow \infty$.

Finally, notice that if we would calculate the magnetic field produced by a volume charge density $\textbf{J}=J_0\hat{x}_3$ with $J_0$ constant, and replaced in our answer $J_0$ by $-\frac{1}{\mu_0}\frac{dB}{dt}$, we'd get exactly the electric field above.

• Thanks. Especially the part about $r\to \infty$ was helpful. However I don't fully understand the symmetry argument. Intuitively in the case of the current in the wire the current is along the $x_3$-axis, which gives you a distinguished axis. However this seems not to be the case in the case of Faradays law. So why not integrate around a circle (inside the magnetic field region) around a point other than the origin? – Julia Jul 10 '14 at 17:21
• Regarding your analogy: The magnetic field in the case of the wire I would calculate via the Biot-Savart law. Is there an analogous way for this in the case of the induction? – Julia Jul 10 '14 at 17:22
• Another point that's not clear is the role of the boundary conditions in this derivation. – Julia Jul 10 '14 at 17:26
• 1. We took $A$ as a disc-shaped area with radius $R$ and center on the $x_3$ axis, hence also here $x_3$ is a distinguished axis, i.e. we have rotational symmetry around this axis. 2. I'm not sure there is, I guess you should be able to get such a law... Maybe Jefimenko's equations? The point is that it's much easier to calculate the magnetic field of the wire via Ampère's law, and then use the analogy. – ScroogeMcDuck Jul 10 '14 at 18:26
• 3. The point is that when using the integral version of Maxwell's law you don't need the boundary conditions, since you're not solving differential equations. When using the differential versions, you can use that if we know the divergence and the rotation of a vector field and the vector field goes to zero sufficiently fast (like $1/r^2$ at least, I think), the vector field is uniquely determined. I think this is called Helmholtz theorem, the (more) exact statement is in Griffiths. – ScroogeMcDuck Jul 10 '14 at 18:29