Applying Ampere's law in situation with non-physical $E$-field?

Question

On an exam I was given this question:

Suppose an electric field in a region with no current $(\textbf{J}=\textbf{0}$) is given by $\textbf{E}(t,x,y,z) = \sin(\omega t)\hat{\textbf{k}}$ and $C$ is the circle of radius $a$ in the $xy$-plane oriented counterclockwise when looking down the $z$-axis. Determine the value of $$ \oint_C \textbf{B}\cdot d\mathbf{\ell} $$ as a function of time.

The intent of the question is obviously to use Ampere's law to find that $$ \oint_C \textbf{B}\cdot d\mathbf{\ell} = \epsilon_0\mu_0\iint_S \frac{\partial\textbf{E}}{\partial t}\cdot d\textbf{S} = \epsilon_0\mu_0\cos(\omega t)\iint_S dS = \epsilon_0\mu_0\pi a^2\cos(\omega t), $$ where $S$ is the disc of radius $a$ in the $xy$-plane centered at the origin. However, if this electric field were to satisfy Maxwell's equations, we should have that the magnetic field is constant, since $$ \frac{\partial\textbf{B}}{\partial t} = -\nabla\times \textbf{E} = 0, $$ and so the circulation should be constant with respect to time as well.

I brought this concern up afterward with the instructor, but they weren't able to provide a satisfying answer to the dilemma.

Is there a way to make sense of this, or should I not bother?

Answer 1

In a region with no current or charge density, the electric and magnetic fields obey the wave equation

$$\left( \frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \nabla^2 \right) \mathbf E(\mathbf r, t) = 0$$

This result follows directly from Maxwell's equations (you are most likely familiar with this derivation; if not, it is straightforward and can be found e.g. here).

Obviously the electric field you were given does not satisfy the wave equation, which means that it is not part of a viable solution to Maxwell's equations. In other words, there is no magnetic field $\mathbf B(\mathbf r,t)$ such that $\mathbf E(\mathbf r,t)$ and $\mathbf B(\mathbf r,t)$ together satisfy all four of Maxwell's equations at the same time, which means that the question (which refers to the magnetic flux) has no meaningful solution.

This is a mistake (or possibly a deliberate omission) on the part of your instructor, likely stemming from a desire to simplify the calculation for you.

A Possible Fix

What s/he could have written is something like

$$\mathbf E(\mathbf r,t) = E_0 \sin(\omega t) \cos(\frac{\omega}{c} y) \hat{\mathbf k}$$

This electric field does satisfy the wave equation, and corresponds to a magnetic field given by

$$\mathbf B(\mathbf r,t) = \frac{E_0}{c} \cos(\omega t) \sin(\frac{\omega}{c}y) \hat{\mathbf i}$$

The trouble with something like this is that it makes that integral rather messy. However, our calculation can be simplified by assuming that the loop size is very small compared to the wavelength of this standing wave, i.e. $\frac{\omega a}{c} \ll 1$. Taylor expanding to second order gives us

$$\mathbf E(\mathbf r,t) = E_0 \sin(\omega t)\left(1 - \frac{\omega^2 y^2}{2c^2}\right) \hat{\mathbf k}$$ and $$\mathbf B(\mathbf r,t) = \frac{E_0}{c} \cos(\omega t) \left(\frac{\omega y}{c}\right)\hat{\mathbf i}$$

Performing the flux integral yields

$$\iint \frac{\partial \mathbf E}{\partial t} \cdot d\mathbf S = \omega E_0\cos(\omega t) \int_0^{2\pi} \int_0^a r \left(1 - \frac{\omega^2}{c^2}r^2 \sin^2(\theta)\right) dr d\theta$$ $$ = \omega E_0\cos(\omega t)\left( \pi a^2 - \frac{\omega^2}{c^2}\pi \frac{a^4}{4}\right)= \pi a^2 \omega E_0 \cos(\omega t)\left( 1 - \left[\frac{\omega a}{c}\right]^2\right) $$

The corresponding circulation in the magnetic field would be

$$ \oint \mathbf B \cdot d\mathbf r = \frac{\omega E_0}{c^2} \cos(\omega t) \int_0^{2\pi} a^2 \sin^2(\theta) d\theta = \frac{\pi a^2 \omega E_0}{c^2}\cos(\omega t)$$

which matches what we expect to lowest order in $\left(\frac{\omega a}{c}\right)$.

Discussion

The key to this is the dimensionless parameter $\epsilon \equiv \frac{\omega a}{c}$. The electric field provided by your instructor can be viewed as the coarsest possible approximation to a real electric field like the one I wrote down. Whether this approximation is sufficient depends on what you want to do with it - as you can see, it correctly gives the lowest order contribution to the electric flux through the loop, but it incorrectly gives that the corresponding magnetic field vanishes everywhere. In order to obtain the lowest order term in $\mathbf B$ from $\mathbf E$, we need to keep at least one higher order term.

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To summarize, your instructor provided you with an approximation to a physical $\mathbf E$ field. The approximation is sufficient to calculate the circulation of the corresponding $\mathbf B$ field to lowest order, but not sufficient to calculate the $\mathbf B$ field itself.

Answer 2

The Ampere-Maxwell Law involves the Partial Derivative of the magnetic induction field:

$$\nabla\times\mathbf{E}(\mathbf{r},t)=-\frac{\partial\mathbf{B}(\mathbf{r},t)}{\partial t}$$

Note the difference between a total derivative and a partial one:

Let $f(x,y,z,t)\equiv f(\mathbf{r},t)$ be a function of position and time, then its total time derivative is : $$ \frac{df(\mathbf{r},t)}{dt}= \frac{\partial f(\mathbf{r},t)}{\partial t}+\dot{\mathbf{r}}\cdot \nabla f(\mathbf{r},t)$$

So, asumming that the magnetic field is of the form $\mathbf{B}(\mathbf{r},t)=(B_x,B_y,B_z)$, the fact that the partial time derivatives vanish:

$$ \frac{\partial B_x}{\partial t}=\frac{\partial B_y}{\partial t}=\frac{\partial B_z}{\partial t}=0$$

does not say that $\mathbf{B}$ is constant in time. In this sense, your exercise is not ill-posed and the correct answer is the one you wrote. Nonetheless, a spatially uniform but time-varying Electric field is not at all physical, as you and some others have pointed out.