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?

  • $\begingroup$ I was missing some minus signs and a derivative (which have been fixed) but that still doesn't solve my dilemma. If $\textbf{B}$ is constant in time, shouldn't the circulation of $\textbf{B}$ be constant in time as well? $\endgroup$
    – newuser188
    Commented Nov 13, 2019 at 19:51
  • 1
    $\begingroup$ The problem seems to be this electric field is not consistent with Maxwell's equation - i.e., this can not be a physical field, since it generates constant B in time (through Faraday's law) and harmonic varying B in time (through Ampère's law). (I deleted the old comment, because I got confused as well). $\endgroup$
    – Woe
    Commented Nov 13, 2019 at 20:37
  • $\begingroup$ Indeed, you can see that this form is not consistent with special relativity either: a change in the field (in time) propagates instantaneosuly to every point in space. $\endgroup$
    – Woe
    Commented Nov 13, 2019 at 20:51

2 Answers 2


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 $\left(\frac{\omega a}{c}\right)$.


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.

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.


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.

  • $\begingroup$ I don't understand your point. $\frac {\partial \mathbf {B}}{\partial t} =0$ implies that $\mathbf {B}(\mathbf {r},t)$ is independent of time. The equation in the yellow box of your answer is connected to the material derivative and I don't think it makes sense in this context. Material derivative is used to study how a macroscopic field $f (\mathbf {r},t) $ of a "material element" changes when subjected to a macroscopic flow/"velocity" field like in the case of fluids. Because what does $\dot {\mathbf {r}} $ mean in this case otherwise? ($\mathbf {r} $ and $t $ are independent parameters) $\endgroup$
    – Ajay Mohan
    Commented Nov 14, 2019 at 2:54
  • $\begingroup$ Also, I believe OP's question is technically inconsistent. Because you can show by differentiating with time both sides of the equation that OP obtained, to get : $\oint_C \frac {\partial \mathbf {B}}{\partial t} \cdot d\mathbf {l} = -\epsilon_0 \mu_0 \omega^2 \pi a^2 sin (\omega t) $. The LHS is zero from Ampere's law but the RHS is non zero. $\endgroup$
    – Ajay Mohan
    Commented Nov 14, 2019 at 3:02
  • $\begingroup$ I am sorry, but $\partial \mathbf{B} / \partial t=0$ does not imply that $\mathbf{B}(\mathbf{r},t)$ is time independent. There is a huge difference between a partial derivative and a total one. Maybe you should check math.stackexchange.com/questions/174270/… $\endgroup$ Commented Nov 14, 2019 at 23:55
  • $\begingroup$ $\mathbf {B}(\mathbf {r},t) $ gives the magnetic field at point $\mathbf {r} $ at time $t $. Here, $\mathbf {r } $ and $t $ are _independent_ parameters (changing t does not affect $\mathbf {r} $ : I think this is where you might be wrong). If $\frac {\partial \mathbf {B}(\mathbf {r},t)} {\partial t}=0$, then it implies that $\mathbf {B}(\mathbf {r},t)$ at each point $\mathbf {r} $ is a constant that doesn't vary with time (I'm not saying that the field is uniform) : In other words, the field is independent of time. $\endgroup$
    – Ajay Mohan
    Commented Nov 15, 2019 at 3:28
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    $\begingroup$ While I admit that I should have used partial derivatives in my questions instead of "total derivatives" (which I've now fixed by editing the question), I think Humberto's answer is not addressing the true question and is rather unhelpful. $\endgroup$
    – newuser188
    Commented Nov 15, 2019 at 19:10

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