Electromagnetism problem: where does the magnetic field come from?

Question

Consider the following problem:

Consider a plane with uniform charge density $\sigma$. Above the said plane, there is a system of conducting wires made up of an U-shaped circuit on which a linear conductor of lenght $d$ can slide with constant velocity $v$. The system as a whole has a rectangular shape and is parallel to the plane. (See the picture). Calculate the line integral of the magnetic field $\bf B$ along the perimeter $L(t)$ of said rectangle as a function of time.

My professor solves this problem using Maxwell's fourth equation in integral form, assuming that the current density $\bf {J} $ is everywhere null, and that the electric field $\bf E$ is the one generated by a uniformly charged plane, i.e. perpendicular to the the plane and of norm $E=\frac{\sigma}{2\epsilon_0}$; thus yielding $$\oint_{L(t)} {\bf B}\cdot dl=\mu_0\epsilon_0\frac{d}{dt}\int_{S(t)} {\bf E}\cdot dS=\mu_0\epsilon_0Edv=0.5\mu_0\sigma dv$$

I think there are some things wrong both with this solution:

1. There should be no magnetic field at all! A uniformly charged plane only produces an electrostatic field. (I know there could be a magnetic field generated by the current inside the wires, but then you couldn't assume that $\bf J$ is null everywhere as my professor did!)
2. Maxwell's fourth equation does not hold in that form if the domains of integration are allowed to vary with time. In fact, by resorting to the differential forms, we find that plugging ${\bf J}=\vec 0$ and $\frac{\partial {\bf E}}{\partial t}=0$, as my professor assumed, yields $rot{\bf B}=0$, and thus the line integral of the magnetic field over any closed curve, at any istant, should be zero by Stokes' theorem!

Therefore, my question is the following.

Are the my professor's assumption ($\bf J$ $=\vec 0$, $\frac{\partial{\bf E}}{\partial t}=\vec 0$) correct, or do both $\bf J$ and $\bf E$ need be modified so as to account for the charges present in the circuit? Is there a current in the circuit at all?

Answer 1

Your insight stated in 2. is correct! In the integral form of the fourth Maxwell equation with time varying integration surface, the time differentiation stays inside the integral: $$\oint_{L(t)} {\bf B}\cdot dl=\mu_0\epsilon_0\int_{S(t)} {\frac{\partial}{\partial t}\bf E}\cdot dS \tag{1}$$ Then from the assumption $\frac{\partial{\bf E}}{\partial t}= 0$ both the left hand and the right hand side should be zero in this case. However, Anton Fetisov has shown in his answer (s. below) that due to induced charges on the moving wire $\frac{\partial{\bf E}}{\partial t}\neq 0$. Therefore, your professor has obviously made mistakes but fortuitously obtained the correct answer.

Addendum following the answer of Anton Fetisov:
In his correct and deep going analysis of the problem, he considers the effects of the finite size of the metallic wire and the electric charges induced on its surface by the homogeneous electric field of the charged plane which are necessary to produce a zero total electric field in the wires. These induced charges and the associated deformation of the electrical field around the wire are moving with velocity $v$ in the $x$-direction.

Thus, from this point of view, there exist currents and time varying electric fields which is inconsistent with two basic assumptions made in the problem, i.e., $\bf J = 0$ and $\frac{\partial{\bf E}}{\partial t}= 0$. The second error is the solution with the wrong integral form of the 4th Maxwell equation for time varying integration surface/contour $$\oint_{L(t)} {\bf B}\cdot dl=\mu_0\epsilon_0\frac{d}{d t}\int_{S(t)} {\bf E}\cdot dS \tag{2}$$ The correct form is equation (1). From the given assumption $\frac{\partial{\bf E}}{\partial t}= 0$ it follows that the right hand side of equation (1) should be zero as I have stated before. This is, however, not correct in this particular case due to the fact that the induced charges on the wire cause a time varying field.

In his detailed analysis, Anton Fetisov has shown, that the right hand side of the correct equation (1) is not zero and that, surprisingly, it is equal to the right hand side of the incorrect equation (2). Thus the solution of the problem found by the professor with the incorrect equation (2) is fortuitously correct. Therefore, I have reduced my original short answer (first paragraph) to the still valid fact, already found by Nicol, that the form of the used Maxwell equation was generally not correct for the time dependent integration surface/contour.

Added simple derivation: For those who are not math virtuosos, I would like to show, on the basis of Anton Fetisov's reasoning, how the right hand side of the correct 4th Maxwell equation (1) can be evaluated for the considered problem in a simple way giving the result quoted in the question of Nicol.

The essential point is the charges on the wire that are electrostatically induced by the homogeneous electric field $E_0=\sigma/\epsilon_0$ of the sheet charge $\sigma$. Only the vertical y-component has to be considered for the the integral. These charges are the sources of an additional electrical field $\epsilon (x)$ in and closely around the wire which exactly cancels $E_0$ inside the wire and reduces it near the wire on a length scale of the wire diameter $2a$. This additional wire field $\epsilon (x)$ has the most negative value at a (flat) minimum $\epsilon _{min}= -E_0$ inside the wire, particularly on its axis. The exact functional form is irrelevant here, as long as its minimum at $x=0$ is $\epsilon (0)=-E_0$ and it is zero a couple of wire diameters horizontally away from the wire axis. The x- and t-dependence of the vertical field in the wire plane of the moving wire can be written as $\epsilon (x,t)=\epsilon (x-vt)$, where the axis of the wire (and field minimum) is located at $x_1=vt$. The total vertical electric field in the wire plane is then given by $$E(x,t)=E_0 + \epsilon (x) + \epsilon (x-vt)$$ (The second term on the RHS is the time-independent field of the left transverse wire.) Thus with $$\frac{\partial{E}}{\partial t}=\frac{\partial{\epsilon (x-vt)}}{\partial t}=\frac{\partial{\epsilon(x-vt)}}{\partial x}(-v)$$ the surface integral of the RHS of equation (1) reduces to $$\int_{S(t)} {\frac{\partial}{\partial t}\bf E}\cdot dS= -vd\int_{x=0}^{x_1=vt} {\frac{\partial \epsilon(x-vt)}{\partial x}} dx =-vd[\epsilon(x-vt)]_{x=0}^{x_1=vt}= vd[\epsilon (-vt)-\epsilon (0)]=vdE_0$$ where it has been assumed that $\epsilon (0)=-E_0$ and $x_1=vt>>2a$ so that $\epsilon (-vt)=0$. This shows that the RHS of equation (1) is indeed $$\frac{\mu_0 v \sigma d}{2}$$ the fortuitously obtained solution quoted by Nicol.

Answer 2

Yes, there is indeed a current in the circuit, however the proposed solution is still valid, even though it requires extra reasoning to justify. I claim that the assumptions $\mathbf J =0$, $\frac{\partial \mathbf E}{\partial t} = 0$ are valid almost everywhere. Specifically, they fail within the conducting contour and in a small vicinity of it, which has the size on the order of the wire diameter. Since we assume wires infinitely thin, macroscopically the assumptions are valid, but it is vital to remember the microscopic details.

I will use Maxwell's equations in Gaussian units, to avoid pesky $\mu_0$'s and $\varepsilon_0$'s. For the reference they look as follows: $$ \begin{eqnarray} \nabla \cdot \mathbf E & = & 4\pi \rho \\ \nabla \cdot \mathbf B & = & 0 \\ \nabla \times \mathbf E & = & -\frac{1}{c}\frac{\partial \mathbf B}{\partial t} \\ \nabla \times \mathbf B & = & \frac{4\pi}{c}\mathbf J + \frac{1}{c}\frac{\partial \mathbf E}{\partial t} \end{eqnarray} $$

This implies (assuming that $S(t)$ is the region of plane bounded by $L(t)$) $$ \begin{eqnarray} \oint_{L(t)} \mathbf B \cdot \mathrm d \mathbf l & = & \iint_{S(t)} (\nabla \times \mathbf B; \mathrm d \mathbf S) \\ & =& \frac{4\pi}{c} \iint_{S(t)} (\mathbf J; \mathrm d \mathbf S) + \frac{1}{c} \iint_{S(t)} \left(\frac{\partial \mathbf E}{\partial t}; \mathrm d \mathbf S \right) \end{eqnarray} $$

However all current flows in the plane $S(t)$, thus its flux through $S(t)$ is $0$.

The term with the partial derivatives is harder to study. First note that macroscopically there are no free charges, in the sense that the macroscopic charge distribution is constant in time. Also the system is quasi-stationary since the speed $v \ll c$ --- this allows us to exclude any EM waves from the problem and only work with charges and currents. This implies that macroscopically $\mathbf E$ is stationary, but if we would assume globally $\frac{\partial \mathbf E}{\partial t} = 0$, then Maxwell's equations would imply that $\mathbf B$ and $\mathbf J$ are also stationary and always $0$. To see that this isn't true we need to consider what happens in the wire itself.

We assume the wire to be an ideal conductor with zero resistance. Ohm's law says that in the wire $\mathbf E = \rho \mathbf J$, if $\rho = 0$ then finite current implies $\mathbf E = 0$ within the wire (EDIT: since we are only interested in the vertical component of $\mathbf E$ and there can be no vertical current, $E_z = 0$ in wire even if $\rho \ne 0$). Thus we see that even before the movement starts the field isn't equal to $\mathbf E_0$ everywhere --- it is $0$ within the wire and has some intermediary value in its vicinity. This also shows that globally $\mathbf E$ isn't stationary --- the movement of the wire causes the movement of the zeroes of $\mathbf E$ and of the shielding charges in the wire. This, in turn, causes the magnetic field and the induced current. If we try to calculate the derivative, then we see that near the wire $E$ changes from $E_0$ to $0$ over an infinitely small interval of time, so the derivative has some delta-function-like form, providing some finite (generally) non-zero value to the integrals.

To calculate the surface integral of $\frac{\partial \mathbf E}{\partial t}$, we need to convert it to a more manageable form, something like a derivative of a continuous function. The general formula for a full derivative of a time-dependent surface integral is $$ \frac{\mathrm d}{\mathrm d t} \iint_{S(t)} (\mathbf F ; \mathrm d \mathbf S) = \iint_{S(t)} \left( \frac{\partial \mathbf F}{\partial t}; \mathrm d \mathbf S \right ) + \frac{1}{\mathrm d t}\iint_{\delta S(t)} (\mathbf F; \mathrm d \mathbf S) $$

Here $\delta S(t)$ is the infinitesimal variation of the surface $S(t)$ and I assume that $S(t)$ varies by adding extra area, like in the problem (i.e. no movement of the interior). This is just the usual product rule for the calculation of derivatives. In our problem $\mathbf F = \mathbf E$ and the surface is chosen so that its boundary passes inside the wire loop. This means that $\mathbf E =0 $ near the boundary of $S(t)$ and thus the integral over the variation of area is $0$, so the second term vanishes and we have $$ \iint_{S(t)} \left( \frac{\partial \mathbf E}{\partial t}; \mathrm d \mathbf S \right ) = \frac{\mathrm d}{\mathrm d t} \iint_{S(t)} (\mathbf E ; \mathrm d \mathbf S) $$

Since $\mathbf E$ is everywhere bounded and almost everywhere equal to $\mathbf E_0$, the answer to your problem follows.

Note that the circulation of $\mathbf B$ doesn't depend on the specific contour passing through the wire, but will change if we move the contour outside the wire. Also note that if we consider an infinitely small loop around a section of the wire, then the circulation of $\mathbf B$ will be finite nonzero if there is a non-zero current passing through the section. This demonstrates that when the problem asks an integral around the perimeter, we must consider exactly the perimeter, even a small variation would give an incorrect answer.

Answer 3

From the 2nd equation (001b) of the Maxwell's equations \begin{align} \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{E} & = -\frac{\partial \mathbf{B}}{\partial t} \tag{001a}\\ \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{B} & = \mu_{0}\mathbf{j}+\frac{1}{c^{2}}\frac{\partial \mathbf{E}}{\partial t} \tag{001b}\\ \nabla \boldsymbol{\cdot} \mathbf{E} & = \frac{\rho}{\epsilon_{0}} \tag{001c}\\ \nabla \boldsymbol{\cdot}\mathbf{B}& = 0 \tag{001d} \end{align} we have

\begin{equation} \int\limits_{S(t)}\left(\boldsymbol{\nabla} \boldsymbol{\times} \mathbf{B}\right)\boldsymbol{\cdot} \mathrm d\mathbf{S}\:=\:\mu_{0}\!\!\int\limits_{S(t)} \mathbf{j}\boldsymbol{\cdot} \mathrm d\mathbf{S}+\frac{1}{c^{2}}\int\limits_{S(t)} \frac{\partial \mathbf{E}}{\partial t}\boldsymbol{\cdot} \mathrm d\mathbf{S} \tag{02} \end{equation} and since with or without current in the wire we have $\:\mathbf{j}\boldsymbol{\cdot} \mathrm d\mathbf{S}\equiv 0\:$ everywhere on the surface $\:S(t)\:$

\begin{equation} \oint\limits_{\partial S(t)}\mathbf{B}\boldsymbol{\cdot} \mathrm d \boldsymbol{\ell}\:=\:\mu_{0}\epsilon_{0}\int\limits_{S(t)} \frac{\partial \mathbf{E}}{\partial t}\boldsymbol{\cdot} \mathrm d\mathbf{S} \tag{03} \end{equation} Now, in order to find the integral in the rhs of above equation I use the Helmholtz transport theorem⁽¹⁾ as suggested by @Michael Seifert in one of his comments. A first form of this theorem is for the flux of a vector field $\:\mathbf{F}\left(\mathbf{x},t\right)\:$ through a surface $\:S(t)\:$ in motion and/or deformation⁽²⁾

\begin{equation} \dfrac{\mathrm d}{\mathrm dt}\int\limits_{S(t)}\mathbf{F}\left(\mathbf{x},t\right)\boldsymbol{\cdot} \mathrm d\mathbf{S}=\int\limits_{S(t)} \left[\dfrac{\partial \mathbf{F}}{\partial t} + \left(\nabla \boldsymbol{\cdot} \boldsymbol{\upsilon}\right)\mathbf{F} + \left(\boldsymbol{\upsilon}\boldsymbol{\cdot}\boldsymbol{\nabla}\right)\mathbf{F} - \left(\mathbf{F}\boldsymbol{\cdot} \boldsymbol{\nabla}\right)\boldsymbol{\upsilon}\right] \boldsymbol{\cdot} \mathrm d\mathbf{S} \tag{04} \end{equation}

expressed here for our purpose as \begin{equation} \dfrac{\mathrm d}{\mathrm dt}\int\limits_{S(t)}\mathbf{F}\left(\mathbf{x},t\right)\boldsymbol{\cdot} \mathrm d\mathbf{S}\:=\:\int\limits_{S(t)} \left[\dfrac{\partial \mathbf{F}}{\partial t} + \left(\nabla \boldsymbol{\cdot} \mathbf{F}\right)\boldsymbol{\upsilon} - \boldsymbol{\nabla} \boldsymbol{\times} \left( \boldsymbol{\upsilon}\boldsymbol{\times} \mathbf{F}\right)\right]\boldsymbol{\cdot} \mathrm d\mathbf{S} \tag{05} \end{equation} For $\:\mathbf{F}\left(\mathbf{x},t\right)\equiv \mathbf{E}\left(\mathbf{x},t\right)\:$ \begin{equation} \dfrac{\mathrm d}{\mathrm dt}\int\limits_{S(t)}\mathbf{E}\left(\mathbf{x},t\right)\boldsymbol{\cdot} \mathrm d\mathbf{S}\:=\:\int\limits_{S(t)} \left[\dfrac{\partial \mathbf{E}}{\partial t} + \left(\nabla \boldsymbol{\cdot} \mathbf{E}\right)\boldsymbol{\upsilon} - \boldsymbol{\nabla} \boldsymbol{\times} \left( \boldsymbol{\upsilon}\boldsymbol{\times} \mathbf{E}\right)\right]\boldsymbol{\cdot} \mathrm d\mathbf{S} \tag{06} \end{equation} So \begin{equation} \int\limits_{S(t)} \frac{\partial \mathbf{E}}{\partial t}\boldsymbol{\cdot} \mathrm d\mathbf{S} =\dfrac{\mathrm d}{\mathrm dt}\int\limits_{S(t)}\mathbf{E}\left(\mathbf{x},t\right)\boldsymbol{\cdot} \mathrm d\mathbf{S}-\int\limits_{S(t)}\left(\nabla \boldsymbol{\cdot} \mathbf{E}\right)\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathrm d\mathbf{S}+\int\limits_{S(t)} \boldsymbol{\nabla} \boldsymbol{\times} \left( \boldsymbol{\upsilon}\boldsymbol{\times} \mathbf{E}\right)\boldsymbol{\cdot} \mathrm d\mathbf{S} \tag{07} \end{equation} But firstly \begin{equation} \dfrac{\mathrm d}{\mathrm dt}\int\limits_{S(t)}\mathbf{E}\left(\mathbf{x},t\right)\boldsymbol{\cdot} \mathrm d\mathbf{S}=\lim_{\Delta t \rightarrow 0}\dfrac{1}{\Delta t}\left[\int\limits_{S(t+\Delta t)}\mathbf{E}\left(\mathbf{x},t\right)\boldsymbol{\cdot} \mathrm d\mathbf{S}-\int\limits_{S(t)}\mathbf{E}\left(\mathbf{x},t\right)\boldsymbol{\cdot} \mathrm d\mathbf{S}\right]=+Ed\upsilon \tag{08} \end{equation} secondly, since $\:\rho\equiv0\:$⁽³⁾ every where on $\:S(t)\:$ \begin{equation} \int\limits_{S(t)}\left(\nabla \boldsymbol{\cdot} \mathbf{E}\right)\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathrm d\mathbf{S}=\int\limits_{S(t)}\frac{\rho}{\epsilon_{0}}\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathrm d\mathbf{S}=0 \tag{09} \end{equation} and thirdly \begin{equation} \int\limits_{S(t)} \boldsymbol{\nabla} \boldsymbol{\times} \left( \boldsymbol{\upsilon}\boldsymbol{\times} \mathbf{E}\right)\boldsymbol{\cdot} \mathrm d\mathbf{S}=\oint\limits_{\partial S(t)}\left( \boldsymbol{\upsilon}\boldsymbol{\times} \mathbf{E}\right)\boldsymbol{\cdot} \mathrm d \boldsymbol{\ell}=-Ed\upsilon \tag{10-wrong, see 10$^{\boldsymbol{\prime}}$} \end{equation} Finally \begin{equation} \oint\limits_{\partial S(t)}\mathbf{B}\boldsymbol{\cdot} \mathrm d \boldsymbol{\ell}\:=\:\mu_{0}\epsilon_{0}\int\limits_{S(t)} \frac{\partial \mathbf{E}}{\partial t}\boldsymbol{\cdot} \mathrm d\mathbf{S}=0 \tag{11-wrong, see 11$^{\boldsymbol{\prime}}$ } \end{equation}

---

EDIT A

As @freecharly commented (Jan'12 2018) :

Your derivations are all perfect. Except for equation (10). As Anton Fetisov has pointed out, the path of the integration runs in the metallic wire where the electric field is zero, which is due to surface charges on the wire induced by the homogeneous field of the positive sheet charge. Therefore, as long as this time dependent path runs in the metallic wire circuit, the integral will be zero \begin{equation} \oint\limits_{\partial S(t)}\left( \boldsymbol{\upsilon}\boldsymbol{\times} \mathbf{E}\right)\boldsymbol{\cdot} \mathrm d \boldsymbol{\ell}=0 \nonumber \end{equation} Therefore also the RHS of equ.(11) is $=Edv$!

Under these suggestions the correct equations and the correct final result are as follows

In place of equation (10) \begin{equation} \oint\limits_{\partial S(t)}\left( \boldsymbol{\upsilon}\boldsymbol{\times} \mathbf{E}\right)\boldsymbol{\cdot} \mathrm d \boldsymbol{\ell}=0 \tag{10$^{\boldsymbol{\prime}}$} \end{equation} and finally \begin{equation} \boxed{\:\:\: \oint\limits_{\partial S(t)}\!\mathbf{B}\!\boldsymbol{\cdot}\! \mathrm d \boldsymbol{\ell}\stackrel{(03)}{=\!\!=}\mu_{0}\epsilon_{0}\!\!\int\limits_{S(t)}\! \frac{\partial \mathbf{E}}{\partial t}\!\boldsymbol{\cdot}\!\mathrm d\mathbf{S}\stackrel{(07),(09),(10^{\boldsymbol{\prime}})}{=\!\!=\!\!=\!\!=\!\!=\!\!=\!\!=\!\!=}\:\mu_{0}\epsilon_{0}\dfrac{\mathrm d}{\mathrm dt}\!\!\int\limits_{S(t)}\!\mathbf{E}\left(\mathbf{x},t\right)\!\boldsymbol{\cdot}\!\mathrm d\mathbf{S}\stackrel{(08)}{=\!\!=}\mu_{0}\epsilon_{0}E\!\cdot\! d\!\cdot\!\upsilon=\frac12\mu_{0}\sigma d \upsilon \vphantom{\oint\limits^{\partial S(t)}_{\partial S(t)}a}\:\:\:} \tag{11$^{\boldsymbol{\prime}}$} \end{equation}

---

^{(1) 'Generalized Vector and Dyadic Analysis', Chen-To Tai, IEEE PRESS, 2nd Edition 1997 equations (6.11),(6.12) page 119.See an excerpt here : Vector Analysis of Transport Theorems.}

---

^{(2) I think there exist many textbooks to find a proof of Helmholtz transport theorem. But you could read my effort to prove this (successfully I want to believe) here : Flux of vector field through movable/deformable surfaces}

---

^{(3) We must distinguish the cases of symbols $\:\rho\:$ and $\:\sigma\:$ used in the question, the answers and the comments \begin{align} \rho & = \begin{cases} \textbf{volume charge density} \\ \textbf{resistivity} \end{cases} \tag{note-01}\\ \sigma & = \begin{cases} \textbf{surface charge density} \\ \textbf{conductivity} \end{cases} \tag{note-02} \end{align}}

---