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Consider the following situation:

enter image description here

We know this is a classic example frequently used when teaching Faraday’s Law, and the voltage/electromotive force (emf) $\mathcal{E}$ induced across terminals of the moving conductor (lighter gray bar in the picture) is

$$\mathcal{E} = B l v$$

where $B$ is the uniform magnetic field perpendicular to the area, $l$ is the length of the conductor and $v$ is the speed of conductor relative to the field. So far so good, but…

How can I deduce the formula by directly solving Maxwell's equations $$\begin{aligned} \nabla \cdot \mathbf{D} &= \rho_\text{f}\\ \nabla \cdot \mathbf{B} &= 0\\ \nabla \times \mathbf{E} &= -\frac{\partial \mathbf{B}} {\partial t}\\ \nabla \times \mathbf{H} &= \mathbf{J}_\text{f} + \frac{\partial \mathbf{D}} {\partial t} \end{aligned}$$ ?

By directly I mean calculate the electric field $\mathbf{E}$ first (by solving Maxwell's equations with proper initial conditions (i.c.) and boundary conditions (b.c.), using techniques taught in partial differential equation (PDE) course), and then calculate emf using the definition

$$\mathcal{E}=\oint_{C} \mathbf{E} \cdot \mathrm{d} \boldsymbol{ l }$$

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3 Answers 3

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Starting from $\vec{\nabla}\times\vec{E} = -\partial \vec{B}/\partial t$, take a surface integral of both sides to find

\begin{align} \iint_{S(t)} \vec{\nabla}\times\vec{E}\cdot d\vec{S} &= \iint_{S(t)} -\frac{\partial \vec{B}}{\partial t}\cdot d\vec{S}\\ \therefore\quad \oint_{\partial S(t)} \vec{E}\cdot d\vec{l} &= -\frac{d}{dt} \iint_{S(t)} \vec{B}\cdot \hat{S}\, dS - \oint_{\partial S(t)}\vec{v}\times\vec{B}\cdot d\vec{l} \end{align}

where the LHS of the second line makes use of Stokes' theorem, while the RHS makes use of the Leibniz integral rule.

$\vec{B}\cdot \hat{S}=-B$ (magnetic field points into the page, while take area vector to point out) and $\iint_S dS = A(t) = l x(t)$ is the area of the (rectangular) surface enclosed by the loop, where $x(t)$ is the length of the sides that is varying with time.

So:

\begin{equation} \mathcal{E} = \oint_{\partial S(t)} (\vec{E}+\vec{v}\times\vec{B})\cdot d\vec{l} = -\frac{d(-Blx(t))}{dt} = Bl\frac{dx(t)}{dt} = Blv \end{equation}

N.B. emf is defined as the work done per unit charge, \begin{equation} \mathcal{E} = \frac{dW}{dq} = \frac{d}{dq}\int \vec{F}\cdot d\vec{l}=\frac{d}{dq}\int q(\vec{E}+\vec{v}\times\vec{B})\cdot d\vec{l} =\int (\vec{E}+\vec{v}\times\vec{B})\cdot d\vec{l} \end{equation} So as Puk has pointed out in the comments, this relies on knowing the Lorentz force. See e.g. Wikipedia for a discussion.

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  • $\begingroup$ @xzczd Note that the definition of emf relies on the Lorentz force law. You could certainly calculate the emf integral without using the Lorentz force law, but it has physical meaning (and in fact is defined that way) because of the Lorentz force law. $\endgroup$
    – Puk
    Nov 3, 2020 at 6:22
  • $\begingroup$ It would be useful to explain why do you put $\vec{v}\times \vec{B}$ into the integral for EMF? External magnetic force can't do work on the electrons. $\endgroup$ Nov 3, 2020 at 18:25
  • $\begingroup$ @xzczd No, the problem is why EMF should include any contribution due to magnetic field. The standard definition of EMF is in terms of work. $\endgroup$ Nov 4, 2020 at 9:54
  • $\begingroup$ @JánLalinský I've edited the answer. $\endgroup$
    – hiccups
    Nov 4, 2020 at 12:10
  • $\begingroup$ @hiccups good but why is there a term $\vec v\times \vec B$ contributing to that work? 1) $\vec v$ is velocity of the rod, not of the mobile charges inside (except for the initial moment where $I=0$) 2) Lorentz force on the mobile charge is $\vec{v}+\vec{v}_d$ where $\vec{v}_d$ is drift velocity due to current. Lorentz force on such particle is $\vec{v}\times \vec{B}+\vec{v}_d\times\vec{B}$ and does no work, since it is perpendicular to total velocity. $\endgroup$ Nov 4, 2020 at 16:51
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The motional EMF in a conductor moving in magnetic field is really due to atomic lattice working on electrons, as these get pushed out of equilibrium into the lattice by external magnetic forces. It is a result of microscopic constraint forces keeping the electrons inside the conductor. So this is all about microscopic forces and Maxwell's equations for field are largely irrelevant - we already know the external magnetic field.

One can try to formulate microscopic theory of lattice and mobile electrons and of their interactions among themselves and with the external field. This would include Lorentz forces between all particles, external magnetic force, and forces of constraint, keeping the electrons bound to conductor. This is a hard problem.

Much easier is to visualize what happens in the inertial frame of the conductor. Using Lorentz transformations on the field tensor $F$, we find out that in the frame of the conductor there is external electric field perpendicular to both the magnetic field and to velocity of the conductor. Its magnitude is $Bv$ and it acts as electromotive agent on the mobile electrons. For conducting rod in the scenario you depicted, net work of the electric force per unit charge transported along the rod is $Blv$.

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  • $\begingroup$ So, if I turn to Lorentz transformations, the speed $v$ will be included in the Maxwell equations, and the integral form of Faraday's law is no longer needed? $\endgroup$
    – xzczd
    Nov 4, 2020 at 2:13
  • $\begingroup$ No, $v$ does not appear in Maxwell's equations. Integral of electric field along the circuit is zero, Maxwell's equations are not relevant here. The EMF is due to rod pushing on electrons and slowing down. Magnitude of this EMF is easy to determine in the co-moving frame. $\endgroup$ Nov 4, 2020 at 9:58
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My answwer is you can't. The maxwell equation $\nabla \times \bf E = -\partial \bf B/\partial t $ must be modified to $\nabla \times \bf E = -\partial \bf B/\partial t + \nabla \times (\bf v \times \bf B) $. (cf. H.H. Skilling, "Fundamentals of Electric Waves", p. 82 eq. 194 ) The last term is based on the Lorentz force $ q \bf v \times \bf B $.

You have to be careful to associate time-varying areas with moving media. Sometimes it doesn't work. I have a WORD file showing such a situation but I'm a newbie at this forum & don't know how to upload that file as part of this answer.

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  • $\begingroup$ You can.And isn't there a rule on stackexchange not to answer or critique others' answers? $\endgroup$
    – rude man
    Nov 5, 2020 at 2:54
  • $\begingroup$ cf. H.H. Skilling, "Fundamentals of Electric Waves", p. 82 eq. 194 which is the exact same equation. $\endgroup$
    – rude man
    Nov 5, 2020 at 3:04
  • $\begingroup$ Not sure about the culture of physics.SE, but it's perfectly fine to comment if one thinks certain answer involves mistake in mathematica.SE. (We don't often cast hasty downvote, though. I haven't casted downvote BTW. ) $\endgroup$
    – xzczd
    Nov 5, 2020 at 3:23
  • $\begingroup$ Not familiar with 'mathematica. SE". If "mathematica. SE" flagged my equation as a "mistake" I need to learn about "mathematica" in a hurry so Ican decide if posting on this site even makes sense. $\endgroup$
    – rude man
    Nov 5, 2020 at 3:31
  • $\begingroup$ After a second look at the integral form of Faraday's law of induction, I think you're right. (Incorrect comment deleted. ) $\endgroup$
    – xzczd
    Nov 5, 2020 at 3:44

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