There are two forms of Faraday's law: The first one: $\nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t}$ and the second one: $\oint \vec{E} \cdot d\vec{s}=-\frac{d\Phi_B}{dt}$. If we have a uniform magnetic field that doesn't change as time passes there is no rotational electric field. Furthermore, if we assume no charge or other sources of it, there will be no electric field whatsoever. However, if we imagine a surface perpendicular to $\vec{B}$ with increasing area we have non-zero change of magnetic flux which implies the existence of field $\vec{E}$ according to Farady's law in integral form. It seems weird, if not totally wrong, that changing properties of some imaginary surface can lead to creation of electric field. How is that even possible? $$\\$$ The second part of my question goes like this: imagine a charged particle that entered a uniform magnetic field with its velocity $\vec{v}$ perpendicular to $\vec{B}$. From the perspective of an observer in the laboratory we see Lorentz force causing change of the direction of $\vec{v}$. How would somebody travelling with the particle explain the change in its motion? Relative to the particle it is not moving at all so Lorentz force equals $\vec{0}$.


This addresses only the first question.

An induced emf in a circuit can arise either through (a) a changing magnetic field or (b) through the boundary of the circuit moving in a magnetic field that doesn't change with time. Both these cases are covered by the equation $$\mathscr E=-\frac {d \Phi}{dt}.$$ This is often called "Faraday's law".

The mechanisms by which the emf is induced are rather different for cases (a) and (b).

Case (a) is the one to which your first equation applies: $$\nabla \times \mathbf E = -\frac{\partial \mathbf B}{\partial t}.$$ This is the Faraday-Maxwell equation. [I think it's best seen not as a changing $\mathbf B$ causing an electric field around it, but as stating a relationship between two aspects of a 'single' electromagnetic field.]

Applying Stokes's theorem, we can find the emf around a stationary closed path:

$$\mathscr E=\int_{\partial S} \mathbf E. d \mathbf l=\int_S (\nabla \times \mathbf E). d \mathbf S=-\int_S \frac{\partial \mathbf B}{\partial t}.d \mathbf S$$

in which the integration is over any surface, S, bounded by the path. If the path is fixed (in our frame of reference), we can swap the order of the differentiation and integration, so $$\mathscr E=-\frac{d}{dt} \int_S \mathbf B.d \mathbf S=-\frac{d \Phi}{dt}.$$

This integrated form of the Faraday-Maxwell equation applies whether or not the path is conducting.

For case (b) we do need a conducting path, a circuit. as the emf arises from the magnetic Lorentz force on the free charges in the circuit as it, or parts of it, move(s), carrying the charges with it. Specifically, the emf is given by a line integral around the circuit: $$\mathscr E=\int_{\partial S} (\mathbf v \times \mathbf B).d \mathbf l.$$ $\mathbf v$ is the velocity of directed element $d \mathbf l$ of the circuit. Using an algebraic result for the scalar triple product, we can write this as $$\mathscr E=-\int_{\partial S} (d \mathbf l \times \mathbf v).\mathbf B.$$ The motion of parts of the circuit results in a changing circuit area. $d \mathbf l \times \mathbf v$ is the (directed) increase in the circuit area per unit time due to the movement of $d\mathbf l$. In fact we can rewrite the equation as $$\mathscr E=-\frac{d}{dt}\int_S \mathbf B. \mathbf dS=-\frac{d \Phi}{dt}.$$

Conclusion: in both case (a) and case (b) we can use the same equation for the emf in terms of rate of change of flux, but (considering the emf in a conducting circuit) in case (a) the emf arise from an induced electric field providing an electric Lorentz force, $q \mathbf E$, acting on the charge carriers, whereas in case (b) it arises from a magnetic Lorentz force, $q (\mathbf v \times \mathbf B)$; a change of flux doesn't imply the existence of an induced electric field.

  • $\begingroup$ This answer is more clarified but otherwise identical to mine. $\endgroup$ – my2cts Jan 16 at 22:40
  • $\begingroup$ This answer is misleading as it implies a charge density (aka a circuit) is required for faraday’s law to apply. No such requirement exists. The path in the integral can be over empty space, which is what the question is about. $\endgroup$ – cms Jan 17 at 1:21
  • $\begingroup$ @ cms I entirely agree and have modified my answer. Do you think that for a path changing its shape in a non-varying field, the equation $$\mathscr E=\int_{\partial S}(\mathbf v \times \mathbf B).d \mathbf l$$ is also applicable to free space, without a conducting path? $\endgroup$ – Philip Wood Jan 17 at 14:43

Answering your 2nd question in more details: In order to go to the frame of the particle, you have to do a Lorentz transformation to be consistent with relativity. E and B field get "mixed" after a Lorentz transformation (see details in this wiki article).

In this scenario, it is true that after you go to the particle's frame, there is no magnetic force as it is not moving in its own frame by definition. However, there is a new electric field in this frame due to the Lorentz transformation of fields, and the electric field provides the force.

In sum, electric and magnetic field are really one single entity that morphs into one another based on which reference frame you are in.

  • $\begingroup$ This is actually the answer to the first question as well. $\endgroup$ – cms Jan 17 at 1:28

Q1. In your example $\nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t}=0$. However if there is a somehow expanding conducting wire encircling the expanding surface there will be a current because light, mobile electrons are moving through B. Note that the ions in the metal will also form a current, if the wire is allowed to rotate but this is much smaller due their much larger mass.

  • $\begingroup$ I'm kind of comfortable with this answer, but I'm not sure whether I fully understand it. So electrons in an expending wire experience Lorentz force due to their motion in magnetic field. However, I always thought that Faraday law, and generally all Maxwell equations, apply no matter if there is a wire or not. Hence my problem - as long as there is a physical conductor electrons move as if there was an electric field when in reality they move because of lorentz force... Is that correct? $\endgroup$ – A. J. Bałaziński Jan 16 at 18:19
  • $\begingroup$ My answer is based on the Maxwell equations, which give you the field, and on the Lorentz force. $\endgroup$ – my2cts Jan 16 at 22:05

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