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The Faraday's Paradox

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The Faraday's Paradox The Faraday's Paradox Yun, Ji-Won and Lee, Dongsub [email protected] [email protected] Key Points of the Faraday paradox • The electromotive force(emf) is induced not because flux passing thru the given circuit is changing in time, but there is the Lorentz force exerted on the charge carriers. • Electromagnetic field can change when the frame is changed. Furthermore, moving magnetic field can be seen as an electric field in some other frame and vice versa. • Even though the Lorentz forces exerted on each charge carriers inside the circuit are non-trivial, whether there is a current or not depends on the net Lorentz force integrated along the wire and even it does not have to be closed circuit. • To determine whether there is non-zero emf or not, a good check point is to consider the work done on the system. If there was a work done on it, there might be emf, but if there is not, there never be non-zero net emf. The reason why the Faraday paradox arose lies on the insensitiveness on both the relativity and existence of the electron. Basically the electromagnetism already has relativistic structure. Thus, we have to be careful on the change of frames and to use covariant coordinate change rather than simple Galilean coordinate transform. Then no matter what frame we use to investigate the system, we will reach to the same answer. Besides, original description of Faraday on the induction states about the emf focuses on the induced current rather than each electrons. However the emf should be defined as line integral along the circuit of the Lorentz force which is felt by each single electrons. Even the flux seems to change in time, if the difference of voltages induced on each segment of the circuit cancel out each other because of their opposite direction, there could not be non-zero total emf or current. In such way, we never confuse whether there is emf or not no matter what frame we use and no matter what system we see. If we focused on the movements of each electrons and carefully change the frame while we keep covariance, we never face the Faraday's paradox. Actually it might be very hard to change frames in covariant way for example, in the cases having rotating object. In such cases, the frame in which the source of the field is stationary would be easier to analyse the system than the other frames. Short Summary • The Faraday paradox implies that the relativistic structure is already embedded in electromagnetism. • Covariant change of frame never occur the Faraday paradox. • If the paradox seems to be left, try to focus on the motion of electrons due to the Lorentz forces along the path you see. Rigorous Approach to the Faraday's law of induction The Faraday's law of induction states that the total derivative, it should be emphasised that it is not a partial derivative, of the magnetic flux ΦB penetrating given circuit generates the emf E on the circuit, i.e, dΦ E = − B : (1) dt Here the flux ΦB is a integral of the magnetic field B~ on the given surface S closed by the circuit, i.e, @S = C with path C specified by the circuit. Shortly, Z 2 ΦB = d ~a · B~ (t): (2) S Note that, even the surface S can change as time goes, i.e, S = S(t) is a set whose elements can change in time. As mentioned above, we will check that the origin of emf is the Lorentz force felt by charge carriers by the following evaluation. Figure 1: Surface S(t) and S(t + δt) with the magnetic field B~ (t). Let's consider the most arbitrary situation where the surface S and the magnetic field B~ both change in time as you can see from the Fig. 1. Then the emf E becomes, "Z Z # 1 2 2 E = − lim d ~a2 · B~ (t + δt) − d ~a1 · B~ (t) : (3) δt!0 δt S(t+δt) S(t) Since we are going to take δt ! 0 limit, we can use that, @ 2 B~ (t + δt) = B~ (t) + B~ δt + O δt : (4) @t t Hence, "Z Z Z # 1 2 2 2 @ E = − lim d ~a1 · B(t) − d ~a1 · B~ (t) + d ~a2 · Bδt~ δt!0 δt S(t+δt) S(t) S(t+δt) @t "Z Z # Z 1 2 2 2 = − lim d ~a2 · B~ (t) − d ~a1 · B~ (t) − d ~a1 · @tB~ (t) (5) δt!0 δt S(t+δt) S(t) S(t) For the first limit calculation, we know that r~ · B~ = 0 from the Maxwell's equations. Therefore, for an arbitrary volume V closed by S(t), S(t + δt) and the strip-like surface σ connecting those two surfaces, i.e, @V = S(t) [ S(t + δt) [ σ, we can say, Z I d3~xr~ · B~ = d2~a · B~ = 0: (6) V @V 2 2 Note that, for the surface S(t + δt), d ~a coincides with d ~a2 but, since the outgoing direction of @V is 2 2 2 different with d ~a1, we have to use d ~a = −d ~a1. Applying this to our picture gives, Z Z Z 2 2 2 0 = d ~a2 · B~ (t) − d ~a1 · B~ (t) + d ~a · B~ (t): (7) S(t+δt) S(t) σ Note that, on the strip-like surface σ, as you can see from the Fig. 2, we can write the d2~a as, d2~a = d~l ^ δd:~ (8) Page 2 Figure 2: Strip-like surface σ and its area element. ~ Furthermore, since the δd corresponds to partial movement of the charge carrier, we can write is as ~vtδt, where the velocity of the charge carrier ~v is divided into two terms as ~v = ~vd + ~vt, with drift velocity along the circuit ~vd and transverse velocity ~vt of them. Therefore, the argument of the limit becomes as follow. Z Z 2 2 d ~a2 · B~ (t) − d ~a1 · B~ (t) S(t+δt) S(t) Z = − d2~a · B~ (t) σ Z ~ = − dl ^ ~vtδt · B~ (t) @S(t) "Z # ~ = −δt dl · ~vt ^ B~ (9) @S(t) Please remind that, based on the Maxwell's equations, @tB~ = −r~ ^ E~ . Therefore we can write the emf as, " Z # Z 1 ~ 2 E = lim δt dl · ~vt ^ B~ + d ~a1 · r~ ^ E~ δt! δt @S(t) S(t) Z ~ = dl · E~ + ~vt ^ B~ : (10) @S(t) One thing should be noted is that, the absence of the drift velocity of charge carriers does not affect the ~ ~ result at all. Since the collective motion of those charge carriers gives ~vd k dl, so the term dl · ~vd ^ B~ will automatically vanish. Anyway, if we change the original expression of the Faraday's law of induction to the eqn. (10), we can see that the emf is given by the line integral along the circuit no matter what surface we started with. Also, as you already now, the Lorentz force can be derived while we keep the covariance from the action S such that Z Z µ S = −m ds + q dx Aµ; (11) which is also covariant. Therefore it no more depends on the choice of frame. Page 3 Application on the Example The Fig. 3 is one represents rotating disc which is conductor in presence of the magnetic field. The disc has sliding contact so that wire can left stationary while the disc rotates. Anyway we can consider 4 different situations from this configuration. The first one is when (i)both the source of magnetic field and the disc are stationary. Also we can consider the case (ii)when only the source rotates or (iii)only the disc rotate. Of course there is a case (iv)when both the source and the disc rotate. You may consider that the case (i) and (iv) are equal description while (ii) and (iii) are also equal. However, as a result, we can observe the non-zero emf when (iii) and (iv) only. Except those two cases, we can not see any current at all. Let's use the result from the above and resolve this paradox. Figure 3: Figure 7.14 of the Example The first case is clear. Since there is no change at all, it is natural 7.4 from the Griffith. that there is no current too. What actually make this 4 cases paradoxical are difference between case (ii) and (iii) and the reason why there is current in (iv). For the case (ii), we should change the frame into the one has stationary magnet source. However it would be very hard because we have to use general relativity, not special relativity, to describe rotating system. So let's consider schematically. In that frame, there must be both magnetic field and electric field. On top of that, since there is translational symmetry along the axis of the rotation, we might expect the same symmetry in transformed coordinates. Thus even there is an electric field after the change of the coordinates, we can expect that field also has a translational invariance. Here note that the emf due to those electric field will be induced at two segments, one on the disc which starts from the rotating axis and ends at the sliding contact and the other one is a segment which is at the bottom of the circuit and has the resistance R on the figure. If those two segments have the same emf, it must be canceled because they are in the opposite direction.
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