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If a wire is moving in a magnetic field $B$, there is a Lorentz force acting on both the postive and negative charges, separating them to create an electric field, which is a great explanation to aid my understanding of how ($- \epsilon$) is induced.

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What if I changed the dynamics of this system, making the wire stationary $$\therefore v = 0$$

And moved the magnetic field source(solenoid,magnet,etc...) in the same direction of the previous case(i.e same $v$). Will that change the direction of the Lorentz force acting on the charges? Or is it the same?

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My initial assumption, is that they are relative, leading to the same results. Yet not so sure when considering Lenz's law it confuses me further.

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3 Answers 3

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The principle of relativity: The laws of physics are the same in all inertial frames of reference. Since the Lorentz force is a valid law of physics, it will not change when we pass from one reference frame to another.

First frame, wire is moving. There is no $\mathbf E$ field. Lorentz force $\mathbf F = q\mathbf v\times\mathbf B$. Apparently, you were OK with this frame.

Second frame, you are moving with wire. If you transform the electromagnetic field to the new frame, you will get: $ \mathbf E' = \mathbf v\times\mathbf B $, where $\mathbf v$ is the velocity of your frame of reference with respect to the first frame of reference (provided $\mathbf v$ is perpendicular to $\mathbf B$). And: $\mathbf B' = \mathbf B$.

Its wise to notice, this transformations of the electromagnetic field from one frame to another are Galilean. They are not relativistic. So, they are only valid when the velocity of your frame $v$ is far less than $c$. If you want, the complete relativistic transformations can be seen here.

Now, apply lorentz force to the fields in your frame: $$ \mathbf F = q(\mathbf E' + \mathbf v'\times\mathbf B') = q\mathbf E' $$

where $\mathbf v'$ is the velocity of the wire with respect your frame of reference (ie, zero. After all, you are moving with the wire in this frame, so your relative velocity is zero).

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  • $\begingroup$ A bit confused with the statement "you are moving with the wire", I understand now they are the same(i.e relative) whether moving the wire while the field is stationary or vice versa, it's just a different reference frame, leading to the same results. Which is mindbogglingly amazing. $\endgroup$
    – Pupil
    Commented Jul 18, 2016 at 23:02
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    $\begingroup$ @XCIX Well... you are the guy making the experiments and taking measurements. First frame, the wire were moving (with respect to you, of course). So, there were non-null relative velocity between you and wire. But in the second frame, there is no relative velocity between you and wire, thus, the wire is stationary to you. Or, looking at yourself now from the first frame, "you are moving with the wire". That's what I meant to say. =D. $\endgroup$ Commented Jul 18, 2016 at 23:07
  • $\begingroup$ I came back to this problem after 2-years to add one more thing; The direction of induced $\varepsilon$! Clearly from the two frame it differs, also applying Lenz's law it supports it(opposes the change therefore the polarity of the two frames are different). $\endgroup$
    – Pupil
    Commented Jan 9, 2019 at 11:13
  • $\begingroup$ @Pupil Haha no problem! And.. clearly opposite? It seems to me that both frames produces the EMF on the same direction. How are you calculating it? Here is how I did, on the moving frame: $\mathbf E' = \mathbf v\times \mathbf B = v \mathbf i\times B\mathbf k = -vB\mathbf j$, that is, it becomes positive down below, which is exactly what the first drawing says in your question. $\endgroup$ Commented Jan 10, 2019 at 14:55
  • $\begingroup$ Thank for responding back, I considered Lenz law to the second frame, if I imagined current to flow(if connected to a loop), for the first frame $-F_L$ will oppose to $+F_{pull}$(same direction as $+v$), in the second frame wouldn't the same logic be applied? The rod's current flow would be opposite to the cause of change, $+F_L$ would be to the right. All in order to resist the change. $\endgroup$
    – Pupil
    Commented Jan 19, 2019 at 19:26
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$\def\bE{{\bf E}} \def\PD#1#2{{\partial#1\over\partial#2}}$ We are in a very interesting and apparently paradoxical situation here. Let me explain. You write

And moved the magnetic field

But this should not be said. You may not "move" a magnetic field. You should not think of it as a beam of lines which can be "displaced" here and there. Field lines are only a mathematical representation of how a vector field is directed in different space points. If your field is uniform, field lines are parallel straight lines, everywhere, and can't be "moved".

To be sure, you have added

(solenoid,magnet,etc...)

And this is OK: field sources can move. In our particular case the field source could be a big, infinitely long solenoid, which generates a uniform field inside. And even if you move the solenoid (transversely, as in your figure) the internal field will not change.

We could also appeal to Maxwell equations. In a uniform magnetic field, with no charges and currents in the region we are interested in, we have $$\nabla\cdot\bE = 0 \qquad \nabla\times\bE = 0 \qquad \PD \bE t = 0.\tag1$$ Apparently these equations imply $\bE=0$, but it is untrue. For instance, a uniform $\bE$ constant in time also satisfies eqs. (1). Partial differential equations do not determine a unique solution unless supplemented with boundary conditions.

In our case boundary conditions shoud be assigned at solenoid's surface. But this doesn't yet solve our paradox. @Physicicist137 showed, from field transformation to wire's frame, that a uniform $\bE$ field must exist in that frame, whereas the solenoid brings no net charge. Or does it?

The answer is yes, it does. It can be shown that a wire carrying a current and electrically neutral in a certain frame, if viewed from another frame will show electrical charges of opposite signs in different parts of the circuit. It is a relativistic effect, most easily proven in the simple case of a rectangular loop, but holds in general.

So our moving solenoid acts as a capacitor, and this explains the electric field inside. And also explains why I spoke, at the beginning, of "a very interesting situation": only by having recourse to SR we can achieve a full understanding of it.

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  • $\begingroup$ I went back to this question, after reviewing SR and this problem, which simply boils down to: They produce the same emf, so long as the relative velocity is below $c$, however, the electric field and it's method of conception differs(yet leading the same result in-terms of induction magnitude, and depending on the system could have the same direction or opposite) either it's by the electric force($B$ source moving) or by the magnetic force (conductor moving and $B$ is stationary). From that overly simplistic summary,I didn't understand your "solenoid acts a capacitor" analogy,a bit confused. $\endgroup$
    – Pupil
    Commented Jan 9, 2019 at 11:10
  • $\begingroup$ "You may not move a magnetic field" is a preposterous dogma that came into physics with Feynman's lectures. In fact, in the absence of a gradient in the scalar potential, the force on a charged particle can be written in terms of the velocity of the particle relative to the field, whether (in any given reference frame) the particle is moving through a stationary magnetic field or the particle is at rest in a moving field, or both are moving. $\endgroup$
    – LTFGD
    Commented Jul 3, 2023 at 1:09
  • $\begingroup$ The expression looks just like the Lorentz force law with $v$ replaced by the vector difference v of particle minus v of field. When a radio transmitter is turned on, electric and magnetic fields move outward at speed $c$. It is a staple of intro physics texts to show that a block of crossed $E$ and $B$ fields can move at speed $c$ with $E/B = c$. $\endgroup$
    – LTFGD
    Commented Jul 3, 2023 at 1:09
  • $\begingroup$ Feynman himself refers to this as a "travelling" field and is reduced to a tortured distinction between "moving" and "travelling." The point here is not that there is something wrong with Lorentz transforming the electromagnetic field tensor from one frame to another, but rather that there's more than one way to model electromagnetic phenomena. The force exerted by a moving magnetic field is intuitively illuminating and computationally rigorous. $\endgroup$
    – LTFGD
    Commented Jul 3, 2023 at 1:09
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The answer to the question is yes, the direction of the induced voltage will be reversed in the second case. The direction of the magnetic force on a conduction charge depends on its velocity relative to the field, whether the charge or the field is moving, or both. The relative velocity in the second case is reversed from the first case, so the force is reversed.

The magnitude of the force and voltage are frame dependent, although the difference is only appreciable at high relative speeds. One way of understanding this, without resorting to transformation of the field tensor, is that the flux of the moving magnet is compressed by Lorentz contraction into a smaller area, increasing B (and thereby the force) by the relativistic factor gamma.

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