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I recently came across a problem in which a conducting rod was moving with constant velocity in a constant magnetic field, arrangement is such that rod, velocity, magnetic field are mutually perpendicular to each other and the question asks to calculate the electric field inside the conductor in steady state.

Now according to me as this conductor is not connected to any other body and so there is no means of energy loss possible for it, so it is isolated and hence the field inside it must be zero. But according to them the field inside is constant, Am I wrong in saying that this conductor is isolated or something different happens in case of magnetic field or is it due to the motion of the rod.

My contradiction is that when we apply gauss law in let say an isolated moving sphere, we say that the opposite charges move to opposite sides to create an electric field that cancels the external field in order to make the electric field inside the sphere zero.

Now applying the above thing my question, let say that the rod is moving and the fields present are- magnetic and induced electric field initially, now after sometime due to magnetic and induced electric force the opposite charges once again would move to opposite sides and again according to me should cancel the induced electric field to make the net field zero.

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  • $\begingroup$ Why is my question downvoted? $\endgroup$ Commented Jul 23, 2022 at 6:32

3 Answers 3

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A free charge carrier inside the rod has non zero velocity in a magnetic field, so a Lorentz force is applied $ \vec{F}=q\vec{v}\times\vec{B}$. So those carriers start to move.

Since charges in motion is, precisely, an electrical current, local Ohm's law applies in a conductor: $\vec{\jmath}=\gamma\vec{E}$. Since $\vec{\jmath}$ is non-zero, so is $\vec{E}$.

Lastly, since $\vec{\jmath}$ is by definition alongside $\vec{v}$ (velocity of the carriers), so is $\vec{E}$.

I can't be more precise for two reasons:

  • You didn't give details about the exact direction of $\vec{B}$ with relation to the rod and its motion. I assume that it's orthogonal to the rod's axis?
  • You didn't specify whether you wanted the end result or what happens at first. The charge carriers's motion is modified by the electric field, since the latter implies an electric force that adds up and changes the carriers' dynamics. You might want to read about Hall effect to see a slightly different but similar problem.
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  • $\begingroup$ Can I say that the rod is isolated and let say I have an conducting rod placed in an electric field now the electrons and protons suffers forces in opposite directions which means charges in motion and their net is non zero so J becomes non zero, and so becomes E but in such cases we say that the electric field in the conductor must be 0. Why is that? $\endgroup$ Commented Jul 23, 2022 at 9:45
  • $\begingroup$ Ohms law is derived under the assumption that the only force acting on the electrons is the electric field. So saying where there is a $\vec{j}$ there must an $\vec{E}$ wouldn't be correct. $\endgroup$ Commented Jul 23, 2022 at 9:56
  • $\begingroup$ Normally E is zero in a conductor, only once the conductor reaches its steady state, Where the electrons oppose the initial E field and completely cancel out. For this situation there is no static steady state since (I think) as there will always he movement due to the magnetic field $\endgroup$ Commented Jul 23, 2022 at 9:59
  • $\begingroup$ @jensenpaull I kept my answer simple, and that's why I mentioned the Hall effect. Local Ohm's law still applies with a magnetic field, there's simply an extra term. It doesn't change the qualitative conclusion. $\endgroup$
    – Miyase
    Commented Jul 23, 2022 at 10:04
  • $\begingroup$ It still changes what fields are present. Which is the basis of OP's entire question. $\endgroup$ Commented Jul 23, 2022 at 10:14
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An other way to look at the problem is to consider how the electric field changes in a change of referential. Special relativity (remember than EM is intrinsically relativist) states that for non relativistic velocity:$$ \overrightarrow{E'}= \overrightarrow{E}+ \overrightarrow{V} \times \overrightarrow{B} $$ Where $ \overrightarrow{E'}$ is the electric field in the moving frame (the rod), $ \overrightarrow{E}$ is the electric field in the static frame and $\overrightarrow{V}$ the velocity of the moving frame. As you can see: $\overrightarrow{E}=\overrightarrow{0}$ does not translate into $\overrightarrow{E'}=\overrightarrow{0}$

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Ignoring the extremely rapid establishment of the steady state, what you then have is two electric fields: the field $ \bf E_m = \bf v \times \bf B $ set up by the emf generated by the motion: emf = Blv, l the bar length and v the velocity.

Then there is the electrostatic field $ \bf E_s $ which internally cancels the $ \bf E_m $ field so the net E field is zero inside the bar as required.

But outside the bar the $ E_s $ field is still present and establishes a voltage = Blv which can be measured by a voltmeter if the meter leads are kept outside the B field. $ E_s $, being electrostatic, its circulation looping inside and outside the bar obeys $ \oint \bf E_s \cdot \bf dl = 0 $ regardless of the path taken, while the circulation $ \oint \bf E_m \cdot \bf dl = Blv $.

BTW "voltage" strictly speaking is the line integral of the $ E_s $ field only, and obeys Kirchhoff's voltage law, while the line integral of the $ E_m $ field does not.

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