4
$\begingroup$

I'm trying to understand how to correctly think about a change of reference frames for a conducting rod moving sideways through a uniform magnetic field.

Thinking of the rod in motion through a uniform B field is easy. The Lorentz force explains the charge separation in the rod and I'm fine with that.

But if you change your point of view to where the rod is stationary, I immediately have an issue with the idea of the magnetic field "moving." Sure, if you picture magnetic field lines drifting one way or the other there's an appearance of movement, but isn't that a wrong way to look at it? Field lines aren't real. It seems to me that 'moving' a uniform B field changes nothing at all--it just stays the same as it always was. I'm tempted to say that a constant B field just cannot be thought of as moving at all. The field is the same as if it was not moving.

Yet clearly there has to be the same charge separation in that rod whether we think of it being in motion or not. Relative motion of the rod and field has to lead to the same effect no matter how you look at it. So where am I going wrong?

I have a vague recollection of something in Griffiths' Electrodynamics book that may have talked about this and a problem with the idea of a uniform B field throughout space violating necessary boundary conditions or something like that. Is that where the trouble is?

Is there a simple way out of my dilemma?

Added: I should have pointed out that my issue was motivated by reading an introductory section from a textbook on Faraday's Law and induced E fields. There the discussion is all about "flux change" as usual. But this argument doesn't work for the rod. Is there some other intuitive way to see the origin of the E field?

More added: I came across this Faraday Paradox today and perhaps it addresses some of my uneasiness. I still would like an intuitive way to see how it all works out though.

$\endgroup$

1 Answer 1

2
$\begingroup$

I immediately have an issue with the idea of the magnetic field "moving."

But you don't need such an idea.

Assume that, in the lab frame, there is a uniform B field pointing 'up' which is to say that at every spatial point in the lab, the value of the B field is

$$\vec B = B\hat z$$

Also, assume that the E field is zero everywhere in the lab frame.

Now, consider the fields in a frame of reference moving uniformly with speed $v$ in the $\hat y$ direction in the lab frame.

In this (primed) frame, there is a uniform electric field given by:

$$\vec E' = \frac{vB}{\sqrt{1 - \frac{v^2}{c^2}}}\hat x$$

and

$$\vec B' = \frac{B}{\sqrt{1 - \frac{v^2}{c^2}}}\hat z$$

Note that both the electric field and magnetic field are not 'moving' in the primed framed, i.e., there is no time dependence.

Moreover, if we assume that the rod extends in the $\hat x$ direction and is stationary in the primed frame, we see that there is no magnetic force on the mobile charge in the rod but there is an electric force in the $\hat x$ direction.

Thus, in the lab frame, there is only a magnetic force on the charge. In the primed frame, there is only an electric force on the charge.

If the rod is moving is also moving in the primed frame with velocity $\vec u$, there is a combination of electric and magnetic force on the mobile charge since

$$\vec F' = q\left(\vec E' + \vec u \times \vec B' \right) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.