Relative Motion of Charges Question

[gibberingmouther]

Homework Statement

What is the B field in component form in the reference frame of the rocket which is moving through a static E field?

Relevant Equations

E = v x B

This problem seems pretty simple but I don't understand how the correct answer is 500,000 smaller than what I calculated.

You have an electric field that has a magnitude of 10^6 V/m in the positive z direction. I assumed (though it was not stated) that the cause of this field was a sheet of charges whose plane is oriented parallel to the xy plane.

A rocket moves in the positive x direction at 10^6 m/s. The question is:

"What are the components of the magnetic field in the reference frame of the rocket?"

The answer is 1.1 * 10^-5 T in the positive y direction. I get the direction from the right hand rule, but I calculated that B in the positive y direction is the electric field divided by the velocity of the rocket which would give you 1 T, not 500,000 times smaller than that.

Can anyone tell me what I'm missing here? Thanks.

 

[haruspex]

Relevant Equations
E = v x B

I calculated that B in the positive y direction is the electric field divided by the velocity of the rocket

I can see how you elected to do that, given your relevant equation, but it is not generally valid to divide by vectors. Are you sure this is the right equation for the context? What answer will it give if the velocity is zero?

Edit: if you cannot find a more appropriate equation, I would model the source of the electric field as a uniform line of charge parallel to the rocket's path. Throw in unknowns for distance and density. In the reference frame of the rocket that gives you a current.
Find the resulting E and B fields. The added unknowns should cancel.

Edit 2: my method does give the book answer.

 

[gibberingmouther]

I can see how you elected to do that, given your relevant equation, but it is not generally valid to divide by vectors. Are you sure this is the right equation for the context? What answer will it give if the velocity is zero?

Edit: if you cannot find a more appropriate equation, I would model the source of the electric field as a uniform line of charge parallel to the rocket's path. Throw in unknowns for distance and density. In the reference frame of the rocket that gives you a current.
Find the resulting E and B fields. The added unknowns should cancel.

Edit 2: my method does give the book answer.

Okay, I'll check it out.

In the problem I thought of the source of the electric field as a sheet of charge rather than a line/current. Maybe that's why my answer was wrong. A lesson to take from this could be to make the minimal amount of assumptions when given a problem in a science class.

 

[haruspex]

Okay, I'll check it out.

In the problem I thought of the source of the electric field as a sheet of charge rather than a line/current. Maybe that's why my answer was wrong. A lesson to take from this could be to make the minimal amount of assumptions when given a problem in a science class.

No, your answer was wrong because you tried to apply an equation that had the wrong cause v. effect relationship. Generally speaking, in a physics vector equation ##\vec z=\vec x\times\vec y## the vectors ##\vec x## and ##\vec y## represent causes and ##\vec z## the result. (Precession is a bit odd in this regard.)
You tried to apply your "relevant equation" in a manner that made ##\vec y## (B) the result, so you found yourself dividing by a vector as though a cross product was just scalar multiplication. That is a no-no.

The safe method is to model the E field with a charge distribution and see what B field results.
I think the flat sheet of charge model should work in principle, but I foresee it leading to infinities that will take some limits logic to handle. The line of charge must be just as valid since it produces a constant E field as far as the rocket is concerned and avoids the infinities.

The answer is surprising and rather elegant.

 

[TSny]

Relevant Equations: E = v x B

Are you familiar with a similar equation except that E and B switch places and, additionally, the speed of light enters the equation? This is just a suggestion for another approach to the problem.

 

[haruspex]

Are you familiar with a similar equation except that E and B switch places and, additionally, the speed of light enters the equation? This is just a suggestion for another approach to the problem.

Yes, the speed of light turns up in the line of charge model. I was not aware of a standard equation.

 

[TSny]

Yes, the speed of light turns up in the line of charge model. I was not aware of a standard equation.

There is a standard set of equations for transforming the fields between frames which is covered in some E & M courses and relativity courses.

@gibberingmouther Forget my suggestion if you haven't covered these transformation equations. But I'm a little curious as to where you got your equation E = v x B.

 

[PeroK]

Yes, the speed of light turns up in the line of charge model. I was not aware of a standard equation.

The speed here is non-relativistic, so you could just take the charge density to be the same in each frame. Relativistically, the density in the frame where the charge is moving would be increased by the ##\gamma## factor, on account of length contraction. This would give you in this case:

##B = \frac{\gamma v E}{c^2}##

Where ##\gamma = \frac{1}{\sqrt{1- v^2/c^2}} = 1## for non-relativistic speeds.

This is how essentially Einstein developed SR in the first place: to explain why the electromagnetic phenomena were independent of the state of motion of the charges, created by a change in reference frame (*).

In this example, a similar argument shows that the electric field in the rocket frame is ##E' = \gamma E## and the force on a charged particle at rest in the original frame would be:

In the original frame:

##F = qE##

And in the rocket frame:

##F' = q(E' - vB') = q\gamma E(1 - \frac{v^2}{c^2}) = \frac{F}{\gamma}##

Which is the relativistic relationship between the force, ##F##, on a particle in its own rest frame and the force ##F'## in a frame where the particle is moving at speed ##v## - in the special case where the force is perpendicular to the motion.

So, yes, it is all extremely elegant!

(*) Note that in the non-relativistic approximation, you have the same E field in both cases but an additional (albeit small) B field in the rocket frame. Hence, the force on the particle is not the same in the two frames, which is unexplained by classical physics.

 

[haruspex]

The speed here is non-relativistic, so you could just take the charge density to be the same in each frame. Relativistically, the density in the frame where the charge is moving would be increased by the ##\gamma## factor, on account of length contraction. This would give you in this case:

##B = \frac{\gamma v E}{c^2}##

Where ##\gamma = \frac{1}{\sqrt{1- v^2/c^2}} = 1## for non-relativistic speeds.

This is how essentially Einstein developed SR in the first place: to explain why the electromagnetic phenomena were independent of the state of motion of the charges, created by a change in reference frame (*).

In this example, a similar argument shows that the electric field in the rocket frame is ##E' = \gamma E## and the force on a charged particle at rest in the original frame would be:

In the original frame:

##F = qE##

And in the rocket frame:

##F' = q(E' - vB') = q\gamma E(1 - \frac{v^2}{c^2}) = \frac{F}{\gamma}##

Which is the relativistic relationship between the force, ##F##, on a particle in its own rest frame and the force ##F'## in a frame where the particle is moving at speed ##v## - in the special case where the force is perpendicular to the motion.

So, yes, it is all extremely elegant!

(*) Note that in the non-relativistic approximation, you have the same E field in both cases but an additional (albeit small) B field in the rocket frame. Hence, the force on the particle is not the same in the two frames, which is unexplained by classical physics.

That's very interesting... so the B field calculated in this problem wouid not happen; it is an artefact of ignoring SR?
As I wrote, my method was to postulate an infinite line of charge at some constant density and distance from the rocket. The relationship between the resulting E and B fields turns out to involve μ₀ε₀, i.e. c².

 

[PeroK]

That's very interesting... so the B field calculated in this problem wouid not happen; it is an artefact of ignoring SR?
As I wrote, my method was to postulate an infinite line of charge at some constant density and distance from the rocket. The relationship between the resulting E and B fields turns out to involve μ₀ε₀, i.e. c².

No, you get both fields in SR in the rocket frame. But the field transformations result in a consistent relationship between the two reference frames. The laws of EM are seen to apply in both frames.

If you use the classical approximation, then the particle is subject to less force in the rocket frame, which means the laws of EM cannot be the same in both frames. Because, of course, force is supposed to be frame invariant.

 

[PeroK]

In summary:

If you calculate the force on the particle in your classical approximation, it is ##F## in the original frame and ##F/\gamma^2## in the rocket frame. Hence EM is not Galilean invariant.

In SR you get a force of ##F/\gamma## in the rocket frame but that is in fact consistent with SR dynamics, where the force is not invariant.