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Relation of Magnetic and Electric Force of particles in vacuum

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Homework Statement


There are two particles side by side in a vacuum with electric charge q and mass m, traveling at distance d apart, with v speed(less than speed of light). What is the ratio of magnitude of magnetic to electric force the particle exert on each other?


Homework Equations



F(mag)/ F(elec) is the ratio

Fm= kq1q2/ r2 and Fe=qvB


The Attempt at a Solution



Fm/ Fe

=(qvB) / (kq1q2/ r2)

= vBr2/ kq

Does this look correct? And would this ratio have no units?
 

Answers and Replies

  • #2
gabbagabbahey
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Homework Statement


There are two particles side by side in a vacuum with electric charge q and mass m, traveling at distance d apart, with v speed(less than speed of light). What is the ratio of magnitude of magnetic to electric force the particle exert on each other?


Homework Equations



F(mag)/ F(elec) is the ratio

Fm= kq1q2/ r2 and Fe=qvB


The Attempt at a Solution



Fm/ Fe

=(qvB) / (kq1q2/ r2)

= vBr2/ kq

Does this look correct? And would this ratio have no units?
As long as the charges are moving much less than the speed of light (as opposed to just "less than the speed of light" as given in your post), then this looks correct (a little confusing with you using "q" for the magnetic force and "q1" & "q2" for the electric force, but still correct. However, you haven't calculated B.

In any case, I'm not sure this is exactly what your professor is looking for here. Consider that for the particles to continue to move along parallel paths (as implied by the statement that they are "traveling at distance d apart, with v speed"), the component of the electric force pushing them apart must be balanced by the component of the magnetic field that pulls them together. This will give a relationship between "v" and "d", and I suspect that "v" will be relativistic.
 
  • #3
cepheid
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I'm going to be bold here and try to apply physical reasoning, even though I myself am not 100% sure of the answer.

If I interpret this problem correctly, both particles are travelling with speed v in the same direction, so that they remain side-by-side at all times, right?

If that's true, you can imagine transforming to a reference frame in which both particles are stationary and just sitting there with separation d. All of the force is electric in this frame, and none of it magnetic, since we have no moving charges.

But, you might protest that the whole point of relativistic electromagnetism is that a phenomenon that appears electric in one reference frame might be magnetic in another. Yes, but my counterargument is this: in both frames, the observed effects have to be the same. So, if the force is entirely electric in one frame, but partially electric and partially magnetic in another frame, the total force has to be the same. Problem: the electric force between the charges depends only on their separation, and their separation d, is the same in both the lab frame and the frame moving along with the charges. So, the lab frame can't also have additional magnetic forces, because then the observed effects would be different than in the other frame. The only conclusion I can come two is that the effects of the magnetic fields of the two moving charges in the lab frame must cancel each other out.

It's not clear to me exactly why this is true, and you made no attempt to address the question of what "B" was in your solution. But if the magnetic field of a single moving point charge is anything similar to the magnetic field of a steady current, then it is in a circle in a plane perpendicular to the direction of motion, and it's direction is given by the right hand rule. So, in between the two charges, the magnetic fields from each one will be the same in magnitude and opposite in direction. In fact, that might even be true everywhere, which would solve our problem. (EDIT 3: no, it cannot be true everywhere)

EDIT: Or am I just being dumb, because it seems like a one charge would be affected only by the other one's magnetic field and not by its own. I must be missing something here.

EDIT 2: Okay, I looked at gabbagabbahey's post, and I am even more confused. If the magnetic force is supposed to be cancelling out the mutual repulsion of the charges, how is this explained in the co-moving frame, in which there is no cause for magnetic fields, and yet the charges appear to be somehow held at separation d, in apparent violation of Coulomb's law? Is there supposed to be an *external* magnetic field in this problem?
 
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  • #4
gabbagabbahey
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So, in between the two charges, the magnetic fields from each one will be the same in magnitude and opposite in direction. In fact, that might even be true everywhere, which would solve our problem. (EDIT 3: no, it cannot be true everywhere)
Actually, I think it is true everywhere (at least as long as v is constant). Working in the lab frame and choosing coordinate axes such that [itex]\mathbf{r}_1=vt \mathbf { \hat{z} }[/itex] and [itex]\mathbf{r}_2=vt \mathbf { \hat{z} } +d \mathbf { \hat{x} }[/itex], and using the results of example 10.4 from Grifitth's Introduction to Electrodynamics 3rd Ed., I get [itex]\mathbf{B_1}=-\mathbf{B_2}[/itex] everywhere.

Edit: Nevermind, I made a mathematical error. [itex]\mathbf{B_1} \neq -\mathbf{B_2}[/itex] everywhere., only in the plane halfway between the two charges.

EDIT 2: Okay, I looked at gabbagabbahey's post, and I am even more confused. If the magnetic force is supposed to be cancelling out the mutual repulsion of the charges, how is this explained in the co-moving frame, in which there is no cause for magnetic fields, and yet the charges appear to be somehow held at separation d, in apparent violation of Coulomb's law? Is there supposed to be an *external* magnetic field in this problem?
Barring an external field/force, I think it is only possible in the limiting cases of [itex]q=0[/itex], or [itex]d \to \infty[/itex], so I think we may have both misinterpreted the question, and we are only to assume that the particles move at speed "v" a distance "d" apart for one instant, unless there is additional info omitted from the OP's problem statement.
 
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  • #5
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Thanks much everyone!
the particles are moving on parallel trajectory, does this have any significance?
And will the ratio be unit less?
 
  • #6
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I'm going to be bold here and try to apply physical reasoning, even though I myself am not 100% sure of the answer.

If I interpret this problem correctly, both particles are travelling with speed v in the same direction, so that they remain side-by-side at all times, right?

If that's true, you can imagine transforming to a reference frame in which both particles are stationary and just sitting there with separation d. All of the force is electric in this frame, and none of it magnetic, since we have no moving charges.

But, you might protest that the whole point of relativistic electromagnetism is that a phenomenon that appears electric in one reference frame might be magnetic in another. Yes, but my counterargument is this: in both frames, the observed effects have to be the same. So, if the force is entirely electric in one frame, but partially electric and partially magnetic in another frame, the total force has to be the same. Problem: the electric force between the charges depends only on their separation, and their separation d, is the same in both the lab frame and the frame moving along with the charges. So, the lab frame can't also have additional magnetic forces, because then the observed effects would be different than in the other frame. The only conclusion I can come two is that the effects of the magnetic fields of the two moving charges in the lab frame must cancel each other out.

It's not clear to me exactly why this is true, and you made no attempt to address the question of what "B" was in your solution. But if the magnetic field of a single moving point charge is anything similar to the magnetic field of a steady current, then it is in a circle in a plane perpendicular to the direction of motion, and it's direction is given by the right hand rule. So, in between the two charges, the magnetic fields from each one will be the same in magnitude and opposite in direction. In fact, that might even be true everywhere, which would solve our problem. (EDIT 3: no, it cannot be true everywhere)

EDIT: Or am I just being dumb, because it seems like a one charge would be affected only by the other one's magnetic field and not by its own. I must be missing something here.

EDIT 2: Okay, I looked at gabbagabbahey's post, and I am even more confused. If the magnetic force is supposed to be cancelling out the mutual repulsion of the charges, how is this explained in the co-moving frame, in which there is no cause for magnetic fields, and yet the charges appear to be somehow held at separation d, in apparent violation of Coulomb's law? Is there supposed to be an *external* magnetic field in this problem?
The entire question is stated as so:

Two particles, each with electric charge q and mass m, are traveling in vacuum side-by-side,
on parallel trajectories a distance d apart, both with speed v (much less than the speed of
light). Calculate the ratio of the magnitude of the magnetic force to that of the electric
force each exerts on the other.
 
  • #7
gabbagabbahey
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The entire question is stated as so:

Two particles, each with electric charge q and mass m, are traveling in vacuum side-by-side,
on parallel trajectories a distance d apart, both with speed v (much less than the speed of
light). Calculate the ratio of the magnitude of the magnetic force to that of the electric
force each exerts on the other.
Okay, in that case there must be some external force/field which you don't need to worry about, and you are almost there. You just need to find the non-relativistic (v<<c) magnetic field due to a moving point charge and plug its magnitude in for B.
 
  • #8
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I used E=vB to solve for be, but then I get:

vEr2/kq and that leave E not solved now?
 
  • #9
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The magnetic fields act oppositely to the electric field so that total force vanishes as v approaches c. The ratio is not unitless. E/H has units of Ohms. You can work out the units of E/B.
 
  • #10
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What is H?
 
  • #11
gabbagabbahey
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I used E=vB to solve for be, but then I get:

vEr2/kq and that leave E not solved now?
You've already calculated E (E=kq^2/d^2). The bigger question is why you are using E=vB?
 

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