Moving Charge in a Magnetic field Problem

  • #1
Hi all. I am alluding to the Moving Charge in a Magnetic field problem by quoting an MIT lecture below.

Scott Huges – Lecture
http://web.mit.edu/sahughes/www/8.022/lec10.pdf
Suppose we now examine this situation from the point of view of the charge (the “charge frame”). From the charge’s point of view, it is sitting perfectly still. If it is sitting still, there can be no magnetic force!

In the quote, the author says, in the charge frame, the charge is at rest therefore there can be no magnetic force (Lorentz force). But this is incorrect since velocity is relative here. If the charge frame appears still, it's because the magnetic field lines (B) are moving at v in the Lorentz force equation. (F = qvB).

Current in a wire, for eg, is not zero if you pick a moving charge's frame, since the protons appear to move in the opposite direction. Therefore, current or magnetic field is never zero in a live wire, no matter which frame one picks: electrons or protons. This seems pretty pedestrian. What is going on here?

PS. Lorentz force: F = qvB, where v is the relative velocity between the charge q and a field line (B).
 
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  • #2
Orodruin
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In the quote, the author says, in the charge frame, the charge is at rest therefore there can be no magnetic force (Lorentz force). But this is incorrect since velocity is relative here. If the charge frame appears still, it's because the magnetic field lines (B) are moving at v in the Lorentz force equation. (F = qvB).
Field lines do not "move" in the way you are imagining it. However, what is a magnetic or an electric field also changes depending on the inertial frame. What is a magnetic force in one frame may very well be an electric force in another. This is a crucial point in understanding the Lorentz force ##\vec F = q(\vec E + \vec v \times \vec B)##.

PS. Lorentz force: F = qvB, where v is the relative velocity between the charge q and a field line (B).
This is incomplete (see above).
 
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  • #3
Nugatory
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The Lorentz force on the particle is given by ##\vec{F}=q\vec{E}+q(\vec{v}\times\vec{B})##. The formula that you're trying to use is the simplified version that works only when there is no electric field present.

Start with a situation in which we can properly use the simplified version: we have a particle moving in a magnetic field and the electric field is zero. The force on the particle is ##q(\vec{v}\times\vec{B})##, with no contribution from the ##q\vec{E}## term because ##\vec{E}## is zero.

Now examine this situation using the frame in which the particle is at rest. Because of the way that electric and magnetic fields transform, in this frame ##\vec{E}## is not zero; because ##\vec{E}## is not zero there's still a force on the particle even though the ##\vec{v}\times\vec{B}## term is now zero. (##\vec{B}## is also different, but we'd need a particle that is moving in this frame to see that effect). Thus, we have the same force on the particle in both frames but we explain it differently - "moving in a magnetic field" versus "subject to an electric field" - in the different frames.

It's generally best to think in terms of a single unified electromagnetic field instead of distinct electrical and magnetic fields. The mathematical description of the electromagnetic field is the same in all frames; the way that we divide this field into electrical and magnetic fields is different in different frames.
 
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  • #4
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moving charge's frame, since the protons appear to move in the opposite direction. Therefore, current or magnetic field is never zero in a live wire
That is right. The electromagnetic field divides into electrical and magnetic fields in different ways in different frames, and depending on the particular setup it may or may not be possible to find a frame in which one of them is zero.

It turns out that a current-carrying wire will have a non-zero ##\vec{B}## field in all frames. Conversely, an isolated charged particle will have a non-zero ##\vec{E}## field in all frames, but we can find a frame (the one in which the particle is at rest) in which its ##\vec{B}## field is zero.
 
  • #5
Now examine this situation using the frame in which the particle is at rest... even though the →v×→Bv→×B→\vec{v}\times\vec{B} term is now zero.
Hi Nugatory, the term qvB cannot be zero, because it is an external magnet field. The field lines (B) move at v relative to q.
 
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  • #6
Orodruin
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Hi Nugatory, the term qvB cannot be zero, because its an external magnet field. The field lines (B) move at v relative to q.
Again, this is wrong. Field lines do not move in the sense you are imagining. You have had two experienced physicists tell you the same thing. Continuing your argumentation is not going to make it any more correct. The v in the Lorentz force is the velocity of the charge. The Lorentz force does not care in the slightest about how the fields are changing - just the value of the fields at that particular time. (That is described by Maxwell’s equations.)
 
  • #8
Field lines do not move in the sense you are imagining.
I am imagining a moving permanent magnet. When the magnet moves at v, its field lines move at the same v, relative to a stationary charge q. I am not sure I understand what you mean. Please let me know if I am mistaken.
 
  • #9
Dale
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Lorentz force: F = qvB, where v is the relative velocity between the charge q and a field line (B).
No, v is the velocity of the charge q in an inertial reference frame. No velocity is ever attributed to field lines.

When the magnet moves at v, its field lines move at the same v, relative to a stationary charge q.
No, fields and field lines do not carry the velocity of their sources. Velocity is not something that is or can be attributed to the fields (although waves in the field can have velocity).

But this is incorrect since velocity is relative here.
Velocity is relative here, but it is the velocity of the charge relative to the reference frame. The fields do not have a velocity, so velocity relative to the fields doesn’t make sense.
 
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  • #10
No, fields and field lines do not carry the velocity of their sources.
At relativistic speeds or at non-relativistic speeds?
 
  • #11
Dale
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At relativistic speeds or at non-relativistic speeds?
At any speed. Velocity is simply not a property attributed to a magnetic field. It is like asking for the Meyer Briggs personality type of gravity. It simply is not a property that it has.
 
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  • #12
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Hey @DJ_Juggernaut, here's a little exercise which might make everything clear.

Let me add that I do actually have some kind of point with this.:smile:

Exercise 1: I have a magnetometer in my left hand, it says the magnetic field is 2 magnetic field strength units and pointing to the left. On my right hand I have a charge of 1 Coulombs. The right hand is moving forwards at speed 10 m/s. Calculate the Lorentz force that the right hand feels.

Exercise 2: Same as exercise 1, but the speed is 0.0001 m/s.
 
  • #13
Orodruin
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No, v is the velocity of the charge q in an inertial reference frame.
Just to clarify: In particular, in the inertial frame where the fields are ##\vec E## and ##\vec B##.


Also, to be clear on the transformation laws: As long as the electron does not travel at relativistic velocities, the electromagnetic force is the same in the electron rest frame as in the lab frame. The relation between the electric field ##\vec E## in the electron rest frame and the magnetic field ##\vec B_0## in the lab frame is given by
$$
\vec E = \vec v \times \vec B_0
$$
(again, for ##v \ll c##) and so the Lorentz force in the electron rest frame is
$$
\vec F = q(\vec E + \vec 0 \times \vec B) = q\vec E = q \vec v \times \vec B_0,
$$
just as the force in the lab frame - but the interpretation in the electron rest frame is in terms of an electric force, not a magnetic one.
 
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  • #14
Velocity is simply not a property attributed to a magnetic field.
It applies to a magnet. By extension (or indirectly) the field lines of the magnet "move" too. One can measure this change in flux. Although at very high speeds, I don't know what happens. My point was that, a moving charged particle in a magnetic field experiences a force given by, F = qvB. If you switch the frame to the charged particle frame, the charged particle q appears to be at rest and the magnet appears to move and (by extension its field lines move too). Therefore, the Lorentz force is still, a magnitude of, qvB. I don't want to repeat myself. So I take it we agree to disagree on this.
 
  • #15
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So I take it we agree to disagree on this.
It's not about disagreeing. What you state is simply plain wrong and you are unwilling to learn facts.
 
  • #16
Nugatory
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Therefore, the Lorentz force is still, a magnitude of, qvB. I don't want to repeat myself. So I take it we agree to disagree on this.
One of the good things about empirical science is that there is no "agree to disagree" - faced with conflicting theories we can run experiments and see which theory is confirmed. A century and a half of increasingly convincing experimental results say that electromagnetism does not behave the way you're saying.
 
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  • #17
Orodruin
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So I take it we agree to disagree on this.
As has already been said, it is not a matter of disagreeing. If you disagree, you are simply wrong.
the magnet appears to move and (by extension its field lines move too)
Again, just wrong. Even in the non-relativistic limit where you can do this, it does not matter because the ##\vec v## that appears in the Lorentz force law is not the velocity of the field lines, it is the velocity of the electron in that frame. Saying anything else does not make it true. What is true is that there is an electric field in the rest frame of the electron and that accounts exactly for the Lorentz force on the electron. This is not an issue that is up for debate, it is how empirical studies have confirmed that electromagnetism works. If you disagree with it, then you disagree with how the world works.
 
  • #18
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[...] One can measure this change in flux[...]
Here is the equation for magnetic flux (the simplest version):

$$Φ_B = BS cos θ $$

where B is the magnitude of the magnetic field and S is the area of the surface, and θ is the angle between the field lines and the normal to S.




Tell me, where is the symbol representing velocity? Note: you won't find a v in any version of that formula. The only v you'll ever find that is remotely related to this is in the Lorentz force law, and it's already been explained to you what that v means (but you "agree to disagree" about it). Field lines don't have velocity, and flux doesn't have anything to do with velocity, either.

Flux can change with respect to time, but that isn't a velocity: it's ##\frac{dΦ_B}{dt}##.
 
  • #19
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Hey @DJ_Juggernaut, here's a little exercise which might make everything clear.

Let me add that I do actually have some kind of point with this.:smile:

Exercise 1: I have a magnetometer in my left hand, it says the magnetic field is 2 magnetic field strength units and pointing to the left. On my right hand I have a charge of 1 Coulombs. The right hand is moving forwards at speed 10 m/s. Calculate the Lorentz force that the right hand feels.

The magnetic force the hand feels is of course zero - as discussed. But the magnetic force that the shoulder feels is 20 Newtons.
 
  • #20
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The magnetic force the hand feels is of course zero - as discussed. But the magnetic force that the shoulder feels is 20 Newtons.
It is not clear to me exactly what situation you are describing, but I'm having trouble coming up with any way of reading your previous post that makes this statement correct.
 
  • #21
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It is not clear to me exactly what situation you are describing, but I'm having trouble coming up with any way of reading your previous post that makes this statement correct.


If 'hand' is the part of the upper limb below the wrist, and said hand is capable of feeling a force by itself, then the force it feels is not a magnetic force, but an electric force, I mean according to the hand. Right?

Well in reality the brain feels the force - and it's correct for the brain to think that the force on the hand is a magnetic force.

So the brain observes that 'there is a magnetic field B in this place' and 'there's a force qvB pulling the hand'

How can this be so complicated:smile:
 
  • #22
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The magnetic field around a straight wire running along the x axis is$$B=\frac{\mu_0I}{2\pi r^2}\left(\begin{array}{c}0\\-z\\y\end{array}\right)$$Transformed into a frame moving in the x direction at speed ##v## it is $$B'=\frac{\gamma\mu_0I}{2\pi r^2}\left(\begin{array}{c}0\\-z'\\y'\end{array}\right)$$In other words, the field intensity changes but the magnetic field lines are identical to those in the original frame (there's also an electric field in this frame). They aren't moving in any sense - they aren't even time varying (although that's a special case when the current and velocity are colinear).
 
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  • #23
Orodruin
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I just reread this thread and I think you should follow your initial approach:
Please let me know if I am mistaken.
rather than the approach you took a few posts later:
I don't want to repeat myself. So I take it we agree to disagree on this.
Essentially you asked people to let you know if you were mistaken. When told that indeed you were - by several experienced physicists - you instead wanted to agree to disagree. This is just a bit of advice, but this change of attitude is not beneficial if you want to acually learn physics.
 
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  • #24
Mister T
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Well in reality the brain feels the force - and it's correct for the brain to think that the force on the hand is a magnetic force.
Certainly the person feels the force, regardless of whether or not the person concludes the force is magnetic in origin.

So the brain observes that 'there is a magnetic field B in this place' and 'there's a force qvB pulling the hand'
And what people are telling you is that that conclusion depends on an arbitrary choice of reference frame. Choose a different frame, one that's in motion relative to the prior choice, and that conclusion may no longer be valid.

How can this be so complicated:smile:
In 1905, when the idea was presented to people who researched this issue as a profession, it was overwhelmingly rejected. It took years of experimental verification for this grand idea to gain acceptance. Why? Apparently it has something to do with a disconnect between our intuition concerning the way Nature behaves and the way it actually behaves. Note that this disconnect is present even when people are presented with the much much older formalism developed by Newton. Einstein of course piled on in a way that very few seemed to have expected.
 
  • #25
Dale
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By extension (or indirectly) the field lines of the magnet "move" too.
Nope. That is incorrect. The field lines do not have velocity, including the velocity of the source. The v in the Lorentz force law is the velocity of the charge relative to the given inertial frame.

One can measure this change in flux.
Certainly. ##d\Phi/dt## can be measured, but ##d\Phi/dt\ne v##

My point was that, a moving charged particle in a magnetic field experiences a force given by, F = qvB. If you switch the frame to the charged particle frame, the charged particle q appears to be at rest and the magnet appears to move and (by extension its field lines move too). Therefore, the Lorentz force is still, a magnitude of, qvB. I don't want to repeat myself. So I take it we agree to disagree on this.
No need to repeat yourself. Just please provide the reference that supports your specific claim here.

I can cite references supporting mine, including the MIT lecture you cited above. The usual reference is Purcell who is famous for this line of reasoning.
 
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