Magnetic field and force and work

[miringains]

Do both magnetic field, and magnetic force not do any work on charged particles, due to perpendicular movement?

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[shaboobly]

Do both magnetic field, and magnetic force not do any work on charged particles, due to perpendicular movement?

the magnetic field results in the force. w=Fdcos(theta). it doesnt depend on the direction of the magnetic field. the force can cause the particle to move and thus work is done.

 

[Eastfolding]

You are correct, there is absolutely no component of magnetic force that is in the the same direction as motion.

 

[V5RED]

the magnetic field results in the force. w=Fdcos(theta). it doesnt depend on the direction of the magnetic field. the force can cause the particle to move and thus work is done.

No it can't. Magnetic fields don't cause charged particles to move, electric fields do.

(edit: this was incorrect of me to say: Magnetic fields can do work on stuff like a magnetic piece of metal(for the metal, they cause work, but the semantics of what is doing the work are debatable) or a current carrying wire), but they can't do work on a charged particle unless the charged particle is also a magnetic piece of metal.

Going with the formula you posted, the angle between the magnetic force and the direction of motion of a charged particle is 90 degrees. Cos90=0 so work is zero.

 

[shaboobly]

No it can't. Magnetic fields don't cause charged particles to move, electric fields do.

Magnetic fields can do work on stuff like a magnetic piece of metal or a current carrying wire, but they can't do work on a charged particle unless the charged particle is also a magnetic piece of metal.

Going with the formula you posted, the angle between the magnetic force and the direction of motion of a charged particle is 90 degrees. Cos90=0 so work is zero.

explain how the magnetic force cant do work. you need to take the angle in the direction the particle is traveling and how do you know the magnetic force doesnt make the particle move in the same direction of the force. that kinda implies that the magnetic force is doing work.

 

[V5RED]

explain how the magnetic force cant do work. you need to take the angle in the direction the particle is traveling and how do you know the magnetic force doesnt make the particle move in the same direction of the force. that kinda implies that the magnetic force is doing work.

I did not say that magnetic fields cannot do work. I said that they can only do work on (edit: this part was incorrect of me to say: a current carrying wire or a magnetic metal(for the metal, they cause work, but the semantics of what is doing the work are debatable).)

The magnetic force on a charged particle with a velocity is the cross product of the velocity and the magnetic field multiplied by the charge on the particle. This means that the force vector is perpendicular to the velocity vector(edit: and the force is perpendicular to the magnetic field...recall right hand rule. This perpendicular force leads to circular motion sort of like an object in orbit).

A particle moves in the direction of its velocity. This means that when you use the work equation you posted for a charged particle in a magnetic field, the force and direction are perpendicular to each other and the work is zero.

You can also look at this in terms of energy. The potential energy of the particle is not altered by the magnetic field, its speed is not affected, and it does not get hotter or colder due to the field, so there is no exchange of energy. If the charged particle were at rest, the magnetic field would have no effect on it.

A magnetic field does change the direction of the moving charged particle, so it does produce a force, but it has no effect on the speed or energy in any way, so it is not a force doing work.

 

[stevenz]

This topic comes up often in physics lectures and I can say with authority that magnetic fields can do NO WORK under ANY circumstances, whether static or changing on any particle/object/current/wire/metal. All work done by a magnetic field is indirect and due to electric fields that it generates via Faraday's Law.

Edit: this is a really common misconception so I'm just going to quote from wiki:

"Because the magnetic force is always perpendicular to the motion, the magnetic field can do no work on an isolated charge. It can only do work indirectly, via the electric field generated by a changing magnetic field. It is often claimed that the magnetic force can do work to a non-elementary magnetic dipole, or to charged particles whose motion is constrained by other forces, but this is incorrect[18] because the work in those cases is performed by the electric forces of the charges deflected by the magnetic field."

V5RED's example of the current carrying wire/metal is a good example of the "fake" work done by a magnetic field. In fact, no work is being done by the magnetic field. I can explain it if you want but it's somewhat technical.

2nd Edit: I thought of a decent "intuition" way of explaining it. When a B field does "work" on a ferromagnetic species, the energy comes from the object itself, in that the spin vectors on the electrons in the metal align, therefore lowering the total energy of the metal itself.

 

[V5RED]

This topic comes up often in physics lectures and I can say with authority that magnetic fields can do NO WORK under ANY circumstances, whether static or changing on any particle/object/current/wire/metal. All work done by a magnetic field is indirect and due to electric fields that it generates via Faraday's Law.

Edit: this is a really common misconception so I'm just going to quote from wiki:

"Because the magnetic force is always perpendicular to the motion, the magnetic field can do no work on an isolated charge. It can only do work indirectly, via the electric field generated by a changing magnetic field. It is often claimed that the magnetic force can do work to a non-elementary magnetic dipole, or to charged particles whose motion is constrained by other forces, but this is incorrect[18] because the work in those cases is performed by the electric forces of the charges deflected by the magnetic field."

V5RED's example of the current carrying wire/metal is a good example of the "fake" work done by a magnetic field. In fact, no work is being done by the magnetic field. I can explain it if you want but it's somewhat technical.

2nd Edit: I thought of a decent "intuition" way of explaining it. When a B field does "work" on a ferromagnetic species, the energy comes from the object itself, in that the spin vectors on the electrons in the metal align, therefore lowering the total energy of the metal itself.

For ferromagnetic materials, I have two links to physicists saying that it is a misconception that magnetic fields do no work on them. They say that the field does no work on a charged particle, but to extend that to examples like a magnet lifting a paperclip is a misconception.

http://www.ilovephysics.com/2009/04/20/common-misconceptions-work-done-by-magnetic-field/

http://www.physlink.com/education/askexperts/ae354.cfm

Aside from your post, I can't find any source that says otherwise. I have to ask though, you say you can say it with authority. Do you have a PhD/MS in physics? If not I would suggest you can't. If you do, then I am surprised you are here.

edit: either way, direct or indirect, the field's net result is work on the metal...I have only taken physics 1 and 2, so explanations beyond what is learned there will be beyond me. I am pretty sure the topic of the semantics of work on magnets by magnets is beyond the MCAT as well. I doubt that the idea that magnetic fields don't do work on charged particles is beyond the MCAT though and in that respect we are in agreement.

 

[stevenz]

I am a fourth year physics student, and I can tell you right now that I don't need to have a PhD to understand the underlying physics as to why it is impossible for a B field to do work. You have referenced two sources which you have probably googled in your defence. I'm not sure if you are joking or trolling or what if you believe that some physics blog and physlink.com can be considered "reliable sources". You should have just gone to wikipedia for "appeal to authority" like I did.

Edit: you should probably actually LOOK at your links before you post them. The first link supports me in that magnetic fields cannot do any work...

This is an incredibly common misconception and there are many, many physicists who agree with me, including Griffith (whose textbook I am looking at right now, which says in bold, no magnetic fields can do any work), and my professor, whom I've had this conversation with before.

In any case, the proof is rather well known and tractable for any 2nd year physics student. I don't need to base my answers on appeal to authority so it is up to you whether or not you believe me. Here is a pretty good discussion of it on physics forums, for your interest, though it may be a bit technical for your understanding.

http://www.physicsforums.com/showthread.php?t=130549

 

[V5RED]

I am a fourth year physics student, and I can tell you right now that I don't need to have a PhD to understand the underlying physics as to why it is impossible for a B field to do work. You have referenced two sources which you have probably googled in your defence. I'm not sure if you are joking or trolling or what if you believe that some physics blog and physlink.com can be considered "reliable sources". You should have just gone to wikipedia for "appeal to authority" like I did.

Edit: you should probably actually LOOK at your links before you post them. The first link supports me in that magnetic fields cannot do any work...

This is an incredibly common misconception and there are many, many physicists who agree with me, including Griffith (whose textbook I am looking at right now, which says in bold, no magnetic fields can do any work), and my professor, whom I've had this conversation with before.

In any case, the proof is rather well known and tractable for any 2nd year physics student. I don't need to base my answers on appeal to authority so it is up to you whether or not you believe me. Here is a pretty good discussion of it on physics forums, for your interest, though it may be a bit technical for your understanding.

http://www.physicsforums.com/showthread.php?t=130549

Did you "LOOK" at the link? It literally says the opposite of what you did. "The magnetic field does NO work on a charged particle. In fact, the only thing a magnet can do work on is another magnet or a ferromagnetic material."

You initially resorted to an appeal to authority by linking to wikipedia and claiming to be an authority(which you are not). The point of my links was to show that people who actually count as authorities disagree with you and the wikipedia entry.

I absolutely could be wrong about magnets or magnetic fields doing work(edit: on ferromagnetic materials), but your argument form has been abysmal and very abrasive which is a horrible way to convince anyone of anything.

edit: I have found an actual good and intuitive explanation for why magnetic fields do not do work on a current carrying wire, so I was mistaken about that. The moving charged particles are having their direction altered by the magnetic field, but not their speed, so even though the wire moves, no work was done by the magnetic field. Also, I was thinking about the force acting on the wire when it is really the magnetic field producing a force on a lot of individual charges which, of course, does no work. I still have yet to see a good explanation for why a ferromagnetic material can't have work done on it by a magnetic field though. I mean if I turn on an electromagnet above some paperclips, they fly up, so obviously something is doing work.

 

[stevenz]

Concerning your link. The first one. Read the last line, the author says:

The magnetic field can do no work on a charge.

Concerning my "appeal to authority". As mentioned in my first post, I would be glad to post the solution, if anybody was interested. I mean, first of all, I think I'm pretty safe to assume that you aren't in physics. The question of why I have to even go to the lengths of writing this out is dumb, since you can tell just by looking at the equation of the magnetic force that it does no work. I mean, its pretty clear right? Magnetic fields only act on moving charges, and they only act perpendicular to their velocity. It's not like this magically changes if you move the charges in weird ways, or put them in different atoms. You say that it does work on ferromagnets - that is one WEIRD exception if its true, and there are rarely exceptions in physics. If this were a true scientific discussion, I should be asking YOU to prove why magnetic fields do work on this one type of material but not others. But since I just hate ignorance, here you are:

Consider a mass m attached to a loop of clockwise oriented current, where the top wire of current is exposed to a magnetic field in the direction of the page. It is clear that if we apply sufficient current into the loop, the force of the magnetic field acting on the top wire (current going left to right) will yield a Lorentz force upwards, giving the appearance of the magnetic field doing work.

In particular, let a be the width of the (rectangular) wire loop, and h be the height that the loop rises.
Then, W_mag = F_mag * h = IBah.

To show this is actually NOT work done by F_mag, consider what happens when the loop starts to rise. The current (previously with I vector pointing to the right now has a vertical component. That is, electrons moving through the wire not only have the electromotive drift vector, call it v, but a perpendicular drift vector, call it u. It is clear that u is precisely the speed that the wire itself moves upwards.

However, since the magnetic field must be perpendicular to the velocity of electrons, it must change. If we evaluate B x u we find that the magnetic field gains a component which opposes the flow of electrons. Call it F_u.

So its clear now that for the magnetic field to lift up the mass, the electromotive force (the thing providing the current in the first place) must DO MORE WORK in order to push more electrons through the circuit. That is - the MAGNETIC FIELD DOES NO WORK. The magnetic field induces a stronger electric field from the EMF, causing work to be done by the ELECTRIC FIELD. You can easily show that this extra emf is equal precisely to IBah:

Let L be the current density in the wire, then the vertical component of current induced by u is L*u.

F_emf = -F_u = L*u*a*B (I'm too lazy to bold this to indicate vectors...)
W_emf = integral(F_emf)dx = L*a*B integral(u*dx) = L*a*B integral(dh/dt*v*dt) = L*v*a*B*h = IBah

as shown. So what's your classical analogy? Well, if you have a ramp, and you are pushing a block up the ramp, you don't say the normal force does any work on the block. YOU do all the work on the block, the ramp just REDIRECTS the work. In the same way, the magnetic field redirects the work done by the electric field.

How does this apply to ferromagnetic materials? Ferromagnetism is a phenomenon in which the spin vectors on the delocalized electrons in metals can be aligned with an external magnetic field. Now, you might think this has nothing to do with currents, but your high school teacher was actually correct for once - you CAN think of electron spin as little tiny current loops in your atoms. Quantum mechanically, spin has an expectation value which can be interpreted as current.

So, it turns out that magnets can be modelled as masses attached to big current loops (composed of tiny quantum mechanical currents). Where is the work coming from? The metal itself (via induced electric fields). NOT the magnetic field.

 

[V5RED]

Concerning your link. The first one. Read the last line, the author says:


Concerning my "appeal to authority". As mentioned in my first post, I would be glad to post the solution, if anybody was interested. I mean, first of all, I think I'm pretty safe to assume that you aren't in physics. The question of why I have to even go to the lengths of writing this out is dumb, since you can tell just by looking at the equation of the magnetic force that it does no work. I mean, its pretty clear right? Magnetic fields only act on moving charges, and they only act perpendicular to their velocity. It's not like this magically changes if you move the charges in weird ways, or put them in different atoms. You say that it does work on ferromagnets - that is one WEIRD exception if its true, and there are rarely exceptions in physics. If this were a true scientific discussion, I should be asking YOU to prove why magnetic fields do work on this one type of material but not others. But since I just hate ignorance, here you are:

Consider a mass m attached to a loop of clockwise oriented current, where the top wire of current is exposed to a magnetic field in the direction of the page. It is clear that if we apply sufficient current into the loop, the force of the magnetic field acting on the top wire (current going left to right) will yield a Lorentz force upwards, giving the appearance of the magnetic field doing work.

In particular, let a be the width of the (rectangular) wire loop, and h be the height that the loop rises.
Then, W_mag = F_mag * h = IBah.

To show this is actually NOT work done by F_mag, consider what happens when the loop starts to rise. The current (previously with I vector pointing to the right now has a vertical component. That is, electrons moving through the wire not only have the electromotive drift vector, call it v, but a perpendicular drift vector, call it u. It is clear that u is precisely the speed that the wire itself moves upwards.

However, since the magnetic field must be perpendicular to the velocity of electrons, it must change. If we evaluate B x u we find that the magnetic field gains a component which opposes the flow of electrons. Call it F_u.

So its clear now that for the magnetic field to lift up the mass, the electromotive force (the thing providing the current in the first place) must DO MORE WORK in order to push more electrons through the circuit. That is - the MAGNETIC FIELD DOES NO WORK. The magnetic field induces a stronger electric field from the EMF, causing work to be done by the ELECTRIC FIELD. You can easily show that this extra emf is equal precisely to IBah:

Let L be the current density in the wire, then the vertical component of current induced by u is L*u.

F_emf = -F_u = L*u*a*B (I'm too lazy to bold this to indicate vectors...)
W_emf = integral(F_emf)dx = L*a*B integral(u*dx) = L*a*B integral(dh/dt*v*dt) = L*v*a*B*h = IBah

as shown. So what's your classical analogy? Well, if you have a ramp, and you are pushing a block up the ramp, you don't say the normal force does any work on the block. YOU do all the work on the block, the ramp just REDIRECTS the work. In the same way, the magnetic field redirects the work done by the electric field.

How does this apply to ferromagnetic materials? Ferromagnetism is a phenomenon in which the spin vectors on the delocalized electrons in metals can be aligned with an external magnetic field. Now, you might think this has nothing to do with currents, but your high school teacher was actually correct for once - you CAN think of electron spin as little tiny current loops in your atoms. Quantum mechanically, spin has an expectation value which can be interpreted as current.

So, it turns out that magnets can be modelled as masses attached to big current loops (composed of tiny quantum mechanical currents). Where is the work coming from? The metal itself (via induced electric fields). NOT the magnetic field.

You gave ****ty explanations, were super arrogant about them, misread the links I posted, and falsely claimed authority. You came off as a total jackass who can't read.

That said, you finally posted a decent explanation that actually makes sense though you worded it like an *******. Imagine if you had done that from the beginning instead of being a dick.

 

[deleted103644]

.

 

[stevenz]

You gave ****ty explanations, were super arrogant about them, misread the links I posted, and falsely claimed authority. You came off as a total jackass who can't read.

That said, you finally posted a decent explanation that actually makes sense though you worded it like an *******. Imagine if you had done that from the beginning instead of being a dick.

You're kind of funny. Sufficient evidence was provided within the first few lines of the thread and is directly evident through the definition of the magnetic force. I offered an authoritative answer since I was 99.9% that I was right, hoping that it would put to rest people's doubts, and even offered to derive a proper proof, if people wanted me to. Somehow you got it in your thick skull that my claiming authority was me being arrogant, and from what I can tell, you did a 10 second google search, looked at the first few links that you thought would support your claim against me, and posted saying:

I have to ask though, you say you can say it with authority. Do you have a PhD/MS in physics? If not I would suggest you can't. If you do, then I am surprised you are here.

I would not claim to have authority if I didn't. It would be idiotic to expect some one to have a PhD to have to answer questions a 2nd year undergraduate physics major could answer.

Now, as for my wording in my explanation. Maybe you should go read a physics textbook or something. Better yet you should read one on electromagnetism. The language is clear and concise, but detailed, which is what I tried to emulate. It's a bit dry, but if you had enough juice in your brain to follow the argument (which I honestly doubt you do), I think you would find it quite helpful.

 

[V5RED]

You're kind of funny. Sufficient evidence was provided within the first few lines of the thread and is directly evident through the definition of the magnetic force. I offered an authoritative answer since I was 99.9% that I was right, hoping that it would put to rest people's doubts, and even offered to derive a proper proof, if people wanted me to. Somehow you got it in your thick skull that my claiming authority was me being arrogant, and from what I can tell, you did a 10 second google search, looked at the first few links that you thought would support your claim against me, and posted saying:



I would not claim to have authority if I didn't. It would be idiotic to expect some one to have a PhD to have to answer questions a 2nd year undergraduate physics major could answer.

Now, as for my wording in my explanation. Maybe you should go read a physics textbook or something. Better yet you should read one on electromagnetism. The language is clear and concise, but detailed, which is what I tried to emulate. It's a bit dry, but if you had enough juice in your brain to follow the argument (which I honestly doubt you do), I think you would find it quite helpful.

Did you get picked on a lot as a kid? You come off as a very insecure child. Woo hoo you knew something about physics, but you had to act like a little punk about it.

 

[stevenz]

Did you get picked on a lot as a kid? You come off as a very insecure child. Woo hoo you knew something about physics, but you had to act like a little punk about it.

I don't really see why the first thing every one does when they try to throw an insult online is to make a bunch of really specific assumptions. I mean, you're taking a big chance here cause if you're wrong I'll just think you're an idiot.

But then again, I already do so I guess you have me beat there.

 

[Pigglyjuff]

I don't really see why the first thing every one does when they try to throw an insult online is to make a bunch of really specific assumptions. I mean, you're taking a big chance here cause if you're wrong I'll just think you're an idiot.

But then again, I already do so I guess you have me beat there.

Threads like this are why I love and hate SDN.

 

[Tatiana3325]

Sorry OP- there isn't a straight answer to your question. It depends entirely on who you ask. It's a conceptual question.

I love that you asked it though time for you to major in physics.

 

[Tatiana3325]

there are rarely exceptions in physics.

lolz haha rofl

 

[deleted103644]

Considering it's the same force once you take a relativistic view, it's kind of a pointless argument.

 

[Chunkle]

If you put that magnetic forces do work on the MCAT, you'll get it wrong..

Didn't read all the posts above, but I don't really care for the explanations, the MCAT thinks they do no work, and that is what I shall put.

 

[Tatiana3325]

If you put that magnetic forces do work on the MCAT, you'll get it wrong..

Didn't read all the posts above, but I don't really care for the explanations, the MCAT thinks they do no work, and that is what I shall put.

Which makes physical science MCAT scores meaningless.

 

[deleted103644]

Which makes physical science MCAT scores meaningless.

Haha, it's an easy 15, don't knock it.

 

[Tatiana3325]

Haha, it's an easy 15, don't knock it.

It's an interesting thought though. Very scandalous. "MCAT asks confusing conceptual physics questions that physicists would argue about using oversimplified analogies. But the MCAT thinks there is a right answer!!!!! ZOMG scandal!!"

Oooooo. So many people would care. Lol.

Oh well.

 

[deleted103644]

It's an interesting thought though. Very scandalous. "MCAT asks confusing conceptual physics questions that physicists would argue about using oversimplified analogies. But the MCAT thinks there is a right answer!!!!! ZOMG scandal!!"

Oooooo. So many people would care. Lol.

Oh well.

I think the questions are generally well-worded enough that there would be a right answer, but I haven't looked at them in a long time.

 

[Tatiana3325]

I think the questions are generally well-worded enough that there would be a right answer, but I haven't looked at them in a long time.

You're probably right.

 

[brownbaglunch]

If they give you a question like this:

"An electromagnet lifts a two thousand kilogram car to a height of 10 meters. How much work was done by the magnet?" I would be uncomfortable picking zero. When the question clearly indicates that work is done, I wouldn't look for a trick answer.

In the case of the OP's question, the force applied to a moving charge by a magnetic field is entirely perpendicular to the velocity, and can not increase the kinetic energy of the charge. I would be suspect of any answer that implies work is done by magnetism in a system like this.

 

[shaboobly]

i still dont understand why you guys think magnetic fields aren't capable of work on anything with a charge.

in my view: charged particle is moving perp to B field, which then causes a force to act on the charged particle which then implies the particle is accelerating because of newtons laws. since the charged particle accelerates it implies there is a velocity associated with the direction of the force, which is the same direction of the acceleration and then implies work is done on the charged particle.

the particle can still be moving in its original direction, but the way i see it it is not the distance the particle was originally travelling that contributes to the work done, but the distance in the direction in which the particle accelerates, due to the magnetic force.

no im not a physics major, im a chemistry major and i just base this off my logical thinking of what's going on...

 

[deleted103644]

Work = integral of F . ds

If F and ds are always perpendicular, W = 0.

This is what is tripping people up.

 

[shaboobly]

Work = integral of F . ds

If F and ds are always perpendicular, W = 0.

This is what is tripping people up.

the force and original displacement caused by the original velocity are perpendicular and so no work is done in the original direction, but the particle is moved by the force in the same direction and thus work is done.

any force acting on anything means the object is accelerating in the same direction as the force, and thus it means there is also velocity in that same direction. in this case the direction of the force and the direction of the distance moved due to the force (acceleration) are in the same direction and thus work is done.

what's tripping people up i think is that there is no work in the original direction the charged particle is moving because the force is perpendicular to this and the magnetic field. in my eyes, however, there is always work done on the charged in the direction of the force because a force causes an acceleration which causes the particle to move in the same direction of the force and thus have a displacement in the same direction; unless of course the displacement turns out to be 0, in which case no work is done.

i dont see how my way of thinking is wrong..

 

[deleted103644]

the force and original displacement caused by the original velocity are perpendicular and so no work is done in the original direction, but the particle is moved by the force in the same direction and thus work is done.

any force acting on anything means the object is accelerating in the same direction as the force, and thus it means there is also velocity in that same direction. in this case the direction of the force and the direction of the distance moved due to the force (acceleration) are in the same direction and thus work is done.

what's tripping people up i think is that there is no work in the original direction the charged particle is moving because the force is perpendicular to this and the magnetic field. in my eyes, however, there is always work done on the charged in the direction of the force because a force causes an acceleration which causes the particle to move in the same direction of the force and thus have a displacement in the same direction; unless of course the displacement turns out to be 0, in which case no work is done.

i dont see how my way of thinking is wrong..

It's relatively easy to have a non-zero force do zero work.

Just look at a ball tied to a string spinning in a circle. Ignoring friction and gravity, this system could continue more or less forever with the same angular momentum.

A constant force is pulling the ball towards the center, but no work is done because the force is always perpendicular to velocity.

The effect of magnetism on a moving charge could be viewed the same way.

 

[shaboobly]

It's relatively easy to have a non-zero force do zero work.

Just look at a ball tied to a string spinning in a circle. Ignoring friction and gravity, this system could continue more or less forever with the same angular momentum.

A constant force is pulling the ball towards the center, but no work is done because the force is always perpendicular to velocity.

The effect of magnetism on a moving charge could be viewed the same way.

this example is kind of irrelevant though because the mass would be in equilibrium and the net displacement is 0 anyway, which means work would be 0 regardless because there is no NET force and no displacement.

the magnetic force, in my eyes, has an acceleration and velocity associated with it in the same direction because F=ma. not only this but it would cause displacement in the same direction and unless the particle doesn't return back to the original location then nonzero work is done.

there is no force in the original direction of the moving particle and also has no work associated with it because it would be travelling at constant velocity, meaning net force is equal to 0 and no work is done in that direction.

the direction of the work i am saying occurs is in the same direction of the magnetic force, unless there is something to cause the net force to be 0 or the displacement to be 0 then the work would be nonzero in this direction.

 

[deleted103644]

this example is kind of irrelevant though because the mass would be in equilibrium and the net displacement is 0 anyway, which means work would be 0 regardless because there is no NET force and no displacement.

the magnetic force, in my eyes, has an acceleration and velocity associated with it in the same direction because F=ma. not only this but it would cause displacement in the same direction and unless the particle doesn't return back to the original location then nonzero work is done.

there is no force in the original direction of the moving particle and also has no work associated with it because it would be travelling at constant velocity, meaning net force is equal to 0 and no work is done in that direction.

the direction of the work i am saying occurs is in the same direction of the magnetic force, unless there is something to cause the net force to be 0 or the displacement to be 0 then the work would be nonzero in this direction.

Do you know anything about vectors? That makes this easier to explain.

W = int{F.ds}
v = ds/dt
F = qvxB

So ...

W = int{(qds/dt x B).ds} = 0

In general,
(X x Y) . Y = 0

 

[shaboobly]

Do you know anything about vectors? That makes this easier to explain.

W = int{F.ds}
v = ds/dt
F = qvxB

So ...

W = int{(qds/dt x B).ds

In general,
(X x Y) . Y = 0

yea i have no idea what you are trying to say. all im saying is that the force produces an acceleration because there is no other force to make a resultant of 0 and thus wont be in equilibrium. v=at, and when there is a force, a is nonzero and there is a velocity. all these variables are vectors pointing in the same direction. this means that the original velocity and the velocity generated by the force will have a component going in the same direction of the force and thus the displacement will be in the same direction. as a direct result, the work is nonzero in this direction.

think of a charged particle between a parallel plate capacitor. it will accelerate due to the electric force and thus have acceleration and velocity in the same direction and work will also be nonzero in this direction.

i still have no idea where my thinking is wrong in this.

 

[deleted103644]

No one was saying the electric force can do no work.

The force exerted by a magnetic field is always perpendicular to the velocity of a moving charge, do it will do no work on that charge.

When dealing with aggregates (eg an electromagnet lifting a car), I think it's silly to say magnets do no work (although you could make that argument).

For point charges though, it's a no brainer - qv x B must always be perpendicular to v, so no work will be done.

 

[shaboobly]

No one was saying the electric force can do no work.

The force exerted by a magnetic field is always perpendicular to the velocity of a moving charge, do it will do no work on that charge.

When dealing with aggregates (eg an electromagnet lifting a car), I think it's silly to say magnets do no work (although you could make that argument).

For point charges though, it's a no brainer - qv x B must always be perpendicular to v, so no work will be done.

Yea.. That's the original velocity of the charged particle and I agree with that if you read my posts. What I'm saying is the a and v in the same direction of the force all as a result of the force causing the particle to move in the same direction as the force and that means work. You don't understand that newtons laws still count and there is a and v and thus d in the same direction of the B force which is the definition of work

 

[V5RED]

Yea.. That's the original velocity of the charged particle and I agree with that if you read my posts. What I'm saying is the a and v in the same direction of the force all as a result of the force causing the particle to move in the same direction as the force and that means work. You don't understand that newtons laws still count and there is a and v and thus d in the same direction of the B force which is the definition of work

For a charged particle in a magnetic field, the direction of the force is always perpendicular to the velocity of the particle and the magnetic field. When the particle changes direction, so does the direction of the force. There is no point at which the particle is moving in the same direction as the magnetic force acting on it.

http://www.physics.sjsu.edu/becker/physics51/images/28_13A_Orbit_in_B_field.jpg

This image shows you the changes in the direction of the magnetic force on a charged particle as it moves in a field.

 

[deleted103644]

Yea.. That's the original velocity of the charged particle and I agree with that if you read my posts. What I'm saying is the a and v in the same direction of the force all as a result of the force causing the particle to move in the same direction as the force and that means work. You don't understand that newtons laws still count and there is a and v and thus d in the same direction of the B force which is the definition of work

No one was denying Newton (except for the points when I brought up relativity).

I think you are just unfamiliar with dealing with motion in more than one dimension.

The example of tension on a rope with a rotating mass is very similar to the magnetic force on a moving point charge - as others have pointed out, the force is always perpendicular to the velocity. So there is never a component of ma that is in the same direction as v (sorry for using vectors again, but it makes things easier - you should really try to learn a bit about them).

 

[shaboobly]

yeah but the particle isnt going to randomly stop moving in its original direction, which implies some sort of trajectory and the force can be broken into component vectors, i think, to participate in the work.

 

[deleted103644]

yeah but the particle isnt going to randomly stop moving in its original direction, which implies some sort of trajectory and the force can be broken into component vectors, i think, to participate in the work.

Would you prefer if I wrote a as dv/dt?

Nothing random about it. At any infinitesimal time period, the acceleration due to the magnetic force on a point charge will be perpendicular to velocity.

Components and basis do not matter, a.v = 0, ie the vectors are orthogonal (perpendicular) so work is zero.

If you don't understand why you're wrong here, you need a better foundation in calculus and vectors. And yes, you are wrong.

 

[shaboobly]

Would you prefer if I wrote a as dv/dt?

Nothing random about it. At any infinitesimal time period, the acceleration due to the magnetic force on a point charge will be perpendicular to velocity.

Components and basis do not matter, a.v = 0, ie the vectors are orthogonal (perpendicular) so work is zero.

If you don't understand why you're wrong here, you need a better foundation in calculus and vectors. And yes, you are wrong.

i dont think you are understanding what i have been trying to say in the last few posts ive made. i am in no way talking about the original velocity which is required to generate the force, but the velocity associated with the acceleration that the force makes.

what i am thinking is that since there is a magnetic force, at this instant, there is an acceleration from the force because of newtons law. this also implies that because the acceleration is non zero, that there is indeed a velocity at this instant and the particle moves, whether it be just a little bit, and the displacement is also nonzero which is nonzero work.

i understand that the second velocity caused by the acceleration will also produce another force that is perpendicular to it, but it would also create a new acceleration and velocity at the instant the new force effect is felt in the same direction.

if you were to continue adding these works and forces together, then indeed the work would be 0 because there is no displacement and the NET force would end up being zero, but you are misunderstanding everything that i say. you bring up the infinitesimal theory in calculus and this is precisely what i am talking about. the work is non zero at this instant and i dont see how the work can be zero at this instant. when the particle reaches the original location, no work could have been done because there is no displacement and the net force is zero, just like i said. instead of being arrogant, how about you consider what i am saying for just a second and explain to me where i am wrong, instead of just saying the same thing over and over again. according my logic, it seems you are wrong in your thinking, but hey im just being arrogant like you.

 

[deleted103644]

I'm sorry if I'm coming across as arrogant, but you're demonstrating a fundamental error in your reasoning.

At every infinitesimal window, there is no work because the force is perpendicular.

Because of this, the sum aka integral is zero.

I'm not talking about total displacements or anything like that. There are plenty of cases where an object returns to its original position and velocity but does work (eg an engine). That's why I'm mentioning integrals.

I know these topics can be conceptually difficult, but you really need to take my word that you're wrong.

 

[shaboobly]

I'm sorry if I'm coming across as arrogant, but you're demonstrating a fundamental error in your reasoning.

At every infinitesimal window, there is no work because the force is perpendicular.

Because of this, the sum aka integral is zero.

I'm not talking about total displacements or anything like that. There are plenty of cases where an object returns to its original position and velocity but does work (eg an engine). That's why I'm mentioning integrals.

I know these topics can be conceptually difficult, but you really need to take my word that you're wrong.

the magnetic force is perpendicular to the ORIGINAL velocity that the particle had. being that there is a nonzero net magnetic force in one direction at the very instant implies that there is acceleration and the particle moves (gradually adding up all the small movements to have 0 displacement). this is the definition of work, it doesnt matter if the magnitude is very small or insignificant, i still dont see how conceptually the work is 0 at this instant. sure a new force is generated from the new velocity caused by the force to make it go in a circle, but i dont see still how at that instant the work is 0.

 

[deleted103644]

Because the magnetic force is always perpendicular to the velocity (v, not v0).

The formula for magnetic force on a point charge is qvB in a direction perpendicular to both v and B (it's a cross product but too lazy for more bold vectors).

You can calculate work as the integral of F.v dt. Since F is a function of v that always produces a perpendicular vector, the dot product is zero, and a definite integral of 0 is also 0.

You could also say this:

dW = q (v x B) . v dt

You can look up a proof on Wikipedia that (m x n).m is always equal to 0 (just look for cross product).

 

[shaboobly]

Because the magnetic force is always perpendicular to the velocity (v, not v0).

The formula for magnetic force on a point charge is qvB in a direction perpendicular to both v and B (it's a cross product but too lazy for more bold vectors).

You can calculate work as the integral of F.v dt. Since F is a function of v that always produces a perpendicular vector, the dot product is zero, and a definite integral of 0 is also 0.

You could also say this:

dW = q (v x B) . v dt

You can look up a proof on Wikipedia that (m x n).m is always equal to 0 (just look for cross product).

are you saying that the magnetic force doesnt cause an acceleration and thus a small velocity able to displace the particle a tiny amount in the same direction? i dont see how it wouldnt. if a is nonzero, it must mean that v is non zero, and im not talking about the force caused by the velocity of the acceleration of the first force, but only after the original force acts with the original velocity. this means there are an F a and v (not the original v because yes it is perpendicular, but the v caused by the a) all in the same direction which should mean at this instant a small amount of work is done.

let me say it again, im not talking about the second force from the velocity caused by the acceleration of the original force.

 

[deleted103644]

Of course it causes acceleration. Just like tension on a rotating ball on a string. That doesn't mean it needs to ever accelerate in the same direction as the instantaneous velocity.

F = ma = qvxB

That's why I keep mentioning it.

 

[shaboobly]

Of course it causes acceleration. Just like tension on a rotating ball on a string. That doesn't mean it needs to ever accelerate in the same direction as the instantaneous velocity.

F = ma = qvxB

That's why I keep mentioning it.

yea but this is a conceptual question that i dont think can really be answered by setting the two force equations equal.

the ball on the string and this situation are different, in that the velocity of the mass on the string and the acceleration are indeed always perpendicular because of the circular motion aspect. im talking about the very instant the force is generated it generates a nonzero a in the same direction and thus v will increase as well in the same direction. yes i agree that another force is generated perpendicular to this velocity, but i am only talking about the very instant after the first force is felt.

 

[deleted103644]

yea but this is a conceptual question that i dont think can really be answered by setting the two force equations equal.

the ball on the string and this situation are different, in that the velocity of the mass on the string and the acceleration are indeed always perpendicular because of the circular motion aspect. im talking about the very instant the force is generated it generates a nonzero a in the same direction and thus v will increase as well in the same direction. yes i agree that another force is generated perpendicular to this velocity, but i am only talking about the very instant after the first force is felt.

You realize that magnetic forces induce circular motion on point charges too, right?

For example, the northern lights is caused by charged particles moving in a helical path around the Earth's magnetic field lines. (Circular motion with additional velocity components parallel to the field - also perpendicular to the magnetic force so no work.)

Magnetic fields cause moving charged particles to travel in curved paths, you can even get circles if you set it up carefully.

There is no work, and the string example is directly relevant.

 

[shaboobly]

You realize that magnetic forces induce circular motion on point charges too, right?

For example, the northern lights is caused by charged particles moving in a helical path around the Earth's magnetic field lines. (Circular motion with additional velocity components parallel to the field - also perpendicular to the magnetic force so no work.)

Magnetic fields cause moving charged particles to travel in curved paths, you can even get circles if you set it up carefully.

There is no work, and the string example is directly relevant.

yea i already said this is all true, but you didnt talk about the instant right after the first force is felt because like i already said and agree with you 100% the force will continue to change directions and induce circular motion and thus will have a type of centripetal acceleration which then implies there is no force in the direction of the displacement.

im talking about the very instant (i thought you would know because it seems you understand vectors and calculus pretty well) right after the first force is felt. it causes the particle to accelerate in the same direction because of newtons law, and once accelerated a velocity will be also in the same direction (like i said before and ill say again, there is another force that occurs once this velocity is initiated, which induces the circular motion aspect) and thus a small displacement will be made in the same direction also; this all implies non zero work at this instant.

here ill say it again: im not talking about the entire trajectory, just the instant after the original force is produced, not the second, not the third, not any other force, just the first force experienced at that very instant.

 

[deleted103644]

It doesn't matter which instant you're choosing.

dW = (qv x B).v dt = 0

Regardless of the value of v or B at that point, the incremental value of work is zero.