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Hope99
 3 years ago
Magnetic force... Conservative or Nonconservative?
Hope99
 3 years ago
Magnetic force... Conservative or Nonconservative?

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experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0magnetic field is a non conservative field http://upload.wikimedia.org/wikipedia/en/math/4/8/b/48bf197f3f2cd48dd50295af72d9c989.png

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1ok, this turns out to be a much more subtle question that I thought mathematically experimentX has shown by definition that it is not technically conservative, though is meets the criteria for a conservative force in many physical situations here are some discussions on the matter http://www.physicsforums.com/archive/index.php/t225524.html

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1now I'm very confused because here's a very different conclusion http://www.physicsforums.com/archive/index.php/t172893.html

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1I honestly am getting lost at this point... for a conservative vector field\[\nabla\times\vec F=0\]but\[\nabla\times\vec B=\mu_0\vec J+\mu_0\epsilon_0\frac{\partial\vec E}{\partial t}\] but if the Efield is not changing, and there is no current flowing (J=0 like in a bar magnet) doesn't this become\[\nabla\times\vec B=0\]???

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0the electrons are revolving around the nucleus, if these loops align in same direction then they give net magnetic fields. Not sure though.

fwizbang
 3 years ago
Best ResponseYou've already chosen the best response.1The magnetic force is \[\vec F = q (\vec v \times \vec B)\] where v is the particle velocity and B is the magnetic FIELD. Just as for any force, you can calculate the power, which is the rate at which the force does work, by \[P= \vec F \cdot\vec v = 0\] So the magnetic force never does any work. Since the definition of a conservative force is that the work it does doesn't depend on the path you choose to take, the magnetic force is conservative. Since the work is always zero, there isn't a useful magnetic potential energy.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1@fwizbang that was my original argument as well, but what do you make of the fact that in many cases\[\nabla\times\vec B\neq0\]which is a requirement of a nonconservative vector field?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1oh but you are saying that\[\nabla\times\vec F_B=0\]perhaps? that could make sense...

fwizbang
 3 years ago
Best ResponseYou've already chosen the best response.1Since the magnetic force is velocity dependent, the curl F =0 condition doesn't apply. It assumes F=F(x,y,z), which isn't true for the magnetic force.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1ok... that makes sense still thinking more deeply though thanks

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2MAGNETIC FIELDS! + Forces truly raises a lot of questions... Still mysterious it is.

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2So magnetic fields are conservative? And will never be able to do work ? @fwizbang I believe that in some cause magnetic field are conservative and some cases there not...

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2What about his cause: "The magnetic force can do work to a magnetic dipole, or to a charged particle whose motion is constrained by other forces." Is that true? @TuringTest @fwizbang @experimentX

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0the force is always perpendicular to velocity => work done = 0

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2Then why do people assume it might be nonconservative?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0can i explain this in detail tomorrow? i have got lot of qualms myself.

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2So magnetic force WILL never be capable of doing work huh? Sure, I'll research this more and more.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0I'm not sure about magnetic dipole. for for lorentz force . F is always perpendicular to the velcity and magnetic field.

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2@experimentX In a motor the magnetic force is able to dor work? "the force is always perpendicular to velocity => work done = 0"? What velocity do you mean? In a magnetic field or what?(Lost here) F = q( v+ E x B)? I'm looking at this law more: F = I x L x B Velocity is out of this order. I think magnetic forces under certain conditions tend to create/do weried things lol...

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0looks like i came to contradiction. can we discuss this later?

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2Sure thing mate. I'll be waiting anyway :) Waiting for @TuringTest & @fwizbang to join us again!

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1I told you in the other post that magnetic fields \(can\) do work on dipoles refer to figure section 8.4 of this page http://ocw.mit.edu/courses/physics/802scphysicsiielectricityandmagnetismfall2010/magneticfield/MIT8_02SC_notes16to18.pdf particularly towards the end but that is due to the configuration of the dipole, and will not act in a closed path

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1\[U=\vec\mu\cdot\vec B\]so the energy will seek to minimize by making \(\vec\mu\) and \(\vec B\) parallel, which means the magnetic field will be doing work

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2Thanks @TuringTest I got lost from a lot of things... Thanks though!

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2The link you gave me @TuringTest is very useful! Satisfies a lot of my questions!

fwizbang
 3 years ago
Best ResponseYou've already chosen the best response.1This is one of those questions where the answer depends on how far down the rabbit hole you want to chase it..... Where to begin? First, it's good to clearly separate the magnetic field B from the magnetic force F. The magnetic field doesn't have a vanishing curl, as has already been mentioned, but that's not relevant to energy conservation, which is what we're really discussing here. At the level of introductory physics courses, the magnetic force is F = q (v x B),which has no power, and so does no work. If the force does no work, then it doesn't "really" matter if its conservative or not at some level, as the potential energy would be a constant, which could be chosen to be zero. Note that this also applies to the force on a current carrying wire, because current is the flux of moving charge, so a velocity is implied, at least in an average sense. (in fact, one derives this force by assuming a collection of moving charges.) The situation gets more complicated when you consider things like dipole moments. If the dipole can be thought of as being produced by moving charges, then clearly the magnetic field does no work, because F=q (v x B) for those charges and the above argument still holds. But if you want the magnetic dipole to stay constant in magnitude(which is what is generally assumed), then you need to keep current flowing around the loop despite the fact that the magnetic force is trying to move the charges in a circle. This is accomplished by electric forces between the charges and the wire, and it is the electric forces that do the work. If you want to talk sensibly about pointlike dipoles, like electrons, then you have to go further down the rabbit hole, where you'll discover that the potential energy resides not "on" the particles/dipoles, but in the electric and magnetic fields themselves. Since charges, currents, and magnetic moments(even the pointlike ones) are all sources of magnetic fields as well as objects that can be acted on by the fields, their presence alters the total field, which changes the total field energy. The change in the field energy that's produced by the particles/dipoles is "the potential energy" that we associate with the particle. Usually, this isn't included in the discussion at the introductory level, because its too complicated. Of course, at this level, it's not reallly clear what is meant by work either, but that's another discussion.

fahminator
 3 years ago
Best ResponseYou've already chosen the best response.0I like this website It tastes good

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2The thing is most people would ask... Would magnetic field do work on a "charged particle".... I believe magnetic field can do work on a current carrying a wire that is in a loop and has a 90 dgree axiel as @TuringTest stated before magnetic "force" can do work. I think for a magnetic field to do ANY work there has to be another force that triggers something or there has to be a factor that makes everything differert... A simple idea of this is a motor. Without the precense of magnetic field as a KEY element no work can be done based on the magnetic force generated on the loop carrying current. Again the law is stated : F = I x L x B A force "magnetic" will be present and work will be done in a certain configuration. However, as a magnetic field being still without any kind of force/energy being applied on it,It can't do anything...(Maybe apply a force on a particle/charged particle but no work is done.). Its truly a mystery, all the natural forces are quite sutting and amazing and truly a time consuming thing we need to study on. Again the main reason I ask that magnetic force can do work or not? Is be cause I look at motors/generators... As @TuringTest stated out before work can be performed under a certain state and in a motor/generator's state I believe so... Work can be done. Since B(Magnetic field) is a main vector in the equation of magnetic force F = I x L x B. If B were to be = 0, No force will be applied and no work will be done... (Correct me if I'm wrong I just feel I hit the tipping point of something here!)

fwizbang
 3 years ago
Best ResponseYou've already chosen the best response.1Certainly, if there is no magnetic field, there is no magnetic force, but that's going to be true whether the magnetic force does work or not. ( If there is zero normal force, the force of kinetic friction vanishes, that doesn't imply that the work done on an eraser sliding across the table is done by the normal force.) The I(L x B) force that you mention comes from adding up all the Lorentz q(vxB) forces on the individual charges q that are moving within the wire. Since the Lorentz forces do no work, neither will their sum, I(Lx B). There are other (electric ) forces that act within the wire to keep the moving charges inside the wire. The reaction forces to these forces(Newton's 3rd Law) act on the rest of the wire in the same direction as the magnetic force acts on the moving charges. The reaction forces are what do the work in this situation. (Note that if you turn off the magnetic field, there is no force required to keep the charges in the wire, and hence no reaction force and no work, just as you describe.)

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2@fwizbang Exactly, Its true that the main KEY and reason for these forces to act on each on other is the electric force that causes current to flow throughout the wire.(Hence the magnetic force can't do work unless in a that certain condition in this cause the presence of the electric force). Without the electric force the Lorentz force can't act upon the electric charge... In a way its kinda systematic if you say so... Like a computer a processer is a crucial part of it but without the "motherboard" it can't process well and so on... Same thing in our cause. Without the electric forces no magnetic force can be applied on it thus no total "work" will be done. Their all dependable as you said "reaction force". Now granted that magnetic fields can't do work... But they exert a force on charged particles that are waiting for another key factor applied on it to do work. Many many system all relie on each other to work properly and its really interesting.

fwizbang
 3 years ago
Best ResponseYou've already chosen the best response.1I have to agree that it is truly wondrous the way all this stuff hangs together sometimes.

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2I wonder though... When I want to calculate the net forces in a motor for me to calculate the total work done, I should calculate both the electric force and the magnetic force? I mean using this formula : F = IL x B, will that be enough for me to find the net force? Or should I calculate both, electric force + magnetic force in that system? I want to know whats the MAIN equation/formula to calculating work based on both force magnetic,electric? Because I truly want to break EVERYTHING down and study each force reaction one by one to understand them. I thought Magnetic force is enough, but now I think electric forces is something else I should consider. :P

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2I'd like to study both force acting on each other. Mainly the magnetic forces acted on a wire carrying current that woud produce a torque and then rotational kinetic energy is presents(I'd like to study the force that causes it). I think F = IL x B is a well rounded formula to study right? Or is there another way I should like at the forces acted on each other? @TuringTest @fwizbang @experimentX & Again thanks for all you're efforts on this question so far!

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1This thread is interesting. Thanks for the notice @TuringTest The distinction between Electric and Magnetic fields and forces, which is quite clear in nonrelativistic electrodynamics, becomes blurred and somewhat artificial when relativity is taken into account. Qualitatively, that gives us a hint that a correct treatment must take into account both Electric and Magnetic effects and their interactions, which can be somewhat tricky. Since it's already been said, I'll just remark that the magnetic field cannot do work, i.e. change the kinetic energy of a moving charged particle. Bearing that in mind, since everything is made up of charged particles, it logically follows that magnetic fields just cannot do work. In rail guns, for instance, we run into murky situations in which the magnetic field exerts a force on a system and ostensibly changes its kinetic energy. I'd like to give such an example. dw:1342214383634:dw

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1In the configuration above, we have a circuit comprised of a resistor, a battery, and wires of zero resistance. One of the wires is laid across the others and may slide without friction. In the presence of the magnetic field B, we note that the sliding wire experiences a force to the right, \[F = BIL \] where L is its length. Therefore, the wire gains kinetic energy at the rate of \[F = BILv\] where v is the velocity of the wire. We may stop here and conclude that the magnetic force does indeed do work on this system, imparting to the wire kinetic energy. However, this would overlook a few subtle details.

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1Let us take note of the emf generated in the circuit as a result of the movement of the wire. Since \[EMF = \frac{d\phi_B}{dt} = B\frac{dA}{dt} = BLv \] We note that the current becomes \[I = \frac{VBLv}{R} \] Note that v is an increasing function of time > It would be straightforward to devise a differential equation and solve for it explicitly but unfortunately I have an engagement right now and must leave.... If nobody's felt the need to do it by tonight, I'll do it then. The eventual outcome is that the velocity asymptotically approaches V/(BL). The point of all this is that if the magnetic force was indeed doing work on the charged particles then new energy unrelated to the battery would be being introduced to the system. The magnetic force indeed diverts the paths of the electrons but does not impart any new kinetic energy to the system. Sorry for the disjoint reply but I lost track of time and am now running late.... cheers until tomorrow :)

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2I got lost in this part "The point of all this is that if the magnetic force was indeed doing work on the charged particles then new energy unrelated to the battery would be being introduced to the system. The magnetic force indeed diverts the paths of the electrons but does not impart any new kinetic energy to the system." Under the "circuit" configuration isn't the magnetic force doing work? And kinetic energy imparting the wire? Or no kinetic energy is present and magnetic force won't do work?(Confused here!) Hope you enjoy the engagement @Jemurray3 and thanks for you're effort I'll be waiting! :)

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2@jemurray3 in motor for example aren't magnet forces doing work?

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2http://hyperphysics.phyastr.gsu.edu/hbase/magnetic/mothow.html#c1 "An electric current in a magnetic field will experience a force" What force is that? Isn't it a magnetic force? = Work being done or already done?!

quarkine
 3 years ago
Best ResponseYou've already chosen the best response.0i quote ""An electric current in a magnetic field will experience a force" What force is that? " for the current wire ,it is simply the same force that any other charge moving in a B field experiences..since this lorentz force is perpendicular,no work is done .. now i have been thinking about conservative forces and was wondering if being conservative or not is really a matter of nature of force or a particular field..for instance, isn't it true that a wire with induced emf has non conservative E field and a normal resistorcircuit has a conservative E field? so how do we say whether Electric force is conservative or not when it can be both??

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2@quarkine :"for the current wire ,it is simply the same force that any other charge moving in a B field experiences..since this lorentz force is perpendicular,no work is done .." What causes work inside the motor... Isn't it the electric current + magnetic force? Without the existence of the electric current nothing would happen right? Ok, if there is no magnetic fields to induce a magnetic force upon the current NO work can be done( As I said without the other there is no net force and no work would be applied like the computer example... One is dependable on the other...). I feel there is something contradicting the other... Maybe we can all agree that magnetic field/force under certain condition like in a motor(Where current is present that produces its own opposing magnetic field) work could possibile be done buy the presence of magnetic field/force + electric current. A net force created by both that would create the total work :D Work can be simply done with one kind of force... In some complicated situations multiple forces can do Work. :) Hope,

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2Again! On a plain simple charged particle I doubt magnetic fields can do any work, they can exert a force true. Still no work has been done and no energy has be transfered. By having multiple force that would add up to a net force and then we would find out the net work done...(Correct me if I'm wrong). Again to all: Magnetic field in the cause of a simple charge particle no work would be done I do agree on that. However, in case of a motor as an example magnetic forces! Acted on the current carrying wire is generated by the presents of both current + magnetic field! @quarkine , @TuringTest , @jemurray3, @fwizbang, @experimentX :)

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1When considering situations in which the force on an object is derivable from the Lorentz force F = q(E + vxB), the magnetic force does no work. Period. There may be induced electric fields created BY the magnetic fields, and THOSE fields might do work, but the magnetic fields themselves are not the direct mechanism by which kinetic energy is imparted to such objects. Permanent magnetic dipoles, i.e. elementary particles, however, are a different story. Since these forces cannot be derived from the Lorentz force, we do not know a priori that they can't do work, and it in fact turns out that they CAN. So we must distinguish between these two "types" of magnetic force when determining whether work can be done or not.

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2Permanent magnetic dipoles... How can they do work? Lets look at that more....

fwizbang
 3 years ago
Best ResponseYou've already chosen the best response.1When you get down to the level of elementary partticles, it seems that the whole concept of work starts to break down. An electron, as far as anybody knows, has no spatial extent, but if it absorbs a photon when its in a magnetic field its magnetic moment can flip and its energy will change. Has work been done? Certainly not in the Fdcos theta sense of classical mechanics, as nothing has moved.

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2Then its safe to say magnetic force/fields can do work on only pure magnetic dipoles like a bar magnet? "Based on the formula, the magnetic force on a charge is qv⃗ ×B⃗ which is identically perpendicular to v⃗ and that's why it does no work. However, forces on magnetic dipoles and more general objects don't have the form v⃗ ×  they're not perpendicular to v⃗ , so they do work in general." (Do you all agree?) A pretty good conclusion I found in another site with the similar topic :D Check it out if you're interested: http://physics.stackexchange.com/questions/10565/doesamagneticfielddoworkonanintrinsicmagneticdipole

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1I find no particular disagreement with that statement, no. That about sums it up.

quarkine
 3 years ago
Best ResponseYou've already chosen the best response.0now that we agree that B fields can do work,the question remains to whether that are conservative or not..due to the symmetry between E and B fields i think that B field may be either conservative or not just like E fields depending on the situation. thinking on the line of above question i got struck at two points: 1.the phenomenon of drift of electrons involve acceleration.so does the current wire emit EM waves? 2.why does the current wire produce B field? what i find weird is that ,as far as i know, a moving charge doesn't create B field. so how can a continuous flow of charges create B field? Anyone??........

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1The electrons "drifting" through a wire in a normal circuit aren't accelerating, they're moving with a particular drift velocity that's proportional to the current. And a moving charge does indeed create a magnetic field.

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2Now, we know that magnets can't do work on electric current right? On a loop for wire force is exerted on it by magnets but work is not done? Although its rotating... And speeding up. There is something that is missing here... Would someone add to this puzzel? (Again related to the motor effect.

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2Just makes no sense... Without the magnetic fields no force can be felt and work can't be done. True the current is present but still... Magnetic force on that wire or loop is confusing me right now!

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2The only explanation that can come to my mind is... The electric charge creates a magnetic field and the permanent magnet has already a magnetic filed and thus opposing forces will be apply forces and work will be done by both the electric charged and the magnetic fileld. As stated by Lorentz law on a current carrying wire: F = IL x B , B = 0 , F = 0 , total work done would equal = zero, simple logic about this matter if you make a small loop and connect it to a battery nothing would happen other then flow of current and a magnetic field is generated by that flow, put a permanent magnet into that state and voila! Motion!

quarkine
 3 years ago
Best ResponseYou've already chosen the best response.0@Hope99 : hmm..till now i've been thinking that since the force on current is due to same lorentz force which involves cross product, it mustn't do any work.but in motor's it is doing work but remember the fact that a current loop is itself an equivalent magnetic dipole so as said above  mag force CAN do work on a magnetic dipole but not on electric charge.. @Jemurray3 : 1. regarding the phenomenon of drift, dont the electron get accelerated for a relaxation time and on average a drift velocity is acquired?? 2. Sorry to ask this again (just confirming) but will an electron moving with a velocity create a B field?

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1Remember that if you have a current loop and then you switch on a magnetic field that torques the loop, the current decreases via the induced electric field, robbing the electrons of kinetic energy as the whole loop starts to rotate. Also, @quarkine I'm not sure what your first statement is saying, but the answer to your second statement is yes.

quarkine
 3 years ago
Best ResponseYou've already chosen the best response.0@Jemurray3 : thanks your ans solved some confusions... It was the flow of current which caused the repulsion and flux change so the induced emf will as a consequence oppose this change... but still the fact remains that the rotation was due to magnetic force and by the formula dw:1342363976778:dw magnetic force did some work.. what i said in my first statement was that in some physics books i read about current, there was a derivation of drift velocity in terms of electron being accelerated from zero to a certain velocity in a time t called the relaxation time..this time was defined as the average time between two consecutive collisions between an electron and metal lattice. this link shows the derivation.. http://physics.stackexchange.com/questions/28630/whichderivationofdriftvelocityiscorrect if this were true then the current wire must emit EM radiations due to this acceleration of electrons.

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1Oh, I see what you mean. Yes that's true, but the electric field present in a wire is small. The power radiated away would be fairly negligible. As far as your first point goes, I would argue that the magnetic force is exerted on the electrons. The paths of the electrons are in general chaotic but the magnetic field would divert the velocities of the electrons, giving them a net component of velocity in the direction of the subsequent rotation of the loop (but not changing their total kinetic energy). Because the electrons are confined to remain inside the loop via electrical forces, it is the reaction of those electrical response forces that imparts rotational kinetic energy to the loop.

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2@quarkine, So a magnetic field generated by a loop is like a permanent bar magnet and can do work? are you sure of that? @Jemurray3, Not sure what you ment by this: "Remember that if you have a current loop and then you switch on a magnetic field that torques the loop, the current decreases via the induced electric field, robbing the electrons of kinetic energy as the whole loop starts to rotate." Why is the current decreasing? Why is ther any loss in general? Could that be a negative effect? + Do you agree on this " magnetic field generated by a loop is like a permanent bar magnet, considered as a permanent dipole?".

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1There's an induced emf in the loop, Faraday's law... and no they're not the same, a dipole can be regarded as a current loop if you increase the current to infinity and decrease the area to zero such that the product of the two, the magnetic moment, remains constant.

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2@Jemurray3 So they can do work on each other as @quarkine said?(Got confused at you're point sorry)

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2Could you explain the "Motor effect" in you're point of view @Jemurray3 and what is exactly doing work? Since I haven't gotten that answer clearly from you yet. I think most of us here agree that magnetic forces or fields can do work in a motor opposing/attracting another magnetic field... So what do you think about this? Because I'd like to sum all you're answers up.

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1To be clear, electromagnetic fields don't interact with one another. That's why you can superpose them. And the "motor effect" seems to be the same as what I said above, "The paths of the electrons are in general chaotic but the magnetic field would divert the velocities of the electrons, giving them a net component of velocity in the direction of the subsequent rotation of the loop (but not changing their total kinetic energy). Because the electrons are confined to remain inside the loop via electrical forces, it is the reaction of those electrical response forces that imparts rotational kinetic energy to the loop. "

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2@Jemurray3, that makes sense more. Now the permanent magnets would only divert the velocities of the electrons? Nothing else? They would not increase the velocity or anything? The repulsion and attraction between the loop and magnets is caused by the simple diversion of velocities by the magnetic field that then would cause the rotation of the loop because the electrons are confined within the wire again by the magnetic filed?(Things started to make more sense now.) If the magnetic field would suddenly was not a constant and its lets say, controllable decreasing and increasing what would be the effect on the electrons? Again I believe more force will be applied.

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2@Jemurray3, I keep reading what you said over and over but still can't get convinced! Mainly because of this law: F = IL x B, I look at this site here and keep making the MOST sense ever: http://hyperphysics.phyastr.gsu.edu/hbase/magnetic/motdc.html When we apply a magnetic field doesn't it do anything but divert the electrons?

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1Try imagining a bunch of electrons traveling in the path traced out by the wire, but not confined to it. What would happen to them? It's not the same picture. Now constrain them to stay in the wire via electrical forces.

quarkine
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1342451522946:dw here the charge ,say, is positive. the charge wants to fly out of wire due to force..the reaction force is actually downward and confines it to the wire.but still the work here is being done by mag field as the direction of motion is along http://hyperphysics.phyastr.gsu.edu/hbase/magnetic/magmom.html let me argue through another a simplified situation.. let a charged particle be confined inside a hollow rod(which is the part of a loop) and given a velocity v inside dw:1342454445470:dw the charge is pushing the wall of rod and making it rotate

quarkine
 3 years ago
Best ResponseYou've already chosen the best response.0sorry for the wrong direction of loop rotation above :( @Jemurray3 : i think what u argued is something like the frictionelectrical or pushelectrical relationship.. when we 'push' a block our force is actually the electrostatic repulsion between surface of our hand and of the block.still we say that our force/push did the work..also it is same as the friction force which is also electrostatic repulsion between two surfaces.but we say the friction is doing work even when the force is actually electrostatic repulsion between the surfaces undergoing relative motion.. are not what you said abstracted as being the 'internal forces?? @Hope99 : i think the current loop may be considered just like a magnet as it has its mag moment and putting it is like putting a magnet in between another mag field. it should work...though i m not sure... i m not very much acquainted with electromagnetism (only till 8.02 lectures).

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1I think that you more or less understand the argument I made.... but the point of the matter is if a particle experiences a force given by \[ \vec{F} = q(\vec{v}\times \vec{B}) \] then it CANNOT do work. It just can't, mathematically. I'm not saying that it won't induce other fields which CAN do work, and I'm not arguing that it's more useful to go this far down the rabbit hole when thinking about it. All I'm saying is that if the question is "can the above force directly do work" (which extends to all of the equations derived from that expression) then the answer is no.

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2@Jemuuray3 I agree with you! But please forget this law F⃗ =q(v⃗ ×B⃗ )... Because thats based on a charged particle(I agree no work is done there) But look at this! F = IL x B"Force on a current carry wire" (In a current loop! its mathematically proven that magnetic fields do work!) Again! That laws is based on "CURRENT carrying wire" that itself is a form of electrical energy, it creates a magnetic field and is temporarily like a magnet! We all agree on that point. Then the permanent magnetic fields apply forces to repel/attract. Again if B = 0, F=0. *KEEP THIS F⃗ =q(v⃗ ×B⃗ ) ASIDE FOR A BIT CONCENTRATE ON THIS F= IL x B*

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1That formula is derived directly from F = q(v x B). I don't understand how this discussion is still going on, frankly.

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1Listen, I'm not going to repeat the same thing a thousand times. I'll give my argument one last shot, and if you guys don't like it then that's certainly your business. The electrons in the loop are initially flowing perpendicularly to the magnetic field. The magnetic field diverts the electrons, and changes the DIRECTION but not the MAGNITUDE of their velocity. The electrons subsequently encounter the edge of the wire, where electrical forces cause them to slow down and keep them from leaving the metal. As the electrons slow down, they impart momentum and energy to the wire, creating a torque. Work is being done, and the amount of work is numerically equal to F = L(I X B)... BUT THE MAGNETIC FIELD IS NOT DOING THE WORK. The magnetic field changes the DIRECTION of the electrons' movement in such a way that NO work is done. The ELECTRIC forces, which CAN do work, subsequently redirect the electrons so that they resume their initial path. It is these internal electrical forces which are directly responsible for the torque placed on the rigid wire, and the subsequent work done on the current loop. What about that explanation is unclear or contradicts your knowledge of physics?

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1I'm sorry, in the middle of that: "Work is being done, and the amount of **force** is numerically equal to...."

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2@Jemurray3, Explanation is clear enough. What source do you recommend studying about this matter more?(Online please.) One thing though: The magnetic fields ONLY change the direction nothing else? If increasing the value of B what would be the result?

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2Because if we increase the values of any vector within this equation F= IL x B The total "F" would be more. Increase the I = higher F Increase the L = higher F Increase the B = higher F Based of a mathematical conclusion :P

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1Yeah..... the force will increase if you increase any of those things, which would change the direction more...

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1And I can't recommend a source because there aren't any, to my knowledge. The fact that F = q(v x B) can do no work is immediately mathematically obvious. There cannot be a debate on that fact. The fact that the magnetic field appears to be the cause of some work being done, like the torque on a current loop or the force on a current carrying wire, is a paradox that is resolved when you recognize the electric fields that are actually performing the work directly.

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1The amount of force that accelerates the wire is numerically equal to F = L(I x B). The fact that it is the electric forces that confine the electrons to the wire that actually accelerates the wire is of no real computational use and often overcomplicates the problem. As such, we use the equation F = L(I x B) and other expressions of the magnetic force without worrying about whether or not the magnetic field itself is doing work or if some other forces are being conscripted to do the work more directly.

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1Whenever the magnetic field appears ostensibly to be doing work, we can always find a hidden electrical force that is the real direct cause of the work being done. It's of course possible that the electric field was induced by the magnetic field, but if work is being done (i.e. kinetic energy being transferred) it is ALWAYS the electrical forces.

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2The more of change in direction the more torque generated hence more force is being produced on the charges. Makes good sense. Magnetic fields in this cause can't do work. But as said before a major influence for this "chain reaction" to occur. Thanks all of you who have participated in this long long question.

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1The only exception to this is the quantum mechanical magnetic dipole moments of elementary particles. They have no internal structure, and as such we cannot pretend that electrons, for instance, are like little current loops. They aren't. We can treat them like current loops in the limit as the current goes to infinity and the area goes to zero in such a way that their product remains constant, but that's not really what they are. In these cases, the magnetic field really CAN do work on these particles. But for classical or macroscopic systems, the magnetic field can only act as a puppetmaster pulling the strings while the electric field does all the work. It's been an interesting discussion, thanks for raising the question.

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2@Jemurray3 " the magnetic field can only act as a puppetmaster pulling the strings while the electric field does all the work." lol, well said there... The magnetic fields are interesting and I will pursue to study them because I believe there is a puzzel that I'm looking forward in solving. I got interested because of a motor... Even more with Lorentz force F = IL x B > A law very useful and key for me to study! Expect more questions about this matter everyone! I'm thirsty for some knowledge! & I think its time to close this question :) Thanks again!

quarkine
 3 years ago
Best ResponseYou've already chosen the best response.0@Jemurray3 thanks for your help from me too, especially about the B field associated with moving charge stuff :) [now that i think of it,it seems obvious that the charge should have a B field because by newton's third law it must apply 'reaction' force back on the source of external B field] @Hope99 : electromagnetism IS really interesting for sure.nice question...

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0well ... http://chat.stackexchange.com/transcript/message/5394715#5394715 i think i agree with this

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0http://physics.stackexchange.com/questions/16326/workdonebythemagneticforce also there's quite relevant material .. a quote from wikipedia article.

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2Wanted to add something to this topic. For those who are still interested check this link out: http://www.physicsforums.com/showthread.php?p=4007012

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1@Hope99 Jemurray has given what I think is the most clear explanation, and I think you should study E&M more deeply (along with some relativity and QM) before you can appreciate everything he has said.

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2@TuringTest I do appreciate everyone's effort on this question, I just wanted to add a useful topic to look at :>

Hope99
 3 years ago
Best ResponseYou've already chosen the best response.2@Jemurray3 I'd like to say after a long time researching I've discovered that magnetic fields do generate electrical fields that eventually do the work. So B = E , E = B and so on. I finally understood what you ment :) Thank you!

Hope99
 2 years ago
Best ResponseYou've already chosen the best response.2@TuringTest @Jemurray3 @experimentX @quarkine @fwizbang When a magnet does work. Is it because of the electric field or magnetic field? Based on Faradays I'd say the magnetic field would generate an electric field that would eventually do work. But generally when two magnet attract/repel they certainly do work on each other.

Hope99
 2 years ago
Best ResponseYou've already chosen the best response.2+ Hope you all are doing well, its been a while! :)
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