Question

Magnetic force... Conservative or Non-conservative?

Answer 1

magnetic field is a non conservative field
http://upload.wikimedia.org/wikipedia/en/math/4/8/b/48bf197f3f2cd48dd50295af72d9c989.png

Answer 2

ok, 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/t-225524.html

Answer 3

now I'm very confused because here's a very different conclusion
http://www.physicsforums.com/archive/index.php/t-172893.html

Answer 4

I 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 E-field is not changing, and there is no current flowing (J=0 like in a bar magnet) doesn't this become\[\nabla\times\vec B=0\]???

Answer 5

@Jemurray3 @JamesJ

Answer 6

the electrons are revolving around the nucleus, if these loops align in same direction then they give net magnetic fields. Not sure though.

Answer 7

The 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.

Answer 8

@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 non-conservative vector field?

Answer 9

oh but you are saying that\[\nabla\times\vec F_B=0\]perhaps?
that could make sense...

Answer 10

Since 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.

Answer 11

ok... that makes sense
still thinking more deeply though
thanks

Answer 12

MAGNETIC FIELDS! + Forces truly raises a lot of questions... Still mysterious it is.

Answer 13

So 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...

Answer 14

What 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

Answer 15

the force is always perpendicular to velocity => work done = 0

Answer 16

Then why do people assume it might be non-conservative?

Answer 17

can i explain this in detail tomorrow? i have got lot of qualms myself.

Answer 18

So magnetic force WILL never be capable of doing work huh?
Sure, I'll research this more and more.

Answer 19

I'm not sure about magnetic dipole. for for lorentz force .
F is always perpendicular to the velcity and magnetic field.

Answer 20

@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...

Answer 21

looks like i came to contradiction. can we discuss this later?

Answer 22

Sure thing mate. I'll be waiting anyway :)

Waiting for @TuringTest & @fwizbang to join us again!

Answer 23

I 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/8-02sc-physics-ii-electricity-and-magnetism-fall-2010/magnetic-field/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

Answer 24

\[U=-\vec\mu\cdot\vec B\]so the energy will seek to minimize by making \(\vec\mu\) and \(\vec B\) parallel, which means the magnetic field will be doing work

Answer 25

Thanks @TuringTest I got lost from a lot of things... Thanks though!

Answer 26

The link you gave me @TuringTest is very useful! Satisfies a lot of my questions!

Answer 27

This 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.

Answer 28

I like this website
It tastes good

Answer 29

The 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!)

Answer 30

@Vaidehi09 Thanks! :)

Answer 31

Certainly, 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.)

Answer 32

@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.

Answer 33

I have to agree that it is truly wondrous the way all this stuff hangs together sometimes.

Answer 34

I 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

Answer 35

I'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!

Answer 36

Any Ideas?

Answer 37

This thread is interesting. Thanks for the notice @TuringTest

The distinction between Electric and Magnetic fields and forces, which is quite clear in non-relativistic 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.