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- Thread starter sridhar_n
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HallsofIvy

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(And please don't double post.)

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Magnetic forces do not work since

F= q (v x B) is a vector perpendicular to the infinitesimal displacement dr, so W= F dr=0

Back to the original question, magnetic forces change the direction of another force. When lifting something, the one doing the work is most likely the generator or motor

Greetings

Javier

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krab

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No. That's the force of a magnetic field on a charged particle. There are other forces as well. Hold two magnets apart; now they have potential energy. Let them go; now the field does work on the magnets and the result is that they accelerate towards one another.eJavier said:I'm afraid I'll have to disagree

Magnetic forces do not work since

F= q (v x B) is a vector perpendicular to the infinitesimal displacement dr, so W= F dr=0

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krab

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I will repeat. A magnetic field does no work on an electric charge. But a magnetic field does work on a magnetic charge. It is true that there are no free magnetic charges, but that is a detail.

Read this thread: https://www.physicsforums.com/showthread.php?t=32547

You did not contribute to that thread, so maybe you missed it. I will not repeat again.

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Because a real magnet is made up of tiny magnetic dipoles which are nothing but tiny currents, which are composed by charges with a certain velocity. If we're talking about forces on magnetic dipoles then said force is F=grad(m B) but this formula comes from the Lorentz force law.

As far as I know, magnetic field exert forces only on moving charges.

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krab

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didn't read the other thread, did you?eJavier said:So you're talking about fiticious magnetic monopoles?

Not so. So you have an isolated neutron. It has a dipole moment. So what are the circulating charges that result in the tiny current loop? Other elementary particles?eJavier said:Because a real magnet is made up of tiny magnetic dipoles which are nothing but tiny currents.

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krab said:Not so. So you have an isolated neutron. It has a dipole moment. So what are the circulating charges that result in the tiny current loop? Other elementary particles?

The neutron isn't an elementary particle; it's made up of quarks. Wouldn't they be the source of the magnetic dipole moment?

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A magnetic field cannot do work on a charged particle. This was stated quite precisley in Griffith even stated this explicitly in his text "Introduction tosridhar_n said:

Electrodynamics - 1st Ed.," on page 179

The magnetic field cannot do work on a current element either, and for the same reason. But for a finite current element such as a current loop I believe the answer to that is yes but I'm not 100% sure.Magnetic forces may alter the 'direction' in which a particle moves, but they cannot speed it up or slow it down. The fact that magnetic forces do no work is an elementary and direct consequence of the Lorentz force law, but there are many situations in which it appears so manifestly false that one's confidence is bound to waver. When a magnetic crane lifts the carcass of a junked car, for instance, 'something' is obviously doing the work, and it seems perverse to deny that the magnetic force is responsible. Well, perverse or not, deny it we must, and it can be a very subtle matter to figure out what agency 'does' deserve the credit in such circumstances.

It only seems like the magnetic field does work. As a worked example of how it appears to do work and the correct explaination please see the AJP article I posted at

www.geocities.com/physics_world/mag.htm

re - "When lifting something, the one doing the work is most likely the generator or motor"

To be precise its the conductors doing work on the charges. A magnet can be modeled as a collection of dioples which in turn can be modeled as a collection of conducting loops of wire. The magnetic field facilitates the work. I.e. Work done by B-field = 0, Work done by wire = Function of B.

re - "fiticious magnetic monopoles" - Undiscovered is a more accurate term. Its not quite correct to call them fictitious if you don't know if they exist or not. That'd be like Newton saying that particles which make up atoms are fictitious because he's never observed them.

Pete

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krab

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A very good question! Before the quark model, it was known that a free neutron decayed to an electron and a proton. It was tried to model neutrons as rotating combinations of these particles, with rotation rate set at whatever gives the correct magnetic dipole moment. But these models went nowhere, and in particular did not help in deriving the currently-accepted quark model. In fact, even earlier with the advent of QM, it was learned that the only bound electron-proton systems are: hydrogen atoms.swansont said:The neutron isn't an elementary particle; it's made up of quarks. Wouldn't they be the source of the magnetic dipole moment?

Another example is the electron by itself. It has never exhibited any behaviour suggesting it is a composite particle. We can investigate a model where its magnetic moment arises from an extended distribution of charge, spinning at the appropriate rate. But we find that we end up with no coherent picture.

Further questions relating to elementary particles should be posted on the Nuclei and Particles forum.

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pmb_phy said:re - "fiticious magnetic monopoles" - Undiscovered is a more accurate term. Its not quite correct to call them fictitious if you don't know if they exist or not. That'd be like Newton saying that particles which make up atoms are fictitious because he's never observed them.

Except that "undiscovered" means you expect to find them, rather than having them be something excluded by Maxwell's equations.

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krab said:A very good question! Before the quark model, it was known that a free neutron decayed to an electron and a proton. It was tried to model neutrons as rotating combinations of these particles, with rotation rate set at whatever gives the correct magnetic dipole moment. But these models went nowhere, and in particular did not help in deriving the currently-accepted quark model. In fact, even earlier with the advent of QM, it was learned that the only bound electron-proton systems are: hydrogen atoms.

But do the motions of the quarks account for the neutron magnetic dipole moment, and if so, isn't that an answer to you question about the presence of circulating charges?

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Maxwell's equations does not exclude monopoles. People just tend to set the monopole density to zero since applications to date don't use them. If they are discovered then the monopole density will be non-zero. But keep that in mind. That they are zero is an assumption and to effect that assumption in Maxwell's equations, the divergence of the magnetic field is set to zero. But it need not. Assuming monopoles exist then you set that divergence to the monopole density in the region that you're interested in.swansont said:Except that "undiscovered" means you expect to find them, rather than having them be something excluded by Maxwell's equations.

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reilly

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sridhar_n said:

The iron filings have a magnetic dipole moment -- among other things because the motions of the charges are highly constrained -- primarily due to the magnetic moments of the atoms. The energy term that results is

-mB where m is the magnetic moment, B is the magnetic field.Magnetic fields also generate torques, which is a key idea behind the galvanometer.

Resnick and Halliday give a good basic presentation of forces, and energetics in a magnetic field. More advanced texts, Panofsky and Phillips, Symthe, Landau and Lihschitz (Electrodynmics of Continuous Media) will tell you more than you ever cared to know about the subject.

Regards, Reilly Atkinson

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krab said:didn't read the other thread, did you?

Not so. So you have an isolated neutron. It has a dipole moment. So what are the circulating charges that result in the tiny current loop? Other elementary particles?

I don't want to get into baryons and the internal structure of neutrons because that's off topic.

All I'm saying is this: the force on a magnetc dipole is derived from the Loretz force law, which means that the magnetic force does not work.

The point I'm trying to make is that Lorentz force law implies no work done by the magnetic force. Reading your posts I get the impression that you're saying that it's only valid for electric charges (ie. Lorentz force law), so I would like you to show me a formula for a force between an electric field and something other than an electric current. I'm honestly asking you, I'm not being a smart-ass.

Best regards, Javier

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krab

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All I'm saying is that's just wrong. If you have say a uniform field [itex]\vec{B}[/itex] and a magnet with dipole moment [itex]\vec{m}[/itex], you have to do work to pull it out of alignment with B. The torque to do this iseJavier said:All I'm saying is this: the force on a magnetc dipole is derived from the Loretz force law, which means that the magnetic force does not work.

[tex]\vec{N}=\vec{m}\times\vec{B}[/tex].

There is a potential energy U:

[tex]U=-\vec{m}\cdot\vec{B}[/tex].

Rotating the dipole out of alignment increases the potential energy. Let it go and it will rotate on its own as the potential energy decreases and the rotational kinetic energy increases. This is known as doing work. The magnetic field is doing work on the dipole, just as gravity does work on a mass that is falling. These formulas are directly from Jackson. Lorentz force is not relevant here. You are mis-applying something else, namely that a magnetic field does no work on a charged particle.

Did you mean magnetic field?The point I'm trying to make is that Lorentz force law implies no work done by the magnetic force. Reading your posts I get the impression that you're saying that it's only valid for electric charges (ie. Lorentz force law), so I would like you to show me a formula for a force between an electric field and something other than an electric current.

Anyway, the force law between two magnetic poles p_1 and p_2 in free space is this

[tex]F={p_1p_2\over r^2}[/tex]

in the direction of the vector r from p_1 to p_2. This can be described as p_2 in the magnetic field

[tex]H=p_1/r^2[/tex]

of p_1. Equally well, describe it by a potential

[tex]\Phi_M=p_1/r[/tex].

So you complain that monopoles don't exist. Fine. They only come in pairs. The potential from an isolated dipole is found be adding the two together:

[tex]\Phi_M=p_1/r_1-p_1/r_2=p_1\delta\cos\theta/r^2[/tex],

where [itex]\delta[/itex] is the separation of the two opposite poles [itex]\vec{\delta}=\vec{r_2}-\vec{r_1}[/itex], and [itex]\theta[/itex] is the angle it makes with the observation point. We call [itex]p_1\delta[/itex] the dipole moment m. So

[tex]\Phi_M=m\cos\theta/r^2=\vec{m}\cdot\vec{r}/r^3[/tex].

This is the same as the formula given by Jackson.

If you have a cylindrical permanent magnet, with magnetization [itex]\vec{M}[/itex], then it acts to a very good approximation as if it has magnetic monopoles on its North surface with a charge density [itex]\sigma_M=|\vec{M}|[/itex], with a surface charge density of exactly opposite sign at the South pole (this is directly from Jackson...). Get two such magnets together and they attract or repel eachother with forces given by applying the formula for F that I gave above to these surfaces and integrating. These forces have potentials and everything, just as in gravity and electrostatics.

Here's a quote from my first post "Hold two magnets apart; now they have potential energy. Let them go; now the field does work on the magnets and the result is that they accelerate towards one another." Sheesh. I'm repeating myself...

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So when a magnet picks up iron filings off a table -- we know work is being done -- but by what?

Note that the end field is different from the start field as more filings will not be picked up. Any explanations Ray.

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krab said:All I'm saying is that's just wrong. If you have say a uniform field [itex]\vec{B}[/itex] and a magnet with dipole moment [itex]\vec{m}[/itex], you have to do work to pull it out of alignment with B. The torque to do this is

[tex]\vec{N}=\vec{m}\times\vec{B}[/tex].

There is a potential energy U:

[tex]U=-\vec{m}\cdot\vec{B}[/tex].

Rotating the dipole out of alignment increases the potential energy. Let it go and it will rotate on its own as the potential energy decreases and the rotational kinetic energy increases. This is known as doing work. The magnetic field is doing work on the dipole, just as gravity does work on a mass that is falling. These formulas are directly from Jackson. Lorentz force is not relevant here. You are mis-applying something else, namely that a magnetic field does no work on a charged particle.

Hi

I have no problem with the above, except that as far as I know [tex]\vec{N}=\vec{m}\times\vec{B}[/tex] comes from the Biot-Savart Law as shown by Jackson's book on page 150.

Furthermore, a magnetic moment is defined for a current distribution J (Jackson p.146)

I think that your point is that magnetic poles aren't formed exclusively by charged particles and/or currents. And in the case when neither charges nor currents are involved, then work is done. Am i getting somewhere?

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magnetic fields act more like kinetic(perm magnetics) energy patterns ...there is an attraction between particles...much like a difference in potential(electrical) ... the electrons flow because of the difference of potential.. similar to mag. fields ..depending on the magintude of the difference charge the objects are there fore attracted to each other

as far as work goes ..its inapproiate to use fluid equations for solid matter... im under the impression that work equations dont work well here....is there work done when electricity flows through a circuit? electrons flow because of thier diff in potential its part of thier physics..is attracting iron filings work ?....as much as electron flow (or hole flow) its just what they do dudes

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To eJ

Therefore electric motors cannot work , or dynamos ???????????

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