How does a permanent magnet lift iron filings

[sridhar_n]

Evryone knows that the work done by a magnetic field is zero. Then how does a permanent magnet lift iron filings.

 

[HallsofIvy]

We do not all know that: many of know that any force field does no work on an object moving perpendicular to its lines of force. Magnetic fields certainly can do work.

(And please don't double post.)

 

[eJavier]

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

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

 

[krab]

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

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]

To bring the magnets together requires changing the magnetic field, and a changing B induces an E. Against this E is the work done. At the beginning and at the end there is no E, but in between there is definitely one.

 

[krab]

You need not change any magnetic field. If big magnets are confusing you, think of 2 elementary particles with no charge, but with magnetic moment. There are magnetic forces between them. These forces are not perpendicular to velocities and so they do work.

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.

 

[eJavier]

So you're talking about fiticious magnetic monopoles?

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.

 

[krab]

So you're talking about fiticious magnetic monopoles?

didn't read the other thread, did you?

Because a real magnet is made up of tiny magnetic dipoles which are nothing but tiny currents.

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?

 

[swansont]

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?

 

[pmb_phy]

Evryone knows that the work done by a magnetic field is zero. Then how does a permanent magnet lift iron filings.

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 to
Electrodynamics - 1st Ed.," on page 179

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.

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.

It only seems like the magnetic field does work. As a worked example of how it appears to do work and the correct explanation 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

 

[krab]

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

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.

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.

 

[swansont]

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.

 

[swansont]

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?

 

[pmb_phy]

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

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.

 

[ArmoSkater87]

Im not an expert on this subject, but a magnetic field can't do work on a charged particle, it can only changed the particle's direction, which is how a particles in particle accelerators are made to go in a loop. But the thing is that like it was said in the biginning, a permanent magnet can pick up iron fillings or another magnet Picking it up takes force, and the force is applied through a distance...soooo, it seems like work is done.

 

[reilly]

Magnetic Moments; References

Evryone knows that the work done by a magnetic field is zero. Then how does a permanent magnet lift iron filings.

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

 

[eJavier]

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.

Javier

 

[krab]

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.

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

$$\vec{N}=\vec{m}\times\vec{B}$$.

There is a potential energy U:

$$U=-\vec{m}\cdot\vec{B}$$.

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.

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.

Did you mean magnetic field?

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

$$F={p_1p_2\over r^2}$$

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

$$H=p_1/r^2$$

of p_1. Equally well, describe it by a potential

$$\Phi_M=p_1/r$$.

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:

$$\Phi_M=p_1/r_1-p_1/r_2=p_1\delta\cos\theta/r^2$$,

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

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

This is the same as the formula given by Jackson.

If you have a cylindrical permanent magnet, with magnetization $\vec{M}$, then it acts to a very good approximation as if it has magnetic monopoles on its North surface with a charge density $\sigma_M=|\vec{M}|$, 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 each other 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...

 

[eJavier]

Thank you for your input krab, I'm still not convinced, but I'll go through your argument with more detail later nonetheless.

 

[rayjohn01]

to anybody
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.

 

[eJavier]

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

$$\vec{N}=\vec{m}\times\vec{B}$$.

There is a potential energy U:

$$U=-\vec{m}\cdot\vec{B}$$.

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 $$\vec{N}=\vec{m}\times\vec{B}$$ 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)

What causes me trouble is that I'm utterly convinced that the magnetic field can't do work on the current density J that provides the magnetic moment m.

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?

 

[esolutions99]

well at the risk of sounding nieve(sp?) yall seem to be looking at this from a skewerd perspective..hehe

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... I am under the impression that work equations don't work well here...is there work done when electricity flows through a circuit? electrons flow because of their diff in potential its part of their physics..is attracting iron filings work ?...as much as electron flow (or hole flow) its just what they do dudes

 

[rayjohn01]

Maybe the misconception ( original) was that a magnetic field does no work on a charged particle applies to a STATIC particle , not one in general.

 

[eJavier]

Magnetic field don't work on any charged particle. This follows from the Lorentz force Law. If you want a proof just ask :D.

 

[rayjohn01]

hence

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

 

[eJavier]

to krab:

I finally found out what's the problem. The formulas you wrote for the energy and work are mechanical work. You still need to add the energy of the electric currents. If you do that then you'll get zero work. This is in one of Feynman's lectures, chapter 13 I think.
I'm kind of busy now but as soon as I get some spare time I'll post here the most important of the aforementioned lecture.

to rayjohn:
A dynamo does not work in the sense that it can't speed up an electron. what it does is this: it takes the work done by the waterfall or wathever and converts it into electric work. So really the one doing the work is the one moving the dynamo.

 

[rayjohn01]

mag work.

To Ejavier
I do not understand your answer, dynamos are quite capable of moving electrons since they generate current , it's clear where the original energy came from, and it's clear that in an ideal set up the field is not dissipating energy in the form of heat but it is transmitting energy and doing so by accellerating electrons.
So I'm not really sure what you mean by not doing work, do you mean producing work as an energy source -- I would agree with that.
But if I compared to a lever lifting a weight ( a transmitter of work) would I not say that the lever is doing work ( but not producing it).
Maybe it's just semantics or there is a different definition being used, but there is no question that magnetic fields accellerate electrons without that effect we'd all fry from the sun's radiation.
Also I believe that a magnetic field can be a source of energy , if you take the case of two bar magnets attracting each other they accellerate collide and dissipate heat and sound. The energy comes from the partial collapse or reorientation of the fields.
In that case the system can be recharged by pulling them apart , but to me that's rather like recharging a battery and were quite happy saying that a battery can do work .

 

[eJavier]

I said that a magnetic field can't SPEED UP electrons and I stand by that. It can deflect them of course.

The fact is that the force that a magnetic field exerts on an electron is always perpendicular to the trajectory (same as a satellite in a circular trajectory in a central force field) so it doesn't work.

In the example you wrote about the bar magnets, there is of course a mechanical work done, but you have to take into account the work done to sustain the currents. It's all explained in the second volume of Feynman's lectures, chapter 13 I think.

 

[hi_rudra]

the field certainly doesnot do anywork.all that happnes is it lifts the iron(or even U may thought of attraction between two magnet),perpendicular to the line of force so that F=vxB is still vaid.In some cases induction also take place and by 3rd law of Maxwell,it is electric field which does the work

 

[rayjohn01]

So then a 'relay' operates via an 'electric' field --- right -- and no work is done in closing it's contacts -- according to this we can do without inductors entirely, and you know you can ( in theory by using an active device -- see switched capacitor filters the phase shift required is 180 dgrees ).
An inductor is the perfect model -- it can have a permanent field and a dynamic one.
It's equivalent circuit is that of pure inductance which is an energy storage device.
It is the stored energy which can do work and if it does the energy must be replenished.
EJaviers example of a rotating vector is like a hammer hitting nothing , but we use the hammer to drive the nail , and we don't say 'my fist did it ' even tho' that's where the energy came from.
Dynamic magnetic fields can speed up electrons which is why the suns storms exist especially due to field collapse.
However if you wish to remove magnetics by talking of charge spin fine .
So according to you electromagnetic energy ( radiowaves ) is only carried by the electric field -- tell Maxwell -- the energy is contained equally in both and the waves can do work if not -- then what the heck have I been listening to all this time on my radio ?

 

[eJavier]

Magnetic fields ain't like a hammer hitting a nail since the hammer does work.

Magnetic fields are like the normal force on a ramp. It doesn't do any work since it's perpendicular to the displacement.

 

[rayjohn01]

time is important.

to EJavier.
Then please explain radiation where both fields are normal to displacement but obviously do work (equally) per the photon. They do work BUT dissappear because the stored energy is used . A purely static and unchanging field does no work , a changing field gives or accepts energy,- like a spring, an accellerating field gives rise to radiation , all of which you could equally say about the electrostatic field.
In a tuned circuit comprising a capacitor and an inductor which is oscillating energy is continually exchanged from one to the other in both potential and kinetic forms -- the introduction of a resistance causes an exponential decrease in oscillation as energy is used -- both fields decrease as work is done producing heat. The cap' and ind' voltages are both normal to the current( in the sinusoidal sense) so they produce no internal heat but they have changing energy, represented by the resistance heat.
But I agree that that is a non-conservative system in that heat is a random motion.
However the photon does not have to produce heat it could release (say) electrons from a metal surface exchanging energy to kinetic of the electron.
The normal force on a ramp does work -- it compresses the object which is stored energy - which can be released if the object falls off the ramp.