How do magnetic fields do no work?

  • Thread starter imaloonru
  • Start date
  • #1
10
0
I remember learning in Electrodynamics last year that magnetic fields don't actually do any work. I don't seem to remember any of the explanation behind that. Can anyone explain why magnetic fields do no work (ie. How they differ from electric and gravitational fields). Thanks!
 

Answers and Replies

  • #2
Born2bwire
Science Advisor
Gold Member
1,779
22
Via the Lorentz force, the force created by a magnetic field acting on a moving charge is always normal to the charge's velocity. Hence, when you take the integral of the applied force over the path of the charge, it will be zero because the magnetic force is normal to the charge's path. The magnetic field instead can only directly change the momentum of the charge.

Often times though, when you transform to the frame of reference as seen by the charge's perspective, the magnetic field may transform into an electric field. This electric field can do work on the charge. This can be seen when analyzing the case of two current carrying wires. So while the magnetic field does no work, that doesn't mean that work can't be done in these situations. It just requires a different perspective that allows the electric field to mediate energy from the magnetic field to the charge.
 
  • #3
10
0
Thank you, Born. It makes sense now!
 
  • #4
5,607
40
Hence it's based on our definition of work....analogous to you pushing as hard as you can against and wall and getting tired....yet no work is done....
 
  • #5
10
0
Indeed. I can see that since
[tex]\vec{F}=q\vec{v} \times \vec{B} \text{ and } W = \int \vec{F} \cdot d\vec{r}[/tex]
then
[tex]W = q\int (\vec{v} \times \vec{B}) \cdot d\vec{r}[/tex]

Since [tex]\vec{v} \times \vec{B}[/tex] is perpendicular to [tex]d\vec{r}[/tex], the dot product, and therefore the work are zero.

From the particle's perspective though, it is being influenced by an electric field instead of a magnetic one, so that
[tex]\vec{F} = q\vec{E} \text{ and } W = q\int \vec{E} \cdot d\vec{r}[/tex]
Since the electron will follow the electric field lines, [tex]\vec{E}[/tex] and [tex]d\vec{r}[/tex] are parallel, making
[tex]W=qE \int dr[/tex]

Thanks for the help guys! This has helped a lot.
 
  • #6
marcusl
Science Advisor
Gold Member
2,795
452
Magnetic fields can do work on magnetic materials, such as other magnets or other objects made of magnetic material. Putting steel shot (a ferromagnetic sphere) in a magnetic field will induce in the sphere a magnetic dipole moment. If the field is non-uniform, the shot will experience a force given by

[tex]\vec{F}=\nabla(\vec{m}\cdot\vec{B})[/tex]

and this force can do work.
 
  • #7
5,441
9
Yes I think you should be careful of the context of this statement.

It may be preparing you for the the difference between real power and reactive power in ac circuits, where inductors may do no work.

As has been pointed out magnetic fields can do work on magnetic materials, including other magnets, but can't do any work on charged objects, just because they are charged
 
  • #8
10
0
Interesting point. Glad you brought that up! Thanks.
 
  • #9
4,662
6
I remember learning in Electrodynamics last year that magnetic fields don't actually do any work. I don't seem to remember any of the explanation behind that. Can anyone explain why magnetic fields do no work (ie. How they differ from electric and gravitational fields).
Magnetic fields could do work on magnetic charges (monopoles), but no one has ever found any. But magnetic fields can do work (torque) on magnetic dipoles, like permanent magnets. Watch a compass needle oscillate about the direction of magnetic North.

Also, if the total energy in a magnetic field is

W = (1/2μ0)∫B2 dVvolume,

then there is a force in the x direction given by

Fx = ∂W/∂x.

and this force can do work.

Bob S
 
  • #10
marcusl
Science Advisor
Gold Member
2,795
452
I'm not sure about the last, Bob S. You identify the field energy as work instead and state that the field has an intrinsic force contained within it. There is no force without an object to work against, however, as in my example.
 
  • #11
4,662
6
I'm not sure about the last, Bob S. You identify the field energy as work instead and state that the field has an intrinsic force contained within it. There is no force without an object to work against, however, as in my example.
If you have two large neodymium magnets 1 cm apart, with a large attractive force between them, you can harvest this magnetic energy by letting the gap decrease to, say, 5 mm. So the stored magnetic field energy can do work.

Bob S
 
  • #12
Born2bwire
Science Advisor
Gold Member
1,779
22
Magnetic fields can do work on magnetic materials, such as other magnets or other objects made of magnetic material. Putting steel shot (a ferromagnetic sphere) in a magnetic field will induce in the sphere a magnetic dipole moment. If the field is non-uniform, the shot will experience a force given by

[tex]\vec{F}=\nabla(\vec{m}\cdot\vec{B})[/tex]

and this force can do work.

Technically though, with the absence of magnetic monopoles, the magnetic dipoles are microscopic loop currents. Saying that magnetic fields do work on magnetized objects is just another shortcut that is taken as the magnetization can be modeled as macroscopic bound currents.
 
  • #13
marcusl
Science Advisor
Gold Member
2,795
452
If you have two large neodymium magnets 1 cm apart, with a large attractive force between them, you can harvest this magnetic energy by letting the gap decrease to, say, 5 mm. So the stored magnetic field energy can do work.

Bob S
In your example, the field from one magnet is doing work on the other, and vice versa. That is how a field can do work. The field from a single magnet in the absence of the second generates no force, and I think your earlier wording was prone to this interpretation.

Technically though, with the absence of magnetic monopoles, the magnetic dipoles are microscopic loop currents. Saying that magnetic fields do work on magnetized objects is just another shortcut that is taken as the magnetization can be modeled as macroscopic bound currents.
True, but such problems are usually solved by referring to the magnetization M or dipole moment m. Calculating the force between a field and the equivalent magnetic surface currents on the solid, or actual microscopic currents within, is an academic exercise only.
 
  • #14
64
0
Born2bwire said:
Via the Lorentz force, the force created by a magnetic field acting on a moving charge is always normal to the charge's velocity. Hence, when you take the integral of the applied force over the path of the charge, it will be zero because the magnetic force is normal to the charge's path. The magnetic field instead can only directly change the momentum of the charge.

Yes, that's where the mistake comes from and why there is a false assumption that magnetic force can do no work. But, if there was no separate displacement due to interaction of magnetic fields we would have never known about this force.

The most important thing here is that fields alone do not make any work in any case, for that you need TWO, you need at least two magnetic fields or there will be no magnetic FORCE, and without the force of course there will be no work done, which is very different from the actual ability to perform the work.


The magnetic field instead can only directly change the momentum of the charge.

Momentum, you mean direction or velocity? Is that not what is force supposed to do and how you measure the work done? Fields on their own can do nothing, you need the force, which means at least TWO of these fields. What is "only" about actually causing a charge to displace?



So, electrons move in one direction in a wire and the displacement is caused by the electric force because there are electric fields interacting along the wire. But you can not expect the wire to be moving sideways because you only have magnetic field around the wire and to have a force, displacement and work done, we therefore need at least one more wire or permanent magnet, something to interact with this magnetic field of a moving charges in our wire.

Sure enough, two parallel current carrying wires attract and if the current goes in opposite directions they will repel due to Lorentz force (Ampere's force), we can measure this displacement and hence the work done by the magnetic force.
 
  • #15
64
0
Hence it's based on our definition of work....analogous to you pushing as hard as you can against and wall and getting tired....yet no work is done....

That is not true, it is yet another popular misconception.

In complex situations you have to decompose displacements, ALL OF THEM, in vectors and find out forces and split them in vector components too, then see what displacements match with what force to find out work done on particular object by a particular force.

When pushing against a wall work is done but the displacement is not obvious as it is in a form of muscle tension, bone friction, blood displacement, hart-rate and all the rest of biological and chemical reactions that go along. Some things are not obvious, but that that does not mean they do not exist.
 
  • #16
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Magnetic fields can do work on magnetic materials, such as other magnets or other objects made of magnetic material. Putting steel shot (a ferromagnetic sphere) in a magnetic field will induce in the sphere a magnetic dipole moment. If the field is non-uniform, the shot will experience a force given by

[tex]\vec{F}=\nabla(\vec{m}\cdot\vec{B})[/tex]

and this force can do work.

The magnetic field isn't actually directly doing any work here. According to The Lorentz Force Law (from which the equation you've posted is derived), magnetic fields never (barring the existence of magnetic monopoles) do any work. The work, in this case , and all cases where it appears that a magnetic field is doing work, is caused by an electric field which is induced by a changing magnetic field. The electric field is what actually directly does the work.
 
  • #17
Born2bwire
Science Advisor
Gold Member
1,779
22
Yes, that's where the mistake comes from and why there is a false assumption that magnetic force can do no work. But, if there was no separate displacement due to interaction of magnetic fields we would have never known about this force.

The most important thing here is that fields alone do not make any work in any case, for that you need TWO, you need at least two magnetic fields or there will be no magnetic FORCE, and without the force of course there will be no work done, which is very different from the actual ability to perform the work.




Momentum, you mean direction or velocity? Is that not what is force supposed to do and how you measure the work done? Fields on their own can do nothing, you need the force, which means at least TWO of these fields. What is "only" about actually causing a charge to displace?



So, electrons move in one direction in a wire and the displacement is caused by the electric force because there are electric fields interacting along the wire. But you can not expect the wire to be moving sideways because you only have magnetic field around the wire and to have a force, displacement and work done, we therefore need at least one more wire or permanent magnet, something to interact with this magnetic field of a moving charges in our wire.

Sure enough, two parallel current carrying wires attract and if the current goes in opposite directions they will repel due to Lorentz force (Ampere's force), we can measure this displacement and hence the work done by the magnetic force.

Ok, I'll post this once more, from Griffith's textbook:

magwork.jpg


There is no stipulation that you need two magnetic fields to create a force. A single magnetic field will result in a force when any charged particle passes through the field.

Force describes the change in momentum. Work is the change in momentum along the object's trajectory. The magnetic field can only change the charge's direction, but not its speed. And again, you do not need two fields, only one field.

In the case of the two current carrying wires, as I explained earlier, the work is done by the electric fields, not the magnetic fields. This is because in the frame of the charges, the magnetic fields transform to electric fields which provide a force that allows for work to be done. This is an example shown in many textbooks such as Griffiths or Grant & Phillips.
 
  • #18
4,254
2
I think we can safely say that the "magnetic fields do no work" is an oversimplification of classical electrodynamics despite born2bwire's text. What's on the following page born2bwire?

Rather, instead, the zeroth derivative of the magnetic field does no work on electric charge according to the Lorentz force equation.

More, the Lorentz force is defined without regard to any specific inertial or non-inertial frame of the observer, where E and B are local fields to the charges in question.

Yet, one simple thought experiment seems to say the Lorentz force is wrong or incomplete (it isn't):

A charge neutral, ring of current is subject to a force in a time invariant magnetic field. To obtain the magnetic field we can use another axially aligned ring of current.

Who can explain the work done by one fixed ring of current, upon on the other, using the Lorentz force equation?
 
Last edited:
  • #19
5,441
9
can be modeled as macroscopic bound currents.

The operative word here is model.
 
  • #20
Born2bwire
Science Advisor
Gold Member
1,779
22
The operative word here is model.

Well we could use the original microscopic atomic loop currents as well, it's no different. In fact I recollect that Griffiths does a treatment of this in his text. We just take the shortcut of superposition to bring it up to macroscopic bound currents. This is the relationship between magnetization and the current densities in Maxwell's equations.
 
  • #21
64
0
Phrak said:
Yet, one simple thought experiment seems to say the Lorentz force is wrong or incomplete (it isn't):

A charge neutral, ring of current is subject to a force in a time invariant magnetic field. To obtain the magnetic field we can use another axially aligned ring of current.

Who can explain the work done by one fixed ring of current, upon on the other, using the Lorentz force equation?

I agree Lorentz force is correct and complete and if what I previously said is true I should be able to explain the work done in any setup involving any em charges, fields or forces.

Can you give full details about these rings of current, what is initial condition in terms of electron velocity in the 'primary' ring and what is observed effect in the 'secondary' ring in terms of electron displacement or velocity. What is their exact relative position, orientation and size.
 
  • #22
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Who can explain the work done by one fixed ring of current, upon on the other, using the Lorentz force equation?

Griffiths' Introduction to Electrodynamics for one. See the discussion immediately prior to eq 6.3 (in the 3rd edittion) along with the outline of problem 6.22 from that text.
 
  • #23
4,254
2
It's a little late for me varga. Besides, I want to see if experts here can rise to the challenge.

gabbagabbahey, I don't have that text. What does it say?
 
  • #24
5,441
9
I agree Lorentz force is correct and complete

Interesting that there is another current thread in this forum and a good many textbooks that suggest the opposite.
 
  • #25
Born2bwire
Science Advisor
Gold Member
1,779
22
It's a little late for me varga. Besides, I want to see if experts here can rise to the challenge.

gabbagabbahey, I don't have that text. What does it say?

Assuming a uniform current, the force on a a loop current is

[tex] \mathbf{F} = I \oint \left(d\mathbf{l}\times\mathbf{B}\right) [/tex]

Which is directly from the Lorentz force by noting that the charge is Idl. In a uniform magnetic field, the net force is zero but not so in a nonuniform field. For certain problems this may not be too difficult. For example, Griffiths' gives the net force if you place a current loop directly above a solenoid such that the solenoid and current loop share the same axial axis. The calculated net force is along the axial direction. Since the charges are flowing in the plane normal to this direction, the initial force direction gives 0 work. This follows from the fact that the force is acting directly on the moving charges, which means that it must be acting along a direction that is not going to contribute to the work. While the initial force comes from the magnetic field alone in the frame of the current loop, once the current loop deflects then there is a time-varying magnetic field and thus a time-varying electric field as well. The same could be said about a single electron in a spatially varying, but time static, magnetic field. From the electron's frame of reference, the spatial variations of the magnetic field appear as a time-varying magnetic field and thus an associated time-varying electric field appears as well.

EDIT: Quick thought occured to me. If we have a system where the lab frame only has a magnetic field, then given a reference frame that is traveling with velocity v, then the electric field in the reference frame is

[tex] \mathbf{E} = \mathbf{v}\times\mathbf{B} [/tex]

Neat huh?

I think a lot of this discussion is mainly a point of pedantics. I don't think anyone is going to seriously try and work every problem from the perspective of electric field work. But I think that the underlying physics that mediate the energy from the magnetic field to a system are rather subtle and seem to cause a lot of confusion. I have not seen in the literature that the magnetic field's force can give rise directly to work. Instead, I have always read that there are subtle mediations that involve field transformations and such.
 
Last edited:
  • #26
64
0
Born2bwire said:
There is no stipulation that you need two magnetic fields to create a force. A single magnetic field will result in a force when any charged particle passes through the field.

According to 2nd and 3rd Newton's law of motion we can not have INTERACTION (force) without there being two of something that can interact. In your example you too have a force only because you have two fields interacting:

1. moving charge is one magnetic field (ignoring spin dipole moment)
2. "(charge) passes through the field" - this describes the SECOND field


Force describes the change in momentum. Work is the change in momentum along the object's trajectory.

Agreed.

The magnetic field can only change the charge's direction, but not its speed. And again, you do not need two fields, only one field.

CHANGE in DIRECTION is ACCELERATION is CHANGE in VELOCITY vector. I'd like to see example where there's force and only one field.


In the case of the two current carrying wires, as I explained earlier, the work is done by the electric fields, not the magnetic fields. This is because in the frame of the charges, the magnetic fields transform to electric fields which provide a force that allows for work to be done. This is an example shown in many textbooks such as Griffiths or Grant & Phillips.

Why are you talking about special relativity and why change reference frame all of a sudden? Why not look at electric field from their reference frame when they exist as magnetic fields then? That is arbitrary and inconsistent, in classical physics we have absolute reference frames and they work just fine for these purposes.

We are talking about classical mechanics and classical electrodynamics, in this framework we can explain these experimental setups completely from our (observer) reference frame, just like we do for all the rest of classical physics, and so there is no reason to involve any other more abstract and complicated explanation, which I find flawed anyway.

We can measure displacement of those parallel current caring wires and it perfectly matches to the magnitude of magnetic force as given by Ampere's force law or Lorentz force equation and Biot-Savart law, so why change anything about it if it works as it is?


Two parallel wires, if electric fields of electrons cancel due to superposition with positive electric fields of protons in these wires, then we are obviously left with some other force here, and what kind of force would be the one that can act in two directions in the same time anyway - the very fact that you have two independent and different displacements suggests you have two unique force vectors at work - two forces.



As for that page, thank you. Unfortunately there are statements there that are wrong according to everything I encountered in my education, and although that was quite some time ago, still I'm talking about basics like vector math and Newton's laws of motion, so I must raise my objections. But I'm not surprised those wired conclusions were reached if the logic of it was derived from Maxwell's equations instead of from Coulomb's law, Biot-Savart law and Lorentz force equations.


- "Magnetic forces may alter the DIRECTION in which particle moves, but they can not speed it up or slow it down."

1. What "speed up" and "speed down" has to do with work done?
2. Is change of direction not displacement either with constant velocity or not?

This is the same as projectile motion or free fall under gravity field. You have to split the movement into VECTOR COMPONENTS. If bullet starts its trajectory horizontal to the ground, what does gravity do? It ALTERS THE DIRECTION. How? By causing displacement in the direction perpendicular to the motion of the bullet, yet we do not say gravity field can ONLY alter the direction, because "altering direction" is what DISPLACEMENT actually is if you look at the vertical component of the motion - vector calculus.

Take the Earth and its gravity field out of the equation and bullet moves straight as its gravity field has nothing to interact with and so we do not expect any gravity force, any change in velocity nor change in direction - Newton's 1st law of motion.
 
  • #27
Born2bwire
Science Advisor
Gold Member
1,779
22
Why are you talking about special relativity and why change reference frame all of a sudden? Why not look at electric field from their reference frame when they exist as magnetic fields then? That is arbitrary and inconsistent, in classical physics we have absolute reference frames and they work just fine for these purposes.

There is no absolute reference frame in classical electrodynamics. Classical electrodynamics by definition satisfies special relativity. Avoiding to properly account for special relativity even when we are discussing non-relativistic velocities can give you incorrect answers because the fields ALWAYS transform using the Lorentz transformations.
 
  • #28
1,477
1
We are talking about classical mechanics and classical electrodynamics, in this framework we can explain these experimental setups completely from our (observer) reference frame, just like we do for all the rest of classical physics, and so there is no reason to involve any other more abstract and complicated explanation, which I find flawed anyway...

You have zero understanding of those "more abstract and complicated explantion," which you continually demonstrate, while loudly making demands and generally being boorish. You endlessly make vague attacks on QM and SR/GR without providing sources or sound arguments. Every thread has you going down the same road regarding Electromagnetism and the same points you refuse to accept.
 
  • #29
marcusl
Science Advisor
Gold Member
2,795
452
There is no stipulation that you need two magnetic fields to create a force. A single magnetic field will result in a force when any charged particle passes through the field.
Your example has two fields--one is the static field while the other is due to the charged particle's motion.

I think we can safely say that the "magnetic fields do no work" is an oversimplification of classical electrodynamics despite born2bwire's text.

Agreed. There is, for example, the work

[tex]W=\frac{1}{2}\int\vec{H}\cdot\vec{B}\kern+2pt dV[/tex]

done in bringing a magnetic field up from zero. As mentioned by others, this equation can be derived by considering the electric potentials that are generated by the time-varying currents needed to create the magnetic field. Note also that, for fields created by currents, this leads directly to the engineer's statement of stored energy in terms of inductance

[tex]U=\frac{1}{2}LI^2,[/tex]

and its extension involving mutual inductance when multiple linked coils are involved. Still, it is most convenient to write the work done in terms of magnetic fields, as above, and not the difficult and detailed equations involving time-varying fields and electric potentials.

If the fields in the equation above for work are from two coupled currents, then the derivative of work gives the force between the currents, as pointed out by Bob S. While much of the subsequent discussion of microscopic currents in a magnetized solid may be valid, it takes us out of the practical realm--in fact, truly correct treatment of microscopic magnetization is not even classical, but requires quantum mechanics. In the classical world, on the other hand, problems involving magnetic matter are best handled through the magnetization M, which properly accounts for their bulk behavior.

In this construct, there is no question that magnetic fields can do work on magnetized solids. To return to two cases already considered, the linear force and the torque on a dipole in a field can be integrated to find the work done as the dipole moves. The general expression for the work done in all such cases is

[tex]W=\frac{1}{2}\int\vec{M}\cdot\vec{B}\kern+2pt dV .[/tex]

If you want references to cross-check Griffiths, take a look at ch. 5 in Jackson or ch. 8 of Smythe.
 
Last edited:
  • #30
96
0
I remember learning in Electrodynamics last year that magnetic fields don't actually do any work. I don't seem to remember any of the explanation behind that. Can anyone explain why magnetic fields do no work (ie. How they differ from electric and gravitational fields). Thanks!

I ran into interesting letter on the subject :

http://academic.csuohio.edu/deissler/PhysRevE_77_036609.pdf

I am planning for some time to clear this thing up in my head. A working solution I don't plan to defend too vigorously is that magnetic field can do work. Within CED answer is no. Reason this isn't so in real world is because field interacts with atomic dipoles which are quantized and can not adjust it's orientation perfectly to cancel working effect of field. Other reason is field interaction with spin which is, of course, not current loop to which classic no-work result applies.
 
  • #31
64
0
You have zero understanding... without providing sources or sound arguments.

Ughh, look at your argument. I wanted to ignore your intrusion as you failed to specify anything in particular and even less make any actual arguments, but since I got some negative points regarding my previous post, as there is apperantly some "misinformation" there, then I decided to respond after all. To you I say, please WHAT is it you actually disagree about? I'll be happy to provide sources and answer all your questions, but what is it you are talking about? And to everyone else, I apologize for any misinformation and any false or ambiguous statements I made here, I did so only because it was written in my books and is my own (mis)understanding, but I will stand corrected as soon as someone actually points out what is it I am wrong about.
 
  • #32
64
0
marcusl said:
While much of the subsequent discussion of microscopic currents in a magnetized solid may be valid, it takes us out of the practical realm--in fact, truly correct treatment of microscopic magnetization is not even classical, but requires quantum mechanics.

Let me say that I'm glad to finally see someone I agree with... at everything but this above. I do not think there is anything in QM that can come even close to describe (explain) orbital magnetic moment.
 
  • #33
marcusl
Science Advisor
Gold Member
2,795
452
I just found this post by Vanadium50 in another thread
https://www.physicsforums.com/showthread.php?t=347539"
that does a better job explaining my point than I am doing:
I don't like teaching the meme "magnetic fields do no work." It is true, but it is not useful.

It's clearly true for a single charged particle: the Lorentz force law has the magnetic force perpendicular to the direction of motion, so the dot product of force and displacement is always zero. It's also clearly true that magnets can do work on each other.

The solution to this apparent contradiction is that complex objects like magnets are made up of many charges, and these charges exhibit both electric and magnetic forces on each other, and if one does the calculation carefully enough, it can be shown that the work actually comes from these (usually internal) electric forces.

So what do you gain by thinking about things this way? To my mind, very little: you're trading a relatively simple calculation - say the torque on a magnetic dipole in a magnetic field - for a very complicated one involving internal electric forces. This seems like a poor trade. Note that I am not arguing that "magnetic fields do no work" is not true. I am arguing that it is not useful. It's (relatively recent) overemphasis is not, in my mind, a good thing.
Thanks V50. I wish I could have put it so clearly.
 
Last edited by a moderator:
  • #34
5,441
9
Here is a challenge to those who say a magnetic field can do no work.

Design a non-magnetic system that will operate my solenoid door bell.

The requirements are that no part of the doorbell may be altered, except the magnetic generator may be removed if desired.

I maintain that there is no known force that can replace the magnetic coupling to transfer energy from the battery to the plunger.
 
  • #35
1,477
1
Here is a challenge to those who say a magnetic field can do no work.

Design a non-magnetic system that will operate my solenoid door bell.

The requirements are that no part of the doorbell may be altered, except the magnetic generator may be removed if desired.

I maintain that there is no known force that can replace the magnetic coupling to transfer energy from the battery to the plunger.

You could reverse the circuit so that the doorbell only goes off when the circuit is broken. In that case you could use a spring and maybe the Strong or Weak force. Granted, you'd need a particle accelerator in your doorknob, but I think that's quite fashionable.

Anyway, you could just design one out of clockwork. :wink: Press the plunger against a piezoelectric element which sparks a tiny gap firing the circuit... etc. Of course, you lose the solenoid...

... Oh wait, nothing can be altered except the removal of a system. How is that conducive to DESIGNING a system?
 

Related Threads on How do magnetic fields do no work?

Replies
84
Views
34K
  • Last Post
Replies
17
Views
9K
Replies
2
Views
851
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
6
Views
165
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
4
Views
4K
Replies
6
Views
1K
  • Last Post
2
Replies
44
Views
47K
Replies
36
Views
2K
Top