# B Magnetic forces -- arguments about doing work

1. Aug 14, 2016

### Conductivity

Well First of all, I noticed tons of thread about this in the forums and tons of the same question being asked in all kind of scientific forums. I am pretty sure everyone here hates this topic, and I have got a bad headache from reading all the comments, but just clarify this to me.

"Most of the people who responded to these threads here stated that magnetic forces dont do work and the work we see comes out from electric forces."

Okay I am alright with that but the question arises a bit why do these electric forces have to equal the magnetic force acquired from The equation $F = qvB$?

Using this you can prove that the EMF induced in a moving conductor is
$\text{Emf} = \frac{qvBL}{q} = BLv$
Which agrees with faraday's law of induction

2. Aug 14, 2016

### Simon Bridge

... do they have to be equal? (By "equal" I am guessing you mean "equal magnitude".)
Is it your assertion that the work done on a charged particle moving a distance d through a constant magnetic field is $W=qvBd$?

I don't think so.
In science you can "prove" things in the sense of demonstrating that Nature acts in congruence with the prediction of a model.
You can also prove that two models are actually the same - which is what I think you are saying when you say you can prove Faradays induction from statements about work and energy: that Faraday's Law can be shown to be a consequence of a more fundamental description or model of Nature. But that does not prove that Faraday's Law is true.

Guessing that is what you mean:
How is the magnetic force used to derive Faraday's Law exactly, and what from? What role does it play? Where is the work happening?
eg. A charge travelling through a constant magnetic field experiences no work, yet it does change direction under an unbalanced force.
Clearly the force contributing to work in this case is not the same strength as the magnetic force the particle experiences.

3. Aug 14, 2016

### Staff: Mentor

The electric force doesn't usually equal the magnetic force.

However, I think I may vaguely understand your underlying question. Are you comfortable with the idea of different reference frames?

4. Aug 14, 2016

### gannordean

Sir,
In electromagnetics courses, the empirical action of electric E and magnetic B fileds on a moving test charge q with velocity v is embebed in the definition of the Lorentz Force:
F=q(E+ v × B)=FE+FB
The magnetic force FB=(q v × B) part of the Lorentz force is responsible for motional electromotive force (or EMF), the phenomenon underlying many electrical generators[1]. Note that here, a wire carrying an electric current is placed in a magnetic field and moved in respect to it. The Laplace Force is an after effect and generated on a macroscopic scale and don't be confused with the definition of the Lorentz force which is intrinsic to every charge in nature.
The work done by the Lorentz Force is by definition zero, as it's orthogonal to the desplacement path. For the Laplace Force, this depend on the geometry of conductor-field configuration and is generally non zero.

So if you and me are okay on that, I want to understand your question more: what do you mean by the E forces ''have to equal'' the B forces? there's no ground for equaling anything with another. Just make sure to understand the whole kinematics of your EM fields configuration and everything will be allright.

Don't panic :), physics is not meant to be hated! It's the epitome of love :p

Hope this help.
Best,
Samir.

[1]: https://en.wikipedia.org/wiki/Lorentz_force#Significance_of_the_Lorentz_force

5. Aug 15, 2016

### Conductivity

Well, The proof upthere isn't mine. It is from my book which sounded weird. Let me try to demonstrate what it proved
Assume you have this setup

What the book said is if you move the rod to the left with velocity v to the left a magnetic force will act upward (across the wire) with a magnitude of qvB.

So It said, the work done is qvBL, However that doesn't make sense because magnetic forces don't do work because of the whole cos 90 = 0
But if you use it to find the Emf which is $Emf = \frac{qvBL}{q} = vBL$ which gives the same result of faraday's law

So it gave the right result, that leaves us with 2 choices either it is a coincidence or there must be a force acting upward that can do work. so I assumed it can be electrostatic force.

Dale, Yes I am familiar with different frame of references. Are you going to talk about special relativity and how distances are going to be shorten from a rest frame of a moving charge which means by coulomb's law it will have a force greater in magnitude?

Another example, Two parallel carrying currents wire in the same direction will result in an attraction. Now Usually we say that the force of attraction is ILB but magnetic forces don't do work, However we could say that magnetic forces change the direction of the elections making them attract the protons and neutron thus moving the whole wire. The energy here is coming from the batteries. But dont we observe that there is a force moving the wire with a magnitude of ILB? So the overall of these electrostatic interactions must be equal to ILB in the same direction

It is really hard to imagine the effects of magnetic forces in small scales

Last edited: Aug 15, 2016
6. Aug 15, 2016

### Simon Bridge

Thats what I thought... I'll get back to you. Sit tight.

7. Aug 15, 2016

### Conductivity

8. Aug 15, 2016

### Staff: Mentor

No, my line of reasoning is non relativistic. If you really use a relativistic framework then the whole question disappears since there is no separate magnetic field. This is basically just an analysis of the E.J term in Poynting's theorem considered in a frame where a conductor is moving and a frame where it is at rest.

The work term $E \cdot J$ includes all the work done on matter, including but not limited to the Ohmic losses. So I wanted to see if any further analysis of that term could be done to pull out the Ohmic losses.

If a piece of material is moving at velocity $v<<c$ then we can make the following transformations:
$E' = E + v \times B$
$J' = J - \rho v$
where the primes indicate values in the reference frame where the material is at rest.

Substituting those in we have:
$E \cdot J = (E' - v \times B) \cdot (J' + \rho v)$
$=E' \cdot J' + E' \cdot \rho v - (v \times B) \cdot J' - (v \times B) \cdot \rho v$
$=E' \cdot J' + \rho v \cdot (E+ v \times B) - (J-\rho v) \cdot (v \times B)$
$=E' \cdot J' + \rho v \cdot E - J \cdot (v \times B)$

So finally we end with
$E \cdot J=E' \cdot J' + v \cdot (\rho E + J \times B)$

Since $E'$ and $J'$ are in the reference frame of the material, I would interpret the term $E' \cdot J'$ as being Ohmic losses. Then $E \cdot J$ contains not only the Ohmic losses, but also the material velocity times the Lorentz force density, which I would call the Lorentz power density.

I don't have a reference for this, so I don't know if someone else has already done this derivation.

9. Aug 16, 2016

### Conductivity

Thank you Dale for the answer, but I am not familiar with the Poynting's theorem. I will read about it soon and get back to you. Thanks again

10. Aug 16, 2016

### Staff: Mentor

Sorry about that, with your question I had assumed you knew Poynting's theorem. It is the one that states that the work done on matter is done by the E field.

From Maxwell's equations you can derive the equation:
$du/dt+\nabla \cdot S + E\cdot J=0$ where $u$ is the energy density (the energy stored in the field), $S$ is the flux density (the energy transported from one location to another in the field), and $E\cdot J$ is the power density (the energy transfered out of the field through work on matter).

So the whole question arises because Poynting's theorem says the work is $E\cdot J$, so there is no $B$ field involved in doing work. And yet we can produce examples like the one you mentioned above, where it certainly looks like the $B$ field does work.

Last edited: Aug 16, 2016
11. Aug 16, 2016

### Conductivity

Oh thanks for the answer,I also found a really good way to prove that magnetic forces don't do work but they might transfer energy.
Assume you have that same wire moving to the left with a velocity of V
Lets assume it has positive charges moving
So a magnetic force will act on the charges causing them to actually move upward in some sort of circular motion (Transient state). It is quite complex to explain how it moves them in the transient state but we can show after reach the steady state how this works.

So you will have the postive ions moving left direction and upward due to the current induced. Inside the wire, because the charges moved initially upward and stacked there we have an electric field and this will end after the two forces of the magnetic and electric field are equal.

Now we have two components of velocity you could combine them and show that the overall magnetic field will be perpendicular to the velocity thus no work, But that doesn't explain why the charges move. Okay now keep the two components of velocity, They will generate two components of magnetic field one in the y direction due to the movement to the left and one in the -x direction due to the movement of the current.

in the y direction we have the force of $q~V_{rod}~B$ and on the -x direction we have a force of $q~V_{drift velocity}~B$
Now we have to prove that the overall work of the magnetic field is zero
lets look at the -x direction, If the rod moves with a velocity of V then we can say that the work done by the force of the magnetic field is
$- q~V_{drift velocity}~B ~ V_{rod} t$ where t is really small

But the in the Y direction we have $q ~ V_{rod} ~ B ~ V_{drift velocity} * t$ and here we can see that the two work are equal. Which means it only transfers energy and Energy is just conserved.

12. Aug 16, 2016

### Simon Bridge

Sorry for the delay - murpheys law in action.
You can have a go seeing what happens when your setup fires a charged pellet along a track through a uniform B field.
Work out the free body diagram on the pellet.
You get qvB pushing the pellets against the side of the track, and a normal force from that side keeping the pellet on the track.
How big does the normal force have to be and where does it come from?

13. Aug 17, 2016

### Conductivity

If you want it to keep on track then the normal force must be equal to qvB
But not sure what you mean with where it comes from, I mean there can be a lot of sources

14. Aug 17, 2016

### Simon Bridge

Do you not know why solid objects do not pass through each other? After all, they are mostly empty space.

15. Aug 18, 2016

### Conductivity

Because of electric forces between the two objects (In atomic scales) ...

16. Aug 18, 2016

### Simon Bridge

.
... OK - so back to the charged pellet: the magnetic force pushes it against one side of the track with force qvB, the track pushes back with normal force magnitude qvB ... the normal force is electric in origin ... therefore the net electric force must be equal in magnitude to the magnetic force. If it didn't, the configuration would not be as described.

Relate back to the charges in a wire.... what constrains the charges to move in the wire?

17. Aug 19, 2016

### Conductivity

The charges on the wire on each side: electrons and protons. Charges act as if they were solid bodies

18. Aug 19, 2016

### Simon Bridge

Well it is the electromagnetic potential due to the fixed charges in the wire ... on this scale it makes no sense to talk of charges acting like solid bodies since the wire is mostly empty space. As an electron (since we are talking conductor here) moves away from the wire, it experiences an unbalanced force from positive charges pulling it back.

In the simplest model, the wire is a line of atoms bonded together with a free electron shared between all the atoms in the wire. This gives a line of net positive charge (atomic nuclei + bound electrons) and a "sea" of negative charge.
A potential difference along the length of the wire makes the electrons move along it - a magnetic field pushes the electrons to the side against the coulomb force from the line of positive charge. It's a bit like pulling a mass on a spring and finding that the equilibrium point is where the spring force is equal and opposite the pulling force then asking "why do they have to be the same?"