# Change in resistance, with respect to magnetic field?

1. Aug 26, 2015

### PhiowPhi

From the following diagram:

Due to the velocity($v$) of the conductor($C$) of resistance($R$) through the uniform magnetic field ($B$), there is an induced $\epsilon$ due to the motion:

$\epsilon = vBL$

This conductor now is a voltage source, when connected to a circuit (and a load) this is the circuit:

My issue, is with the resistance. Now the whole conductor's resistance is $R$, however, when calculating the current in the circuit I should only consider the resistance in the magnetic field ($R_o$)correct? Is the induced voltage all the same in the bottom of the conductor? In the magnetic field, and out like so:

I'm certain about the polarity of the blue(+), what about the rest outside the magnetic field? If we placed it into a circuit or measured via a volt meter?

A comparison/& contrast:

If my issues/confusion is unclear perhaps this circuit might clarify more:

Does this circuit represent the first diagram? Or this:

When focusing on the voltage drop(internal resistance)?
How is it, that the change in the magnetic field region/span by making it smaller at the same strength change things?

Last edited: Aug 26, 2015
2. Aug 27, 2015

### BvU

Hi Phi,

The induced emf is due to the Lorentz force working on the mobile charge carriers in the conductor. The Lorentz force is $\vec F_L = q \; \vec v\times\vec B$ meaning in your case only the "vertical" parts of the wire loop contribute to the induced emf.

But now I have to ask: is the conductor a wire loop or is it a solid rectangle ? Makes a big difference, as you probably can understand.

For a solid rectangle the emf causes short-circuit currents in the area entering the field (and idem exiting) and there won't be much emf left over at all.

If it's a wire loop then the Lorentz force being perpendicular to both B field and velocity vector in your case means only the "vertical" parts of the wire loop contribute to the induced emf (see rightmost picture here) . In your first picture (interpreted as a wire loop) this means that there is a pulse while the righthand side of the loop moves through the area of the B field. Then no emf until the left side enters and a second pulse is induced (with the same emf polarity, so the current is in the other direction) . And asking for the current in the loop is a bit more interesting.

 PS I see form other postings by you that you already have quite some experience and good questions in this area. My guiding principle is that I try to start from the Lorentz force (extended to $\vec F_L = q (\vec E + \vec v \times \vec B\$).

Last edited: Aug 27, 2015
3. Aug 27, 2015

### PhiowPhi

Hello @BvU,

It is a solid rectangle connected to a circuit(it's basically a large wire)a large loop. So, I think it would not short? Because the whole top side and bottom side will be connected to a circuit(and a load).
I think my issue, lies more with the induced EMF and resistance correlation. If you look at the 3rd diagram(not circuits) where the magnetic field region covers the whole conductor Vs. the 1st diagram where the magnetic field covers a smaller region, Since v,B,L are the same only change is the magnetic field "region/span" how does that effect the induced V?

I know it should be the same, but now... we focus on volume of the conductor inside the smaller magnetic field region. Is the resistance($R$) that I use in any calculation, should be of the whole conductor? Or only the part in the magnetic field($R_i$)? That in-terms only effects the current's magnitude not the induced voltage correct?

4. Aug 27, 2015

### BvU

Still not completely in the picture. Is the orange rectangle a slice of a wire ? top side and bottom side are out of and into the screen ? Or are they as in this picture:

And you want a circuit like
?

Then: The magnetic field convention I learned was that a circle with a cross is into the paper (away from you) and a circle with a dot is out of the paper (towards you) -- this in analogy with an arrow the cross is the tail view (the stabilizers) and the dot is the sharp point about to pierce you:

So I interpret your XXXX B field such that
$\vec F_L = q\; (\vec E + \vec v \times \vec B\ )\$ is pointing upwards. Which would make the top surface positive. My mistake or yours ?

5. Aug 27, 2015

### PhiowPhi

This is a proper diagram of the complete "circuit":

The polarity's based on the RHR, the magnetic field going into the page, the velocity to the right, the current should be going down(hence the + on the bottom)?

The terminals(blue rectangles) are stationary as the circuit(excluding the conductor(C)).
The small purple + is the individual EMF of that part of the conductor, while the red + is the complete EMF(with the part outside the magnetic field) because they are all a whole piece(in series) if that makes sense?

6. Aug 27, 2015

### BvU

Check you righthand rule once more. With the field into the page and v to the right, + is up.

then: if C is a conductor, the current won't bother to go through the load. It will go through the conductor.

7. Aug 28, 2015

### PhiowPhi

@BvU I was a bit off switched things you are right.

Wait, how is it that that the current will be limited around the conductor? And not going through the circuit & load?
Would there be any EMF/current in this situation? I feel this is a short circuit or...?

8. Aug 28, 2015

### PhiowPhi

Excluding Eddy currents for a bit, I'd like to understand how current is flowing only in the conductor and not beyond to the circuit and the load, is this a short circuit? Why isn't the induced EMF "added" to the whole circuit?
Could you also compare it to the diagram that had the magnetic field region covering the whole conductor? Why would current flow to a load in that case...?

Greatly confused of the current's behavior in this case, and the EMF's...

9. Aug 29, 2015

### BvU

Good thing the polarity issue is cleared up.

the picture there says it all ! The featuring picture in the current thread is rendering exactly the same situation. Eddy currents and induced voltages are one and the same side of one and the same coin: the Lorentz force.

There is no way to avoid these eddy currents when part of the slab is outside the field.

Which brings us to the diagram that has the magnetic field region covering the whole conductor (picture with header "A comparison/& contrast:" in post #1), where - for a brief moment - all the charge carriers in the slab are being pushed in the same direction. Then emf will force the current through a connected load !

10. Aug 29, 2015

### PhiowPhi

Thank you, that indeed does make sense.

11. Aug 30, 2015

### PhiowPhi

@BvU what if we flipped this, and used a power supply to generate the Lorentz force and $v$ is due to that.
Like so:

Changed the polarity via Len'z law, now how things would work out? I assume the part inside the magnetic field would be: $V_P - V_\epsilon$ ? While the rest outside the magnetic field is simply $V_P$? Making sure my analysis is right.

12. Aug 30, 2015

### BvU

Oh my, let me try to follow this through: You almost short-circuit a power supply to have a current from bottom to top in the moving conductor.
The Lorentz force you generate that way is pointing which way, exactly ?

13. Aug 30, 2015

### PhiowPhi

The moving conductor? Ow, things have changed now. $v$ is not from an exterior source, rather caused from the Lorentz force due to the power source's current flow upwards(as diagrammed) in the magnetic field going into the page, the Lorentz force is point to the right(as $v$'s direction).

Also, why is this a short-circuit when R is a finite value?

14. Aug 31, 2015

### BvU

Funny, my right-hand lets the Lorentz force (current up, B into the 'paper') point opposite to $\vec v$.

"Things have changed now" also mean that it's no longer a conductor ? Pretend I am dumb and give a complete self-contained description of what is done in post #11 and an account of what you expect to happen. To compare, here's what I pick up :

After a small correction
the single arrow to the left designated "Terminals" doesn't make sense. There should be one coming from "Top" and a second one coming from "Bott".​
what I see so far is a conductor short-circuiting the load at the terminals. Not completely because it's not an ideal conductor apparently. But the conductor limits the voltage at the terminals to $I\;R$ with $R^{\;-1} = R_o^{\;-1} + R_i^{\;-1}$ .

The current in the conductor causes a Lorentz force for the part of the slab within the B field; that force pushes the conductor to the left, out of that field. Story ends. ​

15. Aug 31, 2015

### PhiowPhi

My goodness, the right hand rule and Lenz law caused a huge confusion of predicting the force, and the polarity of induced EMF, I had to "Review" that again. Sorry bouncing around that issue.

The red arrow(I think that is what you mean...) is an index, indicating terminals of connection to the circuit and PS.
Alright, so far you got everything right. But I might be confused with the $R^{\;-1}$? Why wouldn't it be $R$? Also, the voltage from the PS is reduced correct? Because we must account for the induced voltage due to motion caused by the Lorentz force.

The topic of short-circuit is not well rounded on my end, so this statement is "unknown" to me, I can't make sense of it because I don't know why there is a short circuit?

16. Sep 1, 2015

### BvU

First things first
I would like to read your account to understand how you look at this and what you expect to happen. There is something of a gap between how we perceive the situation.

Instead you start to comment on, and ask about my account. Now I could respond to that, ask and comment etc etc. Not very effective.
When I type
I do indeed mean the red arrow. What else could I do to identify it uniquely ?
But when you type
I still have no idea what you mean.

--

17. Sep 1, 2015

### PhiowPhi

Alright, edited things I should've done this earlier

The conductor (the orange part) is the only part movable in the system. Hope that clears things.
I know the Lorentz force is equal to $IL \times B$, but my concern is after the conductor starts to move, there is an induced EMF(marked + in white) that resists the change, will the voltage oppose the PS in this expression:

$V = V_P - V_E$

Where $V$ is the net voltage, $V_P$ the voltage from the power supply, and $V_E$ induced EMF of that small area of the conductor where the magnetic field lies.
You commented about shorting the power supply, could you clarify that.

I hope my approach now is "better".

18. Sep 1, 2015

### Staff: Mentor

Last edited: Sep 1, 2015