Exploring EMF in Moving Conductive Bars

[greg_rack]

Hi guys, I have barely started studying EMI, and already have a question about the nature of the current induced in a bar of conductive material traveling through a magnetic field.
The moving charges inside the bar will experience a Lorentz's force, pushing the negatives on one side of the bar and the positives onto the other.
Now, what does the induced EMF come from? Is the separation of charges causing a flow of electrons and thus a current, or is it caused by the final difference in potential between the two "poles" of the bar, in its "final" state?

I hope the question is clear, but I highly doubt it :)

 

[BvU]

the separation of charges causing a flow of electrons

The other way around: mobile charge carriers experience a Lorentz force that causes a (slight) separation and hence a potential difference

 

[etotheipi]

Hey Greg, it's a good question! You can think about a mobile charge in the rod as acted upon by a Lorentz force $q(\mathbf{E} + \mathbf{v} \times \mathbf{B})$. The electric $\mathbf{E}$ field is set up due to charge separation caused by the magnetic $\mathbf{B}$ field you're applying, and soon enough you reach an equilibrium state with zero net force, $q(\mathbf{E} + \mathbf{v} \times \mathbf{B}) = \mathbf{0}$, or in other words $\mathbf{E} = - \mathbf{v} \times \mathbf{B}$. If you set it up nicely so everything is at right angles and you integrate along the length of the rod, this gives rise to a potential difference $\Delta \phi = Blv$. [Edit: @BvU beat me to it ]

If I understand correctly, an EMF ($\mathcal{E}$) is defined for a closed curve, so to talk about EMFs you need to have a complete circuit (e.g. you could attach a voltmeter across the bar). Then, the EMF is just the line integral of the Lorentz force around the closed curve defined by the wire.

 

[greg_rack]

Great! @etotheipi @BvU
But the thing I still don't get is: if the current consists of a flow of electrons through a medium, how could there be a current flowing if all negative charges are "stuck" on one end of the rod, whilst positives on the other?

In this conception, wouldn't there be a current only until all electrons don't get stuck on one end? Therefore until they keep flowing and thus separating.

 

[etotheipi]

I think in the steady state (after the initially reshuffling of mobile charges, whilst the bar is accelerating), for the case of the bar moving at constant velocity through the region with the uniform magnetic field, you don't have any current in the bar. You don't need a current to have a potential difference between two points, you only require that there be a non-zero line integral of the electric field (in this case, the electric field is present due to the charge separation, a bit like a capacitor!) and also that the magnetic field is independent of time.

 

[greg_rack]

I think in the steady state (after the initially reshuffling of mobile charges, whilst the bar is accelerating), for the case of the bar moving at constant velocity through the region with the uniform magnetic field, you don't have any current in the bar. You don't need a current to have a potential difference between two points, you only require that there be a non-zero line integral of the electric field (in this case, the electric field is present due to the charge separation, a bit like a capacitor!) and also that the magnetic field is independent of time.

Mmm okay, so when mobile charges "stop"(because at a certain point the Lorentz's force will be equal in magnitude to the electric one), in the rod we'll have a zero current but of course a non-zero potential difference.
And that's where the EMF comes into place, right?

 

[etotheipi]

I think it's only meaningful to talk about the EMF around a closed curve. Hook the bar up to an external circuit and move the entire configuration through the magnetic field, and then you get a so-called "motional EMF"$$\mathcal{E}_{\text{motional}} = \oint_{\partial \Sigma} (\mathbf{v} \times \mathbf{B}) \cdot d\mathbf{x}$$in addition to the so-called "transformer EMF" due to the electric field$$\mathcal{E}_{\text{transformer}} = \oint_{\partial \Sigma} \mathbf{E} \cdot d\mathbf{x}$$The sum $\mathcal{E} = \mathcal{E}_{\text{motional}} + \mathcal{E}_{\text{transformer}}$ is the (total) EMF around the closed curve. I believe it will depend on the geometry of the external circuit.

Spoiler: Some further details

If the boundary $\partial \Sigma$ is a function of time, i.e. $\partial \Sigma = \partial \Sigma(t)$, you can start with the third Maxwell equation $$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} \iff \oint_{\partial \Sigma(t)} \mathbf{E} \cdot d\mathbf{x} = - \int_{\Sigma(t)} \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{S}$$Because the surface is a function of time you can't just pull the partial derivative outside the integral, you need to use the Reynolds' transport theorem, and you end up with$$\oint_{\partial \Sigma(t)} (\mathbf{E} + \mathbf{v} \times \mathbf{B}) \cdot d\mathbf{x} = - \frac{d}{dt}\int_{\Sigma(t)} \mathbf{B} \cdot d\mathbf{S} = - \frac{d\Phi}{dt}$$That means, in general,$$\frac{d\Phi}{dt} = - \mathcal{E} = -\oint_{\partial \Sigma(t)} (\mathbf{E} + \mathbf{v} \times \mathbf{B}) \cdot d\mathbf{x}$$which is the same as we had before

 

[jartsa]

The other way around: mobile charge carriers experience a Lorentz force that causes a (slight) separation and hence a potential difference

Lorentz force causes an EMF, charge separation causes an EMF, as those two EMFs are equal and opposite, potential difference is zero.

Right?

 

[BvU]

Right?

No.
Lorentz force causes charge separation, charge separation causes an EMF that opposes Lorentz force.

EMF = potential difference. Non-zero.

 

[etotheipi]

charge separation causes an EMF that opposes Lorentz force

I don't know, if I agree with this wording. I would say the charge separation causes an electric force that opposes the magnetic force, and in the steady state you have zero Lorentz force.

 

[Delta2]

The way i see it the "real" EMF in this case, is not the potential difference that is a consequence of charge separation, it is instead the force that causes the charge separation, that is the magnetic part of lorentz force $(\mathbf{v}\times\mathbf{B})q$.
The potential difference due to charge separation creates a conservative electric field E whose line integral along a closed loop is always zero (because it is conservative). The line integral along a closed loop of $(\mathbf{v}\times\mathbf{B})$ is not zero however, and that's indeed the real EMF.

It is how @etotheipi puts it, that is we should always consider the EMF around closed loops to see what's a real EMF. (ok if you don't like the adjective "real" here but I don't know how else to say it).

 

[greg_rack]

I think I agree with @BvU's opinion.
Mobile charges will separate( creating a difference in potential between the two ends of the rod, our "fictitious" EMF, as @Delta2 and @etotheipi pointed out), until Lorentz's force isn't equal in mag and opposite to the electric force.

Therefore, when we want to calculate the EMF($\varepsilon$) induced in an object of conductive material and length $L$, we basically equal Lorentz's and electric forces acting on a charge $q$:
$$\frac{\varepsilon}{L}q=qvB \rightarrow \varepsilon=vBL$$

 

[Delta2]

Therefore, when we want to calculate the EMF($\varepsilon$) induced in an object of conductive material and length $L$, we basically equal Lorentz's and electric forces acting on a charge $q$:
$$\frac{\varepsilon}{L}q=qvB \rightarrow \varepsilon=vBL$$

Ok, this approach works in the case you are interested but for more general problems and cases in order to find the motional EMF we have to calculate the integral of $\mathbf{v}\times \mathbf{B}$ around the closed loop of interest that is $$\mathcal{E}=\int(\mathbf{v}\times \mathbf{B})\cdot d\mathbf{l}$$, where $\mathcal {E}$ is the motional EMF.

 

[jartsa]

Hey guys, do we agree that a good 1.5 V battery always generates an EMF of 1.5 V,
and two 1.5 V batteries connected in series generates an EMF of 3.0 V,
and a 1.5 V battery and a block of wood connected in series generates an EMF of 0 V,
so a block of wood generates an opposing EMF when connected in series with a battery?I mean voltage, or EMF, generated by a battery and a block of wood, connected in series, measured by an ordinary voltmeter, is zero. With an ideal voltmeter with infinite resistance it's a different story. Then of course the voltmeter generates the opposing EMF. Opposing EMF = I*R

 

[artis]

@jartsa Well honestly this is the first time I hear that a block of wood generates anything other than smoke when burnt in a fire place...

An insulator cannot generate opposing EMF, it's just an insulator, it's atomic and molecular structure is such that it doesn't allow for any current to flow whne a voltage is applied to it, instead it's molecules deform under the applied Electric field and so it's dielectric permittivity is higher than that of air typically.

An opposing EMF can normally arise only from a conductor which forms a loop or a circuit that is subjected to a Changing ExB field aka EM field.

 

[artis]

@greg_rack Well if it helps think of it this way, typically we speak about EMF in a circuit/loop, in your mentioned case we have a straight rod conductor that is moving across a B field such that the B field lines cut the conductor perpendicularly.
This is what happens, as you start to move the rod across the field at first the rod isn't polarized and no current flows through it, then as you start to move it due to the atomic structure of conductors the free electrons in the rod are free to move and as soon as they experience a magnetic field they get deflected , this is the Lorentz force. This deflection causes a current in the rod due to electrons being deflected all in a single direction.
This current runs through the rod only as long as enough electrons have reached one end of the rod, after a period of time the current stops and there is a static and steady charge separation between the ends of the rod where one side is positive (less electrons) while the other side is more negative (abundance of electrons).

The reason why the current stops is because your B field is static and it's strength doesn't increase or decrease, your rod velocity is also steady and for a given static B field and velocity there is a fixed force that deflects the electrons, once enough electrons have been deflected and voltage set up between the ends of the rod the deflection force becomes insufficient to push any more electrons.

In order for the current to continue to increase you either need to increase B field strength or continually increase the rod velocity.
Now once the rod is polarized there is an E field within it created from the charge separation. if you would then close the rod ends electrically with a wire creating a circuit you would theoretically have current running through it. But in practice closing the rod ends would not result in a current (only for a short while as the polarization discharges) @greg_rack why do you think that is ?
There is a trick here , can you spot it?

 

[BvU]

I don't know, if I agree with this wording

Good point. I should not call it Lorentz force if I refer to the $q \bf{v}\times \bf B$ part only. Thanks !

 

[Delta2]

and a 1.5 V battery and a block of wood connected in series generates an EMF of 0 V,
so a block of wood generates an opposing EMF when connected in series with a battery?

Sophistic logic that doesn't get us anywhere. The block of wood blocks the flow of current because the wood isn't good conductor, that's why we can't measure the EMF of the voltage source in series. (All instruments and devices need current to flow in them in order to operate.) This doesn't mean of course that the block of wood generates an opposite EMF, that's pure sophistic logic, sorry mate.

 

[jartsa]

An insulator cannot generate opposing EMF, it's just an insulator, it's atomic and molecular structure is such that it doesn't allow for any current to flow whne a voltage is applied to it, instead it's molecules deform under the applied Electric field and so it's dielectric permittivity is higher than that of air typically.

Oh, molecules deform, well, that's charge separation, so clearly an EMF that affects those charges is generated.

What I was thinking was that the few free charges experience non-isotropic collisions, in other words there is an EMF affecting those charges.

 

[vanhees71]

I think there's a big confusion concerning the use of "electromotive force" in this context. I've no clue, where you need it at all. What we consider here is just a metal bar moving through a homogeneous magnetic field. Let's put $\vec{B}=B \vec{e}_z$. The bar is moving with velocity $\vec{v}=v \vec{e}_x$. FAPP we can consider the electrons' velocity in the bar $\vec{v}_e=\vec{v}$ since the drift velocity of the electrons relativive to the bar is negligible (it's about 1mm/s for usual setups).

Let the bar be described by the set of points $(\vec{v} t,\vec{v} t) \times (0,b) \times (0,c)$.

Due to the motion of the bar there's a magnetic force acting on the electrons $$\vec{F}_{\text{mag}}=-e \vec{v} \times \vec{B} = +e v B \vec{e}_y$$. Since the (conduction) electrons are freely movable there'll the electrons drift towards the surface $y=b$ at the bar. In the stationary state there'll be a negative surface charge $-\sigma$ at $y=b$ and a positive surface charge $+\sigma$ at $y=0$. This implies that there's an electric field
$$\vec{E}=\frac{\sigma}{4 \pi \epsilon_0} \vec{e}_y.$$
Since we are in the stationary state, the force on an electron in the interior of the metal must be
$$\vec{F}_{\text{tot}}=\vec{F}_{\text{mag}}-e \vec{E}=(e v B-e E)\vec{e}_y=0 \; \Rightarrow\; E=v B$$
and thus
$$\sigma=4 \pi \epsilon_0 E=4 \pi \epsilon_0 v B.$$
The electric field has an electric potential
$$V=-E y$$
and the voltage at $y=b$ thus is $V_b=-E b=-v b B$.

There's no electromotive force anywhere in this problem, because that's usually defined as
$$\mathcal{E}=\int_C \mathrm{d} \vec{x} \cdot (\vec{E}+\vec{v} \times \vec{B}),$$
where $C$ is a closed curve moving described by the velocity field $\vec{v}(t,\vec{x})$. I don't know, where such a thing should be useful in this physical situation. It usually occurs in the correct integral form of Faraday's Law for moving surfaces:
$$\mathrm{d}_t \int_F \mathrm{d}^2 \vec{f} \cdot \vec{B}=-\int_{\partial F} \mathrm{d} \vec{x}(\vec{E}+\vec{v} \times \vec{B}).$$

 

[etotheipi]

Yes I agree with @vanhees71, it's often mis-stated (even in high-school textbooks) that you get an "EMF induced in the rod", however this is a misleading statement probably arising by confusing potential difference and EMF.

As nicely shown in #20, for this example with the bar, it only makes sense to discuss the potential difference across the bar, and nowhere is the concept of an EMF (a closed curve line integral of the Lorentz force per unit charge) present.