Induced EMF due to motion of a wire perpendicular to a magnetic field

[phymath7]

Homework Statement

A 2 metre long straight wire is being pulled with a speed of 30m/s perpendicular to a magnetic field of $180×10^{-6}T$.What is the amount of induced e.m.f due to this motion?

Relevant Equations

$\epsilon =-\frac{d{ \vec A \cdot \vec B} } {dt} , \text{ } \epsilon=$induced e.m.f ,$\text{ }$$\vec A=$Area vector ,$\text{ } \vec B=$Magnetic field

This question appeared in a university entrance exam.Basically, if magnetic flux passing through a surface of a loop changes over time ,only then e.m.f will be induced to that loop.But here only a straight line is used and there's no chance of forming any area.So by definition there's no chance of induced e.m.f production.Hence the question is wrong.Am I right?

 

[BvU]

Do you know about the Lorentz force ?

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[phymath7]

Do you know about the Lorentz force ?

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Yes I do.But that applies only when both electric and magnetic field are present.But here's no electric field.

 

[vanhees71]

No, there's a force on charges also if there's only a $\vec{B}$-field present.

Hint at your problem: What's most probably meant is to calculate the EMF measured by a Volt meter attached to the ends of the wire.

 

[phymath7]

No, there's a force on charges also if there's only a $\vec{B}$-field present.

But that is when a charge is moving or current is flowing through a war.But neither of them is the case here.Also magnetic force and ordinary force are different.

 

[Gordianus]

Search for "motional emf"

 

[phymath7]

Search for "motional emf"

No match to that one.Perhaps,you haven't read my details.

 

[vanhees71]

Integrate Farday's Law
$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}$$
over a time-dependent surface $A$ with boundary $\partial A$, use Stokes's theorem and take the time derivative out of the integral, not forgetting that the area is time dependent, you get the general Faraday's Law in integral form,
$$\mathrm{d}_t \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B} = -\int_{\partial A} \mathrm{d} \vec{r} \cdot (\vec{E} + \vec{v} \times \vec{B})=-\mathcal{E},$$
where $\vec{v}(t,\vec{r})$ is the velocity field of the the surface (here rather of its boundary).

It's pretty intuitive: Moving your wire through a magnetic field there's a force on the conduction electrons $-e \vec{v} \times \vec{B}$.

Now take the area as spanned by the wire as one side and complete it with an arbitrary rectangle fixed in space to evaluate the electromotive force, $\mathcal{E}$.

 

[Steve4Physics]

Hi @phymath7.

The wire (of length $L$ and speed $v$) sweeps across an area $Lv$ each second: $\frac {dA}{dt} = Lv$.

Note that you don’t need a full circuit or a flowing current to get an emf. Think of the voltage between the plates of a capacitor when the capacitor is connected to a battery and current has stopped flowing.

What happens in the moving wire is very similar to what happens with the capacitor. (But instead of a battery, the conduction electrons are ‘pushed’ by the ‘magnetic force’ (F = Bqv).) We end up with an excess of electrons on one side of the wire and a shortage of electrons on the other side.

 

[vanhees71]

That's also known as the "Hall effect". You can take it into account in the standard constitutive equations by taking into account the magnetic force on the conduction electrons in Ohm's Law.

 

[phymath7]

@Steve4Physics Thanks for your effort. But how can yoy say that $\frac {dA}{dt} = Lv$ ? Both L and v is in the same direction and the wire is one dimensional. Also the magnetic force applied to the wire is perpendicular both to the magnetic field and velocity.So the voltage difference would only occur if the wire had an additional dimension in the direction of magnetic force,making the wire a 'Two dimensional or a plate like conductor'. But here's not the case!

 

[BvU]

Both L and v is in the same direction

Does the exercise wording explicitly say that, or do you make that up ?

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[Steve4Physics]

@Steve4Physics Thanks for your effort. But how can yoy say that $\frac {dA}{dt} = Lv$ ? Both L and v is in the same direction and the wire is one dimensional. Also the magnetic force applied to the wire is perpendicular both to the magnetic field and velocity.So the voltage difference would only occur if the wire had an additional dimension in the direction of magnetic force,making the wire a 'Two dimensional or a plate like conductor'. But here's not the case!

Maybe this is the source of the confusion...

The question (Post #1) is ambiguous. It says the wire is “being pulled”. This could mean the wire’s velocity is ‘along’ the wire. If this is the intention, then you are correct; no flux is cut and the induced emf is zero.

But I’m assuming the wire is being pulled ‘sideways’. For example:
Wire is oriented parallel to x-axis.
Wire’s velocity is parallel to y-axis.
Direction of magnetic field is parallel to z-axis.

Does that help?

 

[phymath7]

But I’m assuming the wire is being pulled ‘sideways’. For example:
Wire is oriented parallel to x-axis.
Wire’s velocity is parallel to y-axis.
Direction of magnetic field is parallel to z-axis.

Does that help?

Yeah that helps! So the induced emf here is the Hall Voltage and the process goes the same as for a plate like conductor. Also there is no electromagnetic induction effect.

The wire (of length L and speed v) sweeps across an area Lv each second

That's why I think the above-mentioned fact doesn't have got to do anything with the question. Am I right?

 

[Steve4Physics]

@phymath7, I assume you are now considering the case where the wire's velocity is perpendicular to the wire (e.g. wire parallel to x-axis and velocity parallel to y-axis).

So the induced emf here is the Hall Voltage and the process goes the same as for a plate like conductor.

It is my understanding that the Hall effect applies to a conductor already carrying a current and placed (not necessarily moving) in a magnetic field. Like this
https://www.researchgate.net/profile/Tomas-Scepka/publication/301549615/figure/fig22/AS:669394711609369@1536607544596/7-Schematic-view-of-the-bar-n-type-Hall-device-The-Hall-voltage-is-generated-due-to.ppm

So in this particular question, I would not describe the emf as a Hall voltage.

(But I am happy to be corrected by anyone more familiar with the terminology.)

Also there is no electromagnetic induction effect.

I would describe the process as electromagnetic induction!

My understanding is that electromagnetic induction can be thought of as occurring when lines of magnetic flux are 'cut' - which is the case here.

(There may be a terminology issue here; again I am happy to be corrected by anyone more familiar with this.)

That's why I think the above-mentioned fact doesn't have got to do anything with the question. Am I right?

No sure which 'above-mentioned fact' you are referring to. However, I hope my comments have helped to resolve the problem.

 

[phymath7]

However, I hope my comments have helped to resolve the problem.

Yes it has.

So in this particular question, I would not describe the emf as a Hall voltage.

I think terminology doesn't affect so much.

My understanding is that electromagnetic induction can be thought of as occurring when lines of magnetic flux are 'cut' - which is the case here.

Can you please explain how lines of magnetic flux are supposed to be cut in this case? It's just a straight wire. I don't see any possibility of that. However, I found the most convincing solution using the concept of magnetic force acting on the conduction electrons creating a voltage between two ends of the wire. And even if you would like to describe it in terms of electromagnetic induction, I didn't find it convincing.

 

[Steve4Physics]

Can you please explain how lines of magnetic flux are supposed to be cut in this case? It's just a straight wire. I don't see any possibility of that. However, I found the most convincing solution using the concept of magnetic force acting on the conduction electrons creating a voltage between two ends of the wire. And even if you would like to describe it in terms of electromagnetic induction, I didn't find it convincing.

Here are a couple of videos which should help:




There are plenty of other explanations available. E.g. look for them using search-terms such as “induced emf cutting flux”.

The “flux cutting” view and the “force on conduction electrons” view are mutually equivalent. Though this may not be immediately obvious because the former is a ‘macroscopic’ view and the latter is a ‘microscopic’ view.

 

[phymath7]

@Steve4Physics Thanks a lot for your consistent help. Now it"s clear to me that the secenario can be explained both by induction and magnetic force.

 

[vanhees71]

Here are a couple of videos which should help:




There are plenty of other explanations available. E.g. look for them using search-terms such as “induced emf cutting flux”.

The “flux cutting” view and the “force on conduction electrons” view are mutually equivalent. Though this may not be immediately obvious because the former is a ‘macroscopic’ view and the latter is a ‘microscopic’ view.

Of course both "views" are equivalent and you can also describe the Lorentz-force-point of view macroscopically, treating the conduction electrons as a quasi-free gas moving in the background of positively charged ions forming the wire.

The consistency of both views follows from the fact that you can derive the Lorentz-force-law by using energy and momentum conservation of the complete closed system consisting of the charged fluid and the electromagnetic field. This is so, because the equations of motion for the fluid are equivalent to the conservation of energy and momentum. For the simpler model of "charged dust", see Sect. 4.4 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf