# Induced EMF of a spinning metal rod [closed]

Hey everyone, I'm currently taking an Eletricity and Magnetism course and I'm having dificulty answering this problem. I've gone to my tutor, but sadly, he hasn't taken E+M for a while now and couldn't recall off the top of his head how to go about solving this. We went through my book together, but couldn't find any examples pertaining to this type of problem. (We did find examples of induced EMF's obviously, but making the leap to a spinning rod seems to be too great a task)

Anyways, here is the problem I'm working on:

A metal rod is spinning on an axis that is located in the middle of the rod, and perpendicular to the rod's length. If the rod is in an external magnetic field that is parallel to the axis the rod is spinning on, how would I go about finding the induced emf?

The rod's thickness is negligible but has length L. It's spinning at 1500 rev/s and the magnetic field is 7.5T

Any pointers would be appreciated!

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– Mark Eichenlaub May 3 '11 at 6:00
@mark Thanks for your reply. As I stated above, I'm having trouble solving for the case of a rod spinning. I've worked through examples where the focus is on a point charge, but not on a rod that is rotating. – Jacob Patel May 3 '11 at 6:18

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Did you read the link I gave you? In general, you should try to ask a question that clarifies some physics point, or some technique of problem solving, etc. It's good to ask about a spinning rod if you have some reason in mind that the situation might be different and require some new physical insight compared to other questions about induction. I don't see how that's the case here. As such, I've tried to give an answer about induction in general. – Mark Eichenlaub May 3 '11 at 6:56
Wait, the axis of rotation is perpendicular to the rod's length? Or parallel? – Andrew May 3 '11 at 11:40

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## 4 Answers

It is zero.

Faraday's law states that the induced EMF around the boundary of a surface is proportional to the time derivative of the flux through a that surface.

Therefore, when the magnetic field is constant in both space and time there is no electromagnetic induction.

There would be a Hall effect. Its magnitude would depend on the material, especially the density of charge carriers.

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how does a homopolar generator work where the flux though it remains constant in space? – John McVirgo May 3 '11 at 13:09
@John There is a Lorentz force on the charge carriers. – Mark Eichenlaub May 3 '11 at 16:46
– Andrew May 3 '11 at 17:52
@Andrew I was considering EMF comes from the term in Maxwell's equation related the curl of E to the time derivative of B. – Mark Eichenlaub May 3 '11 at 17:55
Ok, so we both agree that an electric field is generated. Since the definition is not universal, it's possible that the OP isn't clear on the distinction between voltage and EMF. Heck, I'm not sure I'm clear. – Andrew May 3 '11 at 17:57
show 6 more comments

For a wire moving in a magnetic field,

emf = rate of cutting of magnetic flux

Which is one form of Faraday's Law. This is straight forward to use when all parts of the wire are moving at the same velocity. If not, you can divide the wire up into small parts and imagine each small element having a unique velocity, with an emf across it, and the sum across each element giving the total emf across the ends.

One half of the rod sweeps out an area $\pi r^2$ in a time $1/f$ giving a rate of cutting of magnetic flux and hence emf between the center and r as

$emf = Bf\pi r^2$

A more sophisticated approach is to use the Lorentz force law which needs a deeper understanding of what's going on inside the rod.

$F = e( E + v \times B)$

As the rod moves through the magnetic field, it carries with it electrons and metal ions, the former being of a much smaller mass so their motion is affected much more by electric and magnetic fields.

In your example, let $B$ point into the screen, the rod rotating clockwise about $r = 0$ in the plane of the screen. From the definition of the cross product $\times$, the magnetic force has a magnitude $evB$, and points towards r = 0 along the rod for an electron. The electrons being forced in this direction disrupts the neutrality of the rod, so creating an internal E which prevents further electron movement. So we can say that since $F = 0$ then $E = -vB$ and we've now got an internal electric field $E$ which can do work, unlike the static magnetic field B. To calculate the emf, we need the work done in moving a charge of 1 Coulomb between two points

$emf = -\int_{0}^{r}Edr = \int_{0}^{r}vBdr$

Since the speed of an electron at $r$ is $2 \pi fr$, finally

$emf = Bf\pi r^2$

Which gives the same result as before. As expected, there is an emf between the center and each end, but not between the ends.

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You are asking about a homopolar generator. The excellent UMD physics demo page shows an experimental demonstration and includes references. I'll try to find a free version of that article.

EDIT: Fitzpatrick discusses homopolar generators in his online plasma physics textbook.

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The Fitzpatrick link says "In other words, if the rotating disk is a perfect conductor then dynamo action is impossible". Surprising if true. – John McVirgo May 3 '11 at 21:41
He's talking about dynamo, where the current from the homopolar generator creates the magnetic field used in the homopolar generator. Does his reasoning apply to the case of an externally applied magnetic field? – Andrew May 3 '11 at 22:16

Use `e=B*omega*L^2/2`, where `e` is induced EMF, `omega` is the angular velocity and `L` is the length of the rod. Actually, EMF is induced due to change in area. However between the ends, it is 0.

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