# Induced emf when accelerates in uniform magnetic field

Hi everyone,

This question is bothering me for days and would hope to receive some help in this forum. This question is with regards to Faraday's law on EMI.

Will an emf be induced when a copper rod is accelerated between a uniform magnetic field?

My answer would be no if the rod has an uniform area and moving inside an uniform field at a constant direction (angle) as there will be no change in the flux linkage in the rod as BAN will remain constant even though it is moving with increasing speed.

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Philip Wood
Gold Member
I take it that you mean that the rod is moving in a direction at right angles to itself and cutting lines of magnetic flux. In this case there will be an emf induced in it, given by BLv, in which B is the component of magnetic flux density at right angles to the area swept out by the rod (length L) and v is its speed at any instant. So the formula applies at any instant, whether the rod is moving at constant speed or accelerating.

This formula may be derived by considering the motor effect force on the free charges in the rod, as they move through the field because they're carried along inside the moving rod.

Another approach is to consider the rod to make contact with stationary rails, which are connected together at one end by a resistor. Thus there is a complete circuit, whose area changes as the rod moves. Because the area changes, the flux linking the circuit changes, and we can find the emf by equating it to the rate of change of flux linkage.

We get |emf| = d$\Phi$/dt = d (BA)/dt = B dA/dt = BLv.

Reading your post, I think you may be slightly confused about flux linkage. You apply the idea to a circuit, of which the rod is part, as I've explained, not within the rod by itself.

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Short answer: yes, emf will be induced

Long answer: it depends on the relative orientation of the magnetic field and the conductor and the direction of relative motion between the two...look up "right hand rule".

For as long as there is relative motion, emf will be induced....if is constant acceleration or not or constant speed or not, it will still induce emf, but at different rate.

The cross section of the conductor has no bearing on the initial emf induction...it may come into play later once current start flowing depending on the emf and the conductors electrical resistance (depends on cross sectional area)

After the initial motion, there will be no emf. Charge will build up on both ends, causing an electric field inside the conductor such that E + v x B = 0. Thus, the net force on each charge will be zero. Therefore, the emf will be zero as well since it's the integral of the force per unit charge.

Philip Wood
Gold Member
Would you say, then, that a generator being turned but with no load across its terminals, or, for that matter, a battery with no load connected to it, has zero emf ? I would say that they do have emfs. It's the way I use the term 'emf'. But I don't want to make an issue of this and spoil what I think is a good and useful thread. Calls for a new thread, imo.

The OP has a very explicit question: would emf be induced when moving the conductor?

The answer is 'yes'....whether you take advantage of this induced emf or not, it's another matter.

As Philip says, this happens in generators when they are being turned but there is no load...there exists what is called Open Circuit Voltage.

Hi Everyone,

Thank you for the comments. I would agree with Philip that for a rod moving in an magnetic field, the governing formula would be Blv which would justify that an emf will be induced when it moves at any speed across a magnetic field.

Can I ask then if this applies if it were to be a <b>circular coil</b> of wire that is falling and accelerating through a long magnet with an <b>uniform magnetic field</b>? Will there be a period when there will be no emf induced as using the formula of BAN, the flux linkage will be constant when the coil of wire is within the magnet even though it is accelerating inside the uniform field.

Greatly appreciated. Thank you.

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Philip Wood
Gold Member
Spot-on, gkw. That's exactly right. If the ring is co-axial with the magnet, and dropped from above the magnet, there will be an emf induced as the ring cuts the lines of flux splaying out from the top pole. In the central region of the magnet almost all the flux is contained in the magnet itself, with very little field strength outside the magnet for the coil to cut, but cutting will start to happen again as the ring approaches the bottom of the magnet. The peak emf will then be in the opposite direction (but higher, because the ring is moving faster).

I find it rather simpler to think in terms of flux linking (passing through) the coil. An emf is induced as this flux changes. Maximum rates of change are clearly when the ring is near the top and bottom of the magnet. Once the ring has reached the central region of the magnet, there's lots of flux linking it - as lines of magnetic flux are continuous closed loops which pass through the magnet from South pole to North, sprout out from the North end, and return through the air to the South - but there's no emf because the flux through the ring isn't changing.

Here is another one for you...reverse the role of the two pieces involved...

what I mean, take a magnetic ring ( a magnet in the shape of a ring ) and drop it co-axially to a copper rod....what happens? how fast does the magnet fall?

if you cannot get a hold of a ring magnet...you can also simply get any small magnet and drop inside a copper pipe ...what happens?

Philip Wood
Gold Member
The bar magnet dropping through the copper pipe is essentially the same as the copper ring dropping over the co-axial magnet.

But I see you're now asking about the speed of fall. In both cases the relative motion is affected by a retarding force whenever the lines of flux are cut (i.e when the ring passes the end regions of the magnet, or the magnet passes through the pipe. This retarding force arises from the motor effect: because a current flows in the copper, it is a current-carrying conductor in a magnetic field, and experiences a motor effect (BIL) force. [The magnet experience an equal and opposite force.]

The interesting case is the ring-shaped magnet dropping over the copper cylinder. It's the sort of question that gets physicists arguing! I'm certainly not confident about my answer.

If the magnet is a closed ring, the lines of flux will be closed loops running round the ring and confined inside it. There will be almost no magnetic field outside it.

To predict whether or not an emf will be induced in a circuit the safest concept to use is that of flux linkage. Think of the circuit as a link in a chain. Think of the lines of magnetic flux as another link. If the links don't interconnect, no flux is linking the circuit. If the links pass through each other, then flux is linking the circuit. When the flux linkage changes an emf is induced in the circuit.

The difficulty with your case is that the copper bar doesn't form part of a circuit. [Sometimes currents can find their own paths in a lump of metal, and we have so-called 'eddy currents', but I don't think this will happen here.] Instead, let us imagine the copper rod to be one vertical side of a square loop of conducting material. [The square will be in a vertical plane.] Clearly the horizontal sides will prevent us from dropping the ring-shaped magnet over the copper rod. So we'll have to make them of some fluid material, ionised gas, say, so that the magnet can pass through the top conductor and over the copper conductor and out at through the bottom side.

So there will be a change of flux linking the circuit when the ring magnet drops through the top conductor, and when it drops out out through the bottom conductor, and as this happens there will be an induced emf in the square circuit.

It's my hunch that even without the other three conducting sides, we will have an emf in the copper rod as the ring magnet drops past its ends.

This one could run and run...

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The bar magnet dropping through the copper pipe is essentially the same as the copper ring dropping over the co-axial magnet.
I am not so sure...you said yourself...when the magnet is a rod, there may not be many magnetic lines right outside the magnet and so, a falling ring of wire may not experience any field...let alone a field change...the direction of the possible magnetic line and the direction of motion would be aligned and so, there wouldn't be any change in field nor induction...only at the ends where the magnetic lines come out of the magnet to flair out and start turning around...

On the flip side, though...

If you have a small magnet, where both ends are rather close together, the magnetic lines basically surround the entire magnet...it you drop this in the inside or outside of a copper pipe, a portion of these lines will be horizontal and perpendicular to (vertical) falling direction and always be going by a 'delta' section of the rod/pipe that was not expose to the magnetic field before and, so, as the 'delta' section goes from zero field to some field, induction occurs.

And yes, eddy currents...that's exactly the reason why the falling speed is slower than free falling....no perpetual motion here...the induced eddy current's field is going to oppose the 'source' field and hence, slow down the falling magnet.

Philip Wood
Gold Member
gsal
In the (misguided) interests of brevity I conflated two issues. What I should have said is that (1) the bar magnet dropping through the ring produces the same emf sequence as the ring dropping over the magnet, and retarding forces are produced when the emf is produced, i.e. as the magnet enters and leaves the ring. (2) If the magnet is dropping through a copper pipe, the pipe can be considered as a stack of copper rings, so even while the magnet is well inside the pipe, emfs are being induced in rings near the magnet's poles (as that's where the flux cutting is done), so there's a retarding force throughout the magnet's passage through the pipe.

But what about the ring magnet dropped co-axially over the copper rod?

Philip:

Hhhmmm...I am not sure I agree 100% with your point (1), and I quote:

the bar magnet dropping through the ring produces the same emf sequence as the ring dropping over the magnet, and retarding forces are produced when the emf is produced, i.e. as the magnet enters and leaves the ring.
I did not think these two were the same.

(a) A bar magnet much shorter than the copper pipe will have both ends moving along the inside of the pipe and inducing currents in every one of those "delta" rings of pipe...I can see that just fine.

(b) But a copper ring dropping over a long rod magnet? This is where I am not so sure how many magnetic lines the copper ring is going to see on its way down...I thought the rod magnet by its virtue was going to keep the lines inside of it...if any thing, it would be the returning lines coming back through the outside, but then, like I said before, this lines are parallel to the direction of motion of the copper ring.

But what about the ring magnet dropped co-axially over the copper rod?
To me, this is what is similar to my (a), above, just flipped inside out.

Of course, I am envisioning a rather convenient ring magnet that looks like two washers put together where one of them was purely north magnetic pole and the other one purely south pole and when put together make a magnet. Then, as the magnetic lines try to go from one side of the magnet to the other some would go around through the outer radius but some will hopefully fit through the inner radius....this line will have a piece of them in the horizontal direction and will cut the "delta" disk in the rod and hence induce currents there...this "delta" disk in the rod, actually, will probably behave very much like a pipe and the center will not have much current...I am not sure, it all depends on how thick the skin depth is under those conditions...

how does this sound to you?

Philip Wood
Gold Member
Hallo Gsal.
"Hhhmmm...I am not sure I agree 100% with your point (1), and I quote:

"the bar magnet dropping through the ring produces the same emf sequence as the ring dropping over the magnet, and retarding forces are produced when the emf is produced, i.e. as the magnet enters and leaves the ring.
I did not think these two were the same.""

But surely it's a case of relative motion: the magnet falling through the ring is equivalent to the ring falling co-axially over the magnet?

Your point (b): I think we're both saying the same thing: no cutting when the ring is falling over the central region of the magnet:

" If the ring is co-axial with the magnet, and dropped from above the magnet, there will be an emf induced as the ring cuts the lines of flux splaying out from the top pole. In the central region of the magnet almost all the flux is contained in the magnet itself, with very little field strength outside the magnet for the coil to cut, but cutting will start to happen again as the ring approaches the bottom of the magnet. The peak emf will then be in the opposite direction (but higher, because the ring is moving faster)."(Post 8)

As for the ring -shaped magnet, you have in mind quite a different magnetisation from what I was envisaging! I wonder what the OP had in mind. Yours, I bet!

I see now, and yes, we are saying the same thing. Good.

Hi Philip and gsal,

Thank you for the active discussions from which I have become clearer in concept. Fundamentally, BAN would be the governing formula. To me, the challenge would be to ask and answer whether
1) Is the magnetic field uniform throughout the motion?
2) What is the area of flux linkage?

If there is no change in the flux linkage during the motion (either no linkage or at constant linkage), then no emf will be induced.

Thanks again!

It would be best if you could upload a picture (or point to an existing one) of exactly what configuration you have in mind...

Hi, recently I'm confused whether EMF will be produced by crossing the magnetic field or change in magnetic flux linkage.

My teacher said that if there is a metal ring which is "submerged" completely in uniform magnetic flux, neither it moves vertically nor horizontally will not induce EMF, because the magnetic flux is not changing. But if the metal ring is rotating, EMF will be induced because the magnetic flux is changed. That means EMF will be induced only if the magnetic flux changes.

So, how about if the metal ring is replaced with a straight wire? I mean the straight wire is also "submerged" completely in the uniform magnetic field and is moving at certain velocity which is perpendicular to the magnetic field. Is magnetic flux changed?

For my point of understanding: "When we are crossing a magnetic field with a conductor, we are actually bringing the electrons inside the conductor to cross the magnetic field in which the velocity of the electrons are the same as the velocity of the conductor. Then the direction of magnetic force acted on the electrons can be determined by using the Fleming Left Hand Rule, creating potential difference across the conductor. then current will flow and the induced current will be at the opposite direction of the magnetic force, just like the Hall Effect." Is this statement correct?

If so, Then the metal ring which is crossing the magnetic field will also induce EMF, doesn't it? Same as the moving conductor. If not, then what is actually meant by induced EMF?

Furthermore, if there is current flow in the conductor, that means the electrons inside the conductor are moving at drift velocity in the magnetic field. Then force will be induced and I found that the direction of the force is opposite to the velocity of the moving conductor.

I have to get the concept right, so I can't really understand what is meant by Emf=-(change in magnetic flux linkage). Any suggestions?

Could it be that when replacing the metal ring with a straight wire what you are basically doing is replacing a small totally-embedded ring with very large one that has only a section embedded in the field and the rest of the loop outside the magnetic field and, so, when you move the straight wire, you are actually changing the flux linkage inside the large "ring"? http://www.physics.upenn.edu/courses/gladney/phys151/lectures/lecture_mar_17_2003.shtml" [Broken].

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Philip Wood
Gold Member
Ivy. If I may say so, you already have a better understanding of Electromagnetic Induction than many students ever achieve!

Re your first paragraph... Emf = rate of change of flux linkage always works, whether the change in flux linkage is produced by a conductor cutting lines of magnetic flux, or there is no obvious 'cutting', as when the flux density changes in the core of a transformer, and an emf is induced in the secondary coil. To apply the rate of change of flux idea to a moving conductor you need to think of the conductor as part of a circuit; if the conductor moves in such a way that the area of the circuit changes, the flux linking the circuit changes. See post 2 of this thread (where I discuss a conductor sliding on rails joined by a resistor at one end).

What I've just said addresses (I hope) your third paragraph [which I'd like to quote, but need a lesson!] Return now to your second paragraph... What your tutor says is, of course, perfectly correct. Turning the loop about a diameter does change the flux through it, as flux = BA cos($\psi$) in which $\psi$ is the angle between the field direction and the normal to the plane of the loop. But moving the loop in the field without changing its orientation doesn't change the flux through the loop, as BA cos($\psi$) stays the same, so no emf is induced in it.

What particularly impressed me about your post was the fourth paragraph, where you explain that the emf induced in a moving conductor arises from the motor effect. You are absolutely right. In a single straight piece of wire moving at speed v at right angles to the magnetic field the force on each charge carrier due to the motor effect is Bqv. Charge carriers (electrons, say) will pile up at one end of the conductor, leaving the other end oppositely charged. But this piling up will produce an electric field E, which exerts an opposite force, qE, on the charge carriers from the magnetic one, so the piling up will stop when qE = Bqv, so E = Bv. But E = p.d. across ends of wire/ length of wire = V/L, so the induced emf is BLv. This is exactly the same result as we get by considering the rate of change of flux linking a circuit of which the conductor is part, when the rest of the circuit stays still, so that the circuit area changes when the conductor moves. ['Rails' argument in post 2.]

But if we can understand from the motor effect the emf in a moving conductor, why do we need arguments about flux linkage? Well, it's sometimes easier to think this way, as for the ring moving in the uniform field. Also, as I said earlier, e = - dphi/dt also works when the flux changes due to flux density changing, without obvious cutting of flux. One equation deals with both phenomena. Some would say that this is the main aim of Physics: to explain many phenomena using a minimum number of basic laws.

Oh, I see. So you means that the induced EMF in a moving conductor is actually the same as the change in magnetic flux linkage of the whole closed system, isn't it? I mean BLv=-(change in magnetic flux linkage)? Just a rough assumption as their product are the same, that is induced EMF, or maybe only applicable for certain circumstances, I think.

Like what you said, yup, I'm trying to explain this phenomena by using the minimum number of basic law. I just can't accept that why the Faraday's law made such statement (it seems that the formula is not completed). And of course, Faraday's law are more general to state how EMF is induced in magnetic field.

Anyway, I just need to accept it by now, and I'll study it more deeper afterwards.

Philip Wood
Gold Member
Oh, I see. So you means that the induced EMF in a moving conductor is actually the same as the change in magnetic flux linkage of the whole closed system,
Yes, though you mean rate of change of magnetic flux linkage, of course.

Yes, build the moving conductor into a circuit such that the circuit area changes as a result of the conductor moving, then emf = -dphi/dt.

And, as I said, emf = -dphi/dt also applies even when there are no moving conductors, but a stationary circuit through which the flux changes, as a result of changes in flux density.

Good luck!