Understanding Induced Current and EMF in Electromagnetic Induction

[jsmith613]

It would depend on how long the magnet was. If long enough, the field wouldn't be changing whilst the middle section was passing. Your graph would be right for a short one.

I thought it would be if the tube was long enough because once in the tube there is no change in flux linkage within the tube? The longer the magnet, I would have though, the less time it speds inside the tube so surely the less straight the middle section is??

 

[truesearch]

here is a drawing of the falling magnet as it passes through a coil. as the blue end passesthrough you will get the blue part of your trace.
as the red end passes through you will get the red part of the trace.
The size and width of each part of the trace are to do with the fact that the falling magnet is accelerating.
As the middle section of the magnet passes through there is a flux but it is not changing !
Try to picture the field lines for a bar magnet, in the centre they are parallel

 

[sophiecentaur]

The currents will be different at different parts of the tube over its length, remember.

 

[sophiecentaur]

The currents will be different at different parts of the tube over its length, remember.

 

[sophiecentaur]

The currents will be different at different parts of the tube over its length, remember.

 

[sophiecentaur]

The currents will be different at different parts of the tube over its length, remember.

 

[jsmith613]

The currents will be different at different parts of the tube over its length, remember.

I think people are missing the question I had:

if the tube was A LOT longer and the magnet was v. short would the emf be 0 for all the time the magnet was in the LONG tube?

 

[sophiecentaur]

I think people are missing the question I had:

if the tube was A LOT longer and the magnet was v. short would the emf be 0 for all the time the magnet was in the LONG tube?

What EMF where?

 

[jsmith613]

What EMF where?

the EMF induced in the tube?
if I drop a magnet into the tube for the entire time the magnet is in the tube, won't the net emf be zero and hence our graph would have a curved ends with a long flat bit in between
a little bit (but not exactly like:)

/\_________________
........\/

Ignore the ... they are there just to position the V

(this is an emf-time graph)

 

[cabraham]

Is an induced current (conventional current) in the same direction or in the opposite direction to the induced emf.
I ask this in relation to electromagnetic induction. We can predict the direction of the current using the right hand rule but how do I know the direction of emf?

Another question, related to this, is how could an EMF be induced (and hence a current be induced) in a strip of wire, such as an aerial, if it is not in a complete circuit? Surely the whole point of EM induction is to oppose the change in magnetic flux. If a current cannot be induced, how is this done? For example when a car moves forward, an EMF is induced across the ends of the aerial BUT there is no complete circuit

Thanks a lot guys!

Good question. Is the induced current in the same direction as the induced emf? My answer would be that Ohm's law is always in effect. The emf will be oriented in relation to the current and impedance per Ohm's law. For a resistance, positive current enters the positive terminal of the resistance. The positive current enters the positive emf terminal and exits the negative emf terminal. In a resistor, charge carriers are losing energy due to collisions with the material lattice structure.

In an aerial, current and voltage are induced even though the circuit is not a closed loop. This perplexed investigators in the 19th century until Maxwell added a displacement current term to Ampere's Law and published it in 1873. Though an open circuit, the E & H fields incident on the antenna exert forces on the electrons in the wire. The electrons move through the wire. Before they get to the other end the polarity reverses and the electrons reverse direction. For high enough frequencies, the circuit is not a true open, but an infinitessimal series of resistrors, capacitors, conductors, and inductors.

As long as the wavelength of the E/H fields is short compared with the antenna length, a substantial current can exist. The impedance of the antenna actually looks like a pure resistance. With capacitors, displacement currents exist, and the current leads the voltage by about 90 degrees. In an antenna, displacement current exists with about 0 degrees of phase shift between I & V.

The key to understanding induction is that it only works in the ac domain, i.e. time-changing flux (or motion plus static flux which is equivalent). In the ac domain, there is no such thing as a perfect open or short. A superconducting loop has inductive reactance and an open loop has capacitance. Thus current and voltage are both induced together. You can't have one w/o the other. Make sense? I hope I helped.

Claude

 

[jsmith613]

Make sense? I hope I helped.
Claude

How can the polarity suddenly switch direction.. all previous posts seem to suggest the emf CAN exist without the current??

 

[sophiecentaur]

I'm not sure whether this needs clearing up but, for the original falling magnet situation: The RH rule, as we first come to it, is about a wire with two ends across which you measure an emf. When you have a loop or a tube the situation is a bit different. The direction of the induced emf is around the circumference of the loop (a short circuited ring) rather than along the length of the tube (Motion, EMF and Field are all at right angles). The emf and hence the current around the circumference will vary along the length of the tube - having maxima, clockwise and anticlockwise at the places where the poles happen to be passing. The maximum magnitude of the current will depend on the resistivity of the metal tube.

 

[cabraham]

How can the polarity suddenly switch direction.. all previous posts seem to suggest the emf CAN exist without the current??

If the signal generator driving the antenna is ac sine curve, it switches direction every half period. An antenna only works at ac, and high frequency. At dc and/or too low a frequency it is ineffective.

In ac domain, emf CANNOT exist w/o some amount of current. A displacement current exists if the path is open due to capacitance. If the frequency is very low, the current is very low, nanoamp, picoamp, or even less. As frequency increases so does the displacement current. It's called displacement because the path is open, it doesn't circulate in a full loop.

But during the transit through a length of conductor, it behaves just like conduction current. An H field encircles the wire regardless whether it is displacement or conduction current.

An emf induced by a time-changing field is time-changing itself, unless the magnetic field is an infinite ramp, quite rare. The emf is produced by moving charges through the conductor. Lorentz force provides the nmeans of moving said charges. As the charges move, the emf develops. But charges have to move to generate emf, which defines current. To get an emf you need to displace charges which is a current. Again, if the frequency is really tiny, like 0.003 Hz, the current is very small, but the emf wouldn't be without this tiny current.

In time changing conditions, current and voltage cannot exist separately. Likewise for E & H.

Claude

 

[jsmith613]

...

Claude

it seems you missed the point of the question.
I was not referrering to a radio signal
I was referring to the car moving in a magnetic field where the mag field is CONSTANT
hence the cutting of the flux induces emf...surely in this situation no current would flow but an emf would still exits?

 

[sophiecentaur]

the EMF induced in the tube?
if I drop a magnet into the tube for the entire time the magnet is in the tube, won't the net emf be zero and hence our graph would have a curved ends with a long flat bit in between
a little bit (but not exactly like:)

/\_________________
........\/

Ignore the ... they are there just to position the V

(this is an emf-time graph)

I have just re-read this. What did you mean by "in the tube"? The emf will be different at all points on the tube. The Mean EMF (in time and distance) will be zero, of course - or there would be an endless DC component.

We know there is a braking effect (practical evidence) and this 'proves' that the current must be different at all parts of the tube. Why? Because a field that attracts the N pole must be the opposite to a field that attracts the S pole. As it falls, the magnet must be encountering this arrangement of fields all the way down so the induced field must be 'chasing' the magnet and this will involve currents changing (in time and position) as it falls. The EMF to produce these currents must be just right to ensure this happens.

 

[sophiecentaur]

it seems you missed the point of the question.
I was not referrering to a radio signal
I was referring to the car moving in a magnetic field where the mag field is CONSTANT
hence the cutting of the flux induces emf...surely in this situation no current would flow but an emf would still exits?

This is correct but does not contradict anything that has been written. d∅/dt is constant in this case (as the car moves steadily through the field) so the EMF will be constant and no current would flow (load = ∞Ω).
IFFF you actually tried to connect between the ends of your car aerial, to measure or get power, you would be introducing another conductor with an EMF exactly the same as the Aerial so no current would flow. The 'line integral' around the circuit would be zero so your arrangement cannot be used as a source of DC.

 

[jsmith613]

I have just re-read this. What did you mean by "in the tube"? The emf will be different at all points on the tube. The Mean EMF (in time and distance) will be zero, of course - or there would be an endless DC component.

I mean as it is falling through the tube
two currents, in different directions, will be produced
hence cancelling - net emf is zero so graph should look like:

/\_________________
........\/
right??

 

[jsmith613]

This is correct but does not contradict anything that has been written. d∅/dt is constant in this case (as the car moves steadily through the field) so the EMF will be constant and no current would flow (load = ∞Ω).
IFFF you actually tried to connect between the ends of your car aerial, to measure or get power, you would be introducing another conductor with an EMF exactly the same as the Aerial so no current would flow. The 'line integral' around the circuit would be zero so your arrangement cannot be used as a source of DC.

thanks for this :)

 

[sophiecentaur]

I mean as it is falling through the tube
two currents, in different directions, will be produced
hence cancelling - net emf is zero so graph should look like:

/\_________________
........\/
right??

How are these two currents "cancelling"? They are in different places and at anyone time. It's like saying AC 'cancels' because there are as many positives as negatives. I still don't think you have the right picture of what's going on. Currents are flowing all over this tube in different directions and they are all changing in time. There is no one, single, value of current. You could produce a similar effect if your tube were to be replaced by a series of insulated copper rings, fixed one on top of another.

Have you heard of 'eddy currents' which are induced in iron cores of motors and transformers? They are the same phenomenon as in the tube. By laminating the core, you introduce insulating layers between the leaves of the core and eliminate / reduce these eddy currents. If you milled a slot down one side of the tube, the EMFs would still be induced but currents wouldn't flow so the magnet would fall unimpeded.

 

[jsmith613]

How are these two currents "cancelling"? They are in different places and at anyone time. It's like saying AC 'cancels' because there are as many positives as negatives. I still don't think you have the right picture of what's going on. Currents are flowing all over this tube in different directions and they are all changing in time. There is no one, single, value of current. You could produce a similar effect if your tube were to be replaced by a series of insulated copper rings, fixed one on top of another.

Have you heard of 'eddy currents' which are induced in iron cores of motors and transformers? They are the same phenomenon as in the tube. By laminating the core, you introduce insulating layers between the leaves of the core and eliminate / reduce these eddy currents. If you milled a slot down one side of the tube, the EMFs would still be induced but currents wouldn't flow so the magnet would fall unimpeded.

when the magnet falls let's say it falls with north pole first:

the entire magnet is in the tube so:

the bottom of the magnet is increases the flux for the tube (imagine the tube as circlets of wire) beneath it. so a current is induced to repel the magnet (upward). for this to happen it needs to be a north pole so current = anticlockwise

the top of the magnet is falling away and the tube needs to create a north pole (downard) so the current flows clockwise.
thus the NET current is 0 hence the net emf should be zero too, no?

 

[sophiecentaur]

But what do you mean by "net"?

 

[cabraham]

it seems you missed the point of the question.
I was not referrering to a radio signal
I was referring to the car moving in a magnetic field where the mag field is CONSTANT
hence the cutting of the flux induces emf...surely in this situation no current would flow but an emf would still exits?

A car is a conductor. How can an emf exist across the ends of a conductor, or any portion thereof, without current? In order to rearrange charges to obtain emf, the charges must move, which is current. When the charges accumulate at the conductor-air boundary they present an E field which opposes the generated E field. Current will diminish due to the cancellation of E fields. There is a current when the car enters the magnetic field, then another current of opposite polarity when the car exits.

Since the loop is open, there is no sustained conduction current. The key to understanding is to consider all laws, not just Faraday, or Ampere, or Ohm, Lorentz, or CEL (conservation of energy), but all together. All are in effect and are upheld. In order for there to be an emf on the car, charges moved through a finite resistance. Ohm, Ampere, Faraday, Lorentz & CEL all apply here.

Lorentz has not been covered but it applies. An electron in the car body feels a Lorentz force F = -q*(E + (uXB)), where "-q" is the charge on the electron, u is velocity, and E is electric field intensity, B is magnetic flux density. This force results in the electron being moved. That is current. Electrons transit towards one end of the car, holes accumulate at the other end. A potential difference develops due to this charge motion. Charge motion, is of course, referred to as "current". The developed emf is sustained as long as the car is immersed in the magnetic field. The current is not.

The current takes place while the voltage is changing. When the voltage is static, no current exists. Then upon exiting the magnetic field, the current exists as the voltage is decreasing back towards zero. The car is a capacitor.

Pretty straightforward. Anything I missed?

Claude

 

[sophiecentaur]

You are right in saying that it's all the Laws together that should be considered if you want to cover all cases.
I think that the models being used here are too mixed to come to a conclusion. If we're talking about a wire moving through a field then an EMF, initially, will cause some displacement of charges. (If the resistance of the wire were approaching, the current, however, would be near zero but the EMF would be the same. This emf is the sum of all elemental EMFs along the wire and the amount of current flowing will depend upon the capacity between the 'two halves' of the wire and the resistance. Eventually, current will stop and the PD across the ends will reach a steady state. That is the EMF that is referred to when discussing a transformer, for instance and the EMF is there even though the current has stopped flowing. A battery EMF is specifically defined as the PD with no current being taken.
I think we / you have stated all the facts. It remains to explain just how they apply in each new circumstance. There is a bit of chicken and egg with this topic.

 

[cabraham]

You are right in saying that it's all the Laws together that should be considered if you want to cover all cases.
I think that the models being used here are too mixed to come to a conclusion. If we're talking about a wire moving through a field then an EMF, initially, will cause some displacement of charges. (If the resistance of the wire were approaching, the current, however, would be near zero but the EMF would be the same. This emf is the sum of all elemental EMFs along the wire and the amount of current flowing will depend upon the capacity between the 'two halves' of the wire and the resistance. Eventually, current will stop and the PD across the ends will reach a steady state. That is the EMF that is referred to when discussing a transformer, for instance and the EMF is there even though the current has stopped flowing. A battery EMF is specifically defined as the PD with no current being taken.
I think we / you have stated all the facts. It remains to explain just how they apply in each new circumstance. There is a bit of chicken and egg with this topic.

Refer to bold print. I have trouble believing that the displacement of charges is "caused" by the emf. The emf is initially zero. Let's define what we mean by "emf". I assume we mean the potential difference between the front and back of the car, measured by placing voltmeter probes on the front grill, and back trunk lid.

The Lorentz force begins moving charges while the VM still reads zero. After charges reach the trunk and hood, the VM shows non-zero value, increasing as the car further enters the B field. How can the emf be the cause of the displacement current? The emf builds up as a result of charges being displaced. The car is a capacitor. In a capacitor, I leads V, per "Eli the ice man".

I agree with you about the chickens and eggs part. E & B, I & V, under dynamic conditions, cannot exist separately. Only static conditions allow one to exist w/o the other.

In a transformer, powered by an ac source, current never stops at all. As long as a sine curve voltage is at the primary terminals, an exciting current is in the primary, and a displacement current is in the secondary. Even unloaded the secondary has to carry a current. This displacement current is a cosine if the secondary voltage is a sine. It does not stop. An open secondary still has a capacitance.

It cannot be otherwise.

Claude

 

[jsmith613]

But what do you mean by "net"?

well there will be zero overall emf as per the graph?

so the graph would look like:

 

[jsmith613]

...
Claude

thanks for such a brilliant explanation. would it also, then, be correct to say that if the conductor changed speed there would be a current lasting for a short time before dying down (and the emf reamining constant?)

 

[sophiecentaur]

I guess my point is that, even with no wire there, if you move your observation point through the magnetic field then there is an emf. As the resistance is infinite then there is no current but the emf would still be there because the flux is still being cut by your wire probe. Here's the thing, then. If you had two wires of equal dimensions (same capacity) and different resistivities, instantly starting to move at the same speed through the field, what you are implying, I think, is that the rate of change of measured PD wouldn't just be proportional to resistivity (i.e. not proportional to the induced EMF from dΦ/dt). I am implying that it would just be proportional to RC. This all seem like straightforward bookwork and I can't see where the current is necessarily the primary factor. Don't we always say "induced emf" in these problems?

If we're talking in terms of steady state for this wire / car then the EMF / PD will still be there but the current will have stopped flowing long ago. If the EMF were not there, the charge would leak back into place - so it must still be there. (I think that's a QED, actually)

 

[cabraham]

I guess my point is that, even with no wire there, if you move your observation point through the magnetic field then there is an emf. As the resistance is infinite then there is no current but the emf would still be there because the flux is still being cut by your wire probe. Here's the thing, then. If you had two wires of equal dimensions (same capacity) and different resistivities, instantly starting to move at the same speed through the field, what you are implying, I think, is that the rate of change of measured PD wouldn't just be proportional to resistivity (i.e. not proportional to the induced EMF from dΦ/dt). I am implying that it would just be proportional to RC. This all seem like straightforward bookwork and I can't see where the current is necessarily the primary factor. Don't we always say "induced emf" in these problems?

If we're talking in terms of steady state for this wire / car then the EMF / PD will still be there but the current will have stopped flowing long ago. If the EMF were not there, the charge would leak back into place - so it must still be there. (I think that's a QED, actually)

How do you measure said emf with an instrument such as a VM? If there is no wire, how can there be an emf? There is an E field and B field, but for emf to be defined we must provide a specific path. A conductor is the said path in this example, namely the car. Does the car develop current/emf? Yes it does, and it is impossible to have either w/o the other. For the case of a static B field, the current is transient. As far as the emf continuing after the current stopped, that just happens to be a condition associated with this specific example.

If a superconducting closed loop is immersed in a magnetic field, and moved so that induction takes place, then the field is removed, a steady current remains in the loop, dc, with zero voltage/emf.

For an open, only displacement current can exist, but emf can remain indefinitely. For a short circuit, emf is needed during the transient when current is building up, but does not exist for dc steady state conditions.

The fact that the emf in the car case remains after the current has decayed is consistent with capacitive behavior. In a cap, remember that i(t) = C*dv(t)/dt. A steady unchanging v means that i = 0. For inductance, the reverse is true. The steady unchanging emf on the car does not require current to sustain itself, but did require current to establish itself.

As far as "don't we always say 'induced emf'" goes, well, I say induced current, or induced voltage, or induced current/voltage. I acknowledge that both are equally important. To say that emf is primary while current is secondary is a logical contradiction. The emf could not exist unless the Lorentz force first established a current by forcing charges to move.

Since current is absolutely needed to generate the emf, how can it be "secondary". Neither emf nor current is independent of the other. They are each equally as important, no more, no less than the other. To illustrate this, I suggest drawing a picture of the car entering the B field, locating the charges, drawing the Lorentz force vectors, and tracking the electron motion.

At the front grill and back bumper, the charges accumulate and their E field repulses incoming charges. Although the current has decayed, the Lorentz force is still there. It is Lorentz force that maintains the emf w/o the current. Lorentz force drives charges towards the 2 bumpers, and the E field due to charges built up at the bumpers counteracts said Lorentz force, resulting in equilibrium, per Mother Nature's well known tendency.

Anything else that needs to be covered can be discussed as well. I believe I've included all pertinent facts. Comments welcome.

Claude

 

[sophiecentaur]

I think this boils down to how we choose to see things. I am happy with an emf existing even though there is no current flowing. You are happy with the current needing to flow before the emf can be measured. But EMF has units of Joules/Coulomb which is independent of the number of coulombs that have actually flowed (Current times time). I get the same answer no matter how many Coulombs have flowed but you need some current to flow and are bringing in the microscopic (electrons) into your reasoning. That seems to be going somewhere that's not necessary. (Did Maxwell use electrons or even charges with mass?) I do follow what you say, though, and it's quite reasonable.

I say tomayto and you say tomarto.

 

[jsmith613]

I think this boils down to how we choose to see things. I am happy with an emf existing even though there is no current flowing. You are happy with the current needing to flow before the emf can be measured. But EMF has units of Joules/Coulomb which is independent of the number of coulombs that have actually flowed (Current times time). I get the same answer no matter how many Coulombs have flowed but you need some current to flow and are bringing in the microscopic (electrons) into your reasoning. That seems to be going somewhere that's not necessary. (Did Maxwell use electrons or even charges with mass?) I do follow what you say, though, and it's quite reasonable.

I say tomayto and you say tomarto.

just to go back to an earlier question. if the car speeds up will a current be induced for a fraction of a second before the new emf is established?