Understanding Induced Current and EMF in Electromagnetic Induction

[sophiecentaur]

I disagree. The author presents his claim that E and B are there with or without a conductor. I agree with him so far. Then he writes the closed path line integral of E*dl, which is voltage, and relates it to B and area. But the integral E*dl, requires a specific path to have a value. It is a closed loop integral.

If the voltage is there in an imaginary closed loop in space, will charges in free space circulate in said loop? I don't think so. If a voltmeter, VM, were to have its probes placed in 2 points in empty space, would it read a non-zero voltage in the presence of a non-zero E field? What do you think?

The author correctly points out that E & B are there even in empty space. He then concludes that emf which is the integral of E*dl, must also be non-zero, which makes me wonder. In a physical conductor immersed in said fields, charges would move. In space they move, but not in a closed loop like they would in a conductor. Lorentz force is there, and the charges in free space do indeed move, but the path changes. The emf due to varying fields is path dependent. Since the path taken by free electrons in space differs from that taken in a conductor, the voltages are not equal.

To say that a voltage exists in free space can be supported by Maxwell et al. But I don't think it is the same value as the case w/ a conductor because electrons would move along a different path, and voltage value is path dependent. A CRT is an example. Two parallel plates have a charge, and an electron beam is projected between the plates. The electrons get attracted towards the positive plate and away from the negative plate.

In free space, at every point between the plates, it is safe to say there is indeed a potential. But the E field here is static. I've already stated that under static conditions, current can exist w/o voltage and vice-versa. This does not negate my earlier statement involving dynamic conditions.

Same problem, but the field between the plates is ac, sinusoidal for example. The sine curve plate voltage results in a sine curve E field. There is indeed a potential in between the plates, sine curve in nature. But the plates carry a current to maintain the sine E field. Free electrons in between the plates move back and forth, which is ac current.

Dynamic conditions, i.e. time-chaning, are different than static. No ac voltage exists w/o ac current. That is my point. In free space there can be a voltage w/o current, but only static, not dynamic.

As far as joules per coulomb goes, the voltage across the car, bumperto bumper, is the joules per coulomb of charge transported from bumper to bumper. If a resistor were connected across the bumpers, large in value so that its own current generates a B field too small to cancel the external B field, then the voltage is equal to the joules of energy per coulombs transported through said resistor.

Is this clear? Do I need further clarification? BR.

Claude

I think it would because there's a field there. A probe (infinite imdedance voltmeter) would register the appropriate voltage for the field strength and its length.

Because they have mass they would not follow any resulting curved field. I think introducing electrons is really a bad idea for this reason. It's only when in a metal (with almost zero speed,) that electrons will follow curved E field lines.
But DC is only the limit of decreasing frequency there can hardly be a step change in what happens when the current is not changing (in any case, there is no such thing as real DC because it was switched on at some time and will be switched off)

A resistor would also have an equal and opposite emf induced in it so no current would flow. I made this point before in a different context.

 

[cabraham]

I think it would because there's a field there. A probe (infinite imdedance voltmeter) would register the appropriate voltage for the field strength and its length.

Because they have mass they would not follow any resulting curved field. I think introducing electrons is really a bad idea for this reason. It's only when in a metal (with almost zero speed,) that electrons will follow curved E field lines.
But DC is only the limit of decreasing frequency there can hardly be a step change in what happens when the current is not changing (in any case, there is no such thing as real DC because it was switched on at some time and will be switched off)

A resistor would also have an equal and opposite emf induced in it so no current would flow. I made this point before in a different context.

Well, let's clarify what I mean by "current must exist when a dynamic voltage exists". Fields are by their very nature distributed throughout space. An E field generated by a circuit, extends towards infinity, likewise for H. Likewise, the integral of E*dl can be defined for any path in space.

In order for a dynamic E/H to exist, there must be a current somewhere, but not everywhere. This is important. Take a 2 wire transmission line. An E field exists between the wires, plus some fringing, and the H field encircles each wire, extending towards infinity. An E field exists in the space between the 2 wires. But the current is confined to the interior of the 2 wires.

Thus we can state that even under dynamic conditions, that if we confine our region of examination to a space in between the 2 wires, but excluding both wires, that we have E & H field present in said region. We can integrate around a closed path that does not include either wire or portion thereof. Hence one can say that there is a field and an emf in this narrow spatial region, and at the same time without a current. But remember that fields are omnipresent, but current is confined to a locale.

Outside of the wires, the influence of the current is felt in the form of fields, although the current is contained wholly within the wires. So a clarification is in order. An ac voltage in a region of space, requires the existence of ac current, but not necessarily in the same region in space. Obviously current is not present in every point in the universe, whereas E & H fields are.

Do you see what I'm getting at? In a specific spatial volume we have a dynamic E & H, but there is no current present in said volume. But there is a current nearby in the 2 wires. This current is what makes E & H possible. The emf is the integral of E over a specific closed path. But E cannot be unless current is present somewhere, but not necessarily in the region in question. The current in the 2 wire t-line is what gives rise to the proagation of E & H. That is my point. Since current is confined to a narrow region in space, while E, H, and emf are defined everywhere, most closed regions in space have E, H, & emf, without a current in the enclosed region.

I guess I should state my point as follows. A dynamic emf somewhere in space cannot exist unless there is also a current, but not necessarily in the same spatial region. Hopefully we can all agree on that.

As far as the probe registering the right voltage, I say wait a minute. The voltage value is path dependent. The voltage value displayed on the infinite impedance VM would depend on how the leads are arranged. In free space with dynamic fields present, the emf from point a to point b, is path dependent. Re-arrange the voltmeter leads and the displayed reading changes.

So in order to define the emf, one must configure the leads along a specific path in question. But in doing so we have introduced conductors, i.e. the test probe leads, into the picture. Like I said, attempting to define a free space emf w/o conductors is arbitrary and ambiguous. Frankly, it is pretty arbitrary to say that a region in space has an emf w/o defining a conductor with a specific shape.

I believe I've supported my position with solid facts. Anything unclear can be discussed further.

Claude

 

[sophiecentaur]

I think we are going round in circles here and I wonder how you have arrived at some of your opinions. I am very rusty about a lot of this but, when I re-read textbooks, it usually makes sense.

Regarding my high impedance probe. It takes the form of the shortest 'electric dipole' you can get away with and the smallest meter, in situ. Lead lengths are not considered here, although, connecting to a larger meter with a twisted wire feed at right angles to the conductor shouldn't affect things.

But I'm afraid we have hijacked this thread, which was initially about a more mundane problem, if I remember right.

 

[jsmith613]

I think we are going round in circles here and I wonder how you have arrived at some of your opinions. I am very rusty about a lot of this but, when I re-read textbooks, it usually makes sense.

Regarding my high impedance probe. It takes the form of the shortest 'electric dipole' you can get away with and the smallest meter, in situ. Lead lengths are not considered here, although, connecting to a larger meter with a twisted wire feed at right angles to the conductor shouldn't affect things.

But I'm afraid we have hijacked this thread, which was initially about a more mundane problem, if I remember right.

yes
despite how interesting your discussion is..it goes right over my head
I hate to kill a party but I was wondering if we could return to my question post #97
it is the diagram explaining what i was talking about

although I may have cracked it...I think what I am getting at is actually the NET emf is zero but a current still flows...is this a more accurate description of what is going on?

 

[sophiecentaur]

see diagram

This is your set up (just as I imagined) but I want to know where these currents you refer to are flowing (as you see it). This must involve you putting some arrows with labels on that diagram. I still can't see how you are thinking with this.

 

[BruceW]

But it becomes indeterminate. What is the value? What is the path of integration? When a conductor is present there is a definite path of integration meaning that we can compute a specific value. Without a conductor where is the path? How can we transport a charge along an arbitrary path without a conductor in place? Think about what you're saying.

If we define emf as the closed integral of E*dL, then you can choose whatever path of integration you want. That's the beauty of Maxwell's equations. An emf does not necessarily have to cause a movement of charge. Of course, you can define emf differently. I don't know what is the most commonly-used definition of emf.

 

[jsmith613]

This is your set up (just as I imagined) but I want to know where these currents you refer to are flowing (as you see it). This must involve you putting some arrows with labels on that diagram. I still can't see how you are thinking with this.

here is a labelled diagram
or a better image:

 

[BruceW]

yes, if we assume the magnet goes down smoothly, then the induced currents will circulate horizontally (which doesn't happen in practise, but it might be a useful approximation). I see where you're coming from now. So what was your question about emf? (And I'll assume we're talking about emf over a horizontal path around the tube)

 

[jsmith613]

yes, if we assume the magnet goes down smoothly, then the induced currents will circulate horizontally (which doesn't happen in practise, but it might be a useful approximation). I see where you're coming from now. So what was your question about emf? (And I'll assume we're talking about emf over a horizontal path around the tube)

well will either of them be zero for the time the magnet is in the tube?

 

[BruceW]

Were you wondering (in this idealised motion), if the currents going anticlockwise below the magnet are equal in magnitude to the clockwise currents above the magnet? I see no reason why not. The magnet produces the same amount of magnetic field on either side, so the magnetic flux change on both sides is the same.

 

[jsmith613]

Were you wondering (in this idealised motion), if the currents going anticlockwise below the magnet are equal in magnitude to the clockwise currents above the magnet? I see no reason why not. The magnet produces the same amount of magnetic field on either side, so the magnetic flux change on both sides is the same.

so you would agree then that the net EMF and current is zero?

 

[BruceW]

well will either of them be zero for the time the magnet is in the tube?

either of what be zero?

 

[BruceW]

so you would agree then that the net EMF and current is zero?

you mean that there is the same amount of current going clockwise as there is anti-clockwise? Again, I see no reason why not.

 

[jsmith613]

either of what be zero?

the current or emf

 

[BruceW]

the current or emf

They won't be zero. But as I said in my last post, I would guess that there is as much current going anti-clockwise as there is current going clockwise. The magnet is dipole, so I see no reason to break symmetry of clockwise and anti-clockwise.

 

[jsmith613]

They won't be zero. But as I said in my last post, I would guess that there is as much current going anti-clockwise as there is current going clockwise. The magnet is dipole, so I see no reason to break symmetry of clockwise and anti-clockwise.

but why won't they be zero
1 + (-1) = 0
??

 

[truesearch]

Phew! I wonder what Faraday would make of that lot!
Your falling magnet in the copper tube...the induced currents flow around the tube but the current at the bottom is induced to repel the S pole, the one at the top is induced to attract the N pole.
Both currents are induced to generate S poles... They are in the same direction.
This is the difference between the tube with a falling magnet and the (thin) coil with a magnet falling through.

 

[BruceW]

but why won't they be zero
1 + (-1) = 0
??

the charges above the magnet need to go clockwise and the charges underneath need to go anticlockwise (to slow the descent of the magnet). So there is not the same amount of charges going in either direction at all points in space.

 

[jsmith613]

Phew! I wonder what Faraday would make of that lot!
Your falling magnet in the copper tube...the induced currents flow around the tube but the current at the bottom is induced to repel the S pole, the one at the top is induced to attract the N pole.
Both currents are induced to generate S poles... They are in the same direction.
This is the difference between the tube with a falling magnet and the (thin) coil with a magnet falling through.

no but when produces a south pole up and the other a south pole down?

 

[jsmith613]

the charges above the magnet need to go clockwise and the charges underneath need to go anticlockwise (to slow the descent of the magnet). So there is not the same amount of charges going in either direction at all points in space.

so is this diagram wrong:
http://www.thestudentroom.co.uk/attachment.php?attachmentid=139647&d=1333538964

I don't want to show the image here as it is enourmous

 

[BruceW]

Phew! I wonder what Faraday would make of that lot!
Your falling magnet in the copper tube...the induced currents flow around the tube but the current at the bottom is induced to repel the S pole, the one at the top is induced to attract the N pole.
Both currents are induced to generate S poles... They are in the same direction.
This is the difference between the tube with a falling magnet and the (thin) coil with a magnet falling through.

I think they are in different directions. Because they are acting to decrease the change in magnetic flux.

 

[truesearch]

A S pole is needed at each end of the magnet to cause opposition to motion

 

[jsmith613]

A S pole is needed at each end of the magnet to cause opposition to motion

look at the diagram on the previous page
it shows why currents are in different directions

 

[BruceW]

so is this diagram wrong:
http://www.thestudentroom.co.uk/attachment.php?attachmentid=139647&d=1333538964

I don't want to show the image here as it is enourmous

That is a different situation, it is measuring the emf along the entire coil.

 

[jsmith613]

That is a different situation, it is measuring the emf along the entire coil.

so if I were to measure the emf across the entire tube (with the magnet falling) - assuming it were possible - would the current and emf be measured as zero?

 

[truesearch]

Gnight for now

 

[BruceW]

so if I were to measure the emf across the entire tube (with the magnet falling) - assuming it were possible - would the current and emf be measured as zero?

I'm pretty sure emf between the top and bottom of the tube would be zero. In the case of the coil, an emf is measured when the magnet falls into it because the measured emf is effectively the total emf over the horizontal loops, so it will be non-zero when the magnet falls in (and out). But with the tube, you would be measuring the potential difference between the top of the tube and bottom, which will be zero. Its only if you had some other technology, or way of seeing the charge carriers that you would realize something is going on in the case of the tube.

 

[sophiecentaur]

so you would agree then that the net EMF and current is zero?

You still need to define what you mean by "NET".
The only time I would use that term would be in a situation where, for example, forces on a point would Add Up to a 'net' or 'resultant value'. But these currents are all over the place so I think you must mean 'Mean Value' but Mean Value over what? Just the peak values? It is not as simple as you imply, I think. Decide what you really mean and we can come up with an answer.

 

[jsmith613]

I'm pretty sure emf between the top and bottom of the tube would be zero. In the case of the coil, an emf is measured when the magnet falls into it because the measured emf is effectively the total emf over the horizontal loops, so it will be non-zero when the magnet falls in (and out). But with the tube, you would be measuring the potential difference between the top of the tube and bottom, which will be zero. Its only if you had some other technology, or way of seeing the charge carriers that you would realize something is going on in the case of the tube.

just to absolutely clarify and then I think I have gotten it, the current too will be measured as zero between the ends of BOTH the coil and the tube, right?

 

[jsmith613]

You still need to define what you mean by "NET".
The only time I would use that term would be in a situation where, for example, forces on a point would Add Up to a 'net' or 'resultant value'. But these currents are all over the place so I think you must mean 'Mean Value' but Mean Value over what? Just the peak values? It is not as simple as you imply, I think. Decide what you really mean and we can come up with an answer.

well I mean the EMF/current across the ends of the tube / coil
I think maybe the word net / mean was a poor way of saying this

 

[sophiecentaur]

That diagram of the coil would suggest otherwise, wouldn't it?

 

[jsmith613]

That diagram of the coil would suggest otherwise, wouldn't it?

sorry the diagram of the coil suggests exactly this...

 

[jsmith613]

I am off for tonight
we can carry on tomorrow
I continue to express my graditude for all the help you have been giving me (in fact if this is the first time I have said it I really do emphasise how grateful I am for all your help so far :)

 

[BruceW]

just to absolutely clarify and then I think I have gotten it, the current too will be measured as zero between the ends of BOTH the coil and the tube, right?

For the case of the coil, when the magnet is falling into (or out of) the coil, there will be a flow of charge through the coil, to slow the rate of change of flux. But when the magnet is in the middle of the coil, the charges above the magnet want to go clockwise and those underneath the magnet want to go anticlockwise. But this would cause a charge separation, so when the magnet is in the middle, there is no flow of charge.

For the case of the tube, in reality there are lots of eddy currents that try to slow the rate of change of flux caused by the tumbling magnet. If the magnet went down smoothly as you've suggested, it seems plausible that there would be charges flowing anti-clockwise underneath the magnet and clockwise above the magnet (really they are just very ordered eddy currents). These eddy currents continue even as the magnet is falling through the middle of the tube.

In the coil, it is simple to measure the emf which causes the circular flow of charge, because of the fact that it is set-up as a coil. But in the tube, I can't think of a simple way of measuring the flow of charges, and if you just measured the emf between the top and bottom of the tube, I'm pretty sure you'd get nothing, or in any case it would not tell you about the circular flow of charge which is acting to decrease the magnet's acceleration.

I continue to express my graditude for all the help you have been giving me (in fact if this is the first time I have said it I really do emphasise how grateful I am for all your help so far :)

Ah, I'm glad if I've even been of some help.

Edit: maybe you were just saying this to sophiecentaur... I am so quick to accept a thank you

 

[sophiecentaur]

sorry the diagram of the coil suggests exactly this...

The diagram has an oscilloscope trace in it which shows that the EMF is not Zero at all times. How can the following statement be right? Earlier you wrote this:

just to absolutely clarify and then I think I have gotten it, the current too will be measured as zero between the ends of BOTH the coil and the tube, right?

When is it measured as zero? You seem to have a problem with using the term 'Mean'. This is a bit of a handicap in discussions in which both time and position vary. Your term 'Net' doesn't make it clear what you want to say.

I am sorry but I have just followed an endless and frustrating thread, elsewhere, in which one of the contributors absolutely refused to use the right terms appropriately - mixing up force, strength, power etc. in a totally random way. The thread ended without his getting anywhere. I am very sensitive to this sort of thing at the moment, which is why I think the right terminology is important. Unless we agree on the terms to use and what they mean then we may not get anywhere either.
I am assuming you are familiar with the term?

 

[jsmith613]

I am assuming you are familiar with the term?

what term??

 

[jsmith613]

Edit: maybe you were just saying this to sophiecentaur... I am so quick to accept a thank you

absolutely this was aimed at you and everyone else who has helped me!

 

[jsmith613]

The diagram has an oscilloscope trace in it which shows that the EMF is not Zero at all times. How can the following statement be right? Earlier you wrote this:

Oh I see
the question I was asking was for the entire time the magnet is INSIDE THE TUBE
obviously as it approaches / leaves there will be a change in flux BUT INSIDE THE TUBE the emf on the oscilliscope is zero.
look. the magnet is dropped from ABOVE THE COIL so obviously as it falls the emf will increase but my question was is it correct to assume that WHEN THE EMF IS ZERO, the current will also be zero

When I said MEAN/NET I meant WHILST THE MAGNET WAS IN THE TUBE

 

[sophiecentaur]

Right. Whilst inside the coil there is no induced EMF so the meter reads zero. This IS (and I see what you mean here) the NET instantaneous effect of the passage of the N and S poles along the coil length. They are only unbalanced when one pole is inside and one pole is outside the coil (at the ends).

Whilst the magnet is falling inside the tube there are EMFs being produced at all times which are producing circulating currents. Without the currents, the magnet wouldn't be slowed up, would it? These EMFs (and hence the currents) are in 'Lenz's Law' directions. BUT the Mean of all the EMFs around the magnet will be zero because both ends of the magnet are traveling at the same speed.

There is no reason to think that an EMF would ever be measured over the length of the tube because the axis of the tube (the conductor) is not at right angles to the direction of the motion. EMFs are induced around the circumference.

@Bruce You mention the problem of measuring the currents inside the tube. It would be possible to cut the tube, horizontally, half way down and insert an insulated split ring. You could measure the EMF across the gap in the ring and then this could 'indicate' the value of current, knowing the resistivity of the tube metal. The magnet would need to be not tumbling, of course.

 

[jsmith613]

Right. Whilst inside the coil there is no induced EMF so the meter reads zero. This IS (and I see what you mean here) the NET instantaneous effect of the passage of the N and S poles along the coil length. They are only unbalanced when one pole is inside and one pole is outside the coil (at the ends).

Whilst the magnet is falling inside the tube there are EMFs being produced at all times which are producing circulating currents. Without the currents, the magnet wouldn't be slowed up, would it? These EMFs (and hence the currents) are in 'Lenz's Law' directions. BUT the Mean of all the EMFs around the magnet will be zero because both ends of the magnet are traveling at the same speed.

Ok so I think you are saying that the mean emf will be zero but the mean current WILL NOT BE otherwise the magnetic field would not be produced, right?

 

[sophiecentaur]

Ok so I think you are saying that the mean emf will be zero but the mean current WILL NOT BE otherwise the magnetic field would not be produced, right?

??
As far as I'm concerned, as long as you ignore inductance (which you can), the current flowing in the coil of the tube will be directly proportional to the EMF. Where have I given the impression that they don't precisely track each other?

If current is flowing one way (because the emf is causing it to) in one part of the tube and in another direction in another part of the tube (for the same reason) then the mean current is zero but that doesn't imply that the Force on the magnet will be zero, does it? The currents at either end of the magnet must flow in opposite directions or the effect on one of the poles would be to PUSH the magnet down.

Are you reading elsewhere about Induction, RH and LH rules etc.? It may be a good idea because I think you may be mis-reading what I am writing at times. If you read someone elses way of putting things then perhaps it will make more sense to you.

All this stuff is so basic that it is available to read about all over the net. Have you ever tried the 'Hyperphysics ' pages? They are very good. I usually find something there for pretty well every topic.

 

[jsmith613]

If current is flowing one way (because the emf is causing it to) in one part of the tube and in another direction in another part of the tube (for the same reason) then the mean current is zero but that doesn't imply that the Force on the magnet will be zero, does it? The currents at either end of the magnet must flow in opposite directions or the effect on one of the poles would be to PUSH the magnet down.

ok thanks. I just wanted to check. I figured if the mean emf was zero and the mean current was zero I presumed no opposing magnetic forces would act. apparently this was wrong

Are you reading elsewhere about Induction, RH and LH rules etc.? It may be a good idea because I think you may be mis-reading what I am writing at times. If you read someone elses way of putting things then perhaps it will make more sense to you.

Before I started this discussion I looked endlessly over the internet (including Hyperphysics, Wiki) for relevant infromation but as I am only studying A2 Level physics (and don't really want to study more complicated physics than this) I could not find anything.
The A2 websites were not detailed enough and other websites were too detailed.
its a hard balance

 

[sophiecentaur]

A level is strange these days. They spread the net wide and give you a taste of all sorts of new Physics topics*. This means you are spread much more 'thin' than when I did A level. I think it is bad for you because the course books never go far enough if you happen to get an interest in one particular topic. Have you access to a 'proper' textbook that was published a few years ago? You will probably find much better treatment of topics you find difficult. But you're talking £30+ rather than £10+ for a coursebook.The way to get good grades it just to LEARN every fact and formula in the specification (download it if you haven't yet) and then follow, step by step, the Exam questions, doing exactly what they ask. Needless to say, you have to be absolutely rock solid on the simple algebraic manipulation rules (re-arranging equations and the like) and the basic Trig formulae. That's where a lot of students are shaky.

*On the course that I taught (retired recently) they started AS Physics by talking about Fundamental Particles - loads of brand new ideas: "Particle Zoo" long before they dealt with how an electron behaves in an electric field. None of the students got anything out of it except the feeling that they 'knew about' some sexy particles. Totally fragmented.

 

[jsmith613]

A level is strange these days. They spread the net wide and give you a taste of all sorts of new Physics topics*. This means you are spread much more 'thin' than when I did A level. I think it is bad for you because the course books never go far enough if you happen to get an interest in one particular topic. Have you access to a 'proper' textbook that was published a few years ago? You will probably find much better treatment of topics you find difficult. But you're talking £30+ rather than £10+ for a coursebook.The way to get good grades it just to LEARN every fact and formula in the specification (download it if you haven't yet) and then follow, step by step, the Exam questions, doing exactly what they ask. Needless to say, you have to be absolutely rock solid on the simple algebraic manipulation rules (re-arranging equations and the like) and the basic Trig formulae. That's where a lot of students are shaky.

well I am very good at maths (and do A2 maths).
I despise the way A-levels are structured...although the system is very easy to play with it means I come out not really knowing a lot of physics...i think this is why i dislike physics as a subject...the separate bits are fascinating but the way it is taught at A2 level means I learn some facts and just have to accept them.

anyway I would very much like to thank you and BruceW and everyone else who has helped me with this topic..hopefully if a question comes up I will be able to nail it :)

 

[sophiecentaur]

well I am very good at maths (and do A2 maths).
I despise the way A-levels are structured...although the system is very easy to play with it means I come out not really knowing a lot of physics...i think this is why i dislike physics as a subject...the separate bits are fascinating but the way it is taught at A2 level means I learn some facts and just have to accept them.

anyway I would very much like to thank you and BruceW and everyone else who has helped me with this topic..hopefully if a question comes up I will be able to nail it :)

That's a good start!

That's because the course is aimed at 'bums on seats' and supposed to be attractive. I don't know how you got on at GCSE in Science but I know that very few of the students who chose Physics at A/AS had a clue about what they were letting themselves in for. It 'sounds good' as a subject (and of course it is) but it demands a level of rigour, even at A level, which many students have never come across before. As a consequence of student responses to past exam questions, the structure of the course has been adjusted to be suitable for students who don't particularly 'want to think'. Giving lots of taster units and not concentrating on some serious basics means that, as you say, you are given stuff which you have to accept as a fact and can answer questions adequately by just remembering the what, rather than the how. Way back, the topics we were taught were very conventional (heat, light, sound + elec) but, even in the first year, we got the message about how important the 'book-work' was. We had to be able to derive all the basic equations (of which there were a lot) and to REMEMBER them (not a single A level formula was given on the paper - just the constants). I have always told my students that if they need to look up a formula then they can't actually be sure that they will use it properly - after all, they are all pretty straightforward.
Having had my rant, however, I reckon that, if your Maths is good and you are interested enough to be worrying this particular topic to death (haha), then you should do all right. If and when you get to Uni to do Physics then you will find it is more suited to your attitude, I think.

 

[BruceW]

well I am very good at maths (and do A2 maths).
I despise the way A-levels are structured...although the system is very easy to play with it means I come out not really knowing a lot of physics...i think this is why i dislike physics as a subject...the separate bits are fascinating but the way it is taught at A2 level means I learn some facts and just have to accept them.

anyway I would very much like to thank you and BruceW and everyone else who has helped me with this topic..hopefully if a question comes up I will be able to nail it :)

Thanks, jsmith! I'm from uk too, and I recently finished my undergraduate physics degree. I'd say that A-level physics is a pretty good introduction to undergraduate-level physics (in case you're interested in doing that). The main difference is in how much maths goes on. There is definitely more maths involved in undergraduate physics than in A-level physics. And I think this is one of the main reasons that you must just accept some facts at A-level, because they simply haven't taught you the relevant maths yet. Also, they are trying to give you a broad view of physics, so it would take too long to go into detail.

 

[jsmith613]

??
If current is flowing one way (because the emf is causing it to) in one part of the tube and in another direction in another part of the tube (for the same reason) then the mean current is zero but that doesn't imply that the Force on the magnet will be zero, does it? The currents at either end of the magnet must flow in opposite directions or the effect on one of the poles would be to PUSH the magnet down.

I would very much like to check I have understood this topic:
here is the summary of the way I see everything working in relation to the COIL (I am ignoring the tube as it seems more complicated than needed for A-level)

is this summary correct

the net emf (across the entire coil) is zero when the magnet is in the coil so the current in the coil is also zero...but the current at either end of the magnet (above and below it) is creating an instantaneous magnetic field opposing the motion of the magnet as it falls (due to the currents that are present - i.e: before they cancel). This is for when the oscilliscope is attached - it completes the circuit

if we remove the oscilloscope then no current can flow as there is no complete circuit..hence there will be no opposing magnetic field and the magnet will accelerate due to gravity with no opposing forces AT ALL TIMES..

I really hope this is correct

NOTE: I will not be able to sign in until Sunday as I have a busy weekend so I will not respond to any comments until then. I would still be delighted for people to comment:)

 

[sophiecentaur]

Sort of... but an oscilloscope has 10Mohm resistance so you can say that no current flows!. If you were to short the ends together, you might get enough current to produce a measurable braking effect.

 

[jsmith613]

Sort of... but an oscilloscope has 10Mohm resistance so you can say that no current flows!. If you were to short the ends together, you might get enough current to produce a measurable braking effect.

so it would be more correct to say that no breaking effect occurs inside the coil
but as it enters / leaves we will get current and hence breaking effect?
(inside the tube we DO get breaking effect due to eddy currents though, right)

 

[sophiecentaur]

Thanks, jsmith! I'm from uk too, and I recently finished my undergraduate physics degree. I'd say that A-level physics is a pretty good introduction to undergraduate-level physics (in case you're interested in doing that). The main difference is in how much maths goes on. There is definitely more maths involved in undergraduate physics than in A-level physics. And I think this is one of the main reasons that you must just accept some facts at A-level, because they simply haven't taught you the relevant maths yet. Also, they are trying to give you a broad view of physics, so it would take too long to go into detail.

The problem is that you start your A level course with GCSE Maths and not O level Maths, as I did and many people have just done Science GCSE Double Award. That effect trickles up all the way up to Degree level and this is why people are needing Masters Degrees to get the same cred as a good Hons degree in the dim and distant past. I guess, as all you young people are going to live till 100, an extra year doesn't actually hurt anyone. It's just the Politicians (****rds) who claim that kids are getting brighter as a result of their loony policies. Don't get me started on Synthetic Phonics. Grrrrr!