A Gnome-inal Value for ‘g’

The Gnome Experiment Kit from precision scale manufacturers Kern and Sohn.

. . . setting storms and billows at defiance, and visiting the remotest parts of the terraqueous globe.

Samuel Johnson, The Rambler, 17 April 1750

That an object in free fall will accelerate towards the centre of our terraqueous globe at a rate of 9.81 metres per second per second is, at best, only a partial and parochial truth. It is 9.81 metres per second per second in the United Kingdom, yes; but the value of both acceleration due to free fall and the gravitational field strength vary from place to place across the globe (and in the SI System of measurement, the two quantities are numerically equal and dimensionally equivalent).

For example, according to Hirt et al. (2013) the lowest value for g on the Earth’s surface is atop Mount Huascarán in Peru where g = 9.7639 m s-2 and the highest is at the surface of the Arctic Ocean where g = 9.8337 m s-2.

Why does g vary?

There are three factors which can affect the local value of g.

Firstly, the distribution of mass within the volume of the Earth. The Earth is not of uniform density and volumes of rock within the crust of especially high or low density could affect g at the surface. The density of the rocks comprising the Earth’s crust varies between 2.6 – 2.9 g/cm3 (according to Jones 2007). This is a variation of 10% but the crust only comprises about 1.6% of the Earth’s mass since the density of material in the mantle and core is far higher so the variation in g due this factor is probably of the order of 0.2%.

Secondly, the Earth is not a perfect sphere but rather an oblate spheroid that bulges at the equator so that the equatorial radius is 6378 km but the polar radius is 6357 km. This is a variation of 0.33% but since the gravitational force is proportional to 1/r2 let’s assume that this accounts for a possible variation of the order of 0.7% in the value of g.

Thirdly, the acceleration due to the rotation of the Earth. We will look in detail at the theory underlying this in a moment, but from our rough and ready calculations above, it would seem that this is the major factor accounting for any variation in g: that is to say, g is a minimum at the equator and a maximum at the poles because of the Earth’s rotation.


The Gnome Experiment

In 2012, precision scale manufacturers Kern and Sohn used this well-known variation in the value of g to embark on a highly successful advertising campaign they called the ‘Gnome Experiment’ (see link 1 and link 2).

Whatever units their lying LCD displays show, electronic scales don’t measure mass or even weight: they actually measure the reaction force the scales exert on the item in their top pan. The reading will be affected if the scales are accelerating.

In diagram A, the apple is not accelerating so the resultant upward force on the apple is exactly 0.981 N. The scales show a reading of 0.981/9.81 = 0.100 000 kg = 100.000 g (assuming, of course, that they are calibrated for use in the UK).

In diagram B, the apple and scales are in an elevator that is accelerating upward at 1.00 metres per second per second. The resultant upward force must therefore be larger than the downward weight as shown in the free body diagram. The scales show a reading of 1.081/9.81 – 0.110 194 kg = 110.194 g.

In diagram C, the the apple and scales are in an elevator that is accelerating downwards at 1.00 metres per second per second. The resultant upward force must therefore be smaller than the downward weight as shown in the free body diagram. The scales show a reading of 0.881/9.81 – 0.089 806 kg = 89.806 g.


Never mind the weight, feel the acceleration

Now let’s look at the situation the Kern gnome mentioned above. The gnome was measured to have a ‘mass’ (or ‘reaction force’ calibrated in grams, really) of 309.82 g at the South Pole.

Showing this situation on a diagram:

Looking at the free body diagram for Kern the Gnome at the equator, we see that his reaction force must be less than his weight in order to produce the required centripetal acceleration towards the centre of the Earth. Assuming the scales are calibrated for the UK this would predict a reading on the scales of 3.029/9.81= 0.30875 kg = 308.75 g.

The actual value recorded at the equator during the Gnome Experiment was 307.86 g, a discrepancy of 0.3% which would suggest a contribution from one or both of the first two factors affecting g as discussed at the beginning of this post.

Although the work of Hirt et al. (2013) may seem the definitive scientific word on the gravitational environment close to the Earth’s surface, there is great value in taking measurements that are perhaps more directly understandable to check our comprehension: and that I think explains the emotional resonance that many felt in response to the Kern Gnome Experiment. There is a role for the ‘artificer’ as well as the ‘philosopher’ in the scientific enterprise on which humanity has embarked, but perhaps Samuel Johnson put it more eloquently:

The philosopher may very justly be delighted with the extent of his views, the artificer with the readiness of his hands; but let the one remember, that, without mechanical performances, refined speculation is an empty dream, and the other, that, without theoretical reasoning, dexterity is little more than a brute instinct.

Samuel Johnson, The Rambler, 17 April 1750

References

Hirt, C., Claessens, S., Fecher, T., Kuhn, M., Pail, R., & Rexer, M. (2013). New ultrahigh‐resolution picture of Earth’s gravity fieldGeophysical research letters40(16), 4279-4283.

Jones, F. (2007). Geophysics Foundations: Physical Properties: Density. University of British Columbia website, accessed on 2/5/21.

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Nature Abhors A Change In Flux

Aristotle memorably said that Nature abhors a vacuum: in other words. he thought that a region of space entirely devoid of matter, including air, was logically impossible.

Aristotle turned out to be wrong in that regard, as he was in numerous others (but not quite as many as we – secure and perhaps a little complacent and arrogant as we look down our noses at him from our modern scientific perspective – often like to pretend).

An amusing version which is perhaps more consistent with our current scientific understanding was penned by D. J. Griffiths (2013) when he wrote: Nature abhors a change in flux.

Magnetic flux (represented by the Greek letter phi, Φ) is a useful quantity that takes account of both the strength of the magnetic field and its extent. It is the total ‘magnetic flow’ passing through a given area. You can also think of it as the number of magnetic field lines multiplied by the area they pass through so a strong magnetic field confined to a small area might have the same flux (or ‘effect’) as weaker field spread out over a large area.


Lenz’s Law

Emil Lenz formulated an earlier statement of the Nature abhors a change of flux principle when he stated what I think is the most consistently underrated laws of electromagnetism, at least in terms of developing students’ understanding:

The current induced in a circuit due to a change in a magnetic field is directed to oppose the change in flux and to exert a mechanical force which opposes the motion.

Lenz’s Law (1834)

This is a qualitative rather than a quantitive law since it is about the direction, not the magnitude, of an induced current. Let’s look at its application in the familiar A-level Physics context of dropping a bar magnet through a coil of wire.


Dropping a magnet through a coil in pictures

Picture 1

In picture 1 above, the magnet is approaching the coil with a small velocity v. The magnet is too far away from the coil to produce any magnetic flux in the centre of the coil. (For more on the handy convention I have used to draw the coils and show the current flow, please click on this link.) Since there is no magnetic flux, or more to the point, no change in magnetic flux, then by Faraday’s Law of Electromagnetic Induction there is no induced current in the coil.

Picture 2

in picture 2, the magnet has accelerated to a higher velocity v due to the effect of gravitational force. The magnet is now close enough so that it produces a magnetic flux inside the coil. More to the point, there is an increase in the magnetic flux as the magnet gets closer to the coil: by Faraday’s Law, this produces an induced current in the coil (shown using the dot and cross convention).

To ascertain the direction of the current flow in the coil we can use Lenz’s Law which states that the current will flow in such a way so as to oppose the change in flux producing it. The red circles show the magnetic field lines produced by the induced current. These are in the opposite direction to the purple field lines produced by the bar magnet (highlighted yellow on diagram 2): in effect, they are attempting to cancel out the magnetic flux which produce them!

The direction of current flow in the coil will produce a temporary north magnetic pole at the top of the coil which, of course, will attempt to repel the falling magnet; this is ‘mechanical force which opposes the motion’ mentioned in Lenz’s Law. The upward magnetic force on the falling magnet will make it accelerate downward at a rate less than g as it approaches the coil.

Picture 3

In picture 3, the purple magnetic field lines within the volume of the coil are approximately parallel so that there will be no change of flux while the magnet is in this approximate position. In other words, the number of field lines passing through the cross-sectional area of the coil will be approximately constant. Using Faraday’s Law, there will be no flow of induced current. Since there is no change in flux to oppose, Lenz’s Law does not apply. The magnet will be accelerating downwards at g.

Picture 4

As the magnet emerges from the bottom of the coil, the magnetic flux through the coil decreases. This results in a flow of induced current as per Faraday’s Law. The direction of induced current flow will be as shown so that the red field lines are in the same direction as the purple field lines; Lenz’s Law is now working to oppose the reduction of magnetic flux through the coil!

A temporary north magnetic pole is generated by the induced current at the lower end of the coil. This will produce an upward magnetic force on the falling magnet so that it accelerates downward at a rate less than g. This, again, is the ‘mechanical force which opposes the motion’ mentioned in Lenz’s Law.


Dropping a magnet through a coil in graphical form

This would be one of my desert island graphs since it is such a powerfully concise summary of some beautiful physics.

The graph shows the reversal in the direction of the current as discussed above. Also, the maximum induced emf in region 2 (blue line) is less than that in region 4 (red line) since the magnet is moving more slowly.

What is more, from Faraday’s Law (where ℇ is the induced emf and N is total number of turns of the coil), the blue area is equal to the red area since:

and N and ∆Φ are fixed values for a given coil and bar magnet.

As I said previously, there is so much fascinating physics in this graph that I think it worth exploring in depth with your A level Physics students 🙂

Other news

If you have enjoyed this post, then you may be interested to know that I’ve written a book! Cracking Key Concepts in Secondary Science (co-authored with Adam Boxer and Heena Dave) is due to be published by Corwin in July 2021.

References

Lenz, E. (1834), “Ueber die Bestimmung der Richtung der durch elektodynamische Vertheilung erregten galvanischen Ströme”, Annalen der Physik und Chemie107 (31), pp. 483–494

Griffiths, David (2013). Introduction to Electrodynamics. p. 315.

The Coulomb Train Model Revisited (Part 5)

In this post, we are going to look at series circuits using the Coulomb Train Model.

The Coulomb Train Model (CTM) is a helpful model for both explaining and predicting the behaviour of real electric circuits which I think is useful for KS3 and KS4 students.

Without further ado, here is a a summary.


A circuit with one resistor

Let’s look at a very simple circuit to begin with:

This can be represented on the CTM like this:

The ammeter counts 5 coulombs passing every 10 seconds, so the current I = charge flow Q / time t = 5 coulombs / 10 seconds = 0.5 amperes.

We assume that the cell has a potential difference of 1.5 V so there is a potential difference of 1.5 V across the resistor R1 (that is to say, each coulomb loses 1.5 J of energy as it passes through R1).

The resistor R1 = potential difference V / current I = 1.5 / 0.5 = 3.0 ohms.


A circuit with two resistors in series

Now let’s add a second identical resistor R2 into the circuit.

This can be shown using the CTM like this:

Notice that the current in this example is smaller than in the first circuit; that is to say, fewer coulombs go through the ammeter in the same time. This is because we have added a second resistor and remember that resistance is a property that reduces the current. (Try and avoid talking about a high resistance ‘slowing down’ the current because in many instances such as two conductors in parallel a high current can be modelled with no change in the speed of the coulombs.)

Notice also that the voltmeter is making identical measurements on both the circuit diagram and the CTM animation. It is measuring the total energy change of the coulombs as they pass through both R1 and R2.

The current I = charge flow Q / time t = 5 coulombs / 20 seconds = 0.25 amps. This is half the value of the current in the first circuit.

We have an identical cell of potential difference 1.5 V the voltmeter would measure 1.5 V. We can calculate the total resistance using R = V / I = 1.5 / 0.25 = 6.0 ohms.

This is to be expected since the total resistance R = R1 + R2 and R1 = 3.0 ohms and R2 = 3.0 ohms.


Looking at the resistors individually

The above circuit can be represented using the CTM as follows:

Between A and B, the coulombs are each gaining 1.5 joules since the cell has a potential difference of 1.5 V. (Remember that V = E energy transferred (in joules) / Q charge flow (in coulombs.)

Between B and C the coulombs lose no energy; that is to say, we are assuming that the connecting wires have negligible resistance.

Between C and D the coulombs lose some energy. We can use the familar V = I x R to calculate how much energy is lost from each coulomb, since we know that R1 is 3.0 ohms and I is 0.25 amperes (see previous section).

V = I x R = 0.25 x 3.0 = 0.75 volts.

That is to say, 0.75 joules are removed from each coulomb as they pass through R1 which means that (since 1.5 joules were added to each coulomb by the cell) that 0.75 joules are left in each coulomb.

The coulombs do not lose any energy travelling between D and E because, again, we are assuming negligible resistance in the connecting wire.

0.75 joules is removed from each coulomb between E and F making the potential difference across R2 to be 0.75 volts.

Thus we find that the familiar V = V1 + V2 is a direct consequence of the Principle of Conservation of Energy.


FAQ: ‘How do the coulombs know to drop off only half their energy in R1?’

Simple answer: they don’t.

This may be a valid objection for some donation models of electric circuits (such as the pizza delivery van model) but it doesn’t apply to the CTM because it is a continuous chain model (with the caveat that the CTM applies only to ‘steady state’ circuits where the current is constant).

Let’s look at a numerical argument to support this:

  • The magnitude of the current is controlled by only two factors: the potential difference of the cell and the total resistance of the circuit.
  • In other words, if we increased the value of R1 to (say) 4 ohms and reduced the value of R2 to 2 ohms so that the total resistance was still 6 ohms, the current would still be 0.25 amps.
  • However, in this case the energy dissipated by each coulomb passing through R1 would V = I x R = 0.25 x 4 = 1 volt (or 1 joule per coulomb) and similarly the potential difference across R2 would now be 0.5 volts.
  • The coulombs do not ‘know’ to drop off 1 joule at R1 and 0.5 joules at R2: rather, it is a purely mechanical interaction between the moving coulombs and each resistor.
  • R1 has a bigger proportion of the total resistance of the circuit than R2 so it seems self-evident (at least to me) that the coulombs will lose a larger proportion of their total energy passing through R1.
  • A similar analysis would apply if we made R2 = 4 ohms and R1 = 2 ohms: the coulombs would now lose 0.5 joules passing through R1 and 1 joule passing through R2.

Thus, we see that the current in a series circuit is affected by the ‘global’ or ‘whole circuit’ properties such as the potential difference of the cell and the total resistance of the circuit. The CTM models this property of real circuits by being a continuous chain of mechanically-linked ‘trucks’ so that a change in any one part of the circuit affects the movement of all the coulombs.

However, the proportion of the energy lost by a coulomb travelling through one part of the circuit is affected — not by ‘magic’ or a weird form of ‘coulomb telepathy’ — but only by the ‘local’ properties of that section of the circuit i.e. the electrical resistance of that section.

The CTM analogue of a low resistance section of a circuit (top) and a high resistance section of a circuit (bottom)

(PS You can read more about the CTM and potential divider circuits here.)


Afterword

You may be relieved to hear that this is the last post in my series on ‘The CTM revisited’. My thanks to the readers who have stayed with me through the series (!)

I will close by saying that I have appreciated both the expressions of enthusiasm about CTM and the thoughtful criticisms of it.

Mnemonics for the S.I. Prefixes

The S.I. System of Weights and Measures may be a bit of a dog’s dinner, but at least it’s a dog’s dinner prepped, cooked, served and — more to the point — eaten by scientists.

A brief history of the Système international d’unités

It all began with the métre (“measure”), of course. This was first proposed as a universal measure of distance by the post-Revolutionary French Academy of Sciences in 1791. According to legend (well, not legend precisely — think of it as random speculative gossip, if you prefer), they first proposed that the metre should be one millionth of the distance from the North Pole to the equator.

When that turned out to be a little on the large side, they reputedly shrugged in that inimitable Gallic fashion said: “D’accord, faisons un dix millionième alors, mais c’est ma dernière offre.” (“OK, let’s make it one ten millionths then, but that’s my final offer.”)

Since then, what measurement-barbarians loosely (and egregiously incorrectly) call “the metric system” has been through many iterations and revisions to become the S.I. System. Its full name is the Système international d’unités which pays due honour to France’s pivotal role in developing and sustaining it.

When some of those same measurement-barbarians call for a return to the good old “pragmatic” Britsh Imperial System of inches and poundals, I urge all fair-minded people to tell them, as kindly as possible, that they can’t: not now, not ever.

Since 1930, the inch has been defined as 25.4 millimetres. (It was, so I believe, the accuracy and precision needed to design and build jet engines that led to the redefinition. The older definitions of the inch simply weren’t precise enough.)

You simply cannot replace the S.I. system, you can, however, dress it up a little bit and call a distance of 25.4 millimetres “one inch” if you really wanted to — but, in the end, what would be the point of that?

The Power of Three (well, ten to the third power, anyways)

For human convenience, the S.I. system includes prefixes. So a large distance might measured in kilometres where the prefix kilo- indicates multiplying by a factor of 1000 (or 10 raised to the third power). The distance between the Globe Theatre in London and Slough Station is 38.6 km. Longer distances such as London and New York, NY would be 5.6 megametres (or 5.6 Mm — note capital ‘M’ for mega [one million] to avoid confusion with the prefix milli- ).

The S.I. System has prefixes for all occasions, as shown below.

The ‘big’ SI prefixes.
Note that every one of them, except for kilo, is represented by a capital letter.

Note also that one should convert all prefixes into standard units for calculations e.g. meganewtons should be converted to newtons. The sole exception is kilograms because the base unit is the kilogram not the gram, so a megagram should be converted into kilograms, not grams. I trust that’s clear. (Did I mention the “dog’s dinner” part yet?)

For perspective, the distance between Earth and the nearest star outside our Solar System is 40 petametres, and current age of the universe is estimated to be 0.4 exaseconds (give or take a petasecond or two).

A useful mnemonic for remembering these is Karl Marx Gives The Proletariat Eleven Zeppelins (and one can imagine the proletariat expressing their gratitude by chanting in chorus: “Yo! Ta, Mr Marx!” as they march bravely forward.)

Karl Marx Gives The Proletariat Eleven Zeppelins (“Yo! Ta, Mr. Marx!)

But what about the little prefixes?

Milli- we have already covered above. The diameter of one of your red blood cells in 8 micrometres and the time it takes light to travel a distance equal to the diameter of a hydrogen atom is 300 zeptoseconds.

Again, there is an SI prefix for every occasion:

The ‘little’ SI prefixes.
(Handily, all of them are represented by lower case letters — including micro which is shown the lower case Greek letter ‘mu’)

A useful mnemonic would be: Millie’s Microphone Needs a Platform For Auditioning Zebra Yodellers.

For the record, GCSE Physics students are expected to know the SI prefixes between giga- and pico-, but if you’re in for a pico- then you’re in for anything between a yotta- and a yocto- in my opinion (if you catch my drift).

Very, very, very small to very, very, very big

The mean lifetime of a Z-boson (the particle that carries the Weak force) is 0.26 yoctoseconds.

According to our current understanding of physics, the stars will have stopped shining and all the galaxies will dissipate into dissassociated ions some 315 yottaseconds from now.

Apart from that, happy holidays everyone!

The Coulomb Train Model Revisited (Part 4)

In this post, we will look at parallel circuits.

The Coulomb Train Model (CTM) is a helpful model for both explaining and predicting the behaviour of real electric circuits which I think is useful for KS3 and KS4 students.

Without further ado, here is a a summary.

This is part 4 of a continuing series. (Click to read Part 1, Part 2 or Part 3.)


The ‘Parallel First’ Heresy

I advocate teaching parallel circuits before teaching series circuits. This, I must confess, sometimes makes me feel like Captain Rum from Blackadder Two:

The main reason for this is that parallel circuits are conceptually easier to analyse than series circuits because you can do so using a relatively naive notion of ‘flow’ and gives students an opportunity to explore and apply the recently-introduced concept of ‘flow of charge’ in a straightforward context.

Redish and Kuo (2015: 584) argue that ‘flow’ is an example of embodied cognition in the sense that its meaning is grounded in physical experience:

The thesis of embodied cognition states that ultimately our conceptual system grounded in our interaction with the physical world: How we construe even highly abstract meaning is constrained by and is often derived from our very concrete experiences in the physical world.

Redish and Kuo (2015: 569)

As an aside, I would mention that Redish and Kuo (2015) is an enduringly fascinating paper with a wealth of insights for any teacher of physics and I would strongly recommend that everyone reads it (see link in the Reference section).


Let’s Go Parallel First — but not yet

Let’s start with a very simple circuit.

This is not a parallel circuit (yet) because switch S is open. Resistors R1 and R2 are identical.

This can be represented on the coulomb train model like this:

Five coulombs pass through the ammeter in 20 seconds so the current I = Q/t = 5/20 = 0.25 amperes.

Let’s assume we have a 1.5 V cell so 1.5 joules of energy are added to each coulomb as they pass through the cell. Let’s also assume that we have negligible resistance in the cell and the connecting wires so 1.5 joules of energy will be removed from each coulomb as they pass through the resistor. The voltmeter as shown will read 1.5 volts.

The resistance of the resistor R1 is R=V/I = 1.5/0.25 = 6.0 ohms.


Let’s Go Parallel First — for real this time.

Now let’s close switch S.

This is example of changing an example by continuous conversion which removes the need for multiple ammeters in the circuit. The changed circuit can be represented on the CTM as shown

Now, ten coulombs pass through the ammeter in twenty seconds so I = Q/t = 10/20 = 0.5 amperes (double the reading in the first circuit shown).

Questioning may be useful at this point to reinforce the ‘flow’ paradigm that we hope students will be using:

  • What will be the reading if the ammeter moved to a similar position on the other side? (0.5 amps since current is not ‘used up’.)
  • What would be the reading if the ammeter was placed just before resistor R1? (0.25 amps since only half the current goes through R1.)

To calculate the total resistance of the whole circuit we use R = V/I = 1.5/0.5 = 3.0 ohms– which is half of the value of the circuit with just R1. Adding resistors in parallel has the surprising result of reducing the total resistance of the circuit.

This is a concrete example which helps students understand the concept of resistance as a property which reduces current: the current is larger when a second resistor is added so the total resistance must be smaller. Students often struggle with the idea of inverse relationships (i.e. as x increases y decreases and vice versa) so this is a point well worth emphasising.


Potential Difference and Parallel Circuits (1)

Let’s expand on the primitive ‘flow’ model we have been using until now and adapt the circuit a little bit.

This can be represented on the CTM like this:

Each coulomb passing through R2 loses 1.5 joules of energy so the voltmeter would read 1.5 volts.

One other point worth making is that the resistance of R2 (and R1) individually is still R = V/I = 1.5/0.25 = 6.0 ohms: it is only the combined effect of R1 and R2 together in parallel that reduces the total resistance of the circuit.


Potential Difference and Parallel Circuits (2)

Let’s have one last look at a different aspect of this circuit.

This can be represented on the CTM like this:

Each coulomb passing through the cell from X to Y gains 1.5 joules of energy, so the voltmeter would read 1.5 volts.

However, since we have twice the number of coulombs passing through the cell as when switch S is open, then the cell has to load twice as many coulombs with 1.5 joules in the same time.

This means that, although the potential difference is still 1.5 volts, the cell is working twice as hard.

The result of this is that the cell’s chemical energy store will be depleted more quickly when switch S is closed: parallel circuits will make cells go ‘flat’ in a much shorter time compared with a similar series circuit.

Bulbs in parallel may shine brighter (at least in terms of total brightness rather than individual brightness) but they won’t burn for as long.

To some ways of thinking, a parallel circuit with two bulbs is very much like burning a candle at both ends…


More fun and high jinks with coulomb train model in the next instalment when we will look at series circuits.

You can read part 5 here.


Reference

Redish, E. F., & Kuo, E. (2015). Language of physics, language of math: Disciplinary culture and dynamic epistemologyScience & Education24(5), 561-590.

FIFA and Really Challenging GCSE Physics Calculations

‘FIFA’ in this context has nothing to do with football; rather, it is a mnemonic that helps KS3 and KS4 students from across the attainment range engage productively with calculation questions.

FIFA stands for:

  • Formula
  • Insert values
  • Fine-tune
  • Answer

From personal experience, I can say that FIFA has worked to boost physics outcomes in the schools I have worked in. What is especially gratifying, however, is that a number of fellow teaching professionals have been kind enough to share their experience of using it:


Framing FIFA as a modular approach

Straightforward calculation questions (typically 2 or 3 marks) can be ‘unlocked’ using the original FIFA approach. More challenging questions (typically 4 or 5 marks) can often be handled using the FIFA-one-two approach.

However, what about the most challenging 5 or 6 mark questions that are targeted at Grade 8/9? Can FIFA help in solving these?

I believe it can. But before we dive into that, let’s look at a more traditional, non-FIFA, algebraic approach.


A challenging freezing question: the traditional (non-FIFA) algebraic approach

Note: this is a ‘made up’ question written in the style of the GCSE exam.

A pdf of this question is here. A traditional algebraic approach to solving this problem would look like this:

This approach would be fine for confident students with high previous attainment in physics and mathematics. I will go further and say that it should be positively encouraged for students who possess — in Edward Gibbon’s words — that ‘happy disposition’:

But the power of instruction is seldom of much efficacy, except in those happy dispositions where it is almost superfluous.

Edward Gibbon, The Decline and Fall of the Roman Empire

But what about those students who are more akin to the rest of us, and for whom the ‘power of instruction’ is not a superfluity but rather a necessity on which they depend?


A challenging freezing question: the FIFA-1-2-3 approach

Since this question involves both cooling and freezing it seems reasonable to start with the specific heat capacity formula and then use the specific latent heat formula:

FIFA-one-two isn’t enough. We must resort to FIFA-1-2-3.

What is noteworthy here is that the third FIFA formula isn’t on the formula sheet and is not on the list of formulas that need to be memorised. Instead, it is made by the student based on their understanding of physics and a close reading of the question.

Challenging? Yes, undoubtedly. But students will have unlocked some marks (up to 4 out of 6 by my estimation).

FIFA isn’t a royal road to mathematical mastery (although it certainly is a better bet than the dreaded ‘formula triangle’ that I and many other have used in the past). FIFA is the scaffolding, not the finished product.

Genuine scientific understanding is the clock tower; FIFA is simply some temporary scaffolding that helps students get there.

We complete the FIFA-1-2-3 process as follows:


Conclusion: FIFA fixes it

The FIFA-system was born of the despair engendered when you mark a set of mock exam papers and the majority of pages are blank: students had not even attempted the calculation skills.

In my experience, FIFA fixes that — students are much more willing to start a calculation question. And that means that, even when they cannot successfully navigate to a ‘full mark’ conclusion, they gain at least some marks, and and one does not have to be a particularly perceptive scholar of the human heart to understand that gaining ‘some marks‘ is more motivating than ‘no marks‘.

The Coulomb Train Model Revisited (Part 3)

In this post, we will look at explaining electrical resistance using the Coulomb Train Model.

This is part 3 of a continuing series (click to read part 1 and part 2).

The Coulomb Train Model (CTM) is a helpful model for both explaining and predicting the behaviour of real electric circuits which I think is useful for KS3 and KS4 students.

Without further ado, here is a a summary.

A summary of the Coulomb Train Model

Representing Resistance on the CTM

To measure resistance, we would set up this circuit.

We can represent this same circuit on the CTM as follows:

If we count how many ‘coulombs’ (grey trucks) pass one point in a certain time then on this animation we get 5 coulombs in 20 seconds.
This is equivalent to a current of
5 coulombs / 20 seconds = 0.2 coulombs per second = 0.2 amperes.

This way of thinking about current is consistent with the formula charge flow = current x time or Q=It which can be rearranged to give I=Q/t.

We have used identical labels on the circuit diagram and the CTM animation to encourage students to view them as different representations of a real situation. The ammeter at X would read 0.2 amps. We could place the ammeter at any other point in the circuit and still get a reading of 0.2 amps since ammeters only ‘count coulombs per second’ and don’t make any measurement of energy (represented by the orange substance in the trucks).

However, the voltmeter does make a measurement of energy: it compares the energy difference between a single coulomb at Y and a single coulomb at Z. If (say) 1.5 joules of energy is transferred from each coulomb as it passes through the bulb from Y to Z then the voltmeter will read a potential difference (or ‘voltage’ if you prefer) of 1.5 volts.

This way of thinking about potential difference is consistent with the formula energy transferred = charge flow x potential difference or E=QV which we can rearrange to give V=E/Q.

So as you can see, one volt is really equivalent to an energy change of one joule for every coulomb (!)

We can calculate the resistance of the bulb by using R=V/I so R = 1.5/0.2 = 7.5 ohms.


Resistance is not futile . . .

Students sometimes have difficulty accepting the idea of a ‘resistor’: ‘Why would anyone in their right minds deliberately design something that reduces the flow of electric current?’ It’s important to explain that it is vital to be able to control the flow of electric current and that one of the most common electronic components in your phone or games console is — the humble resistor.

One of many resistors on a circuit board. The colour codes tell us the value of each resistor.

Teachers often default to explaining electric circuits using bulbs as the active component. There is a lot to recommend this practice, not least the fact that changes in the circuit instantaneously affect the brightness of the bulb. However, it vital (especially at GCSE) to allow students to learn about circuits featuring resistors and other components rather than just the pedagogically overused (imho) filament lamp.


Calculating the resistance of a resistor

Consider this circuit where we have a resistor R1.

This can be represented as a coulomb train model like this:

The resistor does not glow with visible light as the bulb does, but it would glow pretty brightly if viewed through an infra red camera since the energy carried by the coulombs is transferred to the thermal energy store of the resistor. The only way we can observe this energy shift without such a special camera is to use a voltmeter.

Let’s begin by analysing this circuit qualitatively.

  • The coulombs are moving faster in this circuit than the previous circuit. This means that the current is larger. (Remember: current is coulombs per second.)
  • Because the current is larger, R1 must have a smaller resistance than the bulb. (Remember: resistance is a quantity that reduces the current.)
  • The energy transferred to each coulomb is the same in each example so the potential difference of the cell is the same in both circuits. (Of course, V can be altered by adding a second cell or turning up the setting on a power supply, but in many circuits V is, loosely speaking, a ‘fixed’ or ‘quasi-constant’ value.)
  • Because the ‘push’ or potential difference is the same size but the resistance of R1 is smaller, then the same cell is able to push a larger current around the circuit.

Now let’s analyse the circuit quantitatively.

  • 5 coulombs pass a single point in 13 seconds so the current is 5/13 = 0.38 coulombs per second = 0.4 amperes. (Double the current in the bulb circuit.)
  • The resistance can be calculated using R=V/I = 1.5/0.4 = 3.75 ohms. (Half the resistance of the bulb.)
  • Each coulomb is being loaded with 1.5 J of energy as it passes through the cell. Since this is happening twice as often in the resistor circuit as the bulb circuit, the cell will ‘go flat’ or ’empty its chemical energy store’ in half the time of the bulb cell.

So there we have it: more fun and high jinks with the CTM.

I hope that I have persuaded a few more teachers that the CTM is useful for getting students to think productively and, more importantly, quantitatively using correct scientific terminology about electric circuits.

In the next installment, we will look at series and parallel circuits.

The Coulomb Train Revisited (Part 2)

In this post, we will look at understanding potential difference (or voltage) using the Coulomb Train Model.

This is part 2 of a continuing series. You can read part 1 here.

The Coulomb Train Model (CTM) is a helpful model for both explaining and predicting the behaviour of real electric circuits which I think is suitable for use with KS3 and KS4 students (that’s 11-16 year olds for non-UK educators).

To summarise what has been discussed so far:


Modelling potential difference using the CTM

Potential difference is the ‘push’ needed to make electric charge move around a closed circuit. On the CTM, we can represent the ‘push’ as a gain in the energy of the coulomb. (This is consistent with the actual definition of the volt V = E/Q, where one volt is a change in energy of one joule per coulomb.)

How can we observe this gain in energy? Simple, we use a voltmeter.

Kudos to https://www.circuit-diagram.org/editor/ for the lovely circuit diagrams

On the CTM, this would look like this:

What the voltmeter does is compare the energy contained by two coulombs: one at A and the other at B. The coulombs at B, having passed through the 1.5 V cell, each have 1.5 joules of energy more than than the coulombs at A. This means that the voltmeter in this position reads 1.5 volts. We would say that the potential difference across the cell is 1.5 V. (Try and avoid talking about the potential difference ‘through’ or ‘of’ any part of the circuit.)


More potential difference measurements using the CTM

Let’s move the voltmeter to a different position.

On the CTM, this would look like this:

Let’s make the very reasonable assumption that the connecting wires have zero resistance. This would mean that the coulombs at C have 1.5 joules of energy and that the coulombs at D have 1.5 joules of energy. They have not lost any energy since they have not passed through any part of the circuit that actually has a resistance. The voltmeter would therefore read 0 volts since it cannot detect any energy difference.

Now let’s move the voltmeter one last time.

On the CTM, this would look like this:

Notice that the coulombs at F have 1.5 fewer joules than the coulombs at E. The coulombs transfer 1.5 joules of energy to the bulb because the bulb has a resistance.

Any part of the circuit that has non-zero resistance will ‘rob’ coulombs of their energy. On this very simple model, we assume that only the bulb has a resistance and so only the bulb will ‘push back’ against the movement of the coulombs and cost them energy.

Also on this simple model, the potential difference across the bulb is identical to the potential difference across the cell — but this is not always the case. For example, if the wires had a small but non-negligible resistance and if the cell had an internal resistance, but these would only come into play at A-level.

The bulb is shown as ‘flashing’ on the CTM to provide a visual cue to help students mentally model the transfer of energy from the coulombs to the bulb. In reality, instead of just one coulomb transferring a largish ‘chunk’ of energy, there would be approximately 1.25 billion billion electrons continuously transferring a tiny fraction of this energy over the course of one second (assuming a d.c. current of 0.2 amps) so we wouldn’t see the bulb ‘flash’ in reality.


How do the coulombs ‘know’ how much energy to drop off?

This section is probably more of interest to specialist physics teachers, but all are welcome.

One frequent criticism of donation models like the CTM is how do the coulombs ‘know’ to drop off all their energy at the bulb?

The response to this, of course, is that they don’t. This criticism is an artefact of an (arguably) over-simplified model whereby we assume that only the bulb has resistance. The energy carried by the coulombs according to this model could be shown as a sketch graph, and let’s be honest it does look a little dodgy…

But, more accurately, of course, the energy loss is a process rather than an event. And the connecting wires actually have a small resistance. This leads to this graph:

Realistically speaking, the coulombs don’t lose all their energy passing through the bulb: they merely lose most of their energy here due to the process of passing through a high resistance part of the circuit.

In part 3 of this series, we’ll look at how resistance can be modelled using the CTM.

You can read part 3 here.

The Coulomb Train Model Revisited (Part 1)

The Coulomb Train Model (CTM) is a helpful model for both explaining and predicting the behaviour of real electric circuits which I think is suitable for use with KS3 and KS4 students (that’s 11-16 year olds for non-UK educators).

I have written about it before (see here and here) but I have recently been experimenting with animated versions of the original diagrams.

Essentially, the CTM is a donation model akin to the famous ‘bread and bakery van’ or even the ‘penguins and ski lift’ models, but to my mind it has some major advantages over these:

  • The trucks (‘coulombs’) in the CTM are linked in a continuous chain. If one ‘coulomb’ stops then they all stop. This helps students grasp why a break anywhere in a circuit will stop all current.
  • The CTM presents a simplified but still quantitatively accurate picture of otherwise abstract entities such as coulombs and energy rather than the more whimsical ‘bread van’ = ‘charge carrier’ and ‘bread’ = ‘energy’ (or ‘penguin’ = ‘charge carrier’ and ‘gpe of penguin’ = ‘energy of charge carrier’) for the other models.
  • Explanations and predictions scripted using the CTM use direct but substantially correct terminiology such as ‘One ampere is one coulomb per second’ rather than the woolier ‘current is proportional to the number of bread vans passing in one second’ or similar.

Modelling current flow using the CTM

The coulombs are the ‘trucks’ travelling clockwise in this animation. This models conventional current.

Charge flow (in coulombs) = current (in amps) x time (in seconds)

So a current of one ampere is one coulomb passing in one second. On the animation, 5 coulombs pass through the ammeter in 25 seconds so this is a current of 0.20 amps.

We have shown two ammeters to emphasise that current is conserved. That is to say, the coulombs are not ‘used up’ as they pass through the bulb.

The ammeters are shown as semi-transparent as a reminder that an ammeter is a ‘low resistance’ instrument.


Modelling ‘a source of potential difference is needed to make current flow’ using the CTM

Energy transferred (in joules) = potential difference (in volts) x charge flow (in coulombs) 

So the potential difference = energy transferred divided by the number of coulombs.

The source of potential difference is the number of joules transferred into each coulomb as it passes through the cell. If it was a 1.5 V cell then 1.5 joules of energy would be transferred into each coulomb.

This is represented as the orange stuff in the coulombs on the animation.

What is this energy? Well, it’s not ‘electrical energy’ for certain as that is not included on the IoP Energy Stores and Pathways model. In a metallic conductor, it would be the sum of the kinetic energy stores and electrical potential energy stores of 6.25 billion billion electrons that make up one coulomb of charge. The sum would be a constant value (assuming zero resistance) but it would be interchanged randomly between the kinetic and potential energy stores.

For the CTM, we can be a good deal less specific: it’s just ‘energy’ and the CTM provides a simplified, concrete picture that allows us to apply the potential difference equation in a way that is consistent with reality.

Or at least, that would be my argument.

The voltmeter is shown connected in parallel and the ‘gloves’ hint at it being a ‘high resistance’ instrument.

More will follow in part 2 (including why I decided to have the bulb flash between bright and dim in the animations).

You can read part 2 here.

Teaching Newton’s Third Law

Newton’s First and Second Laws of Motion are universal: they tell us how any set of forces will affect any object.

The “Push-Me-Pull-You” made famous by Dr Doolittle

If the forces are ‘balanced’ (dread word! — saying ‘total force is zero’ is better, I think) then the object will not accelerate: that is the essence of the First Law. If the sum of the forces is anything other than zero, then the object will accelerate; and what is more, it will accelerate at a rate that is directly proportional to the total force and inversely proportional to the mass of the object; and let’s not forget that it will also accelerate in the direction in which the total force acts. Acceleration is, after all, a vector quantity.

So far, so good. But what about the Third Law? It goes without saying, I hope, that Newton’s Third Law is also universal, but it tells us something different from the first two.

The first two tell us how forces affect objects; the third tells us how objects affect objects: in other words, how objects interact with each other.

The word ‘interact’ can be defined as ‘to act in such a way so as to affect each other’; in other words, how an action produces a reaction. However, the word ‘reaction’ has some unhelpful baggage. For example, you tap my knee (lightly!) with a hammer and my leg jerks. This is a reaction in the biological sense but not in the Newtonian sense; this type of reaction (although involuntary) requires the involvement of an active nervous system and an active muscle system. Because of this, there is a short but unavoidable time delay between the stimulus and the response.

The same is not true of a Newton Third Law reaction: the action and reaction happen simultaneously with zero time delay. The reaction is also entirely passive as the force is generated by the mere fact of the interaction and requires no active ‘participation’ from the ‘acted upon’ object.

I try to avoid the words ‘action’ and ‘reaction’ in statements of Newton’s Third Law for this reason.

If body A exerts a force on body B, then body B exerts an equal and opposite force on body A.

The best version of Newton’s Third Law (imho)

In our universe, body B simply cannot help but affect body A when body A acts on it. Newton’s Third Law is the first step towards understanding that of necessity we exist in an interconnected universe.


Getting the Third Law wrong…

Let’s consider a stationary teapot. (Why not?)

This is NOT a good illustation of Newton’s Third Law…

We can reject this as an appropriate example of Newton’s Third Law for two reasons:

  • Reason 1: Force X and Force Y are acting on a single object. Newton’s Third Law is about the forces produced by an interaction between objects and so cannot be illustrated by a single object.
  • Reason 2: Force X and Force Y are ‘equal’ only in the parochial and limited sense of being merely ‘equal in magnitude’ (8.2 N). They are very different types of force: X is an action-at-distance gravitational force and Y is an electromagnetic contact force. (‘Electromagnetic’ because contact forces are produced by electrons in atoms repelling the electrons in other atoms.) The word ‘equal’ in Newton’s Third Law does some seriously heavy lifting…

Getting the Third Law right…

The Third Law deals with the forces produced by interactions and so cannot be shown using a single diagram. Free body diagrams are the answer here (as they are in a vast range of mechanics problems).

This is a good illustration of Newton’s Third Law

The Earth (body A) pulls the teapot (body B) downwards with the force X so the teapot (body B) pulls the Earth (body A) upwards with the equal but opposite force W. They are both gravitational forces and so are both colour-coded black on the diagram because they are a ‘Newton 3 pair’.

It is worth noting that, applying Newton’s Second Law (F=ma), the downward 8.2 N would produce an acceleration of 9.8 metres per second per second on the teapot if it was allowed to fall. However, the upward 8.2 N would produce an acceleration of only 0.0000000000000000000000014 metres per second per second on the rather more massive planet Earth. Remember that the acceleration produced by the resultant force is inversely proportional to the mass of the object being accelerated.

Similarly, the Earth’s surface pushes upward on the teapot with the force Y and the teapot pushes downward on the Earth’s surface with the force Z. These two forces form a Newton 3 pair and so are colour-coded red on the diagram.

We can summarise this in the form of a table:


Testing understanding

One the best exam questions to test students’ understanding of Newton’s Third Law (at least in my opinion) can be found here. It is a really clever question from the legacy Edexcel specificiation which changed the way I thought about Newton’s Third Law because I was suddenly struck by the thought that the only force that we, as humans, have direct control over is force D on the diagram below. Yes, if D increases then B increases in tandem, but without the weighty presence of the Earth we wouldn’t be able to leap upwards…