Necessity may well be the mother of invention, but teacher desperation is often the mother of new pedagogy.
For an unconscionably long time, I think that I failed to adequately help students understand the so called ‘equations of motion’ (the mathematical descriptions of uniformly accelerated motion using the standard v, u, a, s and t notation) because I suffered from the ‘curse of knowledge’: I did not find the topic hard, so I naturally assumed that students wouldn’t either. This, sadly, proved not to be the case, even when the equation was printed on the equation sheet!
What follows is a summary of a dual coding technique that I have found really helpful in helping students become confident with problems involving the ‘equations of motion’. This is especially true at GCSE, where students encounter formulas such as
for the first time.
A dual coding convention for representing motion
Applying dual coding to an equation of motion problem
EXAMPLE: A car is travelling at 6.0 m/s. As the car passes a lamp-post it accelerates up to a velocity of 14.2 m/s over a distance of 250 m. Calculate a) the acceleration; and b) the time taken for this change.
The problem can be represented using the dual coding convention as shown below.
Note that the arrow for v is longer than the arrow for u since the car has a positive acceleration; that is to say, the car in this example is speeding up. Also, the convention has a different style of arrow for acceleration, emphasising that it is an entirely different type of quantity from velocity.
We can now answer part (a) using the FIFA calculation system.
Part (b) can be answered as shown below.
Visualising Stopping Distance Questions
These can be challenging for many students, as we often seem to grabbing numbers and manipulating them without rhyme or reason. Dual coding helps make our thinking explicit,
EXAMPLE: A driver has a reaction time of 0.7 seconds. The maximum deceleration of the car is 2.6 metres per second squared. Calculate the overall stopping distance when the car is travelling at 13 m/s.
We need to calculate both the thinking distance and the braking distance to solve this problem
The acceleration of the car is zero during the driver’s reaction time, since the brakes have not been applied yet.
We visualise the first part of the problem like this.
Now let’s look at the second part of the question.
There are three things to note:
Since the car comes to a complete stop, the final velocity v is zero.
The acceleration is negative (shown as a backward pointing arrow) since we are talking about a deceleration: in other words, the velocity gradually decreases in size from the initial velocity value of u as the car traverses the distance s.
Since the second part of the question does not involve any consideration of time we have omitted the ‘clock’ symbols for the second part of the journey.
We can now apply FIFA:
Since the acceleration arrow points in the opposite direction to the positive arrow, we enter it as a negative value. When we get to the third line of the Fine Tune stage, we see that a negative 169 divided by negative 4.4 gives positive 38.408 — in other words, the dual coding convention does the hard work of assigning positive and negative values!
And finally, we can see that the overall stopping distance is 9.1 + 38.4 = 47.5 metres,
I have found this form of dual coding extremely useful in helping students understand the easily-missed subtleties of motion problems, and hope other science teachers will give it a try.
If you have enjoyed this post, you might enjoy these also:
Students undoubtedly find electromagnetism tricky, especially at GCSE.
I have found it helpful to start with the F = BIl formula.
This means that we can use the streamlined F-B-I mnemonic — developed by no less a personage than Robert J. Van De Graaff (1901-1967) of Van De Graaff generator fame — instead of the cumbersome “First finger = Field, seCond finger = Current, thuMb = Motion” convention.
Students find the beginning steps of applying Fleming’s Left Hand Rule (FLHR) quite hard to apply, so I print out little 3D “signposts” to help them. You can download file by clicking the link below.
The brain is wider than the sky, For, put them side by side, The one the other will include With ease, and you beside.
Emily Dickinson, ‘The Brain’
Most science teachers find that ‘Space’ is one of the most enduringly fascinating topics for many students: the sense of wonder engendered as our home planet becomes lost in the empty vastness of the Solar System, which then becomes lost in the trackless star-studded immensity of the Milky Way galaxy, is a joy to behold.
But a common question asked by students is: How do we know all this? How do we know the distance to the nearest star to the Sun is 4 light-years? Or how do we know the distance to the Sun? Or the Moon?
I admit, with embarrassment, that I used to answer with a casual and unintentionally-dismissive ‘Oh well, scientists have measured them!’ which (though true) must have sounded more like a confession of faith rather than a sober recounting of empirical fact. Which, to be fair, it probably was; simply because I had not yet made the effort to find out how these measurements were first taken.
The technological resources available to our ancestors would seem primitive and rudimentary to our eyes but, coupled with the deep well of human ingenuity that I like to think is a hallmark of our species, it proved not just ‘world-beating’ but ‘universe-beating’.
I hope you enjoy this whistle stop tour of this little-visited corner of the scientific hinterland, and choose to share some these stories with your students. It is good to know that the brain is indeed ‘wider than the sky’.
I have presented this in a style and format suitable for sharing and discussing with KS3/KS4 students (11-16 year olds).
Mad dogs and Eratosthenes go out in the midday Sun…
To begin at the beginning: the first reliable measurement of the size of the Earth was made in 240 BC and it all began (at least in this re-telling) with the fact that Eratosthenes liked talking to tourists. (‘Err-at-oss-THen-ees’ with the ‘TH’ said as in ‘thermometer’ — never forget that students of all ages often welcome help in learning how to pronounce unfamiliar words)
Alexandria (in present day Egypt) was a thriving city and a tourist magnet. Eratosthenes made a point of speaking to as many visitors as he could. Their stories, taken with a pinch of salt, were an invaluable source of information about the wider world. Eratosthenes was chief librarian of the Library of Alexandria, regarded as one of the Seven Wonders of the World at the time, and considered it his duty to collect, catalogue and classify as much information as he could.
One visitor, present in Alexandria on the longest day of the year (June 21st by our calendar), mentioned something in passing to Eratosthenes that the Librarian found hard to forget: ‘You know,’ said the visitor, ‘at noon on this day, in my home town there are no shadows.’
How could that be? pondered Eratosthenes. There was only one explanation: the Sun was directly overhead at noon on that day in Syene (the tourist’s home town, now known as Aswan).
The same was not true of Alexandria. At noon, there was a small but noticeable shadow. Eratosthenes measured the angle of the shadow at midday on the longest day. It was seven degrees.
No shadows at Syene, but a 7 degree shadow at Alexandria at the exact same time. Again, there was only one explanation: Alexandria was ’tilted’ by 7 degrees with respect to Syene.
Seven degrees of separation
The sphericity of the Earth had been recognised by astronomers from c. 500 BC so this difference was no surprise to Eratosthenes, but what he realised that since he was comparing the length of shadows at two places on the Earth’s surface at the same time then the 7o wasn’t just the angle of the shadow: 7o was the angle subtended at the centre of the Earth by radial lines drawn from both locations.
Eratosthenes paid a person to pace out the distance between Alexandria and Syene. (This was not such an odd request as it sounds to our ears: in the ancient world there were professionals called bematists who were trained to measure distances by counting their steps.)
It took the bematist nearly a month to walk that distance and it turned out to be 5000 stadia or 780 km by our measurements.
Eratosthenes then used a simple ratio method to calculate the circumference of the Earth, C:
The modern value for the radius of the Earth is 6371 km.
Ifs and buts…
There is still some debate as to the actual length of one Greek stadium but Eratosthenes’ measurement is generally agreed to within 1-2% of the modern value.
Sadly, none of the copies of the book where Eratosthenes explained his method called On the measure of the earth have survived from antiquity so the version presented here is a simplified one outlined by Cleomedes in a later book. For further details, readers are directed to the excellent Wikipedia article on Eratosthenes.
Astronomer Carl Sagan also memorably explained this method in his 1980 TV documentary series Cosmos.
You might want to read…
This is part of a series exploring how humans ‘measured the size of the sky’:
Nuclear binding energy and binding energy per nucleon are difficult concepts for A-level physics students to grasp. I have found the ‘pool table analogy’ that follows helpful for students to wrap their heads around these concepts.
Since mass and energy are not independent entities, their separate conservation principles are properly a single one — the principle of conservation of mass-energy. Mass can be created or destroyed , but when this happens, an equivalent amount of energy simultaneously vanishes or comes into being, and vice versa. Mass and energy are different aspects of the same thing.
Beiser 1987: 29
E = mc2
There, I’ve said it. This is the first time I have directly referred to this equation since starting this blog in 2013. I suppose I have been more concerned with the ‘andallthat‘-ness side of things rather than E=mc2. Well, no more! E=mc2 will form the very centre of this post. (And about time too!)
The E is for ‘rest energy’: that is to say, the energy an object of mass m has simply by virtue of being. It is half the energy that would be liberated if it met its antimatter doppelganger and particles and antiparticles annihilated each other. A scientist in a popular novel sternly advised a person witnessing an annihilation event to ‘Shield your eyes!’ because of the flash of electromagnetic radiation that would be produced.
Well, you could if you wanted to, but it wouldn’t do much good since the radiation would be in the form of gamma rays which are to human eyes what the sound waves from a silent dog whistle are to human ears: beyond the frequency range that we can detect.
The main problem is likely to be the amount of energy released since the conversion factor is c2: that is to say, the velocity of light squared. For perspective, it is estimated that the atomic bomb detonated over Hiroshima achieved its devastation by directly converting only 0.0007 kg of matter into energy. (That would be 0.002% of the 38.5 kg of enriched uranium in the bomb.)
Matter contains a lot of energy locked away as ‘rest energy’. But these processes which liberate rest energy are mercifully rare, aren’t they?
No, they’re not. As Arthur Beiser put it in his classic Concepts of Modern Physics:
In fact, processes in which rest energy is liberated are very familiar. It is simply that we do not usually think of them in such terms. In every chemical reaction that evolves energy, a certain amount of matter disappears, but the lost mass is so small a fraction of the total mass of the reacting substances that it is imperceptible. Hence the ‘law’ of conservation of mass in chemistry.
Beiser 1987: 29
Building a helium atom
The constituents of a helium nucleus have a greater mass when separated than they do when they’re joined together.
Here, I’ll prove it to you:
The change in mass due to the loss of energy as the constituents come together is appreciable as a significant fraction of its original mass. Although 0.0293/4.0319*100% = 0.7% may not seem like a lot, it’s enough of a difference to keep the Sun shining.
The loss of energy is called the binding energy and for a helium atom it corresponds to a release of 27 MeV (mega electron volts) or 4.4 x 10-12 joules. Since there are four nucleons (particles that make up a nucleus) then the binding energy per nucleon (which is a guide to the stability of the new nucleus) is some 7 MeV.
But why must systems lose energy in order to become more stable?
The Pool Table Analogy for binding energy
Imagine four balls on a pool table as shown.
The balls have the freedom to move anywhere on the table in their ‘unbound’ configuration.
However, what if they were knocked into the corner pocket?
To enter the ‘bound’ configuration they must lose energy: in the case of the pool balls we are talking about gravitational potential energy, a matter of some 0.30 J per ball or a total energy loss of 4 x 0.30 = 1.2 joules.
The binding energy of a pool table ‘helium nucleus’ is thus some 1.2 joules while the ‘binding energy per nucleon’ is 0.30 J. In other words, we would have to supply 1.2 J of energy to the ‘helium nucleus’ to break the forces binding the particles together so they can move freely apart from each other.
Just as a real helium nucleus, the pool table system becomes more stable when some of its constituents lose energy and less stable when they gain energy.
Beiser, A. (1987). Concepts of modern physics. McGraw-Hill Companies.
. . . 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 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.
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.
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
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.
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.
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.
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 🙂
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.
Lenz, E. (1834), “Ueber die Bestimmung der Richtung der durch elektodynamische Vertheilung erregten galvanischen Ströme”, Annalen der Physik und Chemie, 107 (31), pp. 483–494
Griffiths, David (2013). Introduction to Electrodynamics. p. 315.
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 embodiedcognition 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 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.
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.
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.
A solenoid is an electromagnet made of a wire in the form of a spiral whose length is larger than its diameter.
The word solenoid literally means ‘pipe-thing‘ since it comes from the Greek word ‘solen‘ for ‘pipe’ and ‘-oid‘ for ‘thing’.
And they are such an all-embracingly useful bit of kit that one might imagine an alternate universe where The Troggs might have sang:
Pipe-thing! You make my heart sing! You make everything groovy, pipe-thing!
And pipe-things do indeed make everything groovy: solenoids are at the heart of the magnetic pickups that capture the magnificent guitar riffs of The Troggs at their finest.
The Butterfly Field
Very few minerals are naturally magnetised. Lodestones are pieces of the ore magnetite that can attract iron. (The origin of the name is probably not what you think — it’s named after the region, Magnesia, where it was first found). In ancient times, lodestones were so rare and precious that they were worth more than their weight in gold.
Over many centuries, by patient trial-and-error, humans learned how to magnetise a piece of iron to make a permanent magnet. Permanent magnets now became as cheap as chips.
A permanent bar magnet is wrapped in an invisible evanescent magnetic field that, given sufficient poetic license, can remind one of the soft gossamery wings of a butterfly…
The field lines seem to begin at the north pole and end at the south pole. ‘Seem to’ because magnetic field lines always form closed loops.
This is a consequence of Maxwell’s second equation of Electromagnetism (one of a system of four equations developed by James Clark Maxwell in 1873 that summarise our current understanding of electromagnetism).
Using the elegant differential notation, Maxwell’s second equation is written like this:
This also tells us that magnetic monopoles (that is to say, isolated N and S poles) are impossible. A north-seeking pole is always paired with a south-seeking pole.
Magnetising a solenoid
A current-carrying coil will create a magnetic field as shown below.
The wire is usually insulated (often with a tough, transparent and nearly invisible enamel coating for commercial solenoids), but doesn’t have to be. Insulation prevents annoying ‘short circuits’ if the coils touch. At first sight, we see the familiar ‘butterfly field’ pattern, but we also see a very intense magnetic field in the centre of the solenoid,
For a typical air-cored solenoid used in a school laboratory carrying one ampere of current, the magnetic field in the centre would have a strength of about 84 microtesla. This is of the same order as the Earth’s magnetic field (which has a typical value of about 50 microtesla). This is just strong enough to deflect the needle of a magnetic compass placed a few centimetres away and (probably) make iron filings align to show the magnetic field pattern around the solenoid, but not strong enough to attract even a small steel paper clip. For reference, the strength of a typical school bar magnet is about 10 000 microtesla, so our solenoid is over one hundred times weaker than a bar magnet.
However, we can ‘boost’ the magnetic field by adding an iron core. The relative permeability of a material is a measurement of how ‘transparent’ it is to magnetic field lines. The relative permeability of pure iron is about 1500 (no units since it’s relative permeability and we are comparing its magnetic properties with that of empty space). However, the core material used in the school laboratory is more likely to be steel rather than iron, which has a much more modest relative permeability of 100.
So placing a steel nail in the centre of a solenoid boosts its magnetic field strength by a factor of 100 — which would make the solenoid roughly as strong as a typical bar magnet.
But which end is north…?
The N and S-poles of a solenoid can change depending on the direction of current flow and the geometry of the loops.
The typical methods used to identify the N and S poles are shown below.
To go in reverse order for no particular reason, I don’t like using the second method because it involves a tricky mental rotation of the plane of view by 90 degrees to imagine the current direction as viewed when looking directly at the ends of the magnet. Most students, understandably in my opinion, find this hard.
The first method I dislike because it creates confusion with the ‘proper’ right hand grip rule which tells us the direction of the magnetic field lines around a long straight conductor and which I’ve written about before . . .
The direction of the current in the last diagram is shown using the ‘dot and cross’ convention which, by a strange coincidence, I have also written about before . . .
How a solenoid ‘makes’ its magnetic field . . .
To begin the analysis we imagine the solenoid cut in half: what biologists would call a longitudinal section. Then we show the current directions of each element using the dot and cross convention. Then we consider just two elements, say A and B as shown below.
Continuing this analysis below:
The region inside the solenoid has a very strong and nearly uniform magnetic field. By ‘uniform’ we mean that the field lines are nearly straight and equally spaced meaning that the magnetic field has the same strength at any point.
The region outside the solenoid has a magnetic field which gradually weakens as you move away from the solenoid (indicated by the increased spacing between the field lines); its shape is also nearly identical to the ‘butterfly field’ of a bar magnet as mentioned above.
Since the field lines are emerging from X, we can confidently assert that this is a north-seeking pole, while Y is a south-seeking pole.
Which end is north, using only the ‘proper’ right hand grip rule…
First, look very carefully at the geometry of current flow (1).
Secondly, isolate one current element, such as the one shown in picture (2) above.
Thirdly, establish the direction of the field lines using the standard right hand grip rule (3).
Since the field lines are heading into this end of the solenoid, we can conclude that the right hand side of this solenoid is, in fact, a south-seeking pole.
In my opinion, this is easier and more reliable than using any of the other alternative methods. I hope that readers that have read this far will (eventually) come to agree.