## Applying Lenz’s Law

Nature abhors a change of flux.

D. J. Griffiths’ (2013) genius re-statement of Lenz’s Law, modelled on Aristotle’s historically influential but now debunked aphorism that ‘Nature abhors a vacuum’

A student recently asked for help with this AQA A-level Physics multiple choice question:

This question is, of course, about Lenz’s Law of Electromagnetic Induction. The law can be stated easily enough: ‘An induced current will flow in a direction so that it opposes the change producing it.’ However, it can be hard for students to learn how to apply it.

What follows is my suggested explanatory sequence.

### Step 1: simplify the diagram using the ‘dot and cross’ convention

When the switch is closed, a current I begins to flow in coil P. We can assume that I starts at zero and increases to a maximum value in a very small but not negligible period of time.

### Step 2: consider the magnetic field produced by P

You can read more about a simple method of deducing the direction of the magnetic field produced by a coil or a solenoid here.

### Step 3: apply Faraday’s Law to coil Q

Since Q is experiencing a change in magnetic flux, then an induced current will flow through it.

### Step 4: apply Lenz’s Law to coil Q

The current in coil Q must flow in such a direction so that it opposes the change producing it.

Since P is producing an increasing magnetic flux through Q, then the current in Q must flow in such a way so that it tries to prevent the increase in magnetic flux which is inducing it. The direction of the magnetic field BQ produced by Q must therefore be opposite to the direction of the magnetic field produced by P.

### Step 5: consider the polarity of the magnetic fields of P and Q

We can see the magnetic field lines of coil P produce a north magnetic field on its right hand side. The magnetic field of Q will produce a north magnetic field on its left hand side. Coil P will therefore push coil Q to the right.

It follows that we can eliminate options A and C from the question.

### Step 6: What happens when the magnetic field of P reaches its steady value?

Because the magnetic field produced by coil P has how reached its steady maximum value, this means that the magnetic flux through coil Q also has a constant, unchanging value. Since there is no change in magnetic flux, then this means that no emf is induced across the coil so no induced current flows. Since Q does not have a magnetic field it follows that there is no magnetic interaction between them.

The answer to the question must therefore be D.

### Step 7: check student understanding

For the alternative question, the correct answer of C can be explained by going through a process similar to the one outlined above.

• When the switch is opened, the magnetic flux through Y begins to decrease.
• A changing magnetic flux through Y induces current flow.
• Lenz’s Law predicts that the direction of this current is such that it opposes the change producing it.
• The current through Y will therefore be in the same direction as the current through X to produce a magnetic field in the same direction.
• The coils will attract each other.
• Eventually, the magnetic flux produced by coil X drops to a constant value of zero.
• Since there is no change in magnetic flux through Y, there is no induced current flow through Y and hence no magnetic field.
• There is no magnetic interaction between X and Y and therefore the force on Y is zero.

### Conclusion

I hope teachers find this detailed analysis of a Lenz’s Law question useful! As in much of A-level Physics, the devil is not in the detail but rather in the application of the detail. Students who encounter more examples will have a more secure understanding.

### Reference

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

## The worst circuit in the world (part 2)

What is the worst circuit in the world? Many teachers think it is the one below.

This is the circuit that AQA (2018: 47) strongly suggest should be used to capture the data for plotting IV characteristics (aka current against potential difference graphs) for a fixed resistor, a filament lamp and a diode. The reasons why it is ‘the worst circuit in world’ were outlined in part one; and also some reasons why, nonetheless, schools teaching the 2016 AQA GCSE Physics / Combined Science specifications should (arguably) continue to use it.

The procedure outlined isn’t ‘perfect’ but works well using the equipment we have available and enables students to capture (and plot using a FREE Excel spreadsheet!) the data with only minor troubleshooting from the teacher.

### Step the first: ‘These are the graphs you’re looking for.’

I find this required practical runs more smoothly if students have some awareness of what kind of graphs they are looking for. So, to borrow a phrase, I usually just tell ’em.

You can access an unannotated version of the slides on Google Jamboard and pdf below.

### Step the second: capture the data for the fixed resistor

It is a continual source of amazement to me that students seem to find a photograph of a circuit easier to interpret than a nice, clean, minimalist circuit diagram, so for an easier life I present both.

You can, if you have access to ICT, get the students to plot their results ‘live’ on an Excel spreadsheet (link below). I think this is excellent for helping to manage the cognitive demand on our students (as I have argued before here). Please note that I have not used the automated ‘line of best fit’ tools available on Excel as I think it is important for students to practice drawing lines of best fit — including, especially, curved lines of best fit (sorry, Maths teachers, in science there are such things as curved lines!)

### Step the second: capture the data for the filament lamp

In this circuit, we replaced the previous 0-16 ohm variable resistor with a 0 – 1000 ohm variable resistor paired with 2.5 V, 0.2 A filament lamp because the bulb has a resistance of about 60 ohms when run at 2.5 V and so the 0-16 ohm variable resistor is often ineffective. We allowed a maximum potential difference of just over 3.0 V to ‘over run’ the bulb so as to be sure of obtaining the ‘flattening’ of the graph. The method calls for very small adjustments of the variable resistor to obtain noticeable changes of brightness of the bulb. Note that the cells used in the photograph had seen many years of service with our physics department(!) and so were fairly depleted such that three of them were needed to produce a measly three volts; you would likely only need two ‘fresher’, ‘newer’ cells to achieve the same.

### Resources

And, by popular request, a copy of the PowerPoint below (although, trust me, I think Google Jamboard is superior when using ‘live’ in front of a class)

REFERENCES

## Exploring J. S. Mill’s classification of misconceptions (part 1)

The philosopher John Stuart Mill (1806-1873) offers an intriguing system for classifying misconceptions (or ‘fallacies’ as he terms them) that could be useful for teachers in understanding many of the misconceptions and preconceptions that our students hold.

My own thoughts on this issue have been profoundly shaped by the ‘Resources Framework‘ as presented by authors such as Andrea di Sessa, David Hammer, Edward Redish and others. What follows is not a rejection of this approach but rather an exploration of whether Mill’s work offers some relevant insights. My thought is that it quite possibly might; after all, it has happened before . . .

The authors, however, did not use or refer to Mill’s system of logic in developing the programs or in formulating their theory of instruction. They didn’t discover parallels between their theory of instruction and Mill’s logic until after they had finished writing the bulk of ‘Theory of Instruction’. The discovery occurred when they were writing a chapter on theoretical issues. In their search for literature relevant to their philosophical orientation, they came across Mill’s work and were shocked to discover that they had independently identified all the major patterns that Mill had articulated. ‘Theory of Instruction’ (1982) even had parallel principles to the methods in ‘A System of Logic’ (1843)

Engelmann and Carnine 2013: Chapter 2

## Mill’s system for classifying fallacies

In A System of Logic (1843), Mill argues that

Indifference to truth can not, in and by itself, produce erroneous belief; it operates by preventing the mind from collecting the proper evidences, or from applying to them the test of a legitimate and rigid induction; by which omission it is exposed unprotected to the influence of any species of apparent evidence which offers itself spontaneously, or which is elicited by that smaller quantity of trouble which the mind may be willing to take.

Mill 1843: Book V Chap 1

Mill is saying that we don’t believe false things because we want to, but because there are mechanisms preventing our minds from duly noting and weighing the myriad evidences from which we construct our beliefs about the world by the process of induction.

He suggests that there are five major classes of fallacies:

• A priori fallacies;
• Fallacies of observation;
• Fallacies of generalisation;
• Fallacies of ratiocination; and
• Fallacies of confusion

Erroneous arguments do not admit of such a sharply cut division as valid arguments do. An argument fully stated, with all its steps distinctly set out, in language not susceptible of misunderstanding, must, if it be erroneous, be so in some one of these five modes unequivocally; or indeed of the first four, since the fifth, on such a supposition, would vanish. But it is not in the nature of bad reasoning to express itself thus unambiguously.

Mill 1843: Book V Chap 1

Mill is saying that invalid inferences, by their very nature, are ‘messier’ and harder to classify than correct inferences. However, they must all fit into the five categories outlined above. Actually, they are more likely to fit into the first four categories since clear and unambiguous use of language and terms would tend to eliminate fallacies of confusion as a matter of course.

## What is an a priori fallacy?

In philosophy, a priori means knowledge derived from theoretical deduction rather than from empirical observation or experience.

Mill says that a priori fallacies (which he also calls fallacies of simple observation) are

those in which no actual inference takes place at all; the proposition (it cannot in such cases be called a conclusion) being embraced, not as proved, but as requiring no proof; as a self-evident truth.

Mill 1843: Book V Chap 3

In other words, an a priori fallacy is an idea whose truth is accepted on its face value alone; no evidence or justification of its truth is needed. An example from physics education might be ideas such as ‘heavy objects fall’ or ‘wood floats’. Some students accept these as obvious and self-evident truths: there is no need to consider ideas such as weight and resultant force or density and upthrust because these are ‘brute facts’ about the world that admit of no further explanation. This a case of mislabelling subjective facts as objective facts.

Falling is a location-specific behaviour: objects on Earth will indeed tend to accelerate downwards towards the centre of the Earth: this is a subjective fact which is dependent on the location of the object rather than an objective fact about the behaviour of all objects everywhere (although we could, of course, argue that falling is indeed an objective fact about objects which are subject to the influence of gravitational fields). Equally, floating is not a phenomenon restricted to the interaction between wood and water: many woods will sink in low density oils. ‘Wood floats‘ is not an objective fact about the universe but rather a subjective fact about the interaction of wood with a certain liquid.

This may be why some students are incurious about certain phenomena because they regard them as trivial and obvious rather than manifestations of the inner workings of the universe.

Mill lists many other examples of the a priori fallacy, but his examples are drawn from the history of science and philosophy, and so are of less direct relevance to the science classroom, with the possible exception of the two following examples:

Humans tend to default to the assumption that any phenomenon must necessarily have only a single cause; in other words, we assume that a multiplicity of causes is impossible. We are protected from this version of the a priori fallacy by the guard rail of the scientific method. For a complete understanding of a phenomenon, we look at the effect of one independent variable at a time whilst controlling other possible variables.

There remains one a priori fallacy or natural prejudice, the most deeply-rooted, perhaps, of all which we have enumerated; one which not only reigned supreme in the ancient world, but still possesses almost undisputed dominion over many of the most cultivated minds … This is, that the conditions of a phenomenon must, or at least probably will, resemble the phenomenon itself … the natural prejudice which led people to assimilate the action of bodies upon our senses, and through them upon our minds, to the transfer of a given form from one object to another by actual moulding.

Mill 1843: Book V Chap 3

I think that this tendency might be the one in play with the difficulties that many students have with understanding how images are formed: they think that an image is an evanescent ‘clone’ of the object that is being imaged rather than being an artefact of the light rays reflected or emitted from the object. This also might help explain why students find explaining the colour changes produced by looking at an object through a colour filter or illuminating it with coloured light difficult: they assume that colour is an essential unalterable property that adheres to the object and cannot be changed without changing the object.

We’ll continue this exploration of Mill’s classification of misconceptions in later posts.

### References

Engelmann, S., & Carnine, D. (2013). Could John Stuart Mill Have Saved Our Schools? Attainment Company, Inc.

Mill, J. S. (1843). A System of Logic. Collected Works.

## Should we introduce speed using s×t=d instead of s = d÷t?

It is a truth which is by no means universally acknowledged, but one of which I hope shortly to persuade the reader, that introducing speed to 11-14 year-old students as speed=distance÷time or s=d ÷ t is not the most pedagogically effective approach.

This may initially seem like perverse idea since surely s = d ÷ t and s × t = d are mathematically equivalent expressions? They are, but it is my contention that many students find expressions of the format s = d ÷ t more cognitively demanding that s×t=d. This is because many students struggle with the concept of inverse relationships, particularly those involving multiplication and division.

[Researchers have] suggested that multiplicative concepts may be more difficult to acquire than additive ones, and speculated that although addition and subtraction concepts and procedures extend to multiplication and division, the latter also include unique aspects unrelated to addition and subtraction.

Robinson and LeFevre 2012: 426

In short, many students can handle solving problems such as a + b = c where (say) the numerical values of b and c are known. This can be solved by performing the operation a + b – b = c – b leading to a = c – b and hence a solution to the problem. However, students — and many adults(!) — find solving a similar problem of the format a=b÷c much more problematic, especially in cases when b÷c is not a simple integer.

Compounding students’ inability to utilise multiplicative structures, is their failure to recognise the isomorphism between proportion problems. Another possible reason is that a reluctance or inability to deal with the non-integer relationships (‘avoidance of fractions’), coupled with the high processing loads involved, seems to be the likely cause of this error

Singh 2000: 595

## The problem with the s=d÷t format

In this analysis, we will assume that a direct calculation of s when d and t are known is trivial. The problem with the s=d÷t format is that it may require students to apply two problem solving procedures which, to the novice learner, have highly dissimilar surface features and whose underlying isomorphism is, therefore, hidden from them.

• To find d if s and t are known, they need to multiply both sides by t (see Example 1).
• To find t if s and d are known, they need to divide both sides by s and then multiply both sides by t (see Example 2)

(For more on using the ‘FIFA’ mnemonic for calculations, click on this link.)

## Easing cognitive load with the s x t = d format

As above, we will assume that a direct calculation of d when s and t are known is trivial. What happens when we need to find s and t, given that they are the only unknown quantities?

• If t and d are known, then we can find s by dividing both sides by t (see Example 3).
• If s and t are known, then we can find t by dividing both sides s (see Example 4).

Examples 3 and 4 have highly similar surface features as well as a deeper level isomorphism and allow a commonality of approach which I think is immensely helpful for novice learners.

Robinson and LeFevre (2012: 411) call this type of operation ‘the inversion shortcut’ and argue (for a different context than the one presented here) that:

In three-term problems such as a × b ÷ b, the knowledge that b ÷ b = 1, combined with the associative property of multiplication, allows solvers to implement an inversion shortcut on problems such as 4×24÷24. The computational advantage of using the inversion shortcut is dramatic, resulting in greatly reduced solution times and error rates relative to a left-to-right solution procedure. […] Such knowledge of how inverse operations relate in a variety of circumstances forms the basis for understanding and manipulating algebraic expressions, an important mathematical activity for adolescents

## Conclusion

I think there is a strong case to be made for this mode of presentation to be applied to a wider range of physics contexts for 11-16 year-old students such as:

• Power, so that the definition of power is initially presented as P × t = E or P × t = W; that is to say, we define power as the energy transferred in one second.
• Density, so that ρ × V = m; that is to say, we define density as mass of 1 m3 or 1 cm3.
• Pressure, so that the definition of pressure is initially presented as p × A = F; that is to say, we define pressure as the force exerted on an area of 1 metre squared.
• Acceleration, so that a × t = Δv; that is to say, we define acceleration as the change in velocity produced in one second.

### References

Robinson, K. M., & LeFevre, J. A. (2012). The inverse relation between multiplication and division: Concepts, procedures, and a cognitive framework. Educational Studies in Mathematics79(3), 409-428.

Singh, P. (2000). Understanding the concepts of proportion and ratio among grade nine students in Malaysia. International Journal of Mathematical Education in Science and Technology31(4), 579-599.

## Gears for GCSE Physics

I recently made a bit of a mess of teaching the topic of gears by trying to ‘wing it’ with insufficient preparation. To avoid my — and possibly others’ — future blushes, I thought I would compile a post summarising my interpretation of what students need to know about gears for AQA GCSE Physics.

I am going to include some handy gifs and a clean, un-annotated Google Jamboard (my favoured medium for lessons).

Any continuing errors, omissions or misconceptions are entirely my own fault.

## ‘A simple gear system can be used to transmit the rotational effect of a force’ [AQA 4.5.4]

A gear is a wheel with teeth that can transmit the rotational effect of a force.

For example, in the gear train shown above, the first gear (A) is turned by a motor (green dot shown below). The moment (rotational effect) is passed via the interlocking teeth to gear B and so on down the chain to gear E. It is also worth pointing out that gear A has a clockwise moment but gear B has an anticlockwise moment. The direction alternates as we move down the chain. It takes a gear train of five gears to transmit the clockwise moment from gear A to gear E.

Gears A-E are all equal in size with the same number of teeth and, consequently, the moment does not change in magnitude as it passes down the chain (although, as noted above, it does change direction from clockwise to anticlockwise).

## ‘Students should be able to explain how gears transmit the rotational effect of forces’ [AQA 4.5.4]

### Part 1: A reduction gear arrangement

The driving gear (coloured blue) is smaller and has 6 teeth compared with the large gear’s 18 teeth. This is called a reduction gear arrangement.

A reduction gear arrangement does two things:

1. It slows down the speed of rotation. You may notice that the large gear turns only one for each three turns of the small gear.
2. The larger gear exerts a larger moment than the smaller gear. This is because the distance from the centre to the edge is larger for the grey gear.

The blue gear A exerts a force FA on gear B. By Newton’s Third Law, gear B exerts an equal but opposite force FB on gear A. Let’s take the magnitude of both forces to be F.

The anticlockwise moment exerted by gear A is given by m = F x d. The clockwise moment exerted by gear B is given by M=F x D. Since D > d then M > m.

A reduction gear arrangement is typically used in devices like an electric screwdriver. The electric motor in the device produces only a small rotational moment m but a large moment M is needed to turn the screws. The reduction gear produces the large moment M required.

### Part 2: The overdrive arrangement

What happens when the driver gear is larger and has a greater number of teeth than the driven gear? This is called an overdrive arrangement.

The example we are going to look at is the arrangement of gears on a bicycle.

Here the driver gear (on the left) is linked via a chain to the smaller driven gear on the right. This means that the anticlockwise moment of the first gear is transmitted directly to the second gear as an anticlockwise moment. That is to say, the direction of the moment is not reversed as it is when the two gears are directly linked by interlocking teeth.

In the example shown, the big gear A turns only once for each four turns completed by the smaller gear B. Let’s assume that gear A exerts a force F on the chain so that the chain exerts an identical force F on gear B. Since D > d, this means that M > m so that the arrangement works as a distance multiplier rather than a force multiplier. This is, of course, excellent if we are riding at speed along a horizontal road. However, if we encounter an upward incline we may wish to — using the gear changing arrangement on the bike — swap the small gear B with one with a larger value of d. This would have the happy effect of increasing the magnitude of m so as to make it slightly easier to pedal uphill.

The annotate-able Jamboard is available here.

## Acknowledgements

I used Gear Generator, Gear Generator 2 and EZgif to produce the gear animations.

## Dual coding change of momentum

Rosencrantz (an anguished cry): CONSISTENCY IS ALL I ASK!

Tom Stoppard, Rosencrantz and Guildenstern Are Dead (1966)

I think that dual coding techniques can be extremely helpful in helping students understand the concept of change of momentum.

To engage our students’ physical intuitions, let’s consider a question like: Which would hurt more — being hit by a sandbag or being hit by a rubber ball?

Let’s assume that the sandbag and rubber ball have the same mass m and are travelling at the same initial velocity u. We choose ‘u‘ because it’s the initial velocity and we take ‘v‘ as the final velocity: a very subtle piece of dual coding that can reap rewards if applied consistently — pace Rosencrantz(!) — over a range of disparate examples.

To analyse this problem, let’s use the momentum version of Newton’s Second Law of Motion.

We will use the change = final – initial convention (‘Consistency is all I ask!’)). The initial momentum is pi and the final momentum is pf.

Now let’s work out the change in momentum in each case. We will assume that each item is dropped so that it impacts vertically on a horizontal surface. The velocity just before it hits is u so its initial momentum pi is given by pi = mu; its final velocity is v so its final momentum pf is given by pf = mv. The sandbag does not rebound, so its final velocity v is zero.

The rubber ball rebounds from the surface with a velocity v (we have shown that v < u so we are not assuming a perfectly elastic collision).

We will use the down-is-positive convention so that u is positive and the downward momentum pi are positive in both cases. However, the velocity v of the ball is negative so the momentum pf = mv is negative (upwards).

To add vectors, we simply put them ‘nose to tail’. However, in this case, we need to subtract the vectors, not add them. To do this, we use the operation pf + (-pi,). In other words, we put the vector pf nose to tail with minus pi, or with a vector pointing in the opposite direction to the original vector pi. These are shown in the table.

We can see that the change in momentum Δp is larger in the case of the rubber ball.

Applying Newton Second Law that force = change in momentum / change in time then (assuming the time of each interaction is the same) then we can conclude that the (upward) force exerted by the surface on the ball is larger than the force exerted by the surface on the sandbag.

From Newton’s Third Law (that if an object A exerts a force on object B, then object B exerts an equal and opposite force on object A), we can also conclude that the rubber exerts a larger downward force on the surface. This implies that, if the ball hit (say) your hand, then it would hurt more than the sandbag.

Considering change of momentum problems like this helps students answer questions such as the one shown below:

We can discard options C and D since the change of momentum shown is in the wrong direction: the vertical component of momentum will remain unchanged.

A and B show changes of momentum of the same magnitude in the horizontal direction. However, if we take the horizontal component of the initial momentum as positive then the change of momentum on the gas particle must be negative; this implies that the correct answer is B.

Note also that diagram B shows the pf + (-pi) operation outlined above, with the arrow showing minus pi shown in red (added to the original exam question).

## Gravitational potential: the ‘bottom of a hole’ perspective

April 20, 2112: The sky is flat, the land is flat, and they meet in a circle at infinity. No star shows but the big one, a little bigger than it shows through most of the [asteroid] Belt, but dimmed to red, like the sky. It’s the bottom of a hole, and I must have been crazy to risk it. […] The stars are gone, and the land around me makes no sense. Now I know why they call planet dwellers ‘flatlanders’. I feel like a gnat on a table. I’m sitting here shaking, afraid to step outside. […] I’M AT THE BOTTOM OF A LOUSY HOLE!

Larry Niven, ‘At The Bottom of a Hole’ (1966)

Redish and Kuo (2015: 586) suggest that tapping into our students’ innate physical intuitions can be a very productive teaching strategy. For example, Redish observed some physics instructors teaching non-physics majors how to interpret a potential energy U against the separation r between particles graph (diagram 8(a) below).

The students were finding it difficult to answer the question of whether the particles would attract or repel each other when they had energy E and were at a separation of C. Redish noted that the instructors advised the students to consider the derivative of the curve at C (diagram 8(b) above) and, since it had a positive gradient, to surmise that the force between the particles would therefore be attractive since F=-dU/dr. Redish suggested:

A more effective approach for this population might be to begin with an embodied analogy and implicitly supporting epistemologies valuing physical intuition. Start with treating a potential energy curve as a track or hill and, using the analogy of gravitational potential energy, then place a ball on the hill as shown in Fig. 8c.

Redish and Kuo (2015)

Which way would the ball roll in 8(c) roll? Redish said that the students had no problem deducing that the particles would exert an attractive force on each other at C (and a repulsive force when their energy is E at the smaller value of r) after using this analogy.

### Using students’ physical intuitions to help understand gravitational potential

The episode outlined above reminded me of a science fiction story by Larry Niven that I had read many years ago. In ‘At the Bottom of a Hole’, Niven imagined what landing on a planet would feel like to a ‘Belter’; that is to say, to a human being who had spent their entire life navigating between the small worlds of the asteroid Belt: small planetoid-sized worlds whose shallow gravitational fields required only a low-intensity burn for a spaceship to slip free of their influence forever. An extract from the story is quoted as an introduction to this post: in essence, the ‘Belter’ who has lived his life voyaging between the low mass and low gravity worldlets of the asteroid belt finds it emotionally and psychologically disturbing to find himself at the bottom of a deep gravitational hole.

### Gravitational Fields are always ‘holes’

Gravitational fields are always holes (unlike electric fields, of course, which can be either ‘holes’ or ‘mountains’; this may well form the basis of a later post).

The mass of the Earth produces a much deeper gravitational hole than the much smaller mass of an asteroid.

As a consequence, a spaceship near the Earth’s surface (A) needs to burn a lot more fuel (i.e. do a lot more work) to completely escape the gravitational influence of the Earth (B) then a spaceship near to the surface of an asteroid. The spaceship closer to the asteroid (C) needs a much smaller burn to completely escape its gravitational influence (D).

To a mature space-faring civilisation, living on the surface of a planet could well be likened (and seem as eccentric) as living at the bottom of a spectacularly deep hole.

### Gravitational potential

The gravitational potential of an object of mass M is given by:

Note that the magnitude of V gets larger as r decreases. This allows us to represent a gravitational field in terms of equipotential lines (dotted on the diagram below) as well as field lines (solid).

### Modelling gravitational potential as a three dimensional hole

We can engage our own and our students’ physical intuitions by picturing the equipotential lines as being contour lines indicating the depth of a three dimensional hole.

An object represented as a ball at position A will not tend to roll down into the hole since there is no discernible downhill ‘slope’ at A; in effect, as r tends towards infinity then the object is beyond the effects of M’s gravity. A position outside the gravitational field of a massive object has a gravitational potential of zero.

Let’s think about what happens as r decreases until the object is at B. Here we can intuitively surmise that it will experience a small force tending to make it fall deeper into the hole. How much work will the gravitational field have done moving an object from infinity to this position? The answer is, of course, 0.5 MJ for each kilogram of mass.

How much work will be done by the gravitational field moving the object from B to C? The answer is again an extra 0.5 MJ/kg but note that this happens over a much smaller change in r than before because the gravitational field is becoming more intense. Again, we can intuit that the object will experience a stronger gravitational force at C than at B.

We can go on to argue that a similar pattern of behaviour will also occur at D and E.

But the real value of this representation is, in my opinion, helping students understand how much energy a body needs to escape the influence of a gravitational field.

If we start at B, we would have to do 0.5 MJ/kg of work on it to make it escape. In other words, it needs 0.5 MJ/kg to climb out of the hole.

If we started at C, then we would need 1.0 MJ/kg; and D, 1.5 MJ/kg and so on.

If we were considering a spacecraft operating in the vacuum of space, then transferring 2.0 MJ/kg of kinetic energy would allow ot to completely escape the gravitational influence of M; or, in other words, to reach a value of r such that its gravitational potential is zero.

Near the Earth’s surface where r = 6.38 x 106 m, the gravitational potential can be calculated as follows:

That is to say, a body would need to gain 64.4 MJ of kinetic energy for each kilogram of its mass to completely escape from the influence of the Earth’s gravity.

We can therefore calculate the escape velocity for a body near the Earth surface as follows:

As I mentioned above, I think the real power of this way of tapping into physical intuition for understanding fields comes when we use it to represent electric fields. I will cover that in a later post.

#### Reference

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

## The ‘all-in-a-row’ circuit diagram convention for series and parallel circuits

Circuit diagrams can be seen either as pictures or abstractions but it is clear that pupils often find it hard to recognise the circuits in the practical situation of real equipment. Moreover, Caillot found that students retain from their work with diagrams strong images rather than the principles they are intended to establish. The topological arrangement of a diagram or a drawing presents problems for pupils which are easily overlooked. It seems that pupils’ spatial abilities affect their use of circuit diagrams: they sometimes do not regard as identical several circuits, which, though identical, have been rotated so as to have a different spatial arrangement. […] Niedderer found that pupils, when asked whether a circuit diagram would ‘work’ in practice, more often judged symmetrical diagrams to be functioning than non-symmetric ones.

Driver et al. (1994): 124 [Emphases added]

For the reasons outlined by Driver and others above, I think it’s a good idea to vary the way that we present circuit diagrams to students when teaching electric circuits. If students always see circuit diagrams presented so that (say) the cell is at the ‘top’ and ‘facing’ a certain way; or that they are drawn so that they are symmetrical (which is an aesthetic rather that a scientific choice), then they may well incorrectly infer that these and other ‘accidental’ features of our circuit diagrams are the essential aspects that they should pay the most attention to.

One ‘shake it up’ strategy is to redraw a circuit diagram using the ‘all-in-a-row’ convention.

If you arrange the real components in the ‘all-in-a-row’ arrangement, then a standard digital voltmeter has, what is in my opinion a regrettably underused functionality, that will show:

• ‘positive’ potential differences: that is to say, the energy added to the coulombs as they pass through a cell or the electromotive force; and
• ‘negative’ potential differences: that is to say, the energy removed from each coulomb as they pass through a resistor; these can be usefully referred to as ‘potential drops’

This can be shown on circuit diagrams as shown below/

In other words, the difference between the potential difference across the cell (energy being transferred into the circuit from the chemical energy store of the cell) is explicitly distinguished from the potential difference across the resistor (energy being transferred from the resistor into the thermal energy store of the surroundings). The all-in-a-row convention neatly sidesteps a common misconception that the potential difference across a cell is equal to the potential difference across a resistor: they are not. While they may be numerically equal, they are different in sign, as a consequence of Kirchoff’s Second Law. As I have suggested before, I think that this misconception is due to the ‘hidden rotation‘ built into standard circuit diagrams.

### Potential divider circuits and the all-in-a-row convention

Although I am normally a strong proponent of the ‘parallel first heresy‘, I’ll go with the flow of ‘series circuit first’ in this post.

Diagrams 2 and 3 in the sequence show that the energy supplied to the coulombs (+1.5 V or 1.5 joules per coulomb) by the cell is transferred from the coulombs as they pass through the double resistor combination. Assuming that R1 = R2 then, as diagram 4 shows, 0.75 joules will be transferred out of each coulomb as they pass through R1; as diagram 5 shows, 0.75 joules will be transferred out of each coulomb as they pass through R2.

### Parallel circuits and the all-in-a-row convention

I’ve written about using the all-in-a-row convention to help explain current flow in parallel circuits here, so I will focus on understanding potential difference in parallel circuit in this post.

Again, diagrams 2 and 3 in the sequence show that the positive 3.0 V potential difference supplied by the cell is numerically equal (but opposite in sign) to the negative 3.0 V potential drop across the double resistor combination. It is worth bearing in mind that each coulomb passing through the cell gains 3.0 joules of energy from the chemical energy store of the cell. Diagrams 4 and 5 show that each coulomb passing through either R1 or R1 loses its entire 3.0 joules of energy as it passes through that resistor. The all-in-a-row convention is useful, I think, for showing that each coulomb passes through just one resistor as it makes a single journey around the circuit.

### Acknowledgements

Circuit simulations from the excellent https://phet.colorado.edu/sims/html/circuit-construction-kit-dc/latest/circuit-construction-kit-dc_en.html

Circuit diagrams drawn using https://www.circuit-diagram.org/editor/

### Reference

Driver, R., Squires, A., Rushworth, P., & Wood-Robinson, V. (1994). Making sense of secondary science: Research into children’s ideas. Routledge.

## Electromagnetic induction — using the LEFT hand rule…?

They do observe I grow to infinite purchase,
The left hand way;

John Webster, The Duchess of Malfi

Electromagnetic induction — the fact that moving a conductor inside a magnetic field in a certain direction will generate (or induce) a potential difference across its ends — is one of those rare-in-everyday-life phenomena that students very likely will never have come across before. In their experience, potential differences have heretofore been produced by chemical cells or by power supply units that have to be plugged into the mains supply. Because of this, many of them struggle to integrate electromagnetic induction (EMI) into their physical schema. It just seems such a random, free floating and unconnected fact.

What follows is a suggested teaching sequence that can help GCSE-level students accept the physical reality of EMI without outraging their physical intuition or appealing to a sketchily-explained idea of ‘cutting the field lines’.

### ‘Look, Ma! No electrical cell!’

I think it is immensely helpful for students to see a real example of EMI in the school laboratory, using something like the arrangement shown below.

A length of copper wire used to cut the magnetic field between two Magnadur magnets on a yoke will induce (generate) a small potential difference of about 5 millivolts. What is particularly noteworthy about doing this as a class experiment is how many students ask ‘How can there be a potential difference without a cell or a power supply?’

The point of this experiment is that in this instance the student is the power supply: the faster they plunge the wire between the magnets then the larger the potential difference that will be induced. Their kinetic energy store is being used to generate electrical power instead of the more usual chemical energy store of a cell.

But how to explain this to students?

A common option at this point is to start talking about the conductor cutting magnetic field lines: this is hugely valuable, but I recommend holding fire on this picture for now — at least for novice learners.

What I suggest is that we explain EMI in terms of a topic that students will have recently covered: the motor effect.

This has two big ‘wins’:

• It gives a further opportunity for students to practice and apply their knowledge of the motor effect.
• Students get the chance to explain an initially unknown phenomenon (EMI) in terms of better understood phenomenon (motor effect). The motor effect will hopefully act as the footing (to use a term from the construction industry) for their future understanding of EMI.

### Explaining EMI using the motor effect

The copper conductor contains many free conduction electrons. When the conductor is moved sharply downwards, the electrons are carried downwards as well. In effect, the downward moving conductor can be thought of as a flow of charge; or, more to the point, as an electrical current. However, since electrons are negatively charged, this downward flow of negative charge is equivalent to an upward flow of positive charge. That is to say, the conventional current direction on this diagram is upwards.

Applying Fleming Left Hand Rule (FLHR) to this instance, we find that each electron experiences a small force tugging it to the left — but only while the conductor is being moved downwards.

This results in the left hand side of the conductor becoming negatively charged and the right hand side becoming positively charged: in short, a potential difference builds up across the conductor. This potential difference only happens when the conductor is moving through the magnetic field in such a way that the electrons are tugged towards one end of the conductor. (There is, of course, the Hall Effect in some other instances, but we won’t go into that here.)

As soon as the conductor stops moving, the potential difference is no longer induced as there is no ‘charge flow’ through the magnetic field and, hence, no current and no FLHR motor effect force acting on the electrons.

### Faraday’s model of electromagnetic induction

Michael Faraday (1791-1867) discovered the phenomenon of electromagnetic induction in 1831 and explained it using the idea of a conductor cutting magnetic field lines. This is an immensely valuable model which not only explains EMI but can also generate quantitative predictions and, yes, it should definitely be taught to students — but perhaps the approach outlined above is better to introduce EMI to students.

### The left hand rule not knowing what the right hand rule is doing . . .

We usually apply Fleming’s Right Hand Rule (FRHW) to cases of EMI, Can we replace its use with FLHR? Perhaps, if you wanted to. However, FRHR is a more direct and straightforward shortcut to predicting the direction of conventional current in this type of situation.

## Explaining current flow in conductors (part two)

Do we delve deeply enough into the actual physical mechanism of current flow through electrical conductors using the concepts of charge carriers and electric fields in our treatments for GCSE and A-level Physics? I must reluctantly admit that I am increasingly of the opinion that the answer is no.

In part one we discussed two common misconceptions about the physical mechanism of current flow, namely:

1. The all-the-electrons-in-a-conductor-repel-each-other misconception; and
2. The electric-field-of-the-battery-makes-all-the-charge-carriers-in-the-circuit-move misconception.

What, then, does produce the internal electric field that drives charge carriers through a conductor?

Let’s begin by looking at the properties that such a field should have.

### Current and electric field in an ohmic conductor

(You can see a more rigorous derivation of this result in Duffin 1980: 161.)

We can see that if we consider an ohmic conductor then for a current flow of uniform current density J we need a uniform electric field E acting in the same direction as J.

### What produces the electric field inside a current-carrying conductor?

The electric field that drives charge carriers through a conductor is produced by a gradient of surface charge on the outside of the conductor.

Rings of equal charge density (and the same sign) contribute zero electric field at a location midway between the two rings, whereas rings of unequal charge density (or different sign) contribute a non-zero field at that location.

Sherwood and Chabay (1999): 9

These rings of surface charge produce not only an internal field Enet as shown, but also external fields than can, under the right circumstances, be detected.

### Relationship between surface charge densities and the internal electric field

Picture a large capacity parallel plate capacitor discharging through a length of high resistance wire of uniform cross section so that the capacitor takes a long time to discharge. We will consider a significant period of time (a small fraction of RC) when the circuit is in a quasi-steady state with a current density of constant magnitude J. Since E = J / σ then the internal electric field Enet produced by the rings of surface charge must be as shown below.

In essence, the electric field of the battery polarises the conducting material of the circuit producing a non-uniform arrangement of surface charges. The pattern of surface charges produces an electric field of constant magnitude Enet which drives a current density of constant magnitude J through the circuit.

As Duffin (1980: 167) puts it:

Granted that the currents flowing in wires containing no electromotances [EMFs] are produced by electric fields due to charges, how is it that such a field can follow the tortuous meanderings of typical networks? […] Figure 6.19 shows diagrammatically (1) how a charge density which decreases slowly along the surface of a wire produces an internal E-field along the wire and (2) how a slight excess charge on one side can bend the field into the new direction. Rosser (1970) has shown that no more than an odd electron is needed to bend E around a ninety degree corner in a typical wire.

Rosser suggests that for a current of one amp flowing in a copper wire of cross sectional area of one square millimetre the required charge distribution for a 90 degree turn is 6 x 10-3 positive ions per cm3 which they call a “minute charge distribution”.

### Observing the internal and external electric fields of a current carrying conductor

Jefimenko (1962) commented that at the time

no generally known methods for demonstrating the structure of the electric field of the current-carrying conductors appear to exist, and the diagrams of these fields can usually be found only in the highly specialized literature. This […] frequently causes the student to remain virtually ignorant of the structure and properties of the electric field inside and, especially, outside the current-carrying conductors of even the simplest geometry.

Jefimenko developed a technique involving transparent conductive ink on glass plates and grass seeds (similar to the classic linear Nuffield A-level Physics electrostatic practical!) to show the internal and external electric field lines associated with current-carrying conductors. Dry grass seeds “line up” with electric field lines in a manner analogous to iron filings and magnetic field lines.

### Next post

In part 3, we will analyse the transient processes by which these surface charge distributions are set up.

### References

Duffin, W. J. (1980). Electricity and magnetism (3rd ed.). McGraw Hill Book Co.

Jefimenko, O. (1962). Demonstration of the electric fields of current-carrying conductorsAmerican Journal of Physics30(1), 19-21.

Rosser, W. G. V. (1970). Magnitudes of surface charge distributions associated with electric current flow. American Journal of Physics38(2), 265-266.

Sherwood, B. A., & Chabay, R. W. (1999). A unified treatment of electrostatics and circuits. URL http://cil. andrew. cmu. edu/emi. (Note: this article is dated as 2009 on Google Scholar but the text is internally dated as 1999)