Has a school physics experiment or demonstration ever changed the course of human history?

On 21 April 1820, one such demonstration most definitely did. According to physics lore, Hans Christian Øersted was attempting to demonstrate to his students that, according to the scientific understanding of the day, there was in fact no connection between magnetism and electricity.

To this laudable end, he placed a compass needle near to a wire to show that when the current was switched on, the needle would not be affected.

Except that it was affected. Frequently. Each and every time Øersted switched on the electric current, the needle was deflected from pointing North.

Everybody has heard that wise old saw that ‘If it doesn’t work, it’s physics…” except that in this case ‘It did actually work as it was supposed to but in an unexpected way due to a hitherto-unknown-completely-new-branch-of-physics.’

Øersted, to his eternal credit, did not let it lie there and was a pioneer of the new science of electromagnetism.

Push-me-pull-you: or, two current-carrying conductors

One curious consequence of Øersted’s new science was the realisation that, since electric currents create magnetic fields, two wires carrying electric currents will exert a force on each other.

Let’s consider two long, straight conductors placed parallel to each other as shown.

Screenshot 2020-03-22 at 14.39.37.png

In the diagram above, the magnetic field produced by the current in A is shown by the green lines. Applying Fleming’s Left Hand Rule* to conductor B, we find that a force is produced on B which acts towards conductor A. We could go through a similar process to find the force acting on B, but it’s far easier to apply Newton’s Third Law instead: if body A exerts a force on body B, then body B exerts an equal and opposite force on body A. Hence, conductor A experiences a force which pulls it towards conductor B.

So, two long, straight conductors carrying currents in the same direction will be attracted to each other. By a similar analysis, we find that two long, straight conductors carrying currents in opposite directions will be repelled from each other.

In the past, this phenomenon was used to define the ampere as the unit of current: ‘The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 m apart in vacuum, would produce between these conductors a force equal to 2×10−7 newton per metre of length.‘ However, the 2019 redefinition of the SI system has ditched this and adopted a new definition in terms of the transfer of the elementary charge, e.

Enter Albert Einstein, pursuing an enigma

What is the connection between magnetism and electricity? It was precisely this puzzle that started Albert Einstein on the road to special relativity. It is one of the unsung triumphs of this theory that it lays bare the connection between magnetism and electricity.

In what follows, we’re going to apply Einstein’s analysis to the situation of two long, straight current-carrying conductors. Acknowledgment: I’m going to following a line of argument laid out in Beiser 1988: 19-22.

It’s gotta be perfect (or ‘idealised’, if you prefer)

Let’s consider two idealised conductors A and B both at rest in the inertial reference frame of the laboratory. The flow of charge in both conductors is made up of positive and negative charge carriers moving in opposite directions with a speed v.

None of the charges in A interact with the other charges in A because we are considering an idealised conductor. However, the charges in A will interact with the charges in B.

Screenshot 2020-03-22 at 16.08.06.png
Two conductors viewed from the inertial frame of the laboratory

Flip the inertial reference frame

Now let’s look at the situation from the inertial reference frame of one of the positive charges in A. For simplicity, we can focus on a single positive charge in A since it does not interact with any of the other charges in A.

With reference to this inertial frame, the positive charge in A is stationary and the positive charges in B are also stationary.

However, the inertial frame of the laboratory is moving right-to-left with a speed v and the negative charges are moving right-to-left with a speed of 2v.

Screenshot 2020-03-22 at 16.13.28.png
The same two conductors viewed from the inertial frame of one of the positive charges in conductor A. Note that all the positive charges are now stationary; the laboratory is moving with speed v right to left, and the negative charges are moving with speed 2v right to left

Since the positive charges in B are stationary with respect to the positive charge in A, the distance between them is the same as it was in the laboratory inertial frame. However, since the negative charges in B are moving with speed 2v with respect to positive charge in A, the spacing between is contracted due to relativistic length contraction (see Lottie and Lorentzian Length Contraction).

Because of this, the negative charge density of B increases since they are closer together. However, the positive charge density of B remains the same since they are stationary relative to the positive charge in A so there is no length contraction.

This means that, as far as the positive charge in A is concerned, conductor B has a net negative charge which means the positive charge experiences an attractive Coulomb’s Law electrical force towards B.

A similar analysis applied to electric currents in opposite directions would show that the positive charge in A would experience a repulsive Coulomb’s Law electrical force. The spacing between the positive charges in B would be contracted but the spacing between the negative charges remains unchanged, so conductor B has a net positive charge because the positive charge density has increased but the negative charge density is unchanged.


So what we normally think of as a ‘magnetic’ force in the inertial frame of the laboratory can be explained as a consequence of special relativity altering the charge densities in conductors. Although we have just considered a special case, all magnetic phenomena can be interpreted on the basis of Coulomb’s Law, charge invariance** and special relativity.

For the interested reader, Duffin (1980: 388-390) offers a quantitative analysis where he uses a similar argument to derive the expression for the magnetic field due to a long straight conductor.

Update: I’m also indebted to @sbdugdale who points out the there’s a good treatment of this in the Feynman Lectures on Physics, section 13.6.

Notes and references

* Although you could use a non-FLHR catapult field analysis, of course

** ‘A current-carrying conductor that is electrically neutral in one frame of reference might not be neutral in another frame. How can this observation be reconciled with charge invariance? The answer is that we must consider the entire circuit of which the conductor is a part. Because a circuit must be closed for a current to occur in it, for every current element in one direction that a moving observer find to have, say, a positive charge, there must be another current element in the opposite direction which the same observer finds to have a negative charge. Hence, magnetic forces always act between different parts of the same circuit, even though the circuit as a whole appears electrically neutral to all observers.’ Beiser 1988: 21

Beiser, A. (1988). Concepts of modern physics. Tata McGraw-Hill Education

Duffin, W. J. (1980). Electricity and magnetism. McGraw-Hill.

Electric Motors Without The Left Hand Rule

There is little doubt that students find understanding how an electric motor works hard.

What follows is an approach that neatly sidesteps the need for applying Fleming’s Left Hand Rule (FLHR) by using the idea of the catapult field.

The catapult field is a neat bit of Physics pedagogy that appears to have fallen out of favour in recent years for some unknown reason. I hope to rehabilitate and publicise this valuable approach so that more teachers may try out this electromagnetic ‘road less travelled’.

Screenshot 2020-02-21 at 09.48.42

(Incidentally, if you are teaching FLHR, the mnemonic shown above is not the best way to remember it: try using this approach instead.)

The magnetic field produced by a long straight conductor

Moving electric charges produce magnetic fields. When a current flows through a conductor, it produces a magnetic field in the form of a series of cylinders centred on the wire. This is usually shown on a diagram like this:

Screenshot 2020-02-21 at 09.59.26.png

If we imagine looking down from a point directly above the centre of the conductor (as indicated by the disembodied eye), we would see a plan view like this:

Screenshot 2020-02-21 at 10.01.22.png

We are using the ‘dot and cross‘ convention (where an X represents an arrow heading away from us and a dot represents an arrow heading towards us) to easily render a 3D situation as a 2D diagram.

The direction of the magnetic field lines is found by using the right hand grip rule.

Screenshot 2020-02-21 at 10.38.32.png

The thumb is pointed in the direction of the current. The field lines ‘point’ in the same direction as the fingers on the right hand curl.

3D to 2D

Now let’s think about the interaction between the magnetic field of a current carrying conductor and the uniform magnetic field produced by a pair of magnets.

In the diagrams below, I have tried to make the transition between a 3D and a 2D representation explicit, something that as science teachers I think we skip over too quickly — another example of the ‘curse of knowledge’, I believe.

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Magnetic Field on Magnetic Field

If we place the current carrying conductor inside the magnetic field produced by the permanent magnets, we can show the magnetic fields like this:

Screenshot 2020-02-21 at 11.32.31.png

Note that, in the area shaded green, the both sets of magnetic field lines are in the same direction. This leads a to stronger magnetic field here. However, the opposite is true in the region shaded pink, which leads to a weaker magnetic field in this region.

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The resultant magnetic field produced by the interaction between the two magnetic fields shown above looks like this.

Screenshot 2020-02-21 at 11.37.19.png

Note that the regions where the magnetic field is strong have the magnetic field lines close together, and the regions where it is weak have the field lines far apart.

The Catapult Field

This arrangement of magnetic field lines shown above is unstable and is called a catapult field.

Essentially, the bunched up field lines will push the conductor out of the permanent magnetic field.

If I may wax poetic for a moment: as an oyster will form a opalescent pearl around an irritant, the permanent magnets form a catapult field to expel the symmetry-destroying current-carrying conductor.

Screenshot 2020-02-21 at 11.50.55.png

The conductor is pushed in the direction of the weakened magnetic field. In a highly non-rigorous sense, we can think of the conductor being pushed out of the enfeebled ‘crack’ produced in the magnetic field of the permanent magnets by the magnetic field of the current carrying conductor…

Also, the force shown by the green arrow above is in exactly the same direction as the force predicted by Fleming’s Left Hand Rule, but we have established its direction using only the right hand grip rule and a consideration of the interaction between two magnetic field.

The Catapult Field for an electric motor

First, let’s make sure that students can relate the 3D arrangement for an electric motor to a 2D diagram.

Screenshot 2020-02-21 at 13.42.51.png

The pink highlighted regions show where the field lines due to the current in the conductor (red) are in the opposite direction to the field line produced by the permanent magnet (purple). These regions are where the purple field lines will be weakened, and the clear inference is that the left hand side of the coil will experience an upward force and the right hand side of the coil will experience a downward force. As suggested (perhaps a little fancifully) above, the conductors are being forced into the weakened ‘cracks’ produced in the purple field lines.

The catapult field for the electric motor would look, perhaps, like this:

Screenshot 2020-02-21 at 14.39.36.png

And finally…

On a practical teaching note, I wouldn’t advise dispensing with Fleming’s Left Hand Rule altogether, but hopefully the idea of a catapult field adds another string to your pedagogical bow as far as teaching electric motors is concerned (!)

I have certainly found it useful when teaching students who struggle with applying Fleming’s Left Hand Rule, and it is also useful when introducing the Rule to supply an understandable justification why a force is generated by a current in a magnetic field in the first place.

The catapult field is a ‘road less travelled’ in terms of teaching electromagnetism, but I would urge you to try it nonetheless. It may — just may — make all the difference.

Why does kinetic energy = 1/2mv^2?

Why does kinetic energy Ek=½mv2?

Students and non-specialist teachers alike wonder: whence the half?

This post is intended to be a diagrammatic answer to this question using a Singapore Bar Model approach: so pedants, please avert your eyes.

I am indebted to Ben Rogers’ recent excellent post on showing momentum using the Bar Model approach for starting me thinking along these lines.

Part the First: How to get the *wrong* answer

Imagine pushing an object with a mass m with a constant force F so that it accelerates with a constant acceleration a so that covers a distance s in a time t. The object was initially at rest and ends up moving at velocity v.

Screenshot 2019-03-09 at 14.24.59.png

(On the diagram, I’ve used the SUVAT dual coding conventions that I suggested in a previous post.)

So let’s consider the work done on the object by the force:

Step 1: work done = force x distance moved in the direction of the force

Step 2: Wd = F x s

But remember s = v x t so:

Step 3: Wd = F x vt

And also remember that F = m x a so:

Step 4: Wd = ma x vt

Also remember that a = change in velocity / time, so a = (v – 0) / t = v / t.

Step 5: Wd = m (v / t) x vt

The ts cancel so:

Step 6: Wd = mv2

Since this is the work done on the object by the force, it is equal to the energy transferred to the kinetic energy store of the object. In other words, it is the energy the object has gained because it is moving — its kinetic energy, no less: Ek = mv2.

On a Singapore Bar Model diagram this can be represented as follows:

Screenshot 2019-03-09 at 15.14.17

The kinetic energy is represented by the volume of the bar.

But wait: Ek=mv2!?!?

That’s just wrong: where did the half go?

Houston, we have a problem.

Part the Second: how to get the *right* answer

The problem lies with Step 3 above. We wrongly assumed that the object has a constant velocity over the whole of the distance s.

Screenshot 2019-03-09 at 17.35.43.png

It doesn’t because it is accelerating: it starts off moving slowly and ends up moving at the maximum, final velocity v when it has travelled the total distance s.

So Step 3 should read:

But remember that s = (average velocity) x t.

Because the object is accelerating at a constant rate, the average velocity is (v + u) / 2 and since u = 0 then average velocity is v / 2.

Step 3: Wd= F x (v / 2) t

And also remember that F = m x a so:

Step 4: Wd= ma x (v / 2) t

Also remember that a = change in velocity / time, so a = (v – 0) / t = v / t.

Step 5: Wd = m (v / t) x (v / 2) t

The ts cancel so:

Step 6: Wd= ½mv2

Based on this, of course, Ek = ½mv2
(Phew! Houston, we no longer have a problem.)

Screenshot 2019-03-09 at 17.58.45.png

Using the Bar Model representation, the volume of the bar which is above the blue plane represents the kinetic energy of an object of mass m moving at a velocity v.

Another way of representing the kinetic energy as a solid prism is shown below.

The reason it is half the volume of the bar and not the full volume (as in the incorrect Part the First analysis) is because we are considering the work done by a constant force accelerating an object which is initially at rest; the velocity of the object increases gradually from zero as the force acts upon it. It therefore takes a longer time to cover the distance s than if it was moving at a constant velocity v from the very beginning.

So there we have it, Ek = ½mv2 by a rather circuitous method.

But why go “all around the houses” in this manner? For exactly the same reason as we might choose to go by the path less travelled on some of our other journeys: quite simply, we might find that we enjoy the view.

Dual-coding SUVAT Problems

The theory of dual coding holds that the formation of mental images, in tandem with verbal processing, is often very helpful for learners. In other words, if we support verbal reasoning with visual representations, then better learning happens.

Many years ago, I was taught the dual coding technique outlined below to help with SUVAT problems. Of course, it wasn’t referred to as “dual coding” back then, but dual coding it most definitely is.

I found it a very useful technique at the time and I still find it useful to this day. And what is more, it is in my opinion a pedagogically powerful procedure. I genuinely believe that this technique helps students understand the complexities and nuances of SUVAT because it brings many things which are usually implicit out into the open and makes them explicit.

SUVAT: “Made darker by definition”?

BOSWELL. ‘He says plain things in a formal and abstract way, to be sure: but his method is good: for to have clear notions upon any subject, we must have recourse to analytick arrangement.’

JOHNSON. ‘Sir, it is what every body does, whether they will or no. But sometimes things may be made darker by definition. I see a cow, I define her, Animal quadrupes ruminans cornutum. But a goat ruminates, and a cow may have no horns. Cow is plainer.

— Boswell’s Life of Johnson (1791)

As I see it, the enduring difficulty with SUVAT problems is that such things can indeed be made darker by definition. Students are usually more than willing to accept the formal definitions of s, u, v, a and t and can apply them to straightforward and predictable problems. However, the robotic death-by-algorithm approach fails all too frequently when faced with even minor variations on a theme.

Worse still, students often treat acceleration, displacement and velocity as nearly-synonymous interchangeable quantities: they are all lumped together in that naive “intuitive physics” category called MOVEMENT.

The approach that follows attempts to make students plainly see differences between the SUVAT quantities and, hopefully, as make them as plain as a cow (to borrow Dr Johnson’s colourful phrasing).

Visual Symbols for the Dual-coding of SUVAT problems

Screenshot 2018-12-25 at 12.02.38.png

1.1 Analysing a simple SUVAT problem using dual coding

Problem: a motorcycle accelerates from rest at 0.8 m/s2 for a time of 6.0 seconds. Calculate (a) the distance travelled; and (b) the final velocity.

Screenshot 2018-12-25 at 12.09.42.png

Please note:

  1. We are using the AQA-friendly convention of substituting values before rearrangement. (Some AQA mark schemes award a mark for the correct substitution of values into an expression; however, the mark will not be awarded if the expression is incorrectly rearranged. Weaker students are strongly encouraged to substitute before rearrangement, and this is what I model.)
  2. A later time is indicated by the movement of the hands on the clock.

So far, so blindingly obvious, some might say.

But I hope the following examples will indicate the versatility of the approach.

1.2a Analysing a more complex SUVAT problem using dual coding (Up is positive convention)

Problem: A coin is dropped from rest takes 0.84 s to fall a distance of 3.5 m so that it strikes the water at the bottom of a well. With what speed must it be thrown vertically so that it takes exactly 1.5 s to hit the surface of the water?

Screenshot 2018-12-25 at 14.33.25.png

Another advantage of this method is that it makes assigning positive and negative directions to the SUVAT vectors easy as it becomes a matter of simply comparing the directions of each vector quantity (that is to say, s, u, v and a) with the arbitrarily selected positive direction arrow when we substitute values into the expression.

But what would happen if we’d selected a different positive direction arrow?

1.2b Analysing a more complex SUVAT problem using dual coding (Down is positive convention)

Problem: A well is 3.5 m deep so that a coin dropped from rest takes 0.84 s to strike the surface of the water. With what speed must it be thrown so that it takes exactly 1.5 s to hit the surface of the water?

Screenshot 2018-12-25 at 14.43.42.png

The answer is, of course, numerically equal to the previous answer. However, following the arbitrarily selected down is positive convention, we have a negative answer.

1.3 Analysing a projectile problem using dual coding

Let’s look at this typical problem from AQA.

Screenshot 2018-12-25 at 14.50.12.png

We could annotate the diagram like this:

Screenshot 2019-01-03 at 18.30.09.png

Guiding our students through the calculation:

Screenshot 2019-01-03 at 18.34.19.png

Just Show ‘Em!

Some trad-inclined teachers have embraced the motto: Just tell ’em!

It’s a good motto, to which dual coding can add the welcome corollary: Just show ’em!

The Unreasonable Effectiveness of Mathematics in the Natural Sciences

The famous phrase is, of course, from physicist Eugene Wigner (1960: 2):

My principal aim is to illuminate it from several sides. The first point is that the enormous usefulness of mathematics in the natural sciences is something bordering on the mysterious and that there is no rational explanation for it.

Further exploration of the above problem using dual coding can, I believe, give A-level students a glimpse of the truth of Wigner’s phrase.

This Is The Root You’re Looking For

In the calculation above, we found that when s = -1.8 m, v could have a value of plus or minus 6.90 m/s. Since we were interested in the velocity of the kite boarder at the end of the journey, we concluded that it was the negative root that was significant for our purposes.

But does the positive root have any physical significance? Why yes, it does. It indicates the other possible value of v when s = -1.8 m.

The displacement was -1.8 m at only one point on the real journey. However, if the kite boarder had started their projectile motion from the level of the water surface instead of from the top of the ramp, their vertical velocity at this point would have been +6.9 m/s.

Screenshot 2019-01-04 at 14.24.14.png

The fact that the kite boarder did not start their journey from this point is immaterial. Applying the mathematics not only tells us about their actual journey, but all other possible journeys that are consistent with the stated parameters and the subset of the laws of physics that we are considering in this problem — and that, to me, borders enough on the mysterious to bring home Wigner’s point.

And finally…

Screenshot 2019-01-04 at 15.11.47.png

This information allows us to annotate our final diagram as below (bearing in mind, of course, that the real journey of the kite boarder started from the top of the ramp and not from the water’s surface as shown).

Screenshot 2019-01-04 at 15.14.20.png

Let me end on a more cheerful note. The miracle of the appropriateness of the language of mathematics for the formulation of the laws of physics is a wonderful gift which we neither understand nor deserve. We should be grateful for it and hope that it will remain valid in future research and that it will extend, for better or for worse, to our pleasure, even though perhaps also to our bafflement, to wide branches of learning.

Wigner 1960: 9


Wigner, E. (1960). The Unreasonable Effectiveness of Mathematics in the Natural Sciences. Communications in Pure and Applied Mathematics; Vol. 13, No. 1.