The easiest way to make sense of the induced voltages is to first figure out the how the magnetic flux linkage (of the coil) varies with time. Obviously, the flux linkage is strongest when the magnet is centred with the coil.

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1 Fall Through

(a) magnet approaching coil
(b) magnet is centred with coil
(c) magnet left coil

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2 Bungee Dip

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3 Aligned

(a) magnet at the top
(b) magnet centred with the coil
(c) magnet at the bottom
(d) magnet centred with the coil
(e) magnet at the top

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4 Cross-over

(a) magnet at the top
(b) magnet centred with coil
(c) magnet at the bottom
(d) magnet centred with coil
(e) magnet at the top

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When the magnet was moving slowly, the LEDs did not light up because the magnetic flux linkage was changing too slowly. There was an induced emf, but it was too small (Faraday’s Law: ε = -dΦ/dt) to light up the LEDs. (About 1.5V is required to forward bias the LEDs)

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The LEDs also did not light up when the long magnet was moving vigorously in the coil. This is because since the entire magnetic flux of the magnet was already captured by the coil. So even though the magnet was moving rapidly, the magnetic flux linkage remains constant. Hence no emf nor current was induced in the coil. (Faraday’s Law: ε = -dΦ/dt)

First of all, you probably figured out that the two LEDs were both connected in parallel, directly to the coil, but in opposite directions. This would explain why they do not light up together. (LEDs are diodes, so they allow current to flow in only one direction).

Let’s move on to the EMI part of the explanation.

Firstly, Faraday’s Law says that if the magnetic flux linkage of a coil Φ changes, an emf ε directly proportional to the rate of change of Φ will be induced in the coil. (Faraday’s Law: ε = -dΦ/dt)

When the magnet was pushed into the coil, Φ increased, thus inducing an emf and current in one direction in the coil, lighting up the red LED. When the magnet was pulled out of the coil, Φ decreased, thus inducing an emf and current in the coil in the other direction, thus lighting up the green LED.

To analyze such situations, it is usually useful to sketch the variation of Φ with time. We can then tell the induced emf ε from the gradient of the Φ-t graph. (Faraday’s Law: ε = -dΦ/dt)

We have current flowing across the mercury (from one ring to the other), which is sitting in a magnetic field. So there is a F=BIL magnetic force. Or we can think the moving electrons (that constitute the current) experience a F=Bqv magnetic force. This force is directed perpendicular to the current (and magnetic field), which turns out to be along the circumference of the ring. The mercuy is thus pumped in a merry-go-round fashion.

Why should the metal lattice chase after/get dragged by the electrons?

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While a battery provides the EMF (E) to push current through the external circuit, it also comes with its own internal resistance (r). The amount of EMF spent to push current through its own internal resistance is directly proportional to the current passing through the battery (Ir). So the amount of EMF left to push current through the external circuit, called the terminal potential difference (V_t), decreases whenever the battery is made to pump out larger current.

In this video, as we connect more and more bulbs (in parallel) across the battery, we draw larger and larger current from the battery. This must lead to a larger and larger potential difference across the internal resistance of the battery, which implies a smaller and smaller terminal pd, which dims the bulbs.

There is also a simple explanation why the 2V battery could “outshine” the 3V. The 2V battery may have a lower EMF, but by having a smaller internal resistance, it provided a higher terminal pd to the external circuit.

A bulb with a 12 V 50 W rating is designed to have a resistance of 12²/50 = 2.88 Ω when in operation (P=V²/R). Similarly, a bulb rated at 12 V 35 W rating is designed to have an operating resistance of 12²/35 = 4.11 Ω. The point to note is that a 50 W bulb has a smaller resistance than a 35 W bulb.

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At first both bulbs were individually connected across a 6 V battery. Since the potential differences across both bulbs were the same, to compare the power dissipated in the two bulbs, we should think V²/R. So the 50 W bulb, with a smaller resistance, shone brighter.

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Later the bulbs were connected in series across a 12 V battery. Since the current flowing through both bulbs was the same, to compare the power dissipated in the two bulbs, we should think I²R.  So the 35 W bulb, with larger resistance, shone brighter.

(Do note that when connected in series, the potential difference across each bulb was no longer 6 V each. By the potential divider principle, the 35 W bulb, with its larger resistance, ended up with more than 6 V of potential difference across, while the 50 W bulb had less than 6 V across)

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For the sake of discussion, let’s assume that the globe is positively charged.

When the foil is near the globe, the globe’s electric field causes the electrons in the foil to shift, inducing negative and positive charges on opposite ends of the foil.

Thanks to the inverse-square nature of Coulomb’s forces, the electrical attraction acting on the negatively charged near end of the foil is stronger than the electrical repulsion acting on the positively charged far end of the foil. The foil thus accelerates towards the globe.

When the foil touches the globe, electrons are attracted into the globe, leaving the foil positively charged. The foil is thus quickly repelled away from the globe.

As the foil slowly loses its positive charges to the surrounding air molecules, the cycle repeats again.

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Water waves travelling towards along the water container are reflected off the walls when they reach either ends. Incident and reflected waves thus superpose and interfere with one another. A standing wave of significant amplitude can be formed only when the water waves have certain “correct” wavelengths.

Since water waves reflect off a wall without incurring any phase change, displacement antinodes are formed at the boundaries. What was shown in the video were the 1^st, 2^nd and 3^rd harmonics.

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Sound waves reflect repeated off the two ends of the tube and interfere with one another. At certain frequencies, standing sound waves are formed, resulting in alternating pressure nodes and antinodes along the tube. This must be the basis for the formation of alternating tall and short flames along the tube.

To figure out the wavelength of the sound wave, we don’t really have to know whether the positions of tallest (or shortest) flame correspond to pressure nodes, or antinodes. We just need to know that the distance between two tallest (or shortest) flames must correspond to half a wavelength of the sound wave.

If you’re interested in knowing which is which, you should watch the next demo.

This video shows the standing waves that can be formed on a string that is fixed at one end or loose at the other end.

Any standing wave that forms on this string must have a node at the fixed end, and antinode at the loose end.

The simplest standing wave fulfills these conditions is the NA. (Node-Antinode) The next standing wave that can be formed is the NANA, followed by

1. NA,
2. NANA,
3. NANANA,
4. NANANANA,
5. NANANANANA, and so on.

Notice that each ANA corresponds to 2 quarter-wavelength segment. This means

1. NA packs 1 quarter-wavelength along the length of the string,
2. NANA packs 3 quarter-wavelengths along the length of the string,
3. NANANA packs 5 quarter-wavelengths along the length of the string,
4. NANANANA packs 7 quarter-wavelengths along the length of the string,
5. NANANANANA packs 9 quarter-wavelengths along the length of the string, and so on.

Which means that

1. NA’s wavelength is called the fundamental wavelength,
2. NANA’s wavelength is 3x as short as that of NAN’s,
3. NANANA’s wavelength is 5x as short as that of NAN’s,
4. NANANANA’s wavelength is 7x as short as that of NAN’s,
5. NANANANANA’s wavelength is 9x as short as that of NAN’s, and so on.

Which means that

1. NA’s frequency is called the fundamental frequency, or 1^st
2. NANA’s frequency is 3x that of NAN’s, hence called the 3^rd harmonic,
3. NANANA’s frequency is 5x that of NAN’s, hence called the 5th harmonic,
4. NANANANA’s frequency is 7x that of NAN’s, hence called the 7th harmonic,
5. NANANANANA’s frequency is 9x that of NAN’s, hence called the 9th harmonic, and so on.