Answer key · Unit 13

Magnetism and induction.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 6

Lesson 1 of 6 · About 20 minutes

Magnetic fields

What a magnet does to the space round it: field lines and what they say, two poles that always come as a pair, magnets that attract and repel, what a magnet pulls, and the Earth as a magnet.

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The 5 steps and their answers

Step 1 of 5
Open this step
Field lines
Practice · Choose
Question
The paper shows the magnet’s field lines. What does a field line tell you at a point on it?

Correct answer
- Correct Which way a compass needle would point there
Setup, choices and feedback
The Scene at this step
Magnets
One
Distance apart
9
Show the field
0.9 (Magnets → Field)
Choices
Shown in a random order.
- Correct Which way a compass needle would point there
- Wrong The path along which something flows out of the north pole
  Misconception: Thinks something flows out of the north pole along the field lines, like water in a pipe.
  Feedback from the catalogue
  Nothing flows along a field line, and it is not a path anything travels. The line only sums up the field’s direction: at every point on it, a compass needle would point along it. The arrow on the line is the way the needle’s north pole points.
- Wrong Where the field is: between two lines there is no field at all
  Misconception: Thinks the field exists only on the drawn lines, with none in the gaps between them.
  Feedback from the catalogue
  The lines are a drawing, not threads that exist in space. The field fills all the space round the magnet, the gaps between the lines included: a line could be drawn through any point. How many lines to draw is the artist’s choice; what they tell you is the direction and, by how crowded they are, the strength.
- Wrong The path iron filings travel along to reach the magnet
  Feedback
  The filings do not travel along the lines: each grain stays where it is and turns until it lies along the field, like a compass needle.
Feedback for a right answer
A field line is a drawing of a direction: at every point on it, a compass needle would line up along it, its north pole pointing the way of the arrow. Where the lines crowd together the field is strong; where they spread out it is weak.

Hint
Think of a small compass needle laid on the paper.

Worked solution
1. The iron filings on the paper line up like tiny compass needles, each grain along the field where it lies.
2. A field line joins those directions into one continuous line: at every point it runs the way the field does there.
3. So the line tells you which way a compass needle would point, and its arrow tells you which way the needle’s north pole points: outside the magnet, from the north pole to the south.
Nothing flows along the line, and the field is there between the lines too: the person drawing chooses how many lines, not where the field is.
Step 2 of 5
Open this step
Two poles
Practice · Choose
Question
A bar magnet is cut in two, exactly in the middle. What do you get?

Correct answer
- Correct Two smaller magnets, each with a north pole and a south pole
Setup, choices and feedback
The Scene at this step
Magnets
One
Distance apart
9
Show the field
0.9 (Magnets → Field)
Choices
Shown in a random order.
- Correct Two smaller magnets, each with a north pole and a south pole
- Wrong One piece with only a north pole and one with only a south pole
  Misconception: Thinks cutting a bar magnet in half gives a piece with only a north pole and a piece with only a south pole.
  Feedback from the catalogue
  Every piece of a magnet is a whole magnet, with a north pole and a south pole. The magnetism comes from the atoms themselves, each a tiny magnet, so a cut simply makes new poles on either side of it. Nobody has ever found a lone pole.
- Wrong Two pieces of steel with no magnetism, because the cut destroyed the magnet
  Feedback
  The magnetism is not in the ends but in all of the material: the magnet’s atoms are tiny magnets, and they stay lined up after the cut.
Feedback for a right answer
Each piece is a whole magnet. New poles appear on either side of the cut: a south pole on the north half, a north pole on the south half. You could keep cutting down to the atoms and there would always be two poles.

Hint
The magnetism comes from the magnet’s atoms, not just from its ends.

Worked solution
1. The magnet’s atoms are tiny magnets, all lined up the same way.
2. At one end their north poles add up, and at the other their south poles.
3. After the cut, the atoms in each piece are still lined up, so each piece has a north end and a south end of its own.
If you picked a piece with one pole, you thought of the poles as two separate things sitting at the ends. They are the two sides of one arrangement, so they always come in pairs.
Step 3 of 5
Open this step
North facing north
Practice · Predict, then look
Question
Now the north pole of one magnet faces the south pole of the other, and the lines run straight from one to the other. The second magnet is turned round so that north faces north. What will the lines do in the gap between the poles?

Correct answer
- Correct Bend aside, leaving a point in the middle of the gap with no field
Setup, choices and feedback
The Scene at this step
Magnets
N facing S
Distance apart
9
Show the field
0.9 (Magnets → Field)
Choices
- Correct Bend aside, leaving a point in the middle of the gap with no field
- Wrong Still run from one pole to the other, as before
  Feedback
  A field line leaves a north pole and enters a south pole. Between two north poles there is no south pole for it to enter.
- Wrong Meet in the middle and cross each other
  Misconception: Thinks field lines can cross or meet head-on between two like poles.
  Feedback from the catalogue
  Field lines never cross: at each point the field has only one direction, and where two lines crossed it would have two. Between two north poles the lines bend aside, and in the middle there is a point where the two magnets’ fields cancel.
Once the prediction is locked in, the Scene changes
Magnets
N facing N
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Each magnet’s lines bend away from the other magnet and loop back to its own south pole. In the middle, each magnet’s field points away from it with the same strength, and the two cancel: a compass needle there would not be pulled any way at all.

Worked solution
1. Field lines leave a north pole and enter a south pole.
2. With north facing north, the lines from both poles head towards each other, and there is no south pole between them to take them in. So they bend aside and loop back to their own magnet’s south pole.
3. Exactly in the middle, each magnet’s field points away from it, and the two are equally strong: they cancel, and the field there is zero.
If you chose lines that cross, remember that at each point the field has only one direction. Lines that crossed would give that point two.
Step 4 of 5
Open this step
What a magnet pulls
Practice · Choose
Question
Which of these does a magnet pull? Choose every right answer.

Correct answer
- Correct An iron nail
- Correct A steel paper clip
Setup, choices and feedback
The Scene at this step
Magnets
One
Distance apart
9
Show the field
0.9 (Magnets → Field)
Choices
Shown in a random order. More than one choice is correct: every one must be selected.
- Correct An iron nail
- Correct A steel paper clip
- Wrong An aluminium can
  Misconception: Thinks a magnet attracts every metal, including aluminium and copper.
  Feedback from the catalogue
  A magnet attracts only a few materials: iron, nickel and cobalt, and steel, which is mostly iron. Aluminium, copper, gold and silver are metals, but a magnet does not pull them. What matters is whether the material’s atoms are tiny magnets that can be lined up, not whether it is shiny or conducts electricity.
- Wrong A copper wire
  Misconception: Thinks a magnet attracts every metal, including aluminium and copper.
  Feedback from the catalogue
  A magnet attracts only a few materials: iron, nickel and cobalt, and steel, which is mostly iron. Aluminium, copper, gold and silver are metals, but a magnet does not pull them. What matters is whether the material’s atoms are tiny magnets that can be lined up, not whether it is shiny or conducts electricity.
Feedback for a right answer
Iron, and steel, which is mostly iron. The magnet’s field lines up the tiny magnets in them, and they become magnets themselves while the magnet is near. Aluminium and copper are metals, but there is nothing in them to line up.

Hint
Not every metal responds to a magnet. Only in a few materials are the atoms tiny magnets that can be lined up.

Worked solution
1. A magnet attracts ferromagnetic materials: iron, nickel and cobalt, and alloys of them such as steel.
2. The nail is iron and the clip is steel: both are pulled.
3. Aluminium and copper are not ferromagnetic, so the magnet does not pull them, even though they are metals and conduct electricity well.
If you also chose the can or the wire, you assumed every metal is pulled. The difference is in the atoms, not in shine or conduction.
Step 5 of 5
Open this step
The Earth as a magnet
Practice · Choose
Question
A compass needle’s north pole points roughly to the Earth’s geographic North Pole. What kind of magnetic pole is there?

Correct answer
- Correct A south pole
Setup, choices and feedback
The Scene at this step
Magnets
One
Distance apart
9
Show the field
0.9 (Magnets → Field)
Choices
Shown in a random order.
- Correct A south pole
- Wrong A north pole
  Misconception: Thinks the magnetic pole near the geographic North Pole is a north magnetic pole.
  Feedback from the catalogue
  Opposite poles attract. The needle’s north pole is pulled north, so what waits for it there is a magnetic south pole. The name “magnetic North Pole” describes the place on the map, not the kind of pole.
- Wrong No pole: the needle simply points north
  Feedback
  A compass needle is a magnet, and it lines up with the Earth’s magnetic field. It points north because something in the north pulls its north pole.
Feedback for a right answer
Opposite poles attract, so what pulls the needle’s north pole is a magnetic south pole. The Earth behaves like a giant bar magnet whose south pole lies near the geographic North Pole.

Hint
Which poles attract each other?

Worked solution
1. A compass needle is a small magnet, and its north pole is pulled north.
2. Opposite poles attract, and like poles repel.
3. So near the geographic North Pole there is a magnetic south pole. The Earth’s field lines leave in the southern hemisphere and enter in the north.
If you answered a north pole, you took the place’s name for the kind of pole. “North” on a needle only says which way that end turns.

Lesson 2 of 6 · About 25 minutes

Electromagnets

An electric current makes a magnetic field: the field inside a coil, how to work it out, what an iron core adds and where it stops, and which end of the coil is the north pole.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
A current makes a field
No practice question in this step.
The Scene at this step
Current
0 A
Turns of wire
200 turns
Iron core
on
Show the field
Off
Step 2 of 6
Open this step
The field inside a coil
Practice · Calculate
Question
The coil is 6 cm long, has $200$ ⁠ turns, and carries a current of 1 A. The iron core is out. How strong is the magnetic field inside the coil? ( $\mu_0 = 4\pi \times 10^{-7}\ \mathrm{T \cdot m/A}$ ⁠)

Correct answer
4.189 mT ± 3%
Setup, choices and feedback
The Scene at this step
Current
1 Afixed
Turns of wire
200 turnsfixed
Iron core
offfixed
Show the field
Off
Accepted within
± 3%
Computed from the Scene at this step’s setting
Electromagnet: the field at the centre of the coil, with the core if it is in/nodes/coil/outputs/centreField
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.0419 mT ± 5%
  Misconception: Puts a length into an SI formula in centimetres instead of metres.
  Feedback from the catalogue
  The formulas, and $\mu_0$ ⁠ with them, are written in SI units, so a length goes in in metres: 6 cm = 0.06 m and 4 cm = 0.04 m. A number in centimetres is 100 times bigger, and the answer comes out 100 times off.
- 0.0209 mT ± 5%
  Misconception: Leaves the number of turns N out of a formula for a coil.
  Feedback from the catalogue
  In a coil every turn adds its own share: its own field, its own push or its own voltage, and the turns are joined one after another. So $N$ ⁠ multiplies the result. Without $N$ ⁠ you worked out a single turn.
Feedback for a right answer
$B = \dfrac{4\pi \times 10^{-7} \cdot 200 \cdot 1}{0.06} \approx 4.2 \times 10^{-3}\ \mathrm{T}$ ⁠, about 4.2 mT: close to a hundred times the Earth's field, and still too weak to lift a clip off the board.

Hint
$B = \dfrac{\mu_0 N I}{\ell}$ ⁠, with the length in metres.

Worked solution
1. Givens: $N = 200$ ⁠, $I = 1\ \mathrm{A}$ ⁠, $\ell = 6\ \mathrm{cm} = 0.06\ \mathrm{m}$ ⁠.
2. The field inside a long coil: $B = \dfrac{\mu_0 N I}{\ell}$ ⁠.
3. Substitute: $B = \dfrac{4\pi \times 10^{-7} \cdot 200 \cdot 1}{0.06} = 4.19 \times 10^{-3}\ \mathrm{T}$ ⁠.
4. That is 4.2 mT.
If you got 0.042 mT, you put the length in in centimetres. If you got 0.021 mT, you left out $N$ ⁠ and worked it out as if there were one turn.
Step 3 of 6
Open this step
One product
Practice · Reach the target
Question
With no core, change Current and Turns of wire until the field inside the coil is 10 mT, then press Check.

Goal
- 0.01 T ± 0.0003 T
  At the start of the step: 0.004189 T
Setup, choices and feedback
The Scene at this step
Current
1 A
Turns of wire
200 turns
Iron core
offfixed
Show the field
Off
Read from the Scene at
Electromagnet: the field at the centre of the coil, with the core if it is in/nodes/coil/outputs/centreField
One setting that reaches it
Current
4.7 A
Turns of wire
100 turns
Feedback for a right answer
The field depends only on the product $N I$ ⁠: any pair that gives about $480$ ⁠ ampere-turns works, such as $240$ ⁠ turns and 2 A, or $480$ ⁠ turns and 1 A.

Feedback for a wrong answer
Not yet. If the field is too weak, raise the current or the number of turns: the field grows in proportion to each.

Hint
You need $N I = \dfrac{B \ell}{\mu_0}$ ⁠. Work out the product, then pick a pair of numbers that gives it.

Worked solution
1. The target: $B = 10\ \mathrm{mT} = 0.010\ \mathrm{T}$ ⁠, in a coil with $\ell = 0.06\ \mathrm{m}$ ⁠.
2. From $B = \dfrac{\mu_0 N I}{\ell}$ ⁠: $N I = \dfrac{B \ell}{\mu_0} = \dfrac{0.010 \cdot 0.06}{4\pi \times 10^{-7}} \approx 477$ ⁠ ampere-turns.
3. Pick a pair that gives about $480$ ⁠: $240$ ⁠ turns and 2.0 A give 10.05 mT.
Any pair with the same product will do: what counts is how much current goes round the coil in all, not how it is split between turns and amperes.
Step 4 of 6
Open this step
An iron core
Practice · Predict, then look
Question
The coil is still at 1 A and $200$ ⁠ turns, with 4.2 mT inside it. The iron bolt is slid in. What will happen to the field at the centre of the coil?

Correct answer
- Correct It will grow tens of times
Setup, choices and feedback
The Scene at this step
Current
1 A
Turns of wire
200 turns
Iron core
off
Show the field
Off
Choices
- Correct It will grow tens of times
- Wrong It will grow a little, by some tens of per cent
  Feedback
  The iron does not just add itself to the field: the coil’s field lines up countless tiny magnets in it, and their fields join in. The boost is far bigger.
- Wrong It will not change: only the current and the turns set the field
  Feedback
  The formula $B = \dfrac{\mu_0 N I}{\ell}$ ⁠ holds for an empty coil. Iron inside the coil adds a field of its own.
- Wrong It will weaken, because the iron blocks the field
  Feedback
  Iron does not block a magnetic field: it gathers it and strengthens it.
Once the prediction is locked in, the Scene changes
Iron core
on
The Scene then shows: 0.1046 T · Electromagnet: the field at the centre of the coil, with the core if it is in/nodes/coil/outputs/centreField
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The field leaps to about 0.10 T, roughly $25$ ⁠ times as much. The coil’s field lines up the magnetic domains in the iron, and each domain is a small magnet whose field joins in. That is why almost every working electromagnet is wound round iron.

Worked solution
1. Without a core: $B = \dfrac{\mu_0 N I}{\ell} \approx 4.2\ \mathrm{mT}$ ⁠.
2. Soft iron is made of magnetic domains, small regions that are each a magnet. Usually they point every which way and cancel.
3. The coil's field turns them its way, and their fields add on. In the Scene the iron multiplies the field by $25$ ⁠: $25 \cdot 4.2\ \mathrm{mT} \approx 0.10\ \mathrm{T}$ ⁠.
In real iron the boost can be bigger still, and it depends on the kind of iron and the shape of the core.
Step 5 of 6
Open this step
Saturation
Practice · Calculate
Question
Now the current is 5 A and there are $500$ ⁠ turns. Without the core, the coil’s field would be 52 mT; with the core, the field at the centre is 1.08 T. By what factor does the iron boost the field now?

Correct answer
20.6 ± 3%
Setup, choices and feedback
The Scene at this step
Current
5 Afixed
Turns of wire
500 turnsfixed
Iron core
onfixed
Show the field
Off
Accepted within
± 3%
Computed from the Scene at this step’s setting
Electromagnet: how many times the core multiplies the coil's own field (1 without a core)/nodes/coil/outputs/coreFactor
Wrong answers it recognises
- 25 ± 2%
  Misconception: Thinks an iron core multiplies the field by the same factor however strong the coil's own field is.
  Feedback from the catalogue
  Iron boosts the field because its magnetic domains line up. Once most of them are lined up, the iron saturates, and more current adds only the coil’s own small share. So the boost, about $25$ ⁠ at low current, falls as the field nears saturation, around 1.6 T.
Feedback for a right answer
$\dfrac{1.08}{0.052} \approx 21$ ⁠: less than $25$ ⁠. The iron is nearing saturation, most of its domains are already lined up, and more current no longer adds much.

Hint
Divide the field with the core by the field without it, in the same units.

Worked solution
1. Without the core: $B = \dfrac{4\pi \times 10^{-7} \cdot 500 \cdot 5}{0.06} \approx 0.052\ \mathrm{T}$ ⁠.
2. With the core: $1.08\ \mathrm{T}$ ⁠.
3. The boost: $\dfrac{1.08}{0.052} \approx 21$ ⁠.
If you answered $25$ ⁠, you assumed the iron always multiplies by the same number. At low current that is nearly true, but as the field nears about 1.6 T the iron saturates, and more current adds only the coil’s own small share.
Step 6 of 6
Open this step
Which end is north
Practice · Choose
Question
You look straight at one end of a coil. The current goes round that end anticlockwise. Which pole is it?

Correct answer
- Correct A north pole
Setup, choices and feedback
The Scene at this step
Current
1 A
Turns of wire
200 turns
Iron core
on
Show the field
On
Choices
Shown in a random order.
- Correct A north pole
- Wrong A south pole
  Feedback
  Check again with your right hand: with the fingers curling anticlockwise as you see the end, the thumb points at you, out of that end.
- Wrong Neither: a coil has no poles, only a field inside it
  Feedback
  Outside, a coil’s field looks just like a bar magnet’s: the lines leave one end and enter the other. The ends are poles in every sense, and they attract and repel other magnets.
Feedback for a right answer
With the fingers of your right hand curling anticlockwise as you see it, the thumb points at you: the field comes out of this end, so it is a north pole. Reverse the current and the poles swap.

Hint
Curl the fingers of your right hand the way the current goes round the coil. The thumb points to the north pole.

Worked solution
1. The right-hand grip rule: curl your fingers the way the current runs in the turns, and your thumb points along the field inside the coil, towards the north pole.
2. The current runs anticlockwise as you see the end, so with your fingers curling that way, your thumb points at you.
3. The field leaves the coil at this end, and field lines leave a north pole: this is the north pole.
A way to remember: looking at a north pole, the current goes round it anticlockwise; at a south pole, clockwise.

Lesson 3 of 6 · About 20 minutes

The force on a current

A wire carrying a current in a magnetic field is pushed sideways: which way, how hard, and how two opposite pushes turn the coil of an electric motor.

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The 4 steps and their answers

Step 1 of 4
Open this step
A push sideways
Practice · Choose
Question
A straight, level wire carries a current east, in a level magnetic field pointing north. Which way is the force on the wire?

Correct answer
- Correct Up
Setup, choices and feedback
The Scene at this step
Current
1.5 A
Magnetic field
0.2 T
Show the forces
On
Choices
Shown in a random order.
- Correct Up
- Wrong Down
  Feedback
  The order matters: the fingers go along the current first, then curl towards the field. The other way round gives the opposite direction.
- Wrong North, along the field
  Misconception: Thinks the force on a current in a magnetic field points along the field lines.
  Feedback from the catalogue
  The force on a wire carrying a current is at right angles to both the current and the field, not along either of them. If the current runs east and the field points north, the force is neither north nor east but up or down. The right-hand rule says which.
- Wrong East, along the current
  Feedback
  The force never points along the current: it is at right angles to the wire, so it moves the wire sideways rather than pushing the current along.
Feedback for a right answer
The force is at right angles to both, so it is vertical. The right-hand rule picks between up and down: fingers east, curling north, and the thumb points up.

Hint
The force is at right angles to both the current and the field. Point the fingers of your right hand along the current and curl them towards the field; your thumb shows the force.

Worked solution
1. The force on a wire carrying a current in a magnetic field is at right angles to both the current and the field.
2. The current runs east and the field points north, both level, so the force is vertical: up or down.
3. The right-hand rule: fingers along the current (east), curling towards the field (north), and the thumb points up.
If you chose north, you thought the force follows the field lines, as between two magnets. On a current the force is always at right angles to the lines.
Step 2 of 4
Open this step
How hard
Practice · Calculate
Question
The motor’s coil has $50$ ⁠ turns, and each of its sides lying across the field is 4 cm long. The current is 1.5 A and the field 0.2 T. How hard is each of the two sides pushed?

Correct answer
0.6 N ± 2%
Setup, choices and feedback
The Scene at this step
Current
1.5 Afixed
Magnetic field
0.2 Tfixed
Show the forces
On
Accepted within
± 2%
Computed from the Scene at this step’s setting
The push on each side of the coil, N·B·I·L/nodes/motor/outputs/sideForce
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.012 N ± 5%
  Misconception: Leaves the number of turns N out of a formula for a coil.
  Feedback from the catalogue
  In a coil every turn adds its own share: its own field, its own push or its own voltage, and the turns are joined one after another. So $N$ ⁠ multiplies the result. Without $N$ ⁠ you worked out a single turn.
- 60 N ± 5%
  Misconception: Puts a length into an SI formula in centimetres instead of metres.
  Feedback from the catalogue
  The formulas, and $\mu_0$ ⁠ with them, are written in SI units, so a length goes in in metres: 6 cm = 0.06 m and 4 cm = 0.04 m. A number in centimetres is 100 times bigger, and the answer comes out 100 times off.
Feedback for a right answer
$F = 50 \cdot 0.2 \cdot 1.5 \cdot 0.04 = 0.6\ \mathrm{N}$ ⁠. The two sides carry the current opposite ways, so they are pushed opposite ways, and together they turn the coil.

Hint
In each turn the side is a length $L$ ⁠ of wire in the field, and there are $N$ ⁠ turns: $F = N B I L$ ⁠.

Worked solution
1. Givens: $N = 50$ ⁠, $B = 0.2\ \mathrm{T}$ ⁠, $I = 1.5\ \mathrm{A}$ ⁠, $L = 4\ \mathrm{cm} = 0.04\ \mathrm{m}$ ⁠.
2. A straight wire at right angles to the field feels $F = B I L$ ⁠. A side of the coil is $N$ ⁠ such wires side by side, so $F = N B I L$ ⁠.
3. Substitute: $F = 50 \cdot 0.2 \cdot 1.5 \cdot 0.04 = 0.6\ \mathrm{N}$ ⁠.
If you got 0.012 N, you left out the $50$ ⁠ turns. If you got 60 N, you put the length in in centimetres.
Step 3 of 4
Open this step
Twice as hard
Practice · Reach the target
Question
Change Current and Magnetic field until each side of the coil is pushed with 1.2 N, twice as hard as before, then press Check.

Goal
- 1.2 N ± 0.02 N
  At the start of the step: 0.6 N
Setup, choices and feedback
The Scene at this step
Current
1.5 A
Magnetic field
0.2 T
Show the forces
On
Read from the Scene at
The push on each side of the coil, N·B·I·L/nodes/motor/outputs/sideForce
One setting that reaches it
Current
3 A
Magnetic field
0.2 T
Feedback for a right answer
The force depends on the product $B I$ ⁠: any pair with $B I = 0.6\ \mathrm{T \cdot A}$ ⁠ works, such as 0.2 T and 3 A, or 0.3 T and 2 A. The harder the push, the faster the coil settles to turning.

Feedback for a wrong answer
Not yet. The force grows in proportion to the current and to the field: doubling either doubles the force.

Hint
$F = N B I L$ ⁠, and $N$ ⁠ and $L$ ⁠ are fixed. What must the product $B I$ ⁠ be?

Worked solution
1. The target: $F = 1.2\ \mathrm{N}$ ⁠, with $N = 50$ ⁠ and $L = 0.04\ \mathrm{m}$ ⁠.
2. From $F = N B I L$ ⁠: $B I = \dfrac{F}{N L} = \dfrac{1.2}{50 \cdot 0.04} = 0.6\ \mathrm{T \cdot A}$ ⁠.
3. For instance, leave the field at 0.2 T and raise the current to 3 A: $50 \cdot 0.2 \cdot 3 \cdot 0.04 = 1.2\ \mathrm{N}$ ⁠.
Doubling the current alone, or the field alone, doubles the force, because it is in proportion to each.
Step 4 of 4
Open this step
From a push to a turn
No practice question in this step.
The Scene at this step
Current
1.5 A
Magnetic field
0.2 T
Show the forces
On

Lesson 4 of 6 · About 25 minutes

Charged particles in a magnetic field

A charge moving through a magnetic field feels a force at right angles to its motion, so it moves on a circle: the radius mv/qB, a time per turn that does not depend on speed, the charge-to-mass ratio, and a mass spectrometer that sorts isotopes.

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The 6 steps and their answers

Step 1 of 6
Open this step
A moving charge
Practice · Choose
Question
The neon ion moves on a circle, and the magnetic field points straight up, between the coils. Which way does the magnetic force on the ion point, and what does it do to its speed?

Correct answer
- Correct Towards the centre of the circle, at right angles to the motion and to the field, and it does not change the speed
Setup, choices and feedback
The Scene at this step
Ion
Neon ion (Ne⁺)
Speed
50 km/s
Field strength
0.1 T
Build the spectrometer
Ion beam
Choices
Shown in a random order.
- Correct Towards the centre of the circle, at right angles to the motion and to the field, and it does not change the speed
- Wrong Up, along the field lines
  Misconception: Thinks the force on a current in a magnetic field points along the field lines.
  Feedback from the catalogue
  The force on a wire carrying a current is at right angles to both the current and the field, not along either of them. If the current runs east and the field points north, the force is neither north nor east but up or down. The right-hand rule says which.
- Wrong Forward, along the motion, which is why the ion does not stop
  Misconception: Thinks a magnetic force can speed a charged particle up or slow it down.
  Feedback from the catalogue
  The magnetic force on a moving charge, $F = qvB$ ⁠, is always at right angles to the velocity. A force at right angles to the motion does no work, so it changes only the direction of the velocity, not its size. That is why a particle in a uniform magnetic field moves on a circle at a steady speed.
Feedback for a right answer
The force $F = qvB$ ⁠ is at right angles both to the velocity and to the field, so it points to the centre: it is the centripetal force. A force at right angles to the motion does no work, so the speed does not change, only its direction.

Hint
The ion moves on a circle at a steady speed. What force does such a body need?

Worked solution
1. In the Lesson on the force on a wire we saw that the magnetic force on a current is at right angles to the current and to the field. A moving ion is a tiny current, and the force on it is $F = qvB$ ⁠.
2. The force is always at right angles to the velocity, so it turns it without speeding it up or slowing it down.
3. A force of fixed size that is always at right angles to the motion is exactly what circular motion needs: a centripetal force, towards the centre.
If you chose “along the field lines”, that is the one direction the magnetic force never points.
Step 2 of 6
Open this step
The radius
Practice · Predict, then look
Question
With Field strength at 0.2 T and Speed at 50 km/s, the neon ion circles with a radius of about 5.2 cm. What will the radius be if the speed doubles to 100 km/s?

Correct answer
10.36 cm ± 0.5 cm
A number prediction counts as right within ± 0.5 cm
Setup, choices and feedback
The Scene at this step
Ion
Neon ion (Ne⁺)fixed
Speed
50 km/s
Field strength
0.2 Tfixed
Build the spectrometer
Ion beam
Once the prediction is locked in, the Scene changes
Speed
100 km/s
The Scene then shows: 10.36 cm · Radius of the circle, mv/|q|B/nodes/beam/outputs/radius
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About 10.4 cm: twice as big. The force doubles, but a circle as tight as before at twice the speed would need four times the force, so the circle widens.

Worked solution
1. The magnetic force is the centripetal force: $qvB = \dfrac{mv^2}{r}$ ⁠.
2. So $r = \dfrac{mv}{qB}$ ⁠: the radius is proportional to the speed.
3. Twice the speed, twice the radius: $2 \times 5.2 \approx 10.4$ ⁠ centimetres.
If you predicted four times, you used $v^2$ ⁠ without remembering that the force $qvB$ ⁠ grows with $v$ ⁠ too.
Step 3 of 6
Open this step
The time for a turn
Practice · Predict, then look
Question
At 50 km/s the neon ion completes a turn in about 6.5 µs. What will happen to the time for one turn when the speed doubles to 100 km/s?

Correct answer
- Correct It will stay the same
Setup, choices and feedback
The Scene at this step
Ion
Neon ion (Ne⁺)fixed
Speed
50 km/s
Field strength
0.2 Tfixed
Build the spectrometer
Ion beam
Choices
- Correct It will stay the same
- Wrong It will halve, because the ion is twice as fast
  Misconception: Thinks a faster charged particle takes less time to go round its circle in a magnetic field.
  Feedback from the catalogue
  A faster particle goes round a bigger circle: $r = \dfrac{mv}{qB}$ ⁠. The path grows in the same proportion as the speed, so the time for a turn, $T = \dfrac{2\pi m}{qB}$ ⁠, does not depend on the speed at all.
- Wrong It will double, because the circle is twice as big
  Feedback
  The circle does grow twice as big, but the ion also covers it twice as fast. The two changes cancel.
Once the prediction is locked in, the Scene changes
Speed
100 km/s
The Scene then shows: 0.000006509 s · Time for one turn, 2πm/|q|B/nodes/beam/outputs/period
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The time for a turn, $T = \dfrac{2\pi m}{qB}$ ⁠, does not depend on the speed: a faster ion goes round a bigger circle in the same time. The cyclotron is built on this, speeding particles up with pushes at a steady beat.

Worked solution
1. The time for a turn is the circumference divided by the speed: $T = \dfrac{2\pi r}{v}$ ⁠.
2. With $r = \dfrac{mv}{qB}$ ⁠: $T = \dfrac{2\pi m}{qB}$ ⁠. The speed has cancelled.
3. So at 100 km/s the time for a turn stays about 6.5 µs.
If you chose that it halves, you forgot that the path grows twice as long too.
Step 4 of 6
Open this step
A tighter circle
Practice · Reach the target
Question
The neon ion now circles with a radius of about 10.4 cm. Change Speed or Field strength until the radius halves, to about 5.2 cm, and press Check.

Goal
- 5.18 cm ± 0.2 cm
  At the start of the step: 10.36 cm
Setup, choices and feedback
The Scene at this step
Ion
Neon ion (Ne⁺)fixed
Speed
50 km/s
Field strength
0.1 T
Build the spectrometer
Ion beam
Read from the Scene at
Radius of the circle, mv/|q|B/nodes/beam/outputs/radius
One setting that reaches it
Speed
5 km/s
Field strength
0.02 T
Feedback for a right answer
There are two ways: halve the speed to 25 km/s, or double the field to 0.2 T. Either way $\dfrac{v}{B}$ ⁠ halves.

Feedback for a wrong answer
Not yet. A stronger field or a slower ion tightens the circle.

Hint
$r = \dfrac{mv}{qB}$ ⁠. Which two changes would halve it?

Worked solution
1. $r = \dfrac{mv}{qB}$ ⁠: proportional to $v$ ⁠, inversely to $B$ ⁠.
2. To halve $r$ ⁠: halve $v$ ⁠, from 50 to 25 km/s, or double $B$ ⁠, from 0.1 to 0.2 T.
3. Any other combination that halves $\dfrac{v}{B}$ ⁠ works too.
Step 5 of 6
Open this step
The charge-to-mass ratio
Practice · Calculate
Question
A proton at 50 km/s in a 0.1 T field circles with a radius of about 5.2 mm. An alpha particle has four times the mass and twice the charge. What will its circle’s radius be, at the same speed in the same field?

Correct answer
10.44 mm ± 0.3 mm
Setup, choices and feedback
The Scene at this step
Ion
Proton (H⁺)fixed
Speed
50 km/sfixed
Field strength
0.1 Tfixed
Build the spectrometer
Ion beam
Accepted within
± 0.3 mm
Computed from the Scene at this step’s setting
Radius of the circle, mv/|q|B/nodes/beam/outputs/radius
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 20.88 mm ± 0.5 mm
  Misconception: Works out the radius of a charged particle's circle from its mass alone, leaving out its charge.
  Feedback from the catalogue
  In the radius $r = \dfrac{mv}{qB}$ ⁠ the charge is in the denominator: more charge means a bigger force and a tighter circle. What matters is the ratio of mass to charge, not the mass alone.
- 2.61 mm ± 0.2 mm
  Misconception: Thinks a lighter ion swings wider in a magnetic field than a heavier one with the same charge and speed.
  Feedback from the catalogue
  The same charge and speed give the same force, but a heavier ion has more inertia, and the same force bends it less. So $r = \dfrac{mv}{qB}$ ⁠ grows with the mass: the heavier ion swings wider and lands farther out in the spectrometer.
Feedback for a right answer
About 10.4 mm: twice as big. The mass grows four times, but the charge grows too and pulls twice as hard. What matters is the charge-to-mass ratio, and the alpha’s is half the proton’s.

Hint
$r = \dfrac{mv}{qB}$ ⁠: what happens to the ratio $\dfrac{m}{q}$ ⁠?

Worked solution
1. $r = \dfrac{mv}{qB}$ ⁠, with the same $v$ ⁠ and $B$ ⁠.
2. For the alpha, $\dfrac{m}{q}$ ⁠ is $\dfrac{4}{2} = 2$ ⁠ times the proton’s.
3. So the radius is $2$ ⁠ times bigger: $2 \times 5.2 \approx 10.4$ ⁠ millimetres.
If you answered about 21 mm, you used only the mass and forgot the charge. If you answered about 2.6 mm, you turned the ratio over: a particle that is heavier for its charge bends less.
Step 6 of 6
Open this step
A mass spectrometer
Practice · Predict, then look
Question
Neon has three isotopes: neon-20, neon-21 and neon-22. All have the same charge, and they leave the gun at the same speed. In the spectrometer each ion lands on the plate after half a turn. Which one will land farthest from the gun?

Correct answer
- Correct Neon-22, the heaviest
Setup, choices and feedback
The Scene at this step
Ion
Neon ion (Ne⁺)fixed
Speed
50 km/s
Field strength
0.1 T
Build the spectrometer
Helix
Choices
- Correct Neon-22, the heaviest
- Wrong Neon-20, the lightest
  Misconception: Thinks a lighter ion swings wider in a magnetic field than a heavier one with the same charge and speed.
  Feedback from the catalogue
  The same charge and speed give the same force, but a heavier ion has more inertia, and the same force bends it less. So $r = \dfrac{mv}{qB}$ ⁠ grows with the mass: the heavier ion swings wider and lands farther out in the spectrometer.
- Wrong All in the same place: they have the same charge
  Feedback
  The same charge means the same force, but not the same acceleration. A heavier ion has more inertia, and the same force bends it less.
Once the prediction is locked in, the Scene changes
Build the spectrometer
Mass spectrometer
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Neon-22 circles $10\%$ ⁠ wider than neon-20 and lands about 2 cm farther out. A mass spectrometer sorts isotopes this way, and shows how much there is of each: neon-20 is about $90\%$ ⁠ of natural neon.

Worked solution
1. Each ion lands after half a turn, $2r$ ⁠ from the gun.
2. $r = \dfrac{mv}{qB}$ ⁠: with the same $q$ ⁠, $v$ ⁠ and $B$ ⁠, the radius is proportional to the mass.
3. Neon-22 is $10\%$ ⁠ heavier than neon-20, so it lands $10\%$ ⁠ farther out: about 22.8 cm against about 20.7 cm.

Lesson 5 of 6 · About 30 minutes

Electromagnetic induction

A changing magnetic field induces a voltage: magnetic flux, Faraday's law, why more turns and faster motion give more voltage, Lenz's law, and alternating and effective voltage.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
The change is what counts
Practice · Predict, then look
Question
The magnet is pushed in and out once a second, and the needle swings both ways. Now the magnet is stopped and left inside the coil. What will the meter show?

Correct answer
- Correct Zero: the needle returns to the middle
Setup, choices and feedback
The Scene at this step
Speed
1 per second
Turns of wire
200 turns
Motion
Push in and out
Show field and current
Off
Choices
- Correct Zero: the needle returns to the middle
- Wrong A steady voltage, because the magnet is inside the coil
  Misconception: Thinks a magnet resting inside a coil keeps inducing a current.
  Feedback from the catalogue
  A voltage is induced only while the magnetic flux through the coil is changing. A magnet that stays put, however strong and even inside the coil, sends a steady flux through the turns, so $\Delta\Phi = 0$ ⁠ and $\mathcal{E} = 0$ ⁠. The needle moves only while the magnet moves.
- Wrong More voltage than before, because the magnet is closer to the turns
  Misconception: Thinks a magnet resting inside a coil keeps inducing a current.
  Feedback from the catalogue
  A voltage is induced only while the magnetic flux through the coil is changing. A magnet that stays put, however strong and even inside the coil, sends a steady flux through the turns, so $\Delta\Phi = 0$ ⁠ and $\mathcal{E} = 0$ ⁠. The needle moves only while the magnet moves.
Once the prediction is locked in, the Scene changes
Speed
0 per second
The Scene then shows: 0 V · Generator: the largest voltage the meter swings to/nodes/coil/outputs/peakEmf
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Without motion the flux through the turns does not change, so there is no voltage. What makes the voltage is the change in flux, not the flux itself: a very strong magnet sitting inside the coil induces nothing.

Worked solution
1. By Faraday's law, $\mathcal{E} = -N\,\dfrac{\Delta\Phi}{\Delta t}$ ⁠: the voltage is the rate at which the flux changes.
2. With the magnet at rest, its field through each turn is steady, so $\Delta\Phi = 0$ ⁠.
3. So $\mathcal{E} = 0$ ⁠, and the needle reads zero, even with the magnet inside the coil.
If you chose a steady voltage, you thought the flux itself makes a voltage. The flux is large with the magnet inside, but only a change in flux induces a voltage.
Step 2 of 6
Open this step
Faraday's law
Practice · Calculate
Question
A coil has $200$ ⁠ turns. The magnetic flux through each turn rises steadily from zero to 3.0 mWb in 0.10 s. How large is the induced EMF?

Correct answer
6 V ± 2%
Setup, choices and feedback
The Scene at this step
Speed
1 per second
Turns of wire
200 turns
Motion
Push in and out
Show field and current
Off
Accepted within
± 2%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.03 V ± 5%
  Misconception: Leaves the number of turns N out of a formula for a coil.
  Feedback from the catalogue
  In a coil every turn adds its own share: its own field, its own push or its own voltage, and the turns are joined one after another. So $N$ ⁠ multiplies the result. Without $N$ ⁠ you worked out a single turn.
- 0.06 V ± 5%
  Misconception: Multiplies the change in flux by the time instead of dividing by it: N·ΔΦ·Δt.
  Feedback from the catalogue
  The voltage is the rate at which the flux changes: how much it changes each second, $\mathcal{E} = N\,\dfrac{\Delta\Phi}{\Delta t}$ ⁠. The same change in a shorter time gives a bigger voltage, so you divide by the time. And by units: Wb/s is a volt, and Wb·s is not.
Feedback for a right answer
$\mathcal{E} = 200 \cdot \dfrac{3.0 \times 10^{-3}}{0.10} = 6.0\ \mathrm{V}$ ⁠. The same change in a tenth of the time would give ten times the voltage.

Hint
$\mathcal{E} = N\,\dfrac{\Delta\Phi}{\Delta t}$ ⁠, with the flux in webers: 1 mWb = 10⁻³ Wb.

Worked solution
1. Givens: $N = 200$ ⁠, $\Delta\Phi = 3.0\ \mathrm{mWb} = 3.0 \times 10^{-3}\ \mathrm{Wb}$ ⁠, $\Delta t = 0.10\ \mathrm{s}$ ⁠.
2. Faraday's law, in size: $\mathcal{E} = N\,\dfrac{\Delta\Phi}{\Delta t}$ ⁠.
3. Substitute: $\mathcal{E} = 200 \cdot \dfrac{3.0 \times 10^{-3}}{0.10} = 6.0\ \mathrm{V}$ ⁠.
If you got 0.03 V, you worked out a single turn. If you got 0.06 V, you multiplied by the time instead of dividing by it.
Step 3 of 6
Open this step
Faster, and with more turns
Practice · Calculate
Question
With $200$ ⁠ turns and one push a second, the voltage peaks at 0.5 V. Now there are $400$ ⁠ turns, and the magnet is pushed three times a second with the same stroke. What is the peak voltage?

Correct answer
3 V ± 2%
Setup, choices and feedback
The Scene at this step
Speed
3 per secondfixed
Turns of wire
400 turnsfixed
Motion
Push in and outfixed
Show field and current
Off
Accepted within
± 2%
Computed from the Scene at this step’s setting
Generator: the largest voltage the meter swings to/nodes/coil/outputs/peakEmf
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 1.5 V ± 2%
  Misconception: Leaves the number of turns N out of a formula for a coil.
  Feedback from the catalogue
  In a coil every turn adds its own share: its own field, its own push or its own voltage, and the turns are joined one after another. So $N$ ⁠ multiplies the result. Without $N$ ⁠ you worked out a single turn.
- 1 V ± 2%
  Misconception: Thinks moving the magnet faster only makes the needle swing more often, not further.
  Feedback from the catalogue
  Moving faster changes the same flux in less time, so $\dfrac{\Delta\Phi}{\Delta t}$ ⁠ grows, and so does the voltage. A magnet pushed 3 times as fast gives a peak 3 times as high, and the needle also swings 3 times as often.
Feedback for a right answer
Twice the turns and three times the speed: $0.5 \cdot 2 \cdot 3 = 3\ \mathrm{V}$ ⁠. Faster motion changes the same flux in less time, so the voltage grows, rather than only alternating more often.

Hint
The voltage grows in proportion to the number of turns, and to how fast the flux changes.

Worked solution
1. By $\mathcal{E} = N\,\dfrac{\Delta\Phi}{\Delta t}$ ⁠, the voltage is in proportion to $N$ ⁠: twice the turns, twice the voltage.
2. With the same stroke 3 times as fast, the same change in flux happens in a third of the time, so $\dfrac{\Delta\Phi}{\Delta t}$ ⁠ is 3 times as large.
3. The peak: $0.5 \cdot \dfrac{400}{200} \cdot 3 = 3\ \mathrm{V}$ ⁠.
If you answered 1.5 V, you did not multiply by the turns. If you answered 1 V, you thought speed only changes how often the needle swings, not how far.
Step 4 of 6
Open this step
Reach a peak
Practice · Reach the target
Question
Change Speed and Turns of wire until the peak voltage is 1.5 V, then press Check.

Goal
- 1.5 V ± 0.03 V
  At the start of the step: 0.5 V
Setup, choices and feedback
The Scene at this step
Speed
1 per second
Turns of wire
200 turns
Motion
Push in and outfixed
Show field and current
Off
Read from the Scene at
Generator: the largest voltage the meter swings to/nodes/coil/outputs/peakEmf
One setting that reaches it
Speed
4 per second
Turns of wire
150 turns
Feedback for a right answer
The peak depends on turns times speed: any pair that gives $3$ ⁠ times $200 \times 1$ ⁠ works, such as $300$ ⁠ turns and two pushes a second, or $200$ ⁠ turns and three. The bulb glows brighter too.

Feedback for a wrong answer
Not yet. The peak grows in proportion to both the turns and the speed.

Hint
With $200$ ⁠ turns and one push a second the peak is 0.5 V. You need $3$ ⁠ times that.

Worked solution
1. With $N = 200$ ⁠ and one push a second, the peak is $0.5\ \mathrm{V}$ ⁠.
2. The peak is in proportion to $N$ ⁠ and to the speed, so you need $\dfrac{N}{200} \cdot f = \dfrac{1.5}{0.5} = 3$ ⁠, where $f$ ⁠ is the number of pushes a second.
3. For instance $N = 300$ ⁠ and $f = 2$ ⁠: $0.5 \cdot 1.5 \cdot 2 = 1.5\ \mathrm{V}$ ⁠.
$N = 200$ ⁠ with $f = 3$ ⁠, or $N = 500$ ⁠ with $f = 1.2$ ⁠, work as well: what counts is the product.
Step 5 of 6
Open this step
The coil pushes back
Practice · Choose
Question
The north pole of the magnet is pushed into the coil. While it moves in, what does the end of the coil nearest the magnet become?

Correct answer
- Correct A north pole, which pushes the magnet back
Setup, choices and feedback
The Scene at this step
Speed
1 per second
Turns of wire
200 turns
Motion
Push in and out
Show field and current
On
Choices
Shown in a random order.
- Correct A north pole, which pushes the magnet back
- Wrong A south pole, which pulls the magnet in
  Misconception: Thinks the induced current makes a pole that pulls the moving magnet along, helping the motion.
  Feedback from the catalogue
  If the coil helped the magnet along, the magnet would speed up, the current would grow, and so would the help: energy from nothing. In fact the induced current always opposes the change that made it (Lenz’s law): push a north pole in, and the near end of the coil becomes a north pole that pushes it back. The electrical energy comes from the work you do against that push.
- Wrong No pole: the induced current is too small to make one
  Feedback
  Any current in a coil makes it a magnet, even a small one. The only question is which way round.
Feedback for a right answer
The induced current opposes the change that made it: the magnet comes closer, and the coil pushes it back. Keeping up the push takes work, and that work is the energy the bulb receives. This is Lenz’s law, and it is what the minus sign in Faraday’s law means.

Hint
If the coil helped the magnet in, where would the bulb’s energy come from?

Worked solution
1. As the north pole comes closer, the flux through the coil grows.
2. By Lenz’s law, the induced current makes a field that opposes the change: it tries to stop the flux growing, so it pushes the magnet back.
3. To push back a north pole, the near end of the coil must be a north pole.
When the magnet is pulled out, everything reverses: the end becomes a south pole and tries to hold on to it. If you chose a south pole, the coil would pull the magnet in with nobody doing work, making energy from nothing.
Step 6 of 6
Open this step
Spinning: an alternating voltage
Practice · Calculate
Question
The magnet spins twice a second beside the iron core, and with $200$ ⁠ turns the voltage swings as a sine wave between +1.2 V and −1.2 V. What is the effective voltage, the one the bulb glows by?

Correct answer
0.8485 V ± 2%
Setup, choices and feedback
The Scene at this step
Speed
2 per secondfixed
Turns of wire
200 turnsfixed
Motion
Spinfixed
Show field and current
Off
Accepted within
± 2%
Computed from the Scene at this step’s setting
Generator: the RMS (effective) voltage the bulb glows with/nodes/coil/outputs/rmsEmf
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 1.2 V ± 2%
  Misconception: Takes the peak of an alternating voltage as its effective (RMS) value.
  Feedback from the catalogue
  An alternating voltage is at its peak only for an instant, and most of the time it is less. The effective voltage is the steady voltage that would heat the bulb just as much, and for a sine wave it is $\dfrac{V_0}{\sqrt 2} \approx 0.71\,V_0$ ⁠. The 230 V at a socket is an effective voltage too: the peak is about 325 V.
- 0.6 V ± 2%
  Misconception: Takes the effective (RMS) value of a sine wave as half its peak.
  Feedback from the catalogue
  Heating goes with the square of the voltage, so taking “the middle” is not enough. The average of $\sin^2$ ⁠ over a whole cycle is a half, so the mean square is $\tfrac12 V_0^2$ ⁠ and the effective voltage is $\dfrac{V_0}{\sqrt 2} \approx 0.71\,V_0$ ⁠, more than half the peak.
Feedback for a right answer
$\dfrac{1.2}{\sqrt 2} \approx 0.85\ \mathrm{V}$ ⁠. That is the steady voltage that would heat the bulb just as much. Mains electricity is measured the same way: 230 V effective, with a peak of about 325 V.

Hint
For a sine-wave voltage, $V_{\text{eff}} = \dfrac{V_0}{\sqrt 2}$ ⁠.

Worked solution
1. The peak: $V_0 = 1.2\ \mathrm{V}$ ⁠, and the voltage is a sine wave.
2. The bulb's heating goes with $V^2$ ⁠, and the average of $\sin^2$ ⁠ over a whole cycle is $\tfrac12$ ⁠. So $V_{\text{eff}} = \dfrac{V_0}{\sqrt 2}$ ⁠.
3. Substitute: $V_{\text{eff}} = \dfrac{1.2}{\sqrt 2} \approx 0.85\ \mathrm{V}$ ⁠.
If you answered 1.2 V, you took the peak, which the voltage touches only for an instant. If you answered 0.6 V, you took half the peak, but because of the square the effective voltage is more than that.

Lesson 6 of 6 · About 25 minutes

Transformers and the power grid

Two coils on one core: the turns ratio sets the voltage, the current falls as the voltage rises, a transformer does nothing with direct current, and why power lines run at hundreds of thousands of volts.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
The turns ratio
Practice · Calculate
Question
Right now both coils of the transformer have the same number of turns, and the line runs at 25 kV, the generator’s voltage. If the coil on the line side had $4$ ⁠ times as many turns as the coil on the station side, what voltage would the line run at?

Correct answer
100 kV ± 1%
Setup, choices and feedback
The Scene at this step
Turns ratio
1 ×fixed
Power sent
100 MWfixed
Show current and losses
Off
Accepted within
± 1%
Computed from the Scene at this step’s setting
Voltage between two wires of the line/nodes/grid/outputs/lineVoltage
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 6.25 kV ± 2%
  Misconception: Divides by the turns ratio where it should multiply: a step-up transformer that lowers the voltage.
  Feedback from the catalogue
  Each turn of the secondary coil picks up the same voltage, so the voltage grows with the number of turns: $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$ ⁠. The side with more turns has more voltage. A step-up transformer has more turns on the line side, so you multiply by the ratio.
Feedback for a right answer
$25 \cdot 4 = 100\ \mathrm{kV}$ ⁠. Each turn of the secondary coil picks up the same voltage, so $4$ ⁠ times the turns is $4$ ⁠ times the voltage.

Hint
$\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$ ⁠.

Worked solution
1. Givens: $V_p = 25\ \mathrm{kV}$ ⁠, $\dfrac{N_s}{N_p} = 4$ ⁠.
2. In an ideal transformer $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$ ⁠.
3. So $V_s = 25 \cdot 4 = 100\ \mathrm{kV}$ ⁠.
If you got 6.25 kV, you divided by the ratio. The side with more turns has more voltage.
Step 2 of 6
Open this step
The current comes down
Practice · Calculate
Question
The tag on the station side shows that each busbar leaving the generator carries 2,309 A. If the line side had $16$ ⁠ times as many turns, what current would each wire of the line carry?

Correct answer
144.3 A ± 2%
Setup, choices and feedback
The Scene at this step
Turns ratio
1 ×fixed
Power sent
100 MWfixed
Show current and losses
Off
Accepted within
± 2%
Computed from the Scene at this step’s setting
Current in each wire between the generator and the transformer/nodes/grid/outputs/stationCurrent
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 36950 A ± 2%
  Misconception: Thinks a step-up transformer raises the current along with the voltage.
  Feedback from the catalogue
  A transformer adds no power. If $V_p I_p = V_s I_s$ ⁠ and the voltage went up 16 times, the current must go down 16 times. If the current went up too, the power out would be 256 times the power in.
Feedback for a right answer
$\dfrac{2309}{16} \approx 144\ \mathrm{A}$ ⁠. The voltage goes up 16 times, so the current comes down 16 times, and the power $V I$ ⁠ stays the same.

Hint
The transformer adds no power: $V_p I_p = V_s I_s$ ⁠.

Worked solution
1. Givens: $I_p = 2309\ \mathrm{A}$ ⁠, $\dfrac{N_s}{N_p} = 16$ ⁠, so the voltage goes up 16 times.
2. An ideal transformer keeps the power: $V_p I_p = V_s I_s$ ⁠, so $\dfrac{I_s}{I_p} = \dfrac{V_p}{V_s} = \dfrac{N_p}{N_s}$ ⁠.
3. $I_s = \dfrac{2309}{16} \approx 144\ \mathrm{A}$ ⁠.
If you got about 36,950 A, you multiplied the current too. Then the power out would be 256 times the power in.
Step 3 of 6
Open this step
The same power
Practice · Choose
Question
The step-up transformer raises the voltage 16 times. Take it to be ideal. How does the power it sends into the line compare with the power it takes in from the generator?

Correct answer
- Correct The same power
Setup, choices and feedback
The Scene at this step
Turns ratio
16 ×
Power sent
100 MW
Show current and losses
On
Choices
Shown in a random order.
- Correct The same power
- Wrong $16$ ⁠ times more power
  Misconception: Thinks a step-up transformer gives out more power than it takes in.
  Feedback from the catalogue
  A transformer passes energy on; it does not make it. The voltage goes up and the current goes down in the same ratio, so $P = V I$ ⁠ is the same on both sides, and in a real transformer a little less comes out, because a little turns to heat. If a transformer multiplied power, you could feed its output back into its input and get energy for free.
- Wrong $16$ ⁠ times less power
  Feedback
  An ideal transformer loses no power: the voltage goes up and the current down in the same ratio, and the product $V I$ ⁠ does not change. A real one loses less than one per cent.
Feedback for a right answer
The voltage goes up 16 times and the current comes down 16 times, so $P = V I$ ⁠ is the same on both sides. A real transformer passes on more than 99% of what goes in, and the rest becomes heat in the core and the windings.

Hint
The voltage went up 16 times. What happened to the current?

Worked solution
1. A transformer makes no energy: it passes it from coil to coil through the field in the core.
2. In an ideal transformer $P_{\text{in}} = P_{\text{out}}$ ⁠, that is $V_p I_p = V_s I_s$ ⁠.
3. In the Scene: 25 kV times 2,309 A on the station side, and 400 kV times 144 A on the line side: the same product.
If you chose $16$ ⁠ times more, you forgot that the current comes down. A transformer that multiplied power would be a perpetual motion machine.
Step 4 of 6
Open this step
Alternating current only
Practice · Choose
Question
A 12 V battery is connected to the primary coil of a transformer whose secondary coil has $20$ ⁠ times as many turns. What voltage appears across the secondary coil after the moment of connection?

Correct answer
- Correct Zero
Setup, choices and feedback
The Scene at this step
Turns ratio
16 ×
Power sent
100 MW
Show current and losses
On
Choices
Shown in a random order.
- Correct Zero
- Wrong 240 V
  Misconception: Thinks a transformer steps up the voltage of a battery or any steady (DC) supply.
  Feedback from the catalogue
  A transformer works by induction, and induction needs a changing flux. A direct current makes a steady field in the core, and a steady field induces no voltage in the secondary coil, apart from a brief blip at the moment of connection. That is why the grid uses alternating current: it reverses 50 times a second, so the flux in the core is always changing.
- Wrong 12 V
  Feedback
  There is no electrical connection between the coils: voltage reaches the secondary only through a changing field in the core.
- Wrong 0.6 V
  Feedback
  Even with alternating current this would be the wrong way round: more turns give more voltage. And with direct current there is no voltage at all.
Feedback for a right answer
A battery gives direct current, which makes a steady field in the core. A steady field induces no voltage, so the secondary coil is quiet, apart from a brief blip at the moment of connection, as the field rises from zero. That is why the grid runs on alternating current.

Hint
What does it take to induce a voltage in a coil?

Worked solution
1. A transformer works by induction: the current in the primary makes a flux in the core, and a change in that flux induces a voltage in the secondary.
2. A battery gives a steady direct current, so after the moment of connection the flux is steady: $\Delta\Phi = 0$ ⁠.
3. So $\mathcal{E} = N\,\dfrac{\Delta\Phi}{\Delta t} = 0$ ⁠ in the secondary.
If you chose 240 V, you used the turns ratio, which holds only for alternating current, where the flux changes all the time.
Step 5 of 6
Open this step
Heat in the line
Practice · Calculate
Question
The line now runs at 100 kV, the station sends 100 MW, and each wire of the line has a resistance of 5 Ω. How much power turns to heat in the line? Use $P_L = \dfrac{P^2 R}{V^2}$ ⁠.

Correct answer
5 MW ± 2%
Setup, choices and feedback
The Scene at this step
Turns ratio
4 ×fixed
Power sent
100 MWfixed
Show current and losses
Offfixed
Accepted within
± 2%
Computed from the Scene at this step’s setting
Power turned to heat in the line/nodes/grid/outputs/loss
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 80 MW ± 2%
  Misconception: Works out the line's loss with the generator's voltage instead of the line's.
  Feedback from the catalogue
  The loss is set by the current in the line, and the current in the line is set by the line’s voltage, after the transformer. In $P_L = \dfrac{P^2 R}{V^2}$ ⁠, put in the voltage the line runs at. The generator’s voltage is right only for a line with no step-up.
Feedback for a right answer
$P_L = \dfrac{(10^8)^2 \cdot 5}{(10^5)^2} = 5 \times 10^6\ \mathrm{W}$ ⁠: 5 MW, five per cent of what is sent. The tag over the town shows the rest: 95.0 MW.

Hint
Put in the line’s voltage, after the transformer, in SI units.

Worked solution
1. Givens: $P = 100\ \mathrm{MW} = 10^8\ \mathrm{W}$ ⁠, $V = 100\ \mathrm{kV} = 10^5\ \mathrm{V}$ ⁠, $R = 5\ \Omega$ ⁠.
2. $P_L = \dfrac{P^2 R}{V^2}$ ⁠.
3. Substitute: $P_L = \dfrac{10^{16} \cdot 5}{10^{10}} = 5 \times 10^6\ \mathrm{W} = 5\ \mathrm{MW}$ ⁠.
If you got 80 MW, you put in the generator’s voltage, 25 kV. The loss is set by the voltage of the line itself.
Step 6 of 6
Open this step
Double the voltage
Practice · Predict, then look
Question
The Turns ratio is doubled from $4$ ⁠ to $8$ ⁠, and the line’s voltage rises to 200 kV. What will happen to the heat in the wires?

Correct answer
- Correct It will fall to a quarter
Setup, choices and feedback
The Scene at this step
Turns ratio
4 ×
Power sent
100 MWfixed
Show current and losses
Off
Choices
- Correct It will fall to a quarter
- Wrong It will fall to half
  Misconception: Thinks halving the current halves the heat in the line, forgetting the square in I²R.
  Feedback from the catalogue
  The heat in a wire is $I^2 R$ ⁠: the current is squared. Half the current gives $\left(\tfrac12\right)^2 = \tfrac14$ ⁠ of the heat, not half. So doubling the voltage, which halves the current, cuts the loss to a quarter.
- Wrong It will not change, because the same power is sent
  Feedback
  The power sent has not changed, but the current carrying it is smaller, and the heat depends on the current.
- Wrong It will double, because the voltage is higher
  Feedback
  The heat in a wire is $I^2 R$ ⁠; it does not depend on the line’s voltage directly. A higher voltage means a smaller current for the same power.
Once the prediction is locked in, the Scene changes
Turns ratio
8 ×
Show current and losses
On
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Double the voltage, half the current, a quarter of the heat: from 5 MW to 1.25 MW. The chart shows the loss plunging as the voltage rises.

Worked solution
1. For the same power, the current is in inverse proportion to the voltage: double the voltage gives half the current.
2. The heat in the wires is $I^2 R$ ⁠, so half the current gives $\left(\tfrac12\right)^2 = \tfrac14$ ⁠ of the heat.
3. By the formula: $P_L = \dfrac{10^{16} \cdot 5}{(2 \times 10^5)^2} = 1.25 \times 10^6\ \mathrm{W}$ ⁠, a quarter of $5\ \mathrm{MW}$ ⁠.
If you chose half, you forgot that the current is squared.

Across the unit · 26 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Magnetic fields

Field lines

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Two poles

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

North facing north

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

What a magnet pulls

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

The Earth as a magnet

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Electromagnets

A current makes a field

The field inside a coil

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

One product

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

An iron core

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Saturation

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Which end is north

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

The force on a current

A push sideways

Question

Correct answer

The Scene at this step

Choices

Hint