Physics · Secondary School · Electromagnetism

Electromagnetism

The compass that points north, the motor in the blender, the power station that supplies the city and the phone charger all depend on a close link between electricity and magnetism. We start from magnets, see that currents create magnetic fields and that these fields push on charges and wires, and finish with induction, which lies behind generators and transformers.

1Magnets
2Currents make fields
3Force on charges
4Force on wires
5Induction
6Generators
7Transformers
✓Challenges

STEP 1

Why does a compass point north?

A magnet has two regions where its pull on iron is strongest, the poles. We call the north pole the one that, with the magnet free to turn, swings round to face geographic north, and the other one the south pole. Like poles repel each other, and unlike poles attract.

Something that may surprise us is that the poles never come apart. If we break a magnet in half, each piece becomes a complete magnet, with a north and a south, and this keeps holding for smaller and smaller pieces. To this day nobody has found an isolated magnetic pole.

Around the magnet there is a magnetic field \(\vec B\), which we measure in tesla (T). We represent this field with field lines, which, outside the magnet, leave the north pole and arrive at the south pole and crowd together where the field is stronger. We can think of a compass as a small magnet that turns freely and lines up with these lines, with its north end in the direction of \(\vec B\).

The Earth itself behaves like a large magnet. Since the north end of the needle points to geographic north, and unlike poles attract, we conclude that near the geographic north pole there is a magnetic south pole. The axis of this large magnet is tilted by about 10° from the axis of rotation, and so a compass shows geographic north only approximately.

like poles repelunlike poles attract\(\vec B\) in tesla (T)To get a feel for the values, the Earth's field at the surface lies between \(2 \cdot 10^{-5}\) and \(6 \cdot 10^{-5}\ \text{T}\), a fridge magnet produces a few thousandths of a tesla right next to its surface and a neodymium magnet can go above 1 T. In the simulation, the field of each magnet is worked out as if there were a 'magnetic charge' at each pole, a simple model that describes the field outside the magnet well; inside it, the lines run from the south pole to the north pole and close up.

Let's discuss

Watch the compasses around a single magnet. Where does the red tip point near the magnet's north pole, and near its south pole?
Break the magnet in half and drag the pieces away from each other. How many north poles and how many south poles are there on the table now?
Compare the scene with unlike poles face to face with the one with like poles. What happens to the lines and the compasses in the region between the magnets?
Open the Earth scene. Where do the north ends of the needles point, and which magnetic pole is hidden near the geographic north pole?

Magnets on the table

North poles

South poles

Between the magnets

Exercises for step 1 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A bar magnet is cut in half, at right angles to its length. What do we get? Can we end up with a north pole on one side and a south pole on the other, kept apart?

Show solution
Each half becomes a complete magnet, with a north pole at one end and a south pole at the other, and the same happens if we cut each piece again.
We get two smaller magnets, each with its own two poles, and it is not possible to isolate a pole.
basic
Two bar magnets lie in a line on a table, with the north pole of one facing the north pole of the other. Do they attract or repel each other? And if one of them is turned round?

Show solution
Like poles repel, and unlike poles attract.
After the half turn, a north pole and a south pole face each other.
In the first situation they repel; after the half turn, they attract.
basic
A compass is placed on the table, right in front of the north pole of a bar magnet, in line with the magnet. Where does the north end of the needle point? Justify your answer with the field lines.

Show solution
Outside the magnet, the field lines leave the north pole, and there, in line with the magnet, they point away from it.
The needle lines up with the field, with its north end in the direction of the lines.
The north end of the needle points away from the magnet, and the south end faces the magnet's north pole.
basic
Why do we say that there is a magnetic south pole near the Earth's geographic north pole?

Show solution
The north end of a compass needle points, approximately, to geographic north.
Since unlike poles attract, what attracts the north end of the needle is a south pole.
Near the geographic north pole there is a magnetic south pole, and it is what attracts the north end of compasses.
intermediate
A student wants to know whether a metal bar is a magnet or just a piece of iron. He brings one end of the bar close to the north pole of a compass needle, and the pole is attracted. Does this prove that the bar is a magnet? What test would settle the question?

Show solution
An ordinary piece of iron is also attracted by either pole of a magnet, and attraction on its own decides nothing.
Only a magnet can repel a pole. It is enough to bring both ends of the bar close to both poles of the needle and look for a repulsion.
It does not prove it; if either end of the bar repels one of the poles of the needle, the bar is a magnet.
intermediate
Why do two magnetic field lines never cross?

Show solution
At each point, the magnetic field has a single direction, and the field line through that point is tangent to it.
If two lines crossed, a compass placed at the crossing would have to point in two directions at once.
The lines do not cross because the field has only one direction at each point.
intermediate
In São Paulo, the north end of a compass now points about 21° west of geographic north. A hiker wants to head exactly towards geographic north. Which way should she turn from the direction shown by the needle, and by how many degrees?

Show solution
The needle points in a direction 21° west of geographic north.
To reach geographic north, the hiker needs to turn from the needle, the other way, by the same angle.
She should turn about 21° to the east of the needle's direction.
intermediate
At a point on a table, the field of a magnet is \(2 \cdot 10^{-5}\ \text{T}\) and points east, and the Earth's horizontal field is also \(2 \cdot 10^{-5}\ \text{T}\) and points north. Where does the needle of a compass placed there point, and what is the magnitude of the resultant field?

Show solution
The needle lines up with the vector sum of the two fields, which are perpendicular and of equal magnitude.
The resultant makes 45° with each of them and has magnitude \(B = \sqrt{(2 \cdot 10^{-5})^2 + (2 \cdot 10^{-5})^2}\).
The needle points north-east, and \(B = 2\sqrt{2} \cdot 10^{-5}\ \text{T}\) \(\approx 2.8 \cdot 10^{-5}\ \text{T}\).
challenge
A bar magnet is broken into four equal pieces, which are then placed in a row, touching, in the same order and orientation as before. How many north poles and how many south poles are there in the row? Do the pieces stick to one another? How does the row behave, seen from a distance?

Show solution
Each piece is a complete magnet, and the row has four north poles and four south poles.
At the three joins, the north pole of one piece touches the south pole of the next, and they attract.
Seen from a distance, the row has a free north pole at one end and a free south pole at the other, and it behaves like the original magnet.
There are 4 north poles and 4 south poles, the pieces attract at the joins and, from a distance, the row works like the whole magnet.
challenge
A magnetised needle, stuck through a cork floating in a bowl of water, turns until it points north, but it does not swim off towards the north. Why does the Earth's field make the needle turn without pulling it? What would change if we put a strong magnet a few centimetres from the bowl?

Show solution
On the scale of the bowl, the Earth's field is practically uniform. The force on the needle's north pole and the force on its south pole have the same magnitude and opposite directions.
The two forces cancel, and the needle is not pulled, but they act at different points and form a couple that makes it turn.
Near a magnet, the field changes a lot from one point to another, the forces on the two poles are no longer equal, and a resultant force is left over.
In the uniform field of the Earth the needle only turns; near the magnet, where the field is not uniform, it would also be pulled.

STEP 2

Does an electric current create a magnetic field?

In 1820, the Danish physicist Hans Christian Oersted noticed that a compass needle moved when he switched on the current in a nearby wire. It was the first evidence that electricity and magnetism are linked, and today we know that every electric current creates a magnetic field around it.

Around a long straight wire, the field lines are circles centred on the wire. Their direction comes from the right-hand grip rule. If we imagine the right hand closed round the wire, with the thumb in the direction of the current, the other fingers curl in the direction of the lines. In the drawings, ⊙ marks something coming out of the screen, like the tip of an arrow flying towards us, and ⊗ marks something going into the screen, like the feathers of an arrow flying away.

If we bend the wire into a loop, the fields of all its parts add up at the centre. With many loops side by side, in a solenoid, the field inside becomes nearly uniform and, from outside, the solenoid looks like a bar magnet. With an iron core inside, we have an electromagnet, whose field can become hundreds of times stronger and which we switch off simply by cutting the current.

The formulas contain \(\mu_0 = 4\pi \cdot 10^{-7}\ \text{T} \cdot \text{m/A}\), the magnetic permeability of free space, which we also use for air. For the straight wire, for example, it leads to \(B = 2 \cdot 10^{-7}\,i/d\) in SI units, and we see that, 5 cm from a wire carrying 5 A, the field of the current already matches the Earth's.

\(B = \dfrac{\mu_0\, i}{2\pi d}\)\(B = \dfrac{\mu_0\, i}{2R}\)\(B = \mu_0\,\dfrac{N}{L}\, i\)Long straight wire at a distance \(d\), centre of a loop of radius \(R\) and inside a solenoid with \(N\) turns over a length \(L\), always with \(\mu_0 = 4\pi \cdot 10^{-7}\ \text{T} \cdot \text{m/A}\). In the simulation, seen from above, the Earth's field, \(2 \cdot 10^{-5}\ \text{T}\) horizontal and pointing north, also acts on the compasses. The drawing of the lines for the loop and the solenoid treats each piece of wire that crosses the table as a long wire, which gives the shape of the lines, and the values at the centre come from the formulas in the box.

Let's discuss

Start with the current at zero and raise it slowly. At what current does the big compass, 5 cm from the wire, turn by about 45°? Check with \(B = \mu_0 i/(2\pi d)\) and with the Earth's field.
Reverse the current and watch the compasses around the wire. What happens to the direction of the field lines?
Drag the big compass away from the wire. How does the deflection change with distance, and can we predict this from the formula?
Switch to the solenoid and put in the iron core. Why do scrapyard cranes use electromagnets rather than ordinary magnets?

Current

B from the formula

Big compass: field of the current

Deflection of the big compass

Exercises for step 2 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A long straight wire carries a current of 10 A. What is the magnitude of the magnetic field 5 cm from the wire? Use \(\mu_0 = 4\pi \cdot 10^{-7}\ \text{T} \cdot \text{m/A}\).

Show solution
With \(d = 0.05\ \text{m}\), \(B = \dfrac{\mu_0\, i}{2\pi d} = \dfrac{4\pi \cdot 10^{-7} \cdot 10}{2\pi \cdot 0.05}\).
\(B = 4 \cdot 10^{-5}\ \text{T}\)
basic
What is the magnitude of the magnetic field at the centre of a circular loop of radius 10 cm carrying a current of 5 A? Use \(\mu_0 = 4\pi \cdot 10^{-7}\ \text{T} \cdot \text{m/A}\).

Show solution
At the centre of the loop, \(B = \dfrac{\mu_0\, i}{2R} = \dfrac{4\pi \cdot 10^{-7} \cdot 5}{2 \cdot 0.1}\).
\(B = \pi \cdot 10^{-5}\ \text{T} \approx 3.1 \cdot 10^{-5}\ \text{T}\)
basic
A solenoid has 1000 turns spread over a length of 50 cm and carries 2 A. What is the magnitude of the field inside it? Use \(\mu_0 = 4\pi \cdot 10^{-7}\ \text{T} \cdot \text{m/A}\).

Show solution
The number of turns per metre is \(\dfrac{N}{L} = \dfrac{1000}{0.5} = 2000\ \text{m}^{-1}\), and \(B = \mu_0\,\dfrac{N}{L}\,i\) \(= 4\pi \cdot 10^{-7} \cdot 2000 \cdot 2\).
\(B = 1.6\pi \cdot 10^{-3}\ \text{T}\) \(\approx 5.0 \cdot 10^{-3}\ \text{T}\)
basic
A vertical wire carries a current flowing upwards. Looking from above, do the field lines around it go clockwise or anticlockwise? And if the current flows downwards?

Show solution
With the thumb of the right hand in the direction of the current, the other fingers curl in the direction of the field lines.
Looking from above, the upward current comes towards us, like the ⊙ in the drawings.
With the current flowing upwards, the lines go anticlockwise; with the current flowing downwards, clockwise.
intermediate
At what distance from a long straight wire carrying 20 A does the field of the current have the same magnitude as the horizontal component of the Earth's field, \(2 \cdot 10^{-5}\ \text{T}\)? Use \(\mu_0 = 4\pi \cdot 10^{-7}\ \text{T} \cdot \text{m/A}\).

Show solution
Solving for \(d\) in \(B = \dfrac{\mu_0\, i}{2\pi d}\), we get \(d = \dfrac{\mu_0\, i}{2\pi B} = \dfrac{2 \cdot 10^{-7} \cdot 20}{2 \cdot 10^{-5}}\).
\(d = 0.2\ \text{m}\), or 20 cm
intermediate
A compass sits on a table, 4 cm south of a long vertical wire. With no current, the needle points north, and the horizontal component of the Earth's field is \(2 \cdot 10^{-5}\ \text{T}\). What current, flowing up the wire, makes the needle turn by 45°, and which way does it turn? Use \(\mu_0 = 4\pi \cdot 10^{-7}\ \text{T} \cdot \text{m/A}\).

Show solution
Looking from above, the upward current produces anticlockwise lines, and south of the wire the field of the current points east, at right angles to the Earth's.
The needle turns by 45° when the two fields have the same magnitude, \(\dfrac{\mu_0\, i}{2\pi d} = 2 \cdot 10^{-5}\), and then \(i = \dfrac{2 \cdot 10^{-5} \cdot 0.04}{2 \cdot 10^{-7}}\).
\(i = 4\ \text{A}\), and the needle turns from north towards north-east.
intermediate
Two concentric circular loops, in the same plane, have radii of 10 cm and 20 cm and carry 4 A each, in opposite directions. What is the magnitude of the field at the centre? And if the currents were in the same direction? Use \(\mu_0 = 4\pi \cdot 10^{-7}\ \text{T} \cdot \text{m/A}\).

Show solution
The smaller loop produces at the centre \(B_1 = \dfrac{4\pi \cdot 10^{-7} \cdot 4}{2 \cdot 0.1}\) \(= 8\pi \cdot 10^{-6}\ \text{T}\), and the larger one, \(B_2 = \dfrac{4\pi \cdot 10^{-7} \cdot 4}{2 \cdot 0.2}\) \(= 4\pi \cdot 10^{-6}\ \text{T}\).
With opposite currents, the fields point in opposite directions and subtract; with the same direction, they add.
\(4\pi \cdot 10^{-6} \approx 1.3 \cdot 10^{-5}\ \text{T}\) with opposite directions, and \(12\pi \cdot 10^{-6} \approx 3.8 \cdot 10^{-5}\ \text{T}\) with the same direction.
intermediate
A solenoid 20 cm long must produce a field of \(6.0 \cdot 10^{-3}\ \text{T}\) inside it with a current of 3 A. How many turns does it need? Use \(\mu_0 = 4\pi \cdot 10^{-7}\ \text{T} \cdot \text{m/A}\).

Show solution
Solving for \(N\) in \(B = \mu_0\,\dfrac{N}{L}\,i\), we get \(N = \dfrac{B\,L}{\mu_0\, i} = \dfrac{6.0 \cdot 10^{-3} \cdot 0.2}{4\pi \cdot 10^{-7} \cdot 3}\).
\(N \approx 320\) turns
challenge
Two long straight parallel wires are 10 cm apart and carry currents of 6 A in opposite directions. What is the magnitude of the magnetic field at the midpoint between them? And if the currents were in the same direction? Use \(\mu_0 = 4\pi \cdot 10^{-7}\ \text{T} \cdot \text{m/A}\).

Show solution
Each wire, 5 cm from the midpoint, produces there \(B = \dfrac{\mu_0\, i}{2\pi d}\) \(= \dfrac{2 \cdot 10^{-7} \cdot 6}{0.05}\) \(= 2.4 \cdot 10^{-5}\ \text{T}\).
By the right-hand grip rule, with opposite currents the two fields point the same way at the midpoint and add. With currents in the same direction, they point opposite ways and cancel.
\(4.8 \cdot 10^{-5}\ \text{T}\) with opposite currents, and zero with currents in the same direction.
challenge
The electromagnet of a scrapyard crane is a coil of 500 turns, 25 cm long, carrying 10 A. What would the field inside the coil be without a core? Why is the iron core essential, and why is an electromagnet more practical than a permanent magnet for this job? Use \(\mu_0 = 4\pi \cdot 10^{-7}\ \text{T} \cdot \text{m/A}\).

Show solution
Without a core, \(B = \mu_0\,\dfrac{N}{L}\,i\) \(= 4\pi \cdot 10^{-7} \cdot \dfrac{500}{0.25} \cdot 10\).
The iron of the core becomes magnetised and adds its own field to the coil's, which can become hundreds of times stronger. With the electromagnet, the crane drops its load when we switch off the current, something impossible with a permanent magnet.
Without a core, \(B = 8\pi \cdot 10^{-3}\ \text{T} \approx 2.5 \cdot 10^{-2}\ \text{T}\); the iron multiplies it many times over, and the electromagnet can be switched off to drop the scrap.

STEP 3

What does a magnetic field do to a moving charge?

A charge at rest feels no magnetic force at all. When it moves inside a field \(\vec B\), a force appears with magnitude \(F = |q|\,v\,B\,\sin\theta\), where \(\theta\) is the angle between the velocity and the field. The force is greatest with \(\vec v\) perpendicular to \(\vec B\) and vanishes when the charge travels along the field lines.

The force is perpendicular to both \(\vec v\) and \(\vec B\) at the same time. We will use the flat right-hand rule, in which the thumb points in the direction of \(\vec v\), the other fingers, held straight, point in the direction of \(\vec B\), and the palm pushes in the direction of the force on a positive charge. On a negative charge, the force points the opposite way. Many British books use Fleming's left-hand rule instead, with a different arrangement of the fingers, and it gives the same answer.

Since the force is always perpendicular to the velocity, it does no work and does not change the speed, only the direction of motion. In a uniform field, with \(\vec v\) perpendicular to \(\vec B\), the magnetic force plays the part of the centripetal resultant force from the Circular Motion lesson, and the particle moves in uniform circular motion. Setting \(|q|\,v\,B\) equal to \(m\,v^2/R\), we arrive at the radius and the period, and the period, curiously, does not depend on the speed.

This motion turns up in many places. The mass spectrometer separates ions of different masses by the radius of their path, and the aurorae, perhaps the most beautiful example, appear when charged particles coming from the Sun spiral around the lines of the Earth's field, are guided to the polar regions and make the gases high in the atmosphere glow.

\(F = |q|\,v\,B\,\sin\theta\)\(R = \dfrac{m\,v}{|q|\,B}\)\(T = \dfrac{2\pi m}{|q|\,B}\)The last two hold with \(\vec v\) perpendicular to \(\vec B\), in a uniform field. In the simulation, the mass is in atomic mass units, \(1\ \text{u} = 1.66 \cdot 10^{-27}\ \text{kg}\), and the charge in multiples of \(e = 1.6 \cdot 10^{-19}\ \text{C}\). We neglect gravity, tiny next to the magnetic force, and the scene runs much more slowly than reality, by a factor that does not depend on the speed, so that we can compare the times for one revolution.

velocitymagnetic forcepath

Let's discuss

Launch the proton and follow the red force arrow. Where does it point the whole time? Does the speed change?
Double the speed and launch again. What happens to the radius? And to the time for one revolution, which may surprise us?
Swap the proton for the H⁻ ion, which has a negative charge, without changing anything else. Which way does it turn now? Check with the right-hand rule.
Switch to the spectrometer and compare where the two particles land. Why does the heavier one land further from the slit?

Charge

Mass

Speed

Field

Radius of the path

Period

Magnetic force

Speed

Exercises for step 3 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A proton enters a uniform magnetic field of 0.5 T with a speed of \(2 \cdot 10^6\ \text{m/s}\), perpendicular to the field. What is the magnitude of the magnetic force on it? Use \(e = 1.6 \cdot 10^{-19}\ \text{C}\).

Show solution
With \(\theta = 90^\circ\), \(F = |q|\,v\,B\) \(= 1.6 \cdot 10^{-19} \cdot 2 \cdot 10^6 \cdot 0.5\).
\(F = 1.6 \cdot 10^{-13}\ \text{N}\)
basic
An electron moves along the same line and in the same direction as the lines of a uniform magnetic field. What magnetic force does it feel?

Show solution
The angle between the velocity and the field is zero, and \(\sin 0^\circ = 0\) in \(F = |q|\,v\,B\,\sin\theta\).
The magnetic force is zero, and the electron carries on in a straight line.
basic
A proton moves to the right, on the screen, in a magnetic field going into the screen (⊗). Which way does the magnetic force on it point? And if the particle were an electron?

Show solution
With the flat right hand, the thumb to the right (velocity) and the fingers pointing into the screen (field), the palm pushes upwards.
For the electron, with its negative charge, the force points the opposite way.
On the proton, the force points up the screen; on the electron, down.
basic
Does the magnetic force on a moving charge do work? What does that tell us about the charge's kinetic energy?

Show solution
The magnetic force is always perpendicular to the velocity, and a force perpendicular to the displacement does no work.
With no work, the kinetic energy does not change, and neither does the speed.
It does no work, and the charge's kinetic energy stays constant; only the direction of motion changes.
intermediate
A proton moves in a circle in a uniform magnetic field of 0.1 T, with a speed of \(10^6\ \text{m/s}\). What is the radius of its path? Use \(m_p = 1.67 \cdot 10^{-27}\ \text{kg}\) and \(e = 1.6 \cdot 10^{-19}\ \text{C}\).

Show solution
The magnetic force is the centripetal resultant, \(|q|\,v\,B = \dfrac{m\,v^2}{R}\), and so \(R = \dfrac{m\,v}{|q|\,B}\) \(= \dfrac{1.67 \cdot 10^{-27} \cdot 10^6}{1.6 \cdot 10^{-19} \cdot 0.1}\).
\(R \approx 0.10\ \text{m}\), or about 10 cm
intermediate
An electron, of mass \(9.1 \cdot 10^{-31}\ \text{kg}\), circles in a uniform magnetic field of \(1.0 \cdot 10^{-3}\ \text{T}\). What is the period of the motion? Does it change if the electron's speed doubles? Use \(e = 1.6 \cdot 10^{-19}\ \text{C}\).

Show solution
The period is \(T = \dfrac{2\pi m}{|q|\,B}\) \(= \dfrac{2\pi \cdot 9.1 \cdot 10^{-31}}{1.6 \cdot 10^{-19} \cdot 10^{-3}}\), and the speed does not appear in the formula.
With twice the speed, the radius doubles, and so does the distance round one revolution, but the revolution takes the same time.
\(T \approx 3.6 \cdot 10^{-8}\ \text{s}\), and it does not change with the speed.
intermediate
A particle with a charge of \(3\ \mu\text{C}\) crosses a magnetic field of 0.2 T at 500 m/s, in a direction at 30° to the field lines. What is the magnitude of the magnetic force on it?

Show solution
With \(\sin 30^\circ = 0.5\), \(F = |q|\,v\,B\,\sin\theta\) \(= 3 \cdot 10^{-6} \cdot 500 \cdot 0.2 \cdot 0.5\).
\(F = 1.5 \cdot 10^{-4}\ \text{N}\)
intermediate
A proton and an alpha particle, which has twice the charge and four times the mass of the proton, enter the same uniform magnetic field with the same speed, perpendicular to it. Which of the two moves in the circle of larger radius, and how many times larger?

Show solution
The radius is \(R = \dfrac{m\,v}{|q|\,B}\), proportional to the ratio \(m/|q|\) when \(v\) and \(B\) are the same.
For the alpha particle, \(\dfrac{4m}{2e} = 2\,\dfrac{m}{e}\).
The alpha particle moves in the larger radius, twice the radius of the proton.
challenge
In a mass spectrometer, carbon-12 and carbon-14 ions, both with charge \(+e\), enter a field of 0.2 T with a speed of \(10^5\ \text{m/s}\) and travel along semicircles to a detector. How far apart do they reach the detector? Use \(1\ \text{u} = 1.66 \cdot 10^{-27}\ \text{kg}\) and \(e = 1.6 \cdot 10^{-19}\ \text{C}\).

Show solution
For carbon-12, \(R_{12} = \dfrac{12 \cdot 1.66 \cdot 10^{-27} \cdot 10^5}{1.6 \cdot 10^{-19} \cdot 0.2}\) \(\approx 6.23 \cdot 10^{-2}\ \text{m}\), and for carbon-14, \(R_{14} = \dfrac{14}{12}\,R_{12} \approx 7.26 \cdot 10^{-2}\ \text{m}\).
Each ion reaches the detector one diameter away from the slit, and the distance between the landing points is \(2\,(R_{14} - R_{12})\).
About 2.1 cm
challenge
Aurorae appear near the poles. Explain this with the magnetic force on the charged particles arriving from the Sun, and say what kind of path a particle follows when it enters a uniform field with its velocity at a slant to the field lines.

Show solution
The part of the velocity parallel to the field lines feels no force, and the perpendicular part makes the particle circle around the lines.
The combination of the two motions is a helix, a spiral that moves forward along the field lines.
The particles are trapped on the lines of the Earth's field and guided to the polar regions, where these lines come down into the atmosphere, and there they collide with the gases and make them glow.
The particles spiral along the field lines, in a helix, and reach the atmosphere near the poles, where the lines dive into the Earth.

STEP 4

How does the magnetic force make a motor turn?

A current is made of moving charges, and so a wire carrying a current inside a magnetic field also feels a force. Adding up the forces on the charges in a straight piece of length \(L\), we arrive at \(F = B\,i\,L\,\sin\theta\), where \(\theta\) is the angle between the wire and the field. The direction comes from the same right-hand rule, now with the thumb in the direction of the current.

Two parallel wires carrying currents interact, because each one sits in the magnetic field of the other. With the currents in the same direction they attract and, in opposite directions, they repel, the reverse of what we might expect by comparison with electric charges, where like charges repel.

The electric motor makes use of this force. To understand it, we imagine a loop carrying a current between the poles of a magnet, which receives forces in opposite directions on its two sides, and this pair of forces makes it turn. After half a turn the forces would start to brake the rotation, and a commutator, a split ring that turns with the loop and presses against fixed brushes, reverses the current every half turn so that the rotation carries on in the same direction.

We can see the loudspeaker as a close relative of the motor. A coil attached to a cone sits in the field of a magnet, and the current coming from the amplifier, varying in step with the sound, pushes the coil back and forth, while the cone makes the air vibrate.

\(F = B\,i\,L\,\sin\theta\)\(\dfrac{F}{L} = \dfrac{\mu_0\, i_1\, i_2}{2\pi d}\)The second formula gives the force per metre between two long parallel wires a distance \(d\) apart. In the simulation, the loop has 20 turns of wire, sides of 5 cm perpendicular to the field and a width of 4 cm, and sits in a uniform field of 0.2 T, so that each side feels \(F = N\,B\,i\,L\). We assume friction at the axle proportional to the rate of rotation, and the rotation is much slower than that of a real motor.

Let's discuss

With the commutator on, watch the force arrows on the two sides of the loop. In which position is the torque greatest, and in which is it zero?
Switch off the commutator and wait a little. In which position does the loop end up stopping, and why?
Reverse the current while the motor is turning. What happens to the direction of rotation?
Switch to the two wires and reverse the current in wire 2. Do the wires now attract or repel? Double the current and see how much the force grows.

Current

Distance between the wires

Force on each side

Torque now

Rotation

Current

Exercises for step 4 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A straight wire 50 cm long, carrying 4 A, is perpendicular to a uniform magnetic field of 0.3 T. What is the magnitude of the magnetic force on it?

Show solution
With \(\theta = 90^\circ\), \(F = B\,i\,L = 0.3 \cdot 4 \cdot 0.5\).
\(F = 0.6\ \text{N}\)
basic
A wire 20 cm long, carrying 5 A, makes an angle of 30° with the lines of a uniform magnetic field of 0.4 T. What is the force on it? And if the wire were parallel to the field?

Show solution
With \(\sin 30^\circ = 0.5\), \(F = B\,i\,L\,\sin\theta\) \(= 0.4 \cdot 5 \cdot 0.2 \cdot 0.5\). Parallel to the field, \(\sin 0^\circ = 0\).
\(F = 0.2\ \text{N}\) at 30°, and no force with the wire parallel to the field.
basic
Two parallel wires carry currents in the same direction. Do they attract or repel each other? And with currents in opposite directions?

Show solution
Each wire sits in the magnetic field of the other and feels a force given by the right-hand rule.
With currents in the same direction, the wires attract; with opposite currents, they repel.
basic
In a direct-current electric motor, what is the commutator for?

Show solution
After half a turn, the forces on the sides of the loop would start to turn it the other way, and it would rock back and forth until it stopped.
The commutator, a split ring that turns with the loop and presses against fixed brushes, reverses the current in the loop every half turn.
The commutator reverses the current every half turn so that the torque always keeps the same direction of rotation.
intermediate
Two long straight parallel wires are 2 cm apart and carry 10 A each, in the same direction. What is the force per metre of wire, and is it attractive or repulsive? Use \(\mu_0 = 4\pi \cdot 10^{-7}\ \text{T} \cdot \text{m/A}\).

Show solution
The force per metre is \(\dfrac{F}{L} = \dfrac{\mu_0\, i_1\, i_2}{2\pi d}\) \(= \dfrac{2 \cdot 10^{-7} \cdot 10 \cdot 10}{0.02}\), and currents in the same direction attract.
\(\dfrac{F}{L} = 1 \cdot 10^{-3}\ \text{N/m}\), attractive
intermediate
A conducting rod 20 cm long with a mass of 30 g hangs horizontally, in a region with a horizontal magnetic field of 0.5 T, perpendicular to it. What current must pass through the rod for the magnetic force, pointing upwards, to balance its weight? Use g = 10 m/s².

Show solution
Equilibrium requires \(B\,i\,L = m\,g\), and so \(i = \dfrac{m\,g}{B\,L} = \dfrac{0.03 \cdot 10}{0.5 \cdot 0.2}\).
\(i = 3\ \text{A}\)
intermediate
A rectangular coil measuring 10 cm by 5 cm, with 50 turns of wire and a current of 2 A, sits in a uniform magnetic field of 0.4 T, with its 10 cm sides perpendicular to the field. What is the force on each of these sides? What is the greatest torque the coil can experience?

Show solution
On each 10 cm side, with 50 turns of wire, \(F = N\,B\,i\,L = 50 \cdot 0.4 \cdot 2 \cdot 0.1\).
The torque is greatest when the plane of the coil is parallel to the field, and then the two forces, in opposite directions, are separated by the 5 cm width, \(\tau = F \cdot 0.05\).
4 N on each side, and a maximum torque of \(0.2\ \text{N} \cdot \text{m}\).
intermediate
The coil of a loudspeaker has 5 m of wire, all of it perpendicular to the magnet's field of 1.2 T. What is the force on the coil when the current is 0.5 A? What happens to this force when the current changes direction?

Show solution
Adding up all the wire, \(F = B\,i\,L = 1.2 \cdot 0.5 \cdot 5\). With the current reversed, the force also reverses, and the coil is pushed the other way.
\(F = 3\ \text{N}\), and the direction of the force reverses along with the current, which makes the cone move back and forth.
challenge
Until 2019, the ampere was defined as the current that, kept flowing in two long straight parallel wires 1 m apart in a vacuum, produces between them a force of \(2 \cdot 10^{-7}\ \text{N}\) per metre of wire. Show that this definition is consistent with \(\mu_0 = 4\pi \cdot 10^{-7}\ \text{T} \cdot \text{m/A}\).

Show solution
With \(i_1 = i_2 = 1\ \text{A}\) and \(d = 1\ \text{m}\), \(\dfrac{F}{L} = \dfrac{\mu_0\, i_1\, i_2}{2\pi d}\) \(= \dfrac{4\pi \cdot 10^{-7} \cdot 1 \cdot 1}{2\pi \cdot 1}\).
The calculation gives exactly \(2 \cdot 10^{-7}\ \text{N}\) per metre, and the old definition agrees with this value of \(\mu_0\).
challenge
A long horizontal wire carrying a current of 50 A is fixed to the ceiling. Just below it, 1 cm away, a second parallel wire, with a mass of 10 g per metre, hangs held up by the magnetic force alone. What current does the second wire need, and in which direction relative to the current in the first? Use g = 10 m/s². Use \(\mu_0 = 4\pi \cdot 10^{-7}\ \text{T} \cdot \text{m/A}\).

Show solution
The lower wire needs to be attracted upwards, and so its current must flow in the same direction as the first.
Per metre of wire, the weight is \(0.01 \cdot 10 = 0.1\ \text{N}\), and equilibrium requires \(\dfrac{\mu_0\, i_1\, i_2}{2\pi d} = 0.1\), or \(i_2 = \dfrac{0.1 \cdot 0.01}{2 \cdot 10^{-7} \cdot 50}\).
\(i_2 = 100\ \text{A}\), in the same direction as the current in the upper wire.

STEP 5

Can a moving magnet create a current?

If a current creates a magnetic field, it seems natural to ask whether a magnetic field can create a current. Michael Faraday showed, in 1831, that it can, as long as the field through the circuit changes. A magnet at rest beside a coil produces no current at all, and the same magnet, moved closer or further away, makes a current appear.

To describe this, we use the magnetic flux \(\Phi = B\,A\cos\theta\) through a loop of area \(A\), where \(\theta\) is the angle between \(\vec B\) and the line perpendicular to the plane of the loop. We can picture the flux, roughly, as the number of field lines passing through the loop. Faraday's law says that the induced electromotive force is the rate at which this flux changes, multiplied by the number of turns in the coil.

\(\Phi = B\,A\cos\theta\)\(|\varepsilon| = N\,\dfrac{|\Delta\Phi|}{\Delta t}\)\(i = \dfrac{\varepsilon}{R}\)Flux is measured in webers, \(1\ \text{Wb} = 1\ \text{T} \cdot \text{m}^2\). With the sign, Faraday's law is written \(\varepsilon = -N\,\Delta\Phi/\Delta t\), and the minus sign expresses Lenz's law. In the simulation, the coil has 200 turns of radius 2 cm over a length of 8 cm, the magnet is treated as a small magnetic dipole, and the circuit with the galvanometer has \(5\ \Omega\).

The direction of the current comes from Lenz's law, according to which the induced current creates a field that opposes the change in flux. If we bring the north pole of a magnet closer, the face of the coil facing it becomes a north pole and repels it; if we move it away, the face becomes a south pole and tries to hold it back. Here we see the conservation of energy, since the electrical energy of the current comes from the work we do to move the magnet.

A strong magnet dropped inside a copper or aluminium pipe falls very slowly, even though these metals are not attracted by magnets. The fall makes the flux change in the walls of the pipe, the currents induced there oppose the motion, and the magnet falls at almost constant speed.

Let's discuss

Bring the magnet's north pole closer slowly, and then quickly. What changes on the galvanometer pointer and on the graph?
Stop the magnet inside the coil. Why does the galvanometer go back to zero, if the flux there is as large as it can be?
Push the magnet right through the coil and compare the current on the way in and on the way out. Which pole appears on the face of the coil facing the magnet in each case?
Turn the magnet round and repeat the movement. What happens to the direction of the current?

Flux through the 200 turns

Induced emf

Current

Speed of the magnet

Exercises for step 5 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A loop of area \(0.02\ \text{m}^2\) sits in a uniform magnetic field of 0.5 T. What is the magnetic flux through it when the field is perpendicular to the plane of the loop? And when the field is parallel to the plane?

Show solution
With the field perpendicular to the plane, the angle between \(\vec B\) and the perpendicular to the loop is zero, and \(\Phi = B\,A = 0.5 \cdot 0.02\). With the field parallel to the plane, no line passes through the loop.
\(\Phi = 0.01\ \text{Wb}\) in the first case, and zero in the second.
basic
A magnet is at rest inside a coil connected to a galvanometer. Does the galvanometer show a current? And while the magnet is being pulled out?

Show solution
The induced current depends on the change in flux, and with the magnet at rest the flux, although large, does not change.
With the magnet at rest there is no current; while it comes out, the flux decreases and the galvanometer shows a current.
basic
In a coil of 100 turns, the flux through each turn falls from 0.02 Wb to zero in 0.1 s. What is the average induced emf?

Show solution
By Faraday's law, \(|\varepsilon| = N\,\dfrac{|\Delta\Phi|}{\Delta t}\) \(= 100 \cdot \dfrac{0.02}{0.1}\).
\(|\varepsilon| = 20\ \text{V}\)
basic
We bring the south pole of a magnet towards a coil connected to a lamp. Which pole appears on the face of the coil facing the magnet? Is the magnet attracted or repelled by the coil?

Show solution
By Lenz's law, the induced current opposes the approach, and the coil does this by repelling the magnet.
A south pole appears on the face facing the magnet, which is repelled.
intermediate
A circular loop of radius 10 cm and resistance \(0.5\ \Omega\) sits in a field perpendicular to its plane, which falls steadily from 0.8 T to zero in 0.2 s. What is the induced current? Use \(\pi \approx 3.14\).

Show solution
The area is \(A = \pi R^2 = 3.14 \cdot 0.01\) \(= 0.0314\ \text{m}^2\), and the emf is \(|\varepsilon| = \dfrac{|\Delta\Phi|}{\Delta t}\) \(= \dfrac{0.8 \cdot 0.0314}{0.2} \approx 0.126\ \text{V}\).
The current is \(i = \dfrac{\varepsilon}{R} = \dfrac{0.126}{0.5}\).
\(i \approx 0.25\ \text{A}\)
intermediate
A coil of 50 turns, with an area of \(40\ \text{cm}^2\), has its plane perpendicular to a uniform field of 0.2 T. In 0.05 s, it is turned until its plane is parallel to the field. What is the average induced emf?

Show solution
The flux per turn goes from \(B\,A = 0.2 \cdot 40 \cdot 10^{-4}\) \(= 8 \cdot 10^{-4}\ \text{Wb}\) to zero, and \(|\varepsilon| = N\,\dfrac{|\Delta\Phi|}{\Delta t}\) \(= 50 \cdot \dfrac{8 \cdot 10^{-4}}{0.05}\).
\(|\varepsilon| = 0.8\ \text{V}\)
intermediate
A strong magnet falls inside a vertical copper pipe and takes much longer to reach the floor than an identical magnet dropped beside the pipe. Copper is not attracted by magnets. What slows the magnet down, and where does the gravitational potential energy it loses go?

Show solution
During the fall, the magnetic flux changes in each ring of the pipe. Below the magnet it increases, above it it decreases, and induced currents appear in the walls of the pipe.
By Lenz's law, these currents oppose the motion of the magnet, pushing it from below and pulling it from above.
The currents induced in the copper slow the magnet down, and the potential energy it loses becomes heat in the pipe, through the Joule effect.
intermediate
In a coil of 200 turns, the flux through each turn rises steadily from zero to \(3 \cdot 10^{-4}\ \text{Wb}\) between 0 and 0.1 s, stays constant until 0.3 s and falls steadily to zero between 0.3 s and 0.4 s. What is the induced emf in each interval?

Show solution
Between 0 and 0.1 s, \(|\varepsilon| = 200 \cdot \dfrac{3 \cdot 10^{-4}}{0.1}\). Between 0.1 s and 0.3 s, the flux does not change.
Between 0.3 s and 0.4 s, the change has the same size and takes the same time as the first, with the opposite sign.
0.6 V in the first interval, zero in the second and 0.6 V with the sign reversed in the third.
challenge
A square loop of side 10 cm and resistance \(0.02\ \Omega\) moves out, at a constant speed of 2 m/s, from a region with a magnetic field of 0.5 T perpendicular to its plane, with one side parallel to the edge of the region. While it moves out, what is the induced current? What force is needed to keep the speed constant, and where does the energy dissipated in the loop come from?

Show solution
Every second, the area of the loop inside the field decreases by \(L\,v\), and \(|\varepsilon| = B\,L\,v\) \(= 0.5 \cdot 0.1 \cdot 2 = 0.1\ \text{V}\), with \(i = \dfrac{\varepsilon}{R} = \dfrac{0.1}{0.02}\).
The side still in the field feels \(F = B\,i\,L\), against the motion, as Lenz's law predicts, and whoever pulls the loop needs to exert this same force.
The power of whoever pulls, \(F\,v = B\,i\,L\,v\), equals the power dissipated, \(\varepsilon\, i = (B\,L\,v)\, i\).
\(i = 5\ \text{A}\) and \(F = 0.5 \cdot 5 \cdot 0.1 = 0.25\ \text{N}\), and the energy dissipated comes from the work done by whoever pulls the loop.
challenge
In an induction hob, a coil under the glass carries a high-frequency alternating current. Explain why the bottom of a metal pan heats up, while the glass of the hob hardly heats up by itself, and why a glass pan would not work on this hob.

Show solution
The alternating current in the coil produces a magnetic field that changes very rapidly and passes through the bottom of the pan.
In the metal, a conductor, this changing flux induces currents, which heat the bottom through the Joule effect. The glass of the hob and a glass pan are insulators, and practically no current is induced in them.
Only the metal of the pan carries the induced currents that turn into heat; the glass warms up only through contact with the hot pan, and a glass pan would not be heated.

STEP 6

How does a power station produce electrical energy?

Almost all the electrical energy we use comes from generators that work by induction. In a hydroelectric power station, the water coming down from the dam turns a turbine; in a wind turbine, it is the wind that turns the blades; in thermal and nuclear power stations, it is steam. In all these cases, the turbine makes coils turn in a magnetic field, or magnets turn in front of coils, and the flux changes without stopping.

If a coil of \(N\) turns, each of area \(A\), turns in a uniform field with constant angular velocity \(\omega\), the flux through it rises and falls like a cosine, and the induced emf varies as a sine wave of amplitude \(N\,B\,A\,\omega\). The emf changes sign every half turn, and the current changes direction along with it. This is alternating current, which in Brazil completes 60 cycles per second, a frequency of 60 Hz (on the British national grid, it is 50 Hz).

\(\varepsilon_{\text{max}} = N\,B\,A\,\omega\)\(\omega = 2\pi f\)\(\varepsilon = B\,L\,v\)The first holds for a coil turning in a uniform field, and the last for a rod of length \(L\) sliding with velocity \(v\) perpendicular to the field. In the simulation, \(B = 0.5\ \text{T}\) and each turn has \(20\ \text{cm}^2\). The lamp is treated as a \(12\ \Omega\) resistor that shines normally at 1 W, we neglect the resistance of the coil itself, and the rotation is much slower than the 60 Hz of the mains.

A conducting rod sliding along rails, inside a field, is the simplest generator to analyse. Every second, the area of the circuit grows by \(L\,v\) and the flux grows by \(B\,L\,v\), which is the induced emf. We reach the same result through the magnetic force on the electrons in the rod, which pushes them along it.

At first sight, the generator seems to create energy out of the rotation. When there is a current, the magnetic forces on the coil brake the rotation, as Lenz's law predicts, and the turbine has to do work to keep it turning, so that the electrical energy comes from the energy of the water, the wind or the steam.

Let's discuss

Turn the loop at one revolution per second and watch the graph. In which position of the loop is the emf greatest, and in which is it zero?
Double the rate of rotation. What happens to the amplitude, to the period of the sine wave and to the brightness of the lamp?
With the turbine very slow, the lamp flickers. How many times per revolution does it almost go out, and why?
Swap the slip rings for the commutator. What changes on the graph, and why does the lamp still light up in the same way?

Revolutions per second

Number of turns

Peak emf

Frequency

Period

Average power in the lamp

Exercises for step 6 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A rod 0.5 m long slides at 4 m/s along conducting rails, perpendicular to a uniform magnetic field of 0.2 T. What is the induced emf between the ends of the rod?

Show solution
For a rod cutting the field lines, \(\varepsilon = B\,L\,v = 0.2 \cdot 0.5 \cdot 4\).
\(\varepsilon = 0.4\ \text{V}\)
basic
A coil of 100 turns, each of area \(0.01\ \text{m}^2\), turns at 10 rad/s in a uniform field of 0.5 T. What is the peak emf?

Show solution
The peak emf is \(\varepsilon_{\text{max}} = N\,B\,A\,\omega\) \(= 100 \cdot 0.5 \cdot 0.01 \cdot 10\).
\(\varepsilon_{\text{max}} = 5\ \text{V}\)
basic
The Brazilian mains supply runs at 60 Hz. How many times per second does the current in a lamp plugged into the socket change direction?

Show solution
In each cycle the current flows one way for half a cycle and the other way for the other half, and it changes direction twice.
With 60 cycles per second, the current changes direction 120 times per second.
basic
In a hydroelectric power station, what energy transfers take place, from the still water in the reservoir to the electrical energy leaving the generator?

Show solution
The water high up in the reservoir has gravitational potential energy, which becomes kinetic energy as it flows down through the pipes.
The water turns the turbine, and the rotational kinetic energy passes to the generator, which turns it into electrical energy by induction. Part of the energy is lost as heat, in friction and in the wires.
Gravitational potential → kinetic energy of the water → rotational kinetic energy of the turbine → electrical, with some losses as heat.
intermediate
A bicycle dynamo, turned by the tyre, gives a peak emf of 6 V when the bicycle travels at 15 km/h. What would the peak emf be at 30 km/h? Why does the lamp get dimmer on a slow climb?

Show solution
The tyre turns the dynamo, and the angular velocity of the dynamo is proportional to the speed of the bicycle. With \(N\), \(B\) and \(A\) fixed, \(\varepsilon_{\text{max}} = N\,B\,A\,\omega\) is proportional to \(\omega\).
At twice the speed, the peak emf doubles, and on a slow climb, with a small \(\omega\), the emf and the brightness fall.
At 30 km/h, the peak emf would be 12 V; on the slow climb, the dynamo turns slowly and generates less.
intermediate
A rod 40 cm long slides at 5 m/s along rails joined by a \(2\ \Omega\) resistor, in a field of 0.5 T perpendicular to the plane of the rails. Neglect the other resistances and friction. What is the current in the circuit? What force keeps the rod moving at constant speed?

Show solution
The emf is \(\varepsilon = B\,L\,v = 0.5 \cdot 0.4 \cdot 5 = 1\ \text{V}\), and the current, \(i = \dfrac{\varepsilon}{R} = \dfrac{1}{2}\).
The rod carrying the current feels a magnetic force against the motion, \(F = B\,i\,L = 0.5 \cdot 0.5 \cdot 0.4\), and whoever pulls it has to balance it.
\(i = 0.5\ \text{A}\), and the force needed is 0.1 N.
intermediate
A coil of 200 turns, with an area of \(50\ \text{cm}^2\), turns in a uniform field of 0.4 T. At what frequency, in revolutions per second, must it turn for the peak emf to be 12 V? Use \(\pi \approx 3.14\).

Show solution
Solving for \(\omega\) in \(\varepsilon_{\text{max}} = N\,B\,A\,\omega\), we get \(\omega = \dfrac{12}{200 \cdot 0.4 \cdot 50 \cdot 10^{-4}}\) \(= 30\ \text{rad/s}\), and \(f = \dfrac{\omega}{2\pi} = \dfrac{30}{6.28}\).
\(f \approx 4.8\ \text{Hz}\), almost five revolutions per second
intermediate
The graph of a generator's emf is a sine wave with a peak value of 10 V and a period of 0.02 s. What is the frequency? If the generator's rate of rotation doubles, what will the new peak value and the new period be?

Show solution
The frequency is \(f = \dfrac{1}{T} = \dfrac{1}{0.02}\).
With twice the rate of rotation, \(\omega\) doubles, and with it the peak value \(N\,B\,A\,\omega\), while the period halves.
50 Hz; with the rotation doubled, a peak of 20 V and a period of 0.01 s.
challenge
A student turns the handle of a small generator connected to a 12 V, 24 W lamp. Even neglecting friction, he finds the handle stiffer with the lamp connected than with the circuit open. Explain why, and say what the smallest power is that he needs to supply to keep the lamp lit normally.

Show solution
With the circuit open there is no current, and the turns rotate without feeling any magnetic force.
With the lamp connected, the current induced in the turns feels forces that, by Lenz's law, oppose the rotation, and the student needs to overcome them.
The electrical energy the lamp receives comes from this work, and, in a generator with no losses, he would need to supply exactly the power of the lamp.
The handle gets stiff because the forces on the induced current brake the rotation, and he needs to supply at least 24 W.
challenge
A rod 50 cm long with a mass of 20 g slides, without friction, down vertical rails joined by a \(1\ \Omega\) resistor, in a horizontal field of 0.4 T perpendicular to the plane of the rails. Released from rest, it falls and reaches a terminal velocity. What is this velocity? Use g = 10 m/s².

Show solution
At speed \(v\), the emf is \(B\,L\,v\), the current is \(\dfrac{B\,L\,v}{R}\) and the magnetic force, upwards, is \(B\,i\,L = \dfrac{B^2 L^2 v}{R}\).
The terminal velocity is the one at which this force equals the weight, \(\dfrac{B^2 L^2 v}{R} = m\,g\), and then \(v = \dfrac{m\,g\,R}{B^2 L^2} = \dfrac{0.02 \cdot 10 \cdot 1}{0.4^2 \cdot 0.5^2}\).
\(v = 5\ \text{m/s}\)

STEP 7

Why does energy travel at high voltage?

A transformer has two coils wound on the same iron core. The alternating current in the primary creates a flux in the core that changes without stopping, and this flux passes through the secondary and induces an alternating voltage in it. If we neglect losses, each turn of both coils has the same emf, and the voltages are in the ratio of the numbers of turns.

In an ideal transformer, the power going into the primary is the power coming out of the secondary, \(U_1\, i_1 = U_2\, i_2\), and if the voltage goes up, the current goes down in the same proportion. Real transformers typically have efficiencies above 95% and warm up a little. With direct current, the flux in the core stays constant once the circuit is switched on, and the secondary receives nothing.

\(\dfrac{U_1}{U_2} = \dfrac{N_1}{N_2}\)\(U_1\, i_1 = U_2\, i_2\)\(P_{\text{lost}} = R\,i^2\)The first two hold for the ideal transformer, with no losses. In the transmission panel, the power station sends 5 MW along a \(5\ \Omega\) line, and we compare the loss at 13.8 kV, a typical voltage at the output of the generators, with the loss at higher voltages, assuming ideal transformers.

Let's think about a transmission line of resistance \(R\), which loses the power \(R\,i^2\) as heat. To deliver the same power \(P = U\,i\), a voltage ten times higher calls for a current ten times smaller, and the loss becomes a hundred times smaller. That is why power stations step the voltage up to hundreds of thousands of volts, and transformers near the towns and on the poles step it down to the 127 V or 220 V of the sockets in Brazilian homes, where all the values we use come from, something only possible with alternating current.

Old phone chargers, the heavy ones, had a transformer inside that stepped the mains voltage down to a few volts, followed by a circuit that rectified the current. Today's chargers typically make the current oscillate tens of thousands of times per second, or more, and at this high frequency a much smaller transformer does the job.

Let's discuss

Choose the old charger, from 127 V to 5 V. Which side has the larger current? Check that \(U_1\, i_1 = U_2\, i_2\).
Swap the values of \(N_1\) and \(N_2\). Does the transformer now step the voltage up or down?
Connect the transformer to direct current. What happens in the secondary, and why?
In the transmission panel, go from 13.8 kV to 138 kV. The voltage became ten times higher; how many times smaller did the loss become?

Primary voltage

Turns on the primary

Turns on the secondary

Load resistance

Secondary voltage

Primary current

Secondary current

Power (in = out)

Line current

Loss in the line

Exercises for step 7 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A transformer has 1000 turns on the primary and 50 on the secondary. Connected to 220 V, what voltage does it supply at the secondary?

Show solution
The voltages are in the ratio of the turns, \(U_2 = U_1\,\dfrac{N_2}{N_1} = 220 \cdot \dfrac{50}{1000}\).
\(U_2 = 11\ \text{V}\)
basic
An ideal transformer receives 100 W at the primary, at 127 V. How much does it deliver at the secondary? What is the secondary current, if the output voltage is 12 V?

Show solution
In the ideal transformer there are no losses, and it delivers the same 100 W. The secondary current is \(i_2 = \dfrac{P}{U_2} = \dfrac{100}{12}\).
100 W, with \(i_2 \approx 8.3\ \text{A}\)
basic
Why does a transformer not work if we connect it to a car battery?

Show solution
The battery supplies direct current, and, once the circuit is switched on, the flux in the core stays constant.
With no change in flux, there is no voltage induced in the secondary.
With direct current the flux does not change, and the secondary receives no voltage; the transformer needs alternating current.
basic
In an ideal transformer, the secondary voltage is ten times the primary voltage. How do the currents in the two windings compare?

Show solution
Power is conserved, \(U_1\, i_1 = U_2\, i_2\), and with \(U_2 = 10\,U_1\) we have \(i_2 = \dfrac{i_1}{10}\).
The secondary current is ten times smaller than the primary current.
intermediate
We want to step 127 V down to 9 V with a transformer that has 1270 turns on the primary. How many turns must the secondary have? If the appliance connected to the secondary draws 1.5 A, what is the primary current, assuming the transformer is ideal?

Show solution
From the ratio of the voltages, \(N_2 = N_1\,\dfrac{U_2}{U_1} = 1270 \cdot \dfrac{9}{127}\).
By conservation of power, \(i_1 = \dfrac{U_2\, i_2}{U_1} = \dfrac{9 \cdot 1.5}{127}\).
90 turns on the secondary, and \(i_1 \approx 0.11\ \text{A}\) in the primary
intermediate
A power station sends 10 MW along a \(4\ \Omega\) transmission line. Work out the power lost in the line with transmission at 20 kV and at 200 kV.

Show solution
At 20 kV, \(i = \dfrac{10^7}{2 \cdot 10^4} = 500\ \text{A}\), and the loss is \(R\,i^2 = 4 \cdot 500^2\).
At 200 kV, \(i = 50\ \text{A}\), and the loss is \(R\,i^2 = 4 \cdot 50^2\).
1 MW lost at 20 kV, 10% of the power, and only 10 kW at 200 kV, 0.1% of the power, a hundred times less.
intermediate
The voltage of a transmission line goes from 138 kV to 500 kV, with the same power transmitted and the same line. By what factor does the power lost in the line decrease?

Show solution
With \(P = U\,i\) fixed, the current is inversely proportional to the voltage, and the loss \(R\,i^2\) is inversely proportional to \(U^2\).
The factor is \(\left(\dfrac{500}{138}\right)^2\).
The loss becomes about 13 times smaller.
intermediate
A real transformer takes 2.0 A at 220 V in the primary and delivers 9.0 A at 46 V in the secondary. What is its efficiency? Where does the missing energy go?

Show solution
The input power is \(220 \cdot 2.0 = 440\ \text{W}\), and the output power, \(46 \cdot 9.0 = 414\ \text{W}\). The efficiency is \(\eta = \dfrac{414}{440}\).
\(\eta \approx 94\%\); the missing 26 W become heat in the wires of the windings and in the core.
challenge
A power-station generator produces 100 MW at 13.8 kV, and a transformer, which we will take as ideal, steps the voltage up to 500 kV. What is the ratio between the numbers of turns on the secondary and on the primary? What is the current at the generator output, and what is the current in the line?

Show solution
The ratio of the turns is that of the voltages, \(\dfrac{N_2}{N_1} = \dfrac{500}{13.8}\).
At the generator output, \(i_1 = \dfrac{10^8}{13\,800}\), and in the line, \(i_2 = \dfrac{10^8}{5 \cdot 10^5}\).
\(N_2/N_1 \approx 36\), about 7200 A at the generator and 200 A in the line.
challenge
A small hydroelectric plant on a farm supplies 20 kW, and the house is 2 km away. The wires of the line, out and back, add up to \(1\ \Omega\). In a simple model in which the power \(U\,i\) sent along the line is always 20 kW, compare the fraction lost when transmitting at 220 V and at 2200 V, with ideal transformers at both ends in the second case. Why would the first option be unworkable?

Show solution
At 220 V, \(i = \dfrac{20\,000}{220} \approx 91\ \text{A}\), and the loss is \(R\,i^2 \approx 8.3 \cdot 10^3\ \text{W}\).
At 2200 V, \(i \approx 9.1\ \text{A}\), and \(R\,i^2 \approx 83\ \text{W}\).
At 220 V, besides losing almost half the energy as heat, the line would drop the voltage by \(R\,i \approx 91\ \text{V}\), and the house would receive well under 220 V.
About 41% loss at 220 V and 0.4% at 2200 V; at 220 V, the loss and the voltage drop make the connection unworkable.

WRAP-UP

Challenges

Magnets

Two metal bars look exactly alike, but only one of them is a magnet. With no other object, how can we find out which one is the magnet?

Show solution

We touch the end of bar A to the middle of bar B. If there is a strong attraction, A is the magnet, because the middle of a magnet hardly attracts anything, and it is at the ends that the field is strong.
If there is no attraction, we swap roles, and it is B that has an end able to attract the middle of bar A.

Currents make fields

When a car starts, the battery cable carries about 150 A. What is the magnetic field 3 cm from the cable? Compare it with the Earth's field, \(2 \cdot 10^{-5}\ \text{T}\). (\(\mu_0 = 4\pi \cdot 10^{-7}\ \text{T} \cdot \text{m/A}\))

Show solution

Treating the cable as a long wire, \(B = \dfrac{\mu_0\, i}{2\pi d} = \dfrac{2 \cdot 10^{-7} \cdot 150}{0.03}\), which gives \(10^{-3}\ \text{T}\).
That is 50 times the Earth's field, and a compass close by would be completely thrown off while the car starts.

Force on charges

Show that the frequency at which a proton circles in a uniform field does not depend on its speed, and work out this frequency in a field of 1 T. (\(m_p = 1.67 \cdot 10^{-27}\ \text{kg}\), \(e = 1.6 \cdot 10^{-19}\ \text{C}\))

Show solution

The frequency is the reciprocal of the period, \(f = \dfrac{|q|\,B}{2\pi m}\), and the speed does not appear. For the proton, \(f = \dfrac{1.6 \cdot 10^{-19} \cdot 1}{2\pi \cdot 1.67 \cdot 10^{-27}}\), about \(1.5 \cdot 10^7\ \text{Hz}\).
This independence made it possible to build cyclotrons, which push the particles at the same rhythm while they gain speed and open out the spiral.

Force on wires

A horizontal overhead wire for trolleybuses carries 300 A. At a place where the Earth's field is \(5 \cdot 10^{-5}\ \text{T}\) and perpendicular to the wire, what is the magnetic force on 2 m of wire?

Show solution

The force is \(F = B\,i\,L = 5 \cdot 10^{-5} \cdot 300 \cdot 2\), or 0.03 N.
Even with such a large current, the force from the Earth's field is small, roughly the weight of three grams.

Induction

A coil of 500 turns and area \(10\ \text{cm}^2\) has its axis parallel to a field of 0.2 T. In 0.01 s, the field is reversed. What is the average induced emf?

Show solution

The flux per turn goes from \(+0.2 \cdot 10^{-3}\) to \(-0.2 \cdot 10^{-3}\ \text{Wb}\), a change of \(4 \cdot 10^{-4}\ \text{Wb}\), twice what it would be if the field merely dropped to zero.
The emf is \(|\varepsilon| = 500 \cdot \dfrac{4 \cdot 10^{-4}}{0.01}\), or 20 V.

Generators

An aeroplane with a wingspan of 40 m flies horizontally at 250 m/s at a place where the vertical component of the Earth's field is \(4 \cdot 10^{-5}\ \text{T}\). What is the emf induced between the wingtips?

Show solution

The wings act like a rod cutting the vertical field lines, and \(\varepsilon = B\,L\,v = 4 \cdot 10^{-5} \cdot 40 \cdot 250\), or 0.4 V.
This voltage cannot light anything on board, because any wire run from one wingtip to the other moves along with the wing and has the same emf induced in it.

Transformers

A laptop power supply states an output of 19 V and 3.4 A. Assuming an efficiency of 90% and the supply plugged into 127 V, what current does it draw from the socket?

Show solution

The output power is \(19 \cdot 3.4 = 64.6\ \text{W}\), and the input power, \(\dfrac{64.6}{0.9} \approx 71.8\ \text{W}\).
At the socket, \(i = \dfrac{71.8}{127}\), about 0.57 A.

Think it through, no sums

The current induced in a coil always opposes the change in flux. What would happen to the conservation of energy if it helped the change instead?

Show solution

If the induced current, as we brought a magnet closer, pulled the magnet into the coil, a little nudge would be enough for the magnet to speed up on its own, generating more and more current and more and more energy without anyone doing work.
That would create energy out of nothing. Lenz's law is the form the conservation of energy takes in induction.

ENEM-style questions

The ENEM is Brazil's national secondary-school exam, which most students sit to get into university, and we wrote the five questions below in its format, with an everyday context, a question and five options, each one returning to a different step of the lesson. Further down we list real exam questions on electromagnetism, which may well be the best practice once these are done.

Magnets and magnetic field · Step 1
Rita goes hiking and uses a compass to find her way. At a stop, she left the compass on her rucksack, next to her mobile phone, and noticed that the needle no longer pointed north. When she moved the phone about 30 cm away, the needle went back to normal. Mobile phones typically have small magnets inside, in the loudspeaker, for example.

The needle is deflected near the phone because
1. the radio waves given off by the phone exert a force on the needle and make it turn.
2. the phone's battery charges the needle electrically, and it then becomes attracted to the phone.
3. near the phone, the field of its magnet is stronger than the Earth's and changes the direction of the resultant field, which the needle lines up with.
4. the phone blocks the Earth's magnetic field, and the needle, with no field at all, stops in any position.
5. the phone's magnet reverses the Earth's field, and the needle starts pointing to the geographic south pole.
Show solution

Answer: C.
The needle is a small magnet that lines up with the magnetic field where it is, and this field is the vector sum of the Earth's field and that of any nearby magnet.
The field of a magnet falls off quickly with distance. A few centimetres away, the phone's field can exceed the Earth's, which is weak, about \(2 \cdot 10^{-5}\ \text{T}\) horizontally, and at 30 cm it is already small in comparison.
Option A puts the effect down to radio waves, which do not orient the needle, and B confuses an electric effect with a magnetic one. Option D assumes the phone blocks the Earth's field, when in fact the two fields add, and E assumes a reversal of the Earth's field that a small device cannot produce.
Currents make fields · Step 2
In a practical lesson, Sofia repeats Oersted's experiment. A long vertical wire passes through the table, and a compass sits on the table, 2 cm south of the wire. With no current, the needle points north, set by the horizontal component of the Earth's field, which is \(2 \cdot 10^{-5}\ \text{T}\). When a current of 4 A passes through the wire, the needle turns and settles in a new position. Use \(\mu_0 = 4\pi \cdot 10^{-7}\ \text{T} \cdot \text{m/A}\) and, if needed, \(\tan 27^\circ \approx 0.5\), \(\tan 45^\circ = 1\) and \(\tan 63^\circ \approx 2\).

By roughly how many degrees does the needle turn away from north?
1. 0°
2. 27°
3. 45°
4. 63°
5. 90°
Show solution

Answer: D.
South of the wire, the field of the current points east or west, depending on the direction of the current, and is perpendicular to the Earth's field. Its magnitude is \(B = \dfrac{\mu_0\, i}{2\pi d}\) \(= \dfrac{2 \cdot 10^{-7} \cdot 4}{0.02}\) \(= 4 \cdot 10^{-5}\ \text{T}\).
The needle lines up with the resultant, which makes an angle \(\alpha\) with north such that \(\tan\alpha = \dfrac{4 \cdot 10^{-5}}{2 \cdot 10^{-5}} = 2\), about 63°.
Option B inverts the ratio between the fields, and C assumes the two fields are equal. Option E forgets the Earth's field, and A assumes the current does not act on the needle, contrary to what Oersted observed.

Magnetic force on charges · Step 3

In a mass spectrometer, ions enter through a slit, all with a speed of \(1.0 \cdot 10^5\ \text{m/s}\), perpendicular to a uniform magnetic field of 0.10 T. Each ion travels along half a circle and hits a detector plate. A sample contains the ions in the table. Use \(1\ \text{u} = 1.66 \cdot 10^{-27}\ \text{kg}\) and \(e = 1.6 \cdot 10^{-19}\ \text{C}\).

Ions in the sample
Ion	Mass (u)	Charge
hydrogen, H⁺	1	+e
helium, He²⁺	4	+2e
carbon, C⁺	12	+e
oxygen, O²⁺	16	+2e

Which ion hits the plate furthest from the slit, and at what distance from it?

H⁺, at 2.1 cm
He²⁺, at 4.2 cm
O²⁺, at 16.6 cm
C⁺, at 12.5 cm
C⁺, at 24.9 cm

Show solution

Answer: E.
The radius is \(R = \dfrac{m\,v}{|q|\,B}\), proportional to the ratio of mass to charge. This ratio is 1 for H⁺, 2 for He²⁺, 12 for C⁺ and 8 for O²⁺, and C⁺ follows the largest radius.
For C⁺, \(R = \dfrac{12 \cdot 1.66 \cdot 10^{-27} \cdot 1.0 \cdot 10^5}{1.6 \cdot 10^{-19} \cdot 0.10}\) \(\approx 0.125\ \text{m}\), and the plate is hit one diameter away from the slit, \(2R \approx 24.9\ \text{cm}\).
Option D uses the radius in place of the diameter, and C picks the heaviest ion without taking its double charge into account. Options A and B go for the ions with the smallest ratio of mass to charge.

Electromagnetic induction · Step 5
At a science fair, Otávio moves a magnet near a coil of 100 turns connected to a voltmeter, and a sensor records the magnetic flux through each turn. The graph shows the record over 6 s.

In which interval is the emf induced in the coil greatest, and what is its value?
1. Between 3 s and 4 s, at 0.4 V.
2. Between 2 s and 3 s, at 0.4 V.
3. Between 0 and 2 s, at 0.2 V.
4. Between 3 s and 4 s, at 4 mV.
5. Between 0 and 2 s, at 0.4 V.
Show solution

Answer: A.
The emf depends on how fast the flux changes, \(|\varepsilon| = N\,\dfrac{|\Delta\Phi|}{\Delta t}\), and the steepest slope on the graph is between 3 s and 4 s, when the flux falls by 4 mWb in 1 s.
In this interval, \(|\varepsilon| = 100 \cdot \dfrac{4 \cdot 10^{-3}}{1} = 0.4\ \text{V}\). Between 0 and 2 s, the same change takes twice as long, and the emf is 0.2 V.
Option B picks the interval of maximum flux, in which the flux does not change and the emf is zero. Option C picks the longest interval, D forgets the 100 turns, and E forgets to divide by the time.
Energy transmission · Step 7
A small hydroelectric plant is going to send 2 MW to a nearby town along a line whose wires add up to a resistance of \(6\ \Omega\). The engineers compare two options, transmitting at 13.8 kV, the output voltage of the generator, or installing transformers and transmitting at 69 kV. We will assume the transformers are ideal and the same power goes into the line in both cases.

When transmitting at 69 kV, the power lost as heat in the line is approximately
1. 126 kW, the same as at 13.8 kV.
2. 5 kW.
3. 25 kW.
4. 1 kW.
5. 630 kW.
Show solution

Answer: B.
At 13.8 kV, the current in the line is \(i = \dfrac{P}{U} = \dfrac{2 \cdot 10^6}{13\,800} \approx 145\ \text{A}\), and the loss is \(R\,i^2 \approx 6 \cdot 145^2\) \(\approx 126\ \text{kW}\).
With a voltage 5 times higher, the current becomes 5 times smaller, about 29 A, and the loss, which depends on the square of the current, becomes 25 times smaller, \(6 \cdot 29^2 \approx 5\ \text{kW}\).
Option A ignores the change in current, and C divides the loss by only 5. Option D exaggerates the reduction, and E multiplies the loss instead of dividing it.

Real ENEM questions on this topic

The official papers and answer keys are on the INEP website, where we can look each question up by year, day, booklet colour and number.

ENEM 2011, Day 1, blue booklet, question 56. A guitarist swaps steel strings for nylon ones and the guitar's pickup stops working, and the question asks why, which has to do with the magnetisation of the string and the change in flux through the pickup coil.
ENEM 2013, Day 1, blue booklet, question 85. A rod carrying a current, attached to a spring, sits in a magnetic field and must be pushed by the magnetic force to open a door, and we work out the strength of the field.
ENEM 2014, Day 1, blue booklet, question 72. A magnet and a loop move in opposite directions and induce a current, and we need to find another combination of movements and polarity that produces a current in the same direction.
ENEM 2020, Day 2, blue booklet, question 98. The operator of a power station notices that the voltage across the terminals of the generator coils has gone up, and the question asks for the cause, linked to how fast the flux changes.
ENEM 2022, Day 2, blue booklet, question 95. Three compasses sit around a wire carrying a current, as in Oersted's experiment, and we choose the figure with the needles pointing the right way.

ENEM papers and answer keys on the INEP website (in Portuguese) →

Cheat sheet

Straight wire\(B = \dfrac{\mu_0\, i}{2\pi d}\)

Centre of a loop\(B = \dfrac{\mu_0\, i}{2R}\)

Solenoid\(B = \mu_0\,\dfrac{N}{L}\,i\)

Permeability of free space\(\mu_0 = 4\pi \cdot 10^{-7}\ \text{T} \cdot \text{m/A}\)

Force on a charge\(F = |q|\,v\,B\,\sin\theta\)

Charge in a uniform field\(R = \dfrac{m\,v}{|q|\,B}\), \(T = \dfrac{2\pi m}{|q|\,B}\)

Force on a wire\(F = B\,i\,L\,\sin\theta\)

Parallel wires\(\dfrac{F}{L} = \dfrac{\mu_0\, i_1\, i_2}{2\pi d}\)

Flux\(\Phi = B\,A\cos\theta\)

Faraday and Lenz\(\varepsilon = -N\,\dfrac{\Delta\Phi}{\Delta t}\), rod: \(\varepsilon = B\,L\,v\)

Generator\(\varepsilon_{\text{max}} = N\,B\,A\,\omega\)

Ideal transformer\(\dfrac{U_1}{U_2} = \dfrac{N_1}{N_2}\), \(U_1 i_1 = U_2 i_2\)