Electricity and Magnetism 203-SN2-RE • CEGEP, Montreal

Revision sheet: magnetism and induction (203-SN2-RE)

Magnetism asks you to step out of the page: the force on a charge is perpendicular to the plane formed by its velocity and the field, which defeats any reasoning done flat. The right-hand rule is not an ornament, it is the calculating tool of the chapter, in 203-SN2-RE (formerly 203-NYB).

The second half, induction, punishes another habit: looking at the value of a quantity rather than at its change. A huge but constant field induces nothing, a weak field that changes fast induces a lot. This sheet collects the eight mistakes that cost the most marks on these two fronts.

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The thread of the chapter

Nothing lines up in magnetism: the force is PERPENDICULAR to both the velocity and the field, and induction depends not on the flux but on its CHANGE. The two halves of the chapter punish the same reflex, that of reasoning about magnitudes instead of reasoning about directions and slopes.

This chapter is part of 203-SN2-RE, Electricity and Magnetism

The essentials

Magnets and field lines: from north to south

• Outside a magnet, the field lines leave the NORTH pole and enter by the SOUTH pole; they close up inside the magnet, never starting or ending.
• The field $\vec{B}$ is tangent to the lines, and the stronger the closer they are packed, hence near the poles.
• A compass lines up with $\vec{B}$ : the north tip of its needle shows the direction of the field where it sits.

The lines leave the north, in red, and enter at the south, in blue. Each compass lines up with the line through it, red tip in the direction of

\vec{B}

Cutting a magnet in two gives two complete magnets, each with its north and its south: there is no isolated magnetic pole.

The magnetic force: three directions, never two

• $F=|q|vB\sin\theta$ , where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$ . The force is perpendicular to BOTH at once.
• Velocity PARALLEL to the field: $\sin\theta=0$ , so the force is ZERO. Perpendicular velocity: maximum force.
• Right hand for a POSITIVE charge: fingers along $\vec{v}$ , curl them toward $\vec{B}$ , the thumb gives $\vec{F}$ . For a negative charge, reverse the final result.
• The magnetic force never does WORK: perpendicular to the velocity at every instant, it changes the direction of motion, never the magnitude of the velocity or the kinetic energy.
• On a wire: $F=BIL\sin\theta$ . Field of a long straight wire: $B=\dfrac{\mu_{0}I}{2\pi d}$ with $\mu_{0}=4\pi\times 10^{-7}$ T·m/A.

The force points toward the centre, the velocity stays tangent: this is exactly the definition of uniform circular motion. The magnitude of the velocity never changes, only its direction turns.

Two parallel wires carrying currents in the SAME direction attract each other. This is counterintuitive for anyone thinking of electric charges, and it is an almost systematic question in tests.

Paths: circle, helix, and what does not depend on what

• Charge launched perpendicular to a uniform field: circle of radius $r=\dfrac{mv}{|q|B}$ .
• Period $T=\dfrac{2\pi m}{|q|B}$ : it depends on NEITHER the speed NOR the radius. This is the whole principle of the cyclotron.
• Oblique velocity: the component parallel to the field is unaffected, the perpendicular component turns. The path is a HELIX of radius $\dfrac{mv_{\perp}}{|q|B}$ and pitch $p=v_{\parallel}T$ .
• Velocity selector, crossed fields $\vec{E}$ and $\vec{B}$ : particles pass undeflected if $qE=qvB$ , so $v=\dfrac{E}{B}$ , regardless of charge and mass.
• Mass spectrometer: after the selector, the radius $r=\dfrac{mv}{qB}$ separates the isotopes, because only the mass still varies.

After the selector, all the ions have the same speed: in the inward field, the radius

r=\dfrac{mv}{qB}

depends only on the mass, and the heavy isotope lands farther along the detector.

A proton at $10^{6}$ m/s in $0.50$ T follows a circle of $2.1$ cm in $0.13$ μs. Remembering this order of magnitude lets you spot a lost power of ten at once.

The Laplace force: a wire in a field

• A wire of length $L$ carrying $I$ in a field $\vec{B}$ feels $\vec{F}=I\vec{L}\times\vec{B}$ , of magnitude $F=BIL\sin\theta$ : it is the sum of the magnetic forces on the moving charges.
• Right hand: fingers along the current, curl them toward $\vec{B}$ , the thumb gives $\vec{F}$ . The force is perpendicular to the wire AND to the field.
• Between the poles of a U-shaped magnet, the field goes from north to south; reversing the current or flipping the magnet reverses the force.

Field pointing down, from north to south, current coming out of the page: the force on the wire points to the right, perpendicular to the wire and to the field.

The Laplace rail is the classic demonstration: a rod laid across two powered rails starts to roll, and goes back the other way if you swap the terminals of the generator.

The torque on a loop: the electric compass

• A coil of $N$ turns of area $A$ carrying $I$ has a magnetic moment $\mu=NIA$ , carried along its normal by the right-hand rule.
• In a uniform field, the net force is zero but the torque is $\tau=\mu B\sin\theta$ , where $\theta$ is the angle between the NORMAL and $\vec{B}$ : greatest when the plane of the coil contains the field, zero at alignment.
• The torque always tends to align $\vec{\mu}$ with $\vec{B}$ : the energy $U=-\mu B\cos\theta$ is least at alignment, the stable equilibrium.

Coil seen edge-on: the forces on the two sides are opposite and offset, they form a couple that turns

\vec{n}

toward

\vec{B}

, here clockwise.

The DC motor reverses the current at every half-turn, as the coil passes through alignment, so that the torque keeps the same sense of rotation.

Induction: it is the change that counts

• Flux: $\Phi=BA\cos\theta$ , where $\theta$ is the angle between $\vec{B}$ and the NORMAL to the surface, never the plane of the coil.
• Faraday's law: $\varepsilon=N\dfrac{|\Delta\Phi|}{\Delta t}$ . A constant flux, however intense, induces NOTHING.
• Lenz's law: the induced current opposes the CHANGE in flux, not the flux. If the flux decreases, the induced current reinforces the field instead of fighting it.
• A rod of length $L$ sliding at speed $v$ perpendicular to $\vec{B}$ : $\varepsilon=BLv$ , and the force to overcome is $BIL$ .
• Total charge moved during the change: $q=\dfrac{N|\Delta\Phi|}{R}$ , independent of the DURATION of the motion.

Approaching magnet: the ring presents a north face to it and repels it. Receding magnet: a south face, which holds it back. The induced current always opposes the CHANGE.

Three ways to change a flux, and a single question to ask: is it $B$ , the area $A$ , or the angle $\theta$ that changes? The answer alone picks the formula.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Forgetting that the force is zero when the velocity is parallel to the field

the whole question, 5 marks

What not to write

“The particle moves straight along the field lines, in a field of $2.0$ T: the magnetic force is therefore maximum.”

What to write

“ $\vec{v}$ is parallel to $\vec{B}$ , so $\theta=0$ and $F=|q|vB\sin 0=0$ . The particle keeps moving in a straight line at constant speed, whatever the strength of the field.”

Same field, same speed in magnitude: parallel to

\vec{B}

, the force is exactly zero; perpendicular, it is maximum and points out of the plane, here toward the back of the page.

Why: The magnitude of the field only enters as a factor: it is the ANGLE that decides whether anything happens. A huge field aligned with the velocity does nothing at all.

2. Applying the right-hand rule as is to an electron

the whole question, the path curves the wrong way

What not to write

“The electron moves to the right, the field comes out of the page, the right hand gives a force pointing down.”

What to write

“The right-hand rule holds for a POSITIVE charge. Since the electron has $q<0$ , the force points UP, opposite to what the right hand gives.”

Why: The sign of the charge does not appear in the magnitude formula, where we take $|q|$ : it only enters the DIRECTION. As half the problems of the course are about electrons, forgetting it is systematic.

3. Making the magnetic force accelerate a particle

3 marks, and an energy question entirely wrong

What not to write

“The magnetic force acts on the proton for $10$ ns, so it gives it kinetic energy.”

What to write

“The magnetic force is perpendicular to the velocity AT EVERY INSTANT, so its work is zero: the magnitude of the velocity and the kinetic energy stay constant. Only the direction changes.”

Why: Work is computed with the component of the force ALONG the displacement, and that component is zero by construction. This is why a magnetic field alone can never be used to accelerate a particle: you need an electric field for that.

4. Believing the period of circular motion depends on the speed

the whole question, 4 marks, and the principle of the cyclotron missed

What not to write

“The particle goes twice as fast, so it completes a lap twice as fast.”

What to write

“ $T=\dfrac{2\pi m}{|q|B}$ contains neither $v$ nor $r$ : twice as fast means a circle twice as large, traversed in the SAME time.”

Why: The radius and the speed increase in the same proportion, so the duration of a lap does not move. This is exactly what lets a cyclotron be driven at a fixed frequency while the particles accelerate.

5. Measuring the flux angle from the plane of the coil

3 marks, and a flux overestimated by

73\,\%

What not to write

“The coil is tilted $30^{\circ}$ from the field, so $\Phi=BA\cos 30^{\circ}=0.87\,BA$ .”

What to write

“The flux angle is measured between the field and the NORMAL to the coil. A coil at $30^{\circ}$ to the field has its normal at $60^{\circ}$ : $\Phi=BA\cos 60^{\circ}=0.50\,BA$ .”

Coil tilted

30°

from the field: its normal

\vec{n}

makes

60°

with

\vec{B}

, and this is the angle that enters

\Phi=BA\cos\theta

Why: Flux measures what passes THROUGH the surface: it is greatest when the field is perpendicular to the coil, hence parallel to its normal, and zero when the field lies in the plane of the coil.

6. Expecting an electromotive force from an intense but constant field

the whole question, 4 marks

What not to write

“The coil sits in a field of $1.5$ T: the induced electromotive force is therefore large.”

What to write

“The field does not change, the area does not change, the orientation does not change: $\Delta\Phi=0$ , so $\varepsilon=0$ . No electromotive force, whatever the strength of the field.”

The flux is MAXIMUM between the two instants where it changes, and that is exactly where the induced electromotive force is zero. It is not the value of the flux that induces, it is its slope.

Why: Faraday relates the electromotive force to the SLOPE of the flux over time, not to its value. A huge, flat flux induces nothing; a weak flux that collapses in a millisecond induces a lot.

7. Making the induced current oppose the field instead of the change

the whole direction question, 4 marks

What not to write

“The field is upward, so the induced current always creates a field pointing down.”

What to write

“The induced current opposes the CHANGE. If the upward flux INCREASES, the induced current creates a field pointing down; if it DECREASES, it creates an upward field to hold it back.”

The upward field DECREASES: the induced current creates an upward field too, to hold back the flux that is leaving. It reinforces the field instead of opposing it.

Why: Lenz's law is a consequence of conservation of energy: the system resists change, whichever way it goes. A magnet being pulled away is held back, exactly as a magnet being brought closer is pushed away.

8. Believing that moving a magnet more slowly carries less charge

3 marks, and the principle of the ballistic galvanometer missed

What not to write

“Pulling the magnet out twice as slowly, the electromotive force is half as large, so the charge moved is too.”

What to write

“ $q=\dfrac{N|\Delta\Phi|}{R}$ : the duration drops out. Half the current for twice as long carries exactly the SAME charge.”

Why: The electromotive force is proportional to $\dfrac{1}{\Delta t}$ and the duration of the current to $\Delta t$ : their product depends only on the total change in flux. This is what makes it possible to measure a flux with a simple charge counter.

Which method to choose

Which path, depending on the angle between the velocity and the field

The angle between $\vec{v}$ and $\vec{B}$ , and the possible presence of an electric field

If $\vec{v}$ is PARALLEL to $\vec{B}$ → no force: uniform straight-line motion

Example: particle launched along the axis of a solenoid
If $\vec{v}$ is PERPENDICULAR to $\vec{B}$ → circle of radius $r=\dfrac{mv}{|q|B}$ , period $T=\dfrac{2\pi m}{|q|B}$

Example: proton at $10^{6}$ m/s in $0.50$ T: $r=2.1$ cm
If $\vec{v}$ is OBLIQUE → helix: resolve into a free $v_{\parallel}$ and a turning $v_{\perp}$

Example: pitch of the helix $p=v_{\parallel}T$ , with the same $T$ as above
If crossed fields $\vec{E}$ and $\vec{B}$ , and the beam is not deflected → velocity selector: $v=\dfrac{E}{B}$ , with neither mass nor charge

Example: the entrance stage of a mass spectrometer
If you are asked to SEPARATE isotopes → spectrometer: $r=\dfrac{mv}{qB}$ , so the radius is proportional to the mass

Example: at the same speed, the heavy isotope traces the large circle
If the particle must GAIN energy → only an electric field can do it; the magnetic field only bends the path

Example: the D-shaped electrodes of the cyclotron accelerate, the magnet guides

None of these branches requires an integration: all are handled with Newton's second law and the centripetal acceleration $\dfrac{v^{2}}{r}$ . Writing $|q|vB=\dfrac{mv^{2}}{r}$ gives the radius back in one line, with nothing to memorize.

An induction question: which law, depending on what changes

The single quantity that changes over time

If the FIELD changes and the coil is at rest → $\varepsilon=NA\dfrac{|\Delta B|}{\Delta t}$

Example: coil in an electromagnet that is switched off
If the AREA of the circuit changes, rod sliding on rails → $\varepsilon=BLv$ , then $I=\dfrac{\varepsilon}{R}$ and $F=BIL$

Example: $B=0.50$ T, $L=0.30$ m, $v=4.0$ m/s give $0.60$ V
If the ORIENTATION changes, coil turning → alternator: $\varepsilon=NBA\omega\sin\omega t$ , rms value $\dfrac{\varepsilon_{max}}{\sqrt{2}}$

Example: the emf is greatest when the flux is zero, and vice versa
If you are asked the DIRECTION of the induced current → Lenz's law: opposition to the CHANGE, then right hand

Example: an increasing inward flux gives a counterclockwise current
If you are asked the total CHARGE moved → $q=\dfrac{N|\Delta\Phi|}{R}$ , with no duration at all

Example: magnet withdrawn slowly or fast: same charge
If nothing changes → $\varepsilon=0$ , and you have to say so explicitly

Example: coil at rest in a constant field, even a very intense one

A single quantity changes in almost every problem, and identifying it is already half the answer. When two change at the same time, go back to $\Phi=BA\cos\theta$ and differentiate the product.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Determining the direction of an induced current

When to use it: The problem asks for the direction of the current in a loop, or the direction of the force on a moving magnet.

1 Name the initial flux: direction of the field through the loop, and the sense of crossing, inward or outward.
2 Say whether this flux INCREASES or DECREASES, and why: the magnet approaches, the loop leaves the field, the area shrinks.
3 Apply Lenz: the induced current creates a field that opposes this change, so in the opposite direction if the flux increases, in the same direction if it decreases.
4 Translate this wanted field into a current direction with the right hand: thumb along the field created, fingers curled along the current.
5 Conclude with one sentence containing the three elements: the change, the induced field, the direction of the current.

Concluding sentence

“The inward flux through the loop INCREASES as the magnet approaches. By Lenz's law, the induced current creates an OUTWARD field inside the loop to oppose it, which corresponds, by the right-hand rule, to a COUNTERCLOCKWISE current seen from the front.”

The trap: Answering with the direction that opposes the field rather than its change. For a magnet being pulled AWAY, the right answer is the direction that reinforces the field, and this half of the cases is almost always missed.

Marking: Typically 1 mark for the direction of the flux, 1 for its direction of change, 1 for the induced field, 1 for the direction of the current stated unambiguously.

The rod on rails, from the motion to the power balance

When to use it: A conducting bar slides on two rails joined by a resistor, in a field perpendicular to the plane.

1 Electromotive force: $\varepsilon=BLv$ , with $L$ the EFFECTIVE length, that is, the spacing of the rails, not the length of the rod.
2 Current: $I=\dfrac{\varepsilon}{R}$ , stating its direction by Lenz's law.
3 Magnetic force on the rod: $F=BIL$ , always OPPOSED to the motion, which is Lenz again.
4 For a constant speed, the force applied by the operator is exactly equal and opposite to this magnetic force.
5 Close with the balance: the mechanical power $Fv$ must equal the electrical power $\varepsilon I=RI^{2}$ .

Concluding sentence

“The rod induces $\varepsilon=BLv=0.60$ V, so $I=0.30$ A. The magnetic force that brakes it is $BIL=0.045$ N, so $0.045$ N must be applied to keep the speed constant. The power supplied, $Fv=0.18$ W, is entirely dissipated in the resistor, $RI^{2}=0.18$ W.”

The trap: Forgetting that the magnetic force always brakes. A problem asking for “the force to apply to maintain the speed” expects this braking force, not zero on the pretext that the motion is uniform.

Marking: In general 2 marks for the emf, 1 for the current and its direction, 2 for the force, 1 for the power balance.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Perpendicularity, checked on the diagram

Go over every magnetic force drawn and check that it makes a right angle with the velocity AND with the field. A force along the velocity is necessarily wrong.

A force drawn along the path: that would be a force that accelerates, and the magnetic field never produces one.
The speed unchanged in magnitude

In any purely magnetic problem, compare the final speed with the initial speed: they must be EQUAL in magnitude, and the kinetic energy conserved.

A speed going from $1.0\times 10^{6}$ to $1.4\times 10^{6}$ m/s with no electric field: impossible.
The order of magnitude of the radius

A proton at a few $10^{6}$ m/s in a field of about a tesla follows a circle of a few centimetres. An electron, a thousand times lighter, follows a circle a thousand times smaller.

A radius of $21$ m for a proton in $0.50$ T: three powers of ten have been lost.
The direction of the induced current, reread through energy

Check that the induced current BRAKES the motion that creates it. If it accelerated it, the system would make energy all by itself, which is enough to conclude that the direction found is wrong.

A magnetic force in the direction of the rod's displacement: perpetual motion, so a sign error.
The mechanical and electrical balance

On any induction question with motion, compare $Fv$ and $RI^{2}$ : the two must coincide when the speed is constant.

$0.18$ W on each side: the problem is consistent, no force has been forgotten.

The typical problem, taken apart

The rod on rails: from muscle to current

Two horizontal rails $L=0.30$ m apart are joined at one end by a resistor $R=2.0$ $\Omega$ . A conducting rod slides without friction on these rails at a constant speed $v=4.0$ m/s.

The whole setup sits in a uniform magnetic field $B=0.50$ T, perpendicular to the plane of the rails and going into the page. Find the induced electromotive force, the current, the force to apply to the rod, then check the power balance.

The area of the circuit INCREASES as the rod advances to the right: this change, and this alone, induces the current. Its direction and the force that brakes the rod are drawn at the force step.

Step 1

Induced electromotive force: $\varepsilon=BLv=(0.50)(0.30)(4.0)=0.60$ V.

Why

The formula $BLv$ is only a shortcut for Faraday: in advancing by $\Delta x$ , the rod adds an area $L\,\Delta x$ to the circuit, hence a flux $BL\,\Delta x$ , and dividing by $\Delta t$ makes the speed appear.

Step 2

Induced current: $I=\dfrac{\varepsilon}{R}=\dfrac{0.60}{2.0}=0.30$ A. The inward flux increases, so by Lenz the induced current creates an outward field inside the circuit: it flows COUNTERCLOCKWISE.

Why

The direction is not a decorative detail: it is what will give the direction of the force on the rod at the next step, and the problem almost always asks for it explicitly.

Step 3

Magnetic force on the rod carrying this current: $F=BIL=(0.50)(0.30)(0.30)=0.045$ N, directed OPPOSITE to $\vec{v}$ .

Why

The opposite direction does not need to be found with the right hand: Lenz's law guarantees it. A force that accelerated the rod would create energy out of nothing.

Step 4

Force to apply to keep the speed constant: the sum of the forces is zero, so $F_{applied}=0.045$ N, in the direction of motion.

Why

Constant speed does not mean zero force, but zero NET force. This is the step that separates a paper that has understood from one that applies formulas.

Step 5

Mechanical power supplied: $P_{mech}=F v=(0.045)(4.0)=0.18$ W.

Why

This is the energy the operator spends every second. It does not go into kinetic energy, since the speed does not change: it must therefore turn up somewhere else.

Step 6

Power dissipated in the resistor: $P_{elec}=RI^{2}=(2.0)(0.30)^{2}=0.18$ W, or equally $\varepsilon I=(0.60)(0.30)=0.18$ W.

Why

The two powers coincide exactly: all the muscle energy is converted to heat in the resistor. This is the strongest check of the problem, and it uses none of the intermediate steps.

Step 7

Limiting-case check: if the rod stops, $\varepsilon=0$ , so $I=0$ , so $F=0$ . If the speed is doubled, $\varepsilon$ , $I$ and $F$ double, and the power is multiplied by four.

Why

The power going as $v^{2}$ explains why magnetic braking is very effective at high speed and becomes ineffective at low speed: this is the principle of train brakes, which cannot bring the train to a complete stop.

The conclusion, written out

“The rod induces an electromotive force $\varepsilon=BLv=0.60$ V, hence a current $I=0.30$ A flowing counterclockwise. The magnetic force on it is $BIL=0.045$ N and opposes the motion: $0.045$ N must therefore be applied to keep the speed constant. The power supplied, $0.18$ W, is entirely dissipated in the resistor.”

The classic mistake on this problem: Answering that the force to apply is zero because the speed is constant. With no applied force, the magnetic force would brake the rod to a halt, and no current would remain.

Learn by heart

• $F=|q|vB\sin\theta$ : zero if $\vec{v}$ is parallel to $\vec{B}$ , maximum if it is perpendicular to it.
• The magnetic force NEVER does work: the magnitude of the velocity is conserved.
• Right hand for $q>0$ ; for an electron, reverse the result.
• $r=\dfrac{mv}{|q|B}$ but $T=\dfrac{2\pi m}{|q|B}$ : the period depends on neither $v$ nor $r$ .
• Velocity selector: $v=\dfrac{E}{B}$ , with neither mass nor charge.
• $\Phi=BA\cos\theta$ with $\theta$ measured from the NORMAL to the surface.
• $\varepsilon=N\dfrac{|\Delta\Phi|}{\Delta t}$ : a constant flux induces nothing, however intense.
• Lenz: opposition to the CHANGE in flux, hence reinforcement when the flux decreases.
• Rod on rails: $\varepsilon=BLv$ , $F=BIL$ opposed to the motion, and $Fv=RI^{2}$ .
• Total induced charge: $q=\dfrac{N|\Delta\Phi|}{R}$ , independent of the duration.

Frequently asked questions

Why does the magnetic force never accelerate a particle?

Because it is perpendicular to the velocity at every instant. Work is computed with the component of the force along the displacement, and that component is permanently zero. The force therefore changes the direction of the motion without ever changing the magnitude of the velocity or the kinetic energy.

How do you apply the right-hand rule to an electron?

First apply it as for a positive charge, fingers along the velocity and curled toward the field, the thumb giving the force. Then reverse the result, because the charge of the electron is negative. The real force therefore points exactly opposite to what the thumb shows.

Does the period of circular motion depend on the speed?

No. It equals two pi times the mass divided by the product of the charge and the field, and so contains neither the speed nor the radius. A particle twice as fast traces a circle twice as large in the same time. This is what lets a cyclotron be driven at a fixed frequency.

From what is the angle measured in the magnetic flux?

From the normal to the surface, that is, the perpendicular to the plane of the coil, never from the plane itself. The flux is greatest when the field passes straight through the coil, and zero when the field lies in the plane of the coil.

Does a very intense magnetic field induce a current?

Not if it is constant. Faraday's law relates the electromotive force to the change in flux over time, not to its value. A coil at rest in a field of several teslas produces nothing at all, whereas a weak field that drops to zero in a millisecond produces a large voltage.

What exactly does Lenz's law say?

That the induced current opposes the change in flux, not the flux itself. If the flux increases, the induced current creates an opposing field to slow it. If it decreases, the induced current creates a field in the same direction to hold it back. It is a direct consequence of conservation of energy.

Practise it

Corrected exercises: Magnetism and induction, 203-SN2-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

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Revision sheet: magnetism and induction (203-SN2-RE)

The essentials

Magnets and field lines: from north to south

The magnetic force: three directions, never two

Paths: circle, helix, and what does not depend on what

The Laplace force: a wire in a field

The torque on a loop: the electric compass

Induction: it is the change that counts

The mistakes that cost marks

1. Forgetting that the force is zero when the velocity is parallel to the field

2. Applying the right-hand rule as is to an electron

3. Making the magnetic force accelerate a particle

4. Believing the period of circular motion depends on the speed

5. Measuring the flux angle from the plane of the coil

6. Expecting an electromotive force from an intense but constant field

7. Making the induced current oppose the field instead of the change

8. Believing that moving a magnet more slowly carries less charge

Which method to choose

Which path, depending on the angle between the velocity and the field

An induction question: which law, depending on what changes

How the answer is expected to be written

Determining the direction of an induced current

The rod on rails, from the motion to the power balance

Check before you hand in

The typical problem, taken apart

The rod on rails: from muscle to current

Learn by heart

Frequently asked questions

Why does the magnetic force never accelerate a particle?

How do you apply the right-hand rule to an electron?

Does the period of circular motion depend on the speed?

From what is the angle measured in the magnetic flux?

Does a very intense magnetic field induce a current?

What exactly does Lenz's law say?

Corrected exercises: Magnetism and induction, 203-SN2-RE

See also

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