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

Revision sheet: electromagnetic induction (203-SN2-RE)

This sheet does not redo the course on induction, it lists what costs marks on test day. Two errors dominate the whole chapter and are paid for on every question: believing that a strong field induces a lot, when only its CHANGE counts, and believing that the induced current opposes the field, when it opposes the CHANGE in flux. One word of difference, and an answer that is wrong half the time.

It is written for CEGEP students in the Natural Science program (Sciences de la nature) taking 203-SN2-RE (formerly 203-NYB), where induction comes right before the study of RL circuits and the transformer.

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The thread of the chapter

Nothing is induced without a CHANGE of flux: it is never the value of the field that produces an emf, it is its slope, and the induced current opposes the CHANGE, never the field itself.

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

The essentials

Flux: the angle is taken with the normal

• Flux through a flat loop of area $A$ in a uniform field: $\Phi=BA\cos\theta$ , in webers, where $\theta$ is the angle between $\vec{B}$ and the NORMAL $\vec{n}$ to the loop.
• Field perpendicular to the plane of the loop: $\theta=0$ , maximum flux $BA$ . Field in the plane of the loop: $\theta=90°$ , zero flux.
• For a coil of $N$ turns, it is the flux linkage $N\Phi$ that enters Faraday's law.

Loop seen edge-on: the normal

\vec{n}

makes

30°

with

\vec{B}

, the plane of the loop

60°

. It is the cosine of

30°

that enters the flux.

A problem that gives the angle with the PLANE of the loop requires taking the complement: $30°$ with the plane is $60°$ with the normal.

The emf is a SLOPE, not a value

• $\varepsilon=-N\dfrac{d\Phi}{dt}$ , where $\Phi=BA\cos\theta$ is the flux through ONE turn and $\theta$ the angle between $\vec{B}$ and the NORMAL, never between $\vec{B}$ and the surface.
• The flux can only change in three ways: $\dfrac{dB}{dt}\ne 0$ with a fixed circuit, a changing AREA, or a changing ORIENTATION. Faced with a problem, the first question is: which of the three?
• Lenz's law, contained in the minus sign: the induced current opposes the CHANGE in flux. A decreasing flux gives rise to a current that REINFORCES the existing field, which comes as a surprise on a first attempt.
• Motional emf: $\varepsilon=B\ell v$ in translation, but $\varepsilon=\dfrac{1}{2}B\omega\ell^{2}$ for a rod that ROTATES, because each point has its own speed.

The emf vanishes exactly at the instant when the field is MAXIMUM, at

t=2.00

s: this is the definition of a maximum, a horizontal tangent, therefore a zero derivative.

Lenz: the induced current brakes whatever creates it

• The flux increases: the induced current creates a field OPPOSITE to the one passing through the loop. The flux decreases: it creates a field in the SAME direction, to hold it back.
• The direction of the current is then read with the right-hand rule: thumb along the induced field, the curled fingers give the direction of the current in the loop.
• The face through which the induced field emerges is a north pole: facing an approaching magnet, the loop presents the same pole and repels it.

The north pole of the magnet approaches: the flux toward the right increases, the ring presents a north face to the magnet and is repelled, in the direction of

\vec{F}

The check that never fails: the effect of the induced current always opposes the motion, otherwise the magnet would accelerate on its own.

The rod on rails: motional emf and the braking force

• A rod of length $\ell$ sliding at speed $v$ on two rails, perpendicular to $\vec{B}$ , sweeps the area $\ell v$ every second: $\varepsilon=B\ell v$ .
• The induced current $I=\dfrac{\varepsilon}{R}$ flows through the rod, which experiences the Laplace force $F=BI\ell$ , always opposite to $\vec{v}$ .
• To keep the speed constant, the operator supplies $F_{op}=BI\ell$ ; the power $F_{op}v$ comes out entirely as heat, $RI^{2}$ .

Inward field, rod moving to the right: the flux increases, the induced current flows counterclockwise. In the rod it goes up, and the force

\vec{F}

brakes; the operator supplies

\vec{F}_{op}

Released, the rod slows down without ever moving backward: the braking force decreases with the speed, which decays exponentially.

The alternator: the emf is maximum when the flux is zero

• A coil of $N$ turns rotating at $\omega$ in a uniform field sees the flux $\Phi=BA\cos\omega t$ ; its emf is $\varepsilon=NBA\omega\sin\omega t$ .
• Maximum flux, normal aligned with $\vec{B}$ : it no longer varies, $\varepsilon=0$ . Zero flux, plane of the coil parallel to $\vec{B}$ : it varies fastest, $\varepsilon$ is maximum.
• RMS value: $\varepsilon_{\text{rms}}=\dfrac{NBA\omega}{\sqrt{2}}$ . The current delivered brakes the coil, and it is the turbine that supplies the energy.

Two instants of the same revolution: the emf is not maximum when the flux is, but when it vanishes, because that is where it varies fastest.

The same setup, supplied instead of delivering, is a motor: a coil carrying a current in a field experiences a torque.

What follows: coupling, induced field and transformer

• Mutual inductance: $\varepsilon_{2}=-M\dfrac{dI_{1}}{dt}$ with $M=\dfrac{N_{2}\Phi_{2}}{I_{1}}$ . Reciprocity: $M_{12}=M_{21}$ always, so calculate $M$ in the most CONVENIENT direction.
• Coupling $k=\dfrac{M}{\sqrt{L_{1}L_{2}}}$ , between 0 and 1. It equals 1 only if the two windings occupy exactly the same volume.
• Induced electric field: $\displaystyle\oint\vec{E}\cdot d\vec{\ell}=-\dfrac{d\Phi}{dt}$ . This field is NOT conservative: the voltage between two points depends on the path, and the potential no longer makes sense.
• Ideal transformer: $\dfrac{U_{2}}{U_{1}}=\dfrac{N_{2}}{N_{1}}$ , $\dfrac{I_{2}}{I_{1}}=\dfrac{N_{1}}{N_{2}}$ , and a load $Z$ seen from the primary is $\left(\dfrac{N_{1}}{N_{2}}\right)^{2}Z$ .

Step-down transformer with a ratio of 12: the voltage is divided by 12, the current multiplied by 12, and the power

UI=20.0

W is the same on both sides.

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. A coil is placed in the gap of a perfectly steady $10$ -tesla superconducting magnet.

2 marks, and a misunderstanding of the whole chapter

What not to write

The field is enormous, so the emf induced in the coil is enormous.

What to write

$\dfrac{d\Phi}{dt}=0$ , so $\varepsilon=0$ exactly. No emf, whatever the strength of the field.

Why: Faraday's law involves the DERIVATIVE of the flux, never the flux. A steady $10$ T field induces nothing, whereas a $10$ mT field cut off in a millisecond induces far more. This is exactly why a medical imaging scanner induces nothing with its main magnet, and everything with its switched gradients, a thousand times weaker.

2. A magnet is moved AWAY from a coil, north pole first. In which direction does the induced current flow?

3 marks, systematically, on every direction question

What not to write

The induced current always creates a field opposite to that of the magnet, so it opposes the north field.

What to write

The flux DECREASES, so the induced current creates a field in the SAME direction as that of the magnet, to try to maintain it. The coil attracts the magnet.

The north pole is moved away: the flux decreases, the induced current creates a field in the SAME direction as that of the magnet. The ring presents a south face and attracts the magnet.

Why: The false statement is missing a word, and that word changes everything: the induced current opposes the CHANGE in flux, not the flux. The physical check that never fails: the effect always brakes the motion, otherwise the magnet would accelerate on its own and we would have perpetual motion. On approach it is repelled, on withdrawal it is attracted.

3. A coil in an open circuit is placed in a changing field.

2 marks, and the inability to handle any inductive sensor

What not to write

No current can flow, so there is no emf: induction does not take place.

What to write

The emf $-N\dfrac{d\Phi}{dt}$ exists and can be measured across the open terminals. What requires a closed circuit is the CURRENT, and therefore also the Lenz force and the dissipation.

Why: The emf is work per unit charge, produced by the induced electric field: it requires no closed conductor. This is the normal operating mode of almost all inductive sensors, whose open-circuit voltage is read. The slit ring of the classic demonstration does not jump precisely because it cannot carry any current, although the emf is indeed present in it.

4. Calculate the flux through a rectangular frame placed next to a straight wire carrying a current.

3 marks, the whole question

What not to write

$\Phi=BA$ with $B=\dfrac{\mu_{0}I}{2\pi r}$ taken at the middle of the frame.

What to write

The field varies as $\dfrac{1}{r}$ over the frame, so you must integrate: $\Phi=\dfrac{\mu_{0}IL}{2\pi}\ln\dfrac{b}{a}$ .

The field of the wire falls off as

\dfrac{1}{r}

from one edge of the frame to the other: the formula

BA

cannot be used, and the integration makes

\ln\dfrac{b}{a}

appear.

Why: The formula $BA$ assumes the field is UNIFORM over the surface, which is false as soon as the frame is close to the wire: at $2.00$ cm and $7.00$ cm, the field is already three and a half times stronger from one edge to the other. The signature of this geometry is the logarithm, and you must learn to recognize it: any field in $\dfrac{1}{r}$ crossing a flat surface makes one appear.

5. A loop surrounds a region where the magnetic field varies, but the field is zero on the loop itself.

3 marks on the reasoning question

What not to write

The magnetic field being zero at every point of the loop, the induced electric field is zero there too.

What to write

The circulation is $-\dfrac{d\Phi}{dt}$ with the ENCLOSED flux, so $E=\dfrac{R^{2}}{2r}\dfrac{dB}{dt}$ outside: nonzero, although $B$ is zero there.

Beyond

r=5

cm the magnetic field is zero and the induced electric field is not: it falls off as

\dfrac{1}{r}

because the contour still encloses the same changing flux.

Why: Faraday's law in integral form is NONLOCAL: it relates the circulation around a contour to the flux enclosed by that contour, not to the magnetic field where you stand. It is the same phenomenon that lets a transformer work although its windings sit in a weak leakage field. The most unsettling consequence is that the potential ceases to exist: two voltmeters connected to the same terminals by different paths display different values.

6. A step-down transformer, $2400$ to $200$ turns, supplies a $5.00$ Ω resistor. What does the source see on the primary side?

2 marks, and the whole impedance-matching question

What not to write

It sees $5.00$ Ω, or at best $12$ times that value, that is $60.0$ Ω.

What to write

It sees $\left(\dfrac{2400}{200}\right)^{2}\times 5.00=720$ Ω. The turns ratio enters SQUARED.

Why: The transformer divides the voltage by twelve AND multiplies the current by twelve, so it divides the impedance by one hundred and forty-four. Direct check: $\dfrac{120}{0.167}=720$ Ω. This is the property that lets you drive an eight-ohm loudspeaker with a stage that wants to see several kilohms, and it is also why a transformer connected to DIRECT current burns out: without induction, only the ohmic resistance of the winding remains.

Which method to choose

Which of the three causes makes the flux change

Read the problem and spot what moves: the field, the area of the circuit, or its orientation.

If The FIELD varies, circuit at rest → $\varepsilon=-NA\dfrac{dB}{dt}$ : differentiate the expression for $B(t)$ and multiply by $NA$ . The emf vanishes at the extrema of $B$ , never where $B$ vanishes.

Example: $B(t)=0.0800+0.240t-0.0600t^{2}$ with $NA=0.500$ m $^{2}$ : $\varepsilon=-0.120+0.0600t$ .
If The AREA varies, through the translation of a rod or a deformation → Motional emf: $\varepsilon=B\ell v$ for a translation, $\varepsilon=\dfrac{1}{2}B\omega\ell^{2}$ for a rotation about one end.

Example: A $25.0$ cm rod at $60.0$ rad/s in $0.350$ T: $\varepsilon=0.656$ V, and not double.
If The ORIENTATION varies, loop rotating in a fixed field → $\Phi=NBA\cos(\omega t)$ , so $\varepsilon=NBA\omega\sin(\omega t)$ : a sinusoidal emf of amplitude $NBA\omega$ , and of rms value $\dfrac{NBA\omega}{\sqrt{2}}$ .

Example: This is the principle of the alternator, and the reason why a $120$ V rms outlet reaches $170$ V at its peak.
If Two CIRCUITS are coupled and one sees its current vary → $\varepsilon_{2}=-M\dfrac{dI_{1}}{dt}$ , with $M$ calculated in the most convenient direction thanks to the reciprocity $M_{12}=M_{21}$ .

Example: Solenoid of $1200$ turns per metre, secondary of $300$ turns: $M=0.568$ mH, and $250$ A/s give $0.142$ V.
If The problem asks for an ELECTRIC FIELD rather than a current → Integral form: $E(2\pi r)=-\dfrac{d\Phi_{\text{enclosed}}}{dt}$ . Careful, beyond the region where $B$ varies, it is the radius of the REGION that enters the flux, not that of the contour.

Example: Region of $5.00$ cm with $\dfrac{dB}{dt}=0.150$ T/s: $E=3.75$ mV/m at the edge, $1.875$ mV/m at $10.0$ cm.

Once the cause is identified, the SIGN is always decided by Lenz's law and never by the calculation: the induced current opposes the change, so it brakes whatever produces it.

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.

Handling an induction question, from the cause to the sign

When to use it: As soon as a problem makes anything vary around a circuit and asks for an emf, a current or a force.

1 Name the cause of the change in flux among the three possible ones, in one sentence.
2 Write the flux $\Phi(t)$ through ONE turn, with the angle taken relative to the normal.
3 Differentiate and multiply by $N$ to get $|\varepsilon|$ , giving the numerical value.
4 Determine the DIRECTION of the induced current by Lenz's law, stating explicitly which change it opposes.
5 If a force or a power is asked for, check that it BRAKES the imposed motion: this is conservation of energy.

Concluding sentence

“The flux through the coil increases, so the induced current flows in the direction that creates an opposite field inside the loop, that is, clockwise as seen by the reader.”

The trap: Giving the direction of the current without ever naming the change it opposes: the marker cannot tell a reasoned answer from a guess, and half the papers land on the right answer by chance.

Marking: 1 mark for the cause identified, 1 mark for the flux, 1 mark for the differentiation, 1 mark for the justified direction, 1 mark for the energy check.

Check before you hand in

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

The effect always brakes whatever produces it

After finding the direction of the induced current, check that the resulting force opposes the imposed motion. If it accelerates it, the sign is wrong.

An approaching magnet is repelled, a receding magnet is attracted. The reverse would give perpetual motion.
The emf vanishes at the extrema of the flux

On the curve of $\Phi(t)$ or $B(t)$ , locate the points with a horizontal tangent: the emf must be zero there, and change sign as it passes through them.

$B$ maximum at $t=2.00$ s and $\varepsilon=0$ at the same instant: consistent. A maximum emf at the top of the field would be wrong.
The induced charge depends only on the end states

If the problem asks for a total charge, use $q=\dfrac{N\,\Delta\Phi}{R}$ : the path followed between the two states plays no part, unlike the heat dissipated.

A flux that rises and then falls back to its initial value gives a ZERO net charge, although the coil has heated by $1.60$ mJ.
The coupling coefficient is less than 1

After calculating $M$ , $L_{1}$ and $L_{2}$ , check that $\dfrac{M}{\sqrt{L_{1}L_{2}}}$ falls between 0 and 1. A $k$ greater than 1 is mathematically impossible.

$M=0.568$ mH with $L_{1}=0.910$ mH and $L_{2}=2.84$ mH give $k=0.354$ , as it should.

The typical problem, taken apart

A coil in a field that rises and then falls

A flat coil of $N=200$ turns, of area $A=25.0$ cm $^{2}$ and resistance $R=12.0$ Ω, is perpendicular to a field $B(t)=0.0800+0.240\,t-0.0600\,t^{2}$ in teslas, $t$ in seconds. Find the emf, the current, the total charge and the heat dissipated between $t=0$ and $t=4.00$ s.

Step 1

$NA=200\times 25.0\times 10^{-4}=0.500$ m $^{2}$ and $\dfrac{dB}{dt}=0.240-0.120\,t$

Why

We factor out the product $NA$ once and for all: it is constant, and keeping it grouped avoids losing it along the way. Here it is exactly $0.500$ , which simplifies all the following calculations.

Step 2

$\varepsilon(t)=-NA\dfrac{dB}{dt}=-0.120+0.0600\,t$ , so $\varepsilon(0)=-0.120$ V, $\varepsilon(2.00)=0$ , $\varepsilon(4.00)=+0.120$ V

Why

The emf is an AFFINE function of time because $B$ is a second-degree polynomial. It vanishes at $t=2.00$ s and changes sign: the induced current reverses at that instant.

Step 3

$B$ is maximum when $\dfrac{dB}{dt}=0$ , that is $t=\dfrac{0.240}{0.120}=2.00$ s, where $B=0.320$ T

Why

This is the connection that makes the whole exercise: the emf vanishes exactly at the maximum of the field. A problem could just as well offer a constant field of ten teslas, and the emf would be zero. It is never the value that induces, it is the slope.

Step 4

$I(0)=\dfrac{0.120}{12.0}=1.00\times 10^{-2}$ A, that is $10.0$ mA, decreasing to zero and then starting up again in the opposite direction

Why

During the first two seconds the flux increases and the current opposes it; afterwards the flux decreases and the current reverses to try to maintain it. The reversal coincides with the maximum of the field.

Step 5

$q=\dfrac{NA\,\Delta B}{R}$ with $\Delta B=B(4.00)-B(0)=0.0800-0.0800=0$ , so $q=0$

Why

The net charge is strictly zero although a current flowed for four seconds: the two halves cancel exactly. The induced charge depends only on the initial and final states of the flux, never on the path.

Step 6

$W=\displaystyle\int_{0}^{4}\dfrac{\varepsilon^{2}}{R}dt=\dfrac{1}{12.0}\cdot\dfrac{2(0.120)^{3}}{3\times 0.0600}=1.60\times 10^{-3}$ J

Why

The heat, for its part, depends on the WHOLE path: the square never cancels. The contrast with the zero charge is the lesson of the exercise, and this energy is supplied by the device that makes the field vary, against the field of the induced current.

The conclusion, written out

$\varepsilon(t)=-0.120+0.0600\,t$ volts, initial current $10.0$ mA reversing at $t=2.00$ s, zero net charge, and $1.60$ mJ dissipated. The emf vanishes exactly at the maximum of the field.

The classic mistake on this problem: Looking for the instant when the field vanishes instead of the instant when its DERIVATIVE vanishes, or concluding from a zero charge that no current flowed: these are the two confusions this exercise is designed to expose.

Learn by heart

• $\varepsilon=-N\dfrac{d\Phi}{dt}$ and $\Phi=BA\cos\theta$ , the angle being taken with the NORMAL to the surface.
• Lenz's law: the induced current opposes the CHANGE in flux, never the flux. A decreasing flux gives rise to a current that reinforces the field.
• Three possible causes, never four: $B$ varies, the area varies, the orientation varies.
• Motional emf: $B\ell v$ in translation, $\dfrac{1}{2}B\omega\ell^{2}$ in rotation about one end.
• Field of a wire: $B=\dfrac{\mu_{0}I}{2\pi r}$ with $\mu_{0}=4\pi\times 10^{-7}$ . Solenoid: $B=\mu_{0}nI$ .
• Induced charge: $q=\dfrac{N\,\Delta\Phi}{R}$ , independent of the path. The heat, on the other hand, depends on the whole path.
• Transformer: voltages follow $\dfrac{N_{2}}{N_{1}}$ , currents the inverse, and impedance the SQUARE of the ratio.

Frequently asked questions

Does a strong magnetic field always produce a large emf?

No. Faraday's law involves the derivative of the flux with respect to time, not the flux itself. A perfectly steady ten-tesla field induces exactly zero, whereas a field a thousand times weaker cut off in a millisecond induces far more. Only the change counts, never the strength.

Does the induced current oppose the magnetic field?

No, it opposes the CHANGE in flux, and that is very different. When the flux decreases, the induced current creates a field in the same direction as the existing field, to try to maintain it: it reinforces it instead of opposing it. This is why a magnet pulled away from a coil is attracted by it, whereas a magnet brought toward it is repelled.

Is there an emf in an open-circuit coil?

Yes. The emf exists as soon as the flux changes, and it can be measured across the open terminals. What requires a closed circuit is the current, and therefore also the Lenz force and the heating. Most inductive sensors work precisely this way, reading an open-circuit voltage without letting any current flow.

Why does a transformer not work on direct current?

Because a direct current produces a constant flux in the core, so a zero derivative, so no emf in the secondary. Worse, the primary then presents only the ohmic resistance of its winding, a few ohms instead of several hundred: the current drawn explodes and the winding burns out within seconds.

How do you calculate the flux when the field is not uniform?

You must integrate, cutting the surface into strips over which the field is constant. For a rectangular frame placed next to a straight wire, the field falls off as the inverse of the distance, and the integration makes a logarithm of the ratio of the two distances appear. The formula field times area then cannot be used.

Can there be an induced electric field where the magnetic field is zero?

Yes, and this is one of the most unsettling results of the chapter. Faraday's law relates the circulation of the electric field around a contour to the magnetic flux ENCLOSED by that contour, not to the local magnetic field. A contour surrounding a region of changing field therefore carries a nonzero circulation, even if the field is zero everywhere on it.

Practise it

Corrected exercises: Electromagnetic 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

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-phys2-electromagnetic-induction. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: electromagnetic induction (203-SN2-RE)

The essentials

Flux: the angle is taken with the normal

The emf is a SLOPE, not a value

Lenz: the induced current brakes whatever creates it

The rod on rails: motional emf and the braking force

The alternator: the emf is maximum when the flux is zero

What follows: coupling, induced field and transformer

The mistakes that cost marks

1. A coil is placed in the gap of a perfectly steady 101010-tesla superconducting magnet.

2. A magnet is moved AWAY from a coil, north pole first. In which direction does the induced current flow?

3. A coil in an open circuit is placed in a changing field.

4. Calculate the flux through a rectangular frame placed next to a straight wire carrying a current.

5. A loop surrounds a region where the magnetic field varies, but the field is zero on the loop itself.

6. A step-down transformer, 240024002400 to 200200200 turns, supplies a 5.005.005.00 Ω resistor. What does the source see on the primary side?

Which method to choose

Which of the three causes makes the flux change

How the answer is expected to be written

Handling an induction question, from the cause to the sign

Check before you hand in

The typical problem, taken apart

A coil in a field that rises and then falls

Learn by heart

Frequently asked questions

Does a strong magnetic field always produce a large emf?

Does the induced current oppose the magnetic field?

Is there an emf in an open-circuit coil?

Why does a transformer not work on direct current?

How do you calculate the flux when the field is not uniform?

Can there be an induced electric field where the magnetic field is zero?

Corrected exercises: Electromagnetic induction, 203-SN2-RE

See also

Looking for a 203-SN2-RE tutor in Montreal?

1. A coil is placed in the gap of a perfectly steady $10$ -tesla superconducting magnet.

6. A step-down transformer, $2400$ to $200$ turns, supplies a $5.00$ Ω resistor. What does the source see on the primary side?