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D.4

Induction

Theme D · Fields · HL only

Everything on this page comes from one sentence: a changing magnetic flux drives a current. The rest is working out how fast it is changing, and which way the current goes.

🎯What you need to be able to do

  • Calculate magnetic flux \( \Phi = BA\cos\theta \) and flux linkage \( N\Phi \), measuring \(\theta\) from the normal.
  • Apply Faraday’s law to find an induced emf, and name the three things that can change the flux.
  • Apply Lenz’s law to find the direction of an induced current, and argue why it must follow from conservation of energy.
  • Use \( \varepsilon = BvL \), and \( \varepsilon = BvLN \) for a coil, and explain the opposing force that appears once current flows.
  • Use Fleming’s right-hand rule to find the direction of an induced current.
  • Explain the AC generator, use \( \varepsilon = NBA\omega\sin\omega t \), and sketch how the output changes with \(B\) and \(\omega\).
  • Use rms values for alternating current and voltage, and relate average power to peak power.
  • Explain self-induction and back emf, and why a current takes time to build up in a coil.
  • Explain eddy currents, and the everyday effects they produce.

🧭Flux, and the angle everyone measures from the wrong place

\[ \Phi = BA\cos\theta \]

Flux measures how much field passes through a loop. The angle \(\theta\) is between the field and the normal to the area — the line sticking out perpendicular to the loop — and not the plane of the loop. Measure it from the plane instead and every cosine in your working becomes a sine.

Three orientations of a loop in a uniform magnetic field, with the normal to the loop drawn as a dashed line in each. In the first the field is along the normal, at zero degrees, and the flux is a maximum equal to B A. In the second the field is at sixty degrees to the normal and the flux is half of that. In the third the field lies in the plane of the loop, at ninety degrees to the normal, and no field lines pass through the loop at all, so the flux is zero. A warning states that the angle is measured from the normal and not from the plane of the loop, and that measuring it from the plane turns every cosine into a sine.
Flux is greatest when the field is perpendicular to the loop’s plane and zero when the field lies in that plane — which is the opposite of what “90 degrees” makes most people expect. Sketch the normal every time.

Flux linkage is \( N\Phi \) for a coil of \(N\) turns, since each turn links the same flux. Both are measured in webers.

✏️Worked example 1 — flux and flux linkage

A coil of 200 turns and area \( 4.0\times10^{-3} \) m\(^{2}\) sits in a uniform field of 0.25 T. Find the flux and the flux linkage when the field is at 60° to the normal, and when it is along the normal.

At 60° to the normal: \( \Phi = BA\cos\theta = 0.25 \times 4.0\times10^{-3} \times \cos 60^\circ = 5.0\times10^{-4} \) Wb, and \( N\Phi = 200 \times 5.0\times10^{-4} = 0.10 \) Wb.

Along the normal (\(\theta = 0\)): \( \Phi = 1.0\times10^{-3} \) Wb and \( N\Phi = 0.20 \) Wb — the maximum.

\(\cos 60^\circ = 0.5\), so the flux halves. If a question says the field is at 60° to the plane of the coil instead, that is 30° to the normal and the factor is \(\cos 30^\circ = 0.866\). Read that phrase carefully — it is the single most reliable way to lose marks on this topic, and the two answers differ by 73%.

⚡Faraday’s law: only change counts

\[ \varepsilon = -N\frac{\Delta\Phi}{\Delta t} \]

The induced emf depends on the rate of change of flux linkage, not on the flux itself. A coil sitting in the strongest field you can build has no emf in it at all while nothing changes. Something must vary, and there are only three candidates.

Three ways to change the flux through a coil, drawn side by side. First, change the field strength, shown by a magnet moving towards the coil so the field through it grows. Second, change the area, shown by a conducting rod sliding along rails so that the circuit it encloses gets larger. Third, change the orientation, shown by a coil rotating in a fixed field so that the angle between the field and the normal changes. Beneath, a note states that the emf depends on how fast the flux changes and not on how large it is, so a coil in a huge but steady field has no emf induced in it at all.
Change \(B\), change \(A\), or change \(\theta\) — there is nothing else in \( \Phi = BA\cos\theta \) to change. Every induction question in the syllabus is one of these three, and identifying which is usually the first mark.

✏️Worked example 2 — rotating a coil

The same coil (200 turns, \( 4.0\times10^{-3} \) m\(^{2}\), 0.25 T) is turned from “field along the normal” to “field in the plane of the coil” in 0.050 s. Find the average emf induced.

The flux linkage goes from its maximum to zero:

\[ \Delta(N\Phi) = 0.20 - 0 = 0.20\ \text{Wb} \]
\[ \varepsilon = \frac{\Delta(N\Phi)}{\Delta t} = \frac{0.20}{0.050} = 4.0\ \text{V} \]
“Average” is doing real work in that question. Faraday’s law with \(\Delta\) gives the average emf over the interval; the instantaneous emf varies throughout the turn and is largest when the coil sweeps through the position where the field lies in its plane. Note also that turning it faster gives a bigger emf from the same magnet — the flux change is identical, but it happens in less time.

🚫Lenz’s law: the induced current always fights you

That minus sign in Faraday’s law is Lenz’s law: the induced current flows in whatever direction opposes the change that produced it.

Two panels showing a bar magnet and a coil. In the first the magnet is pushed north-pole-first towards the coil, and the induced current flows so that the coil's near face becomes a north pole, which repels the approaching magnet. In the second the magnet is pulled away, and the current reverses so that the near face becomes a south pole, which attracts the retreating magnet. In both cases the coil opposes what is being done to it. A note states that you must therefore do work to move the magnet either way, and that this work is exactly the electrical energy that appears in the coil.
Push it in and the coil pushes back; pull it out and the coil hangs on. Either way you have to work to move the magnet, and that work is precisely the electrical energy that appears.

Why it has to be that way is a standard extended-response question, and the argument is short. Suppose the induced current helped the motion instead.

A flow diagram of the impossible alternative. If the induced current helped the magnet's motion rather than opposing it, the magnet would accelerate, which would make the flux change faster, which would induce a larger current, which would push the magnet harder still, and so on round the loop without limit. The diagram is drawn as a closed cycle with a cross through it and is labelled as producing unlimited energy from nothing, which is forbidden. A conclusion states that Lenz's law is conservation of energy written in the language of induction.
A perpetual motion machine, and a rapidly accelerating one. Lenz’s law is not an extra rule bolted on to Faraday’s — it is conservation of energy, written in the language of induction.

📏A rod moving through a field

The simplest case of all. A rod of length \(L\) moving at speed \(v\) perpendicular to a field \(B\) sweeps out area at a rate \(Lv\), so the flux changes at a rate \(BLv\):

\[ \varepsilon = BvL \]
A conducting rod sliding to the right along two horizontal rails in a magnetic field directed into the page, with the rails joined at the left by a resistor to complete the circuit. The area enclosed by the circuit is shaded and grows as the rod moves, and the rate at which area is swept out is marked as L times v. The induced current is shown flowing round the circuit, and a force B I L is drawn on the rod pointing backwards, opposing its motion. A note states that this opposing force is Lenz's law again, and that the work done pushing against it is exactly the electrical energy dissipated in the resistor.
Once the circuit is complete a current flows, and that current in that field puts a force \(BIL\) on the rod — backwards. Lenz again. Push harder and you get more power out, which is exactly the bargain a generator makes.

Which way does the induced current go? This is what Fleming’s right-hand rule is for, and it is the rule to reach for whenever the motion is what you are given: first finger along the Field, thuMb along the Motion, and the seCond finger then points along the induced Current. Use the left hand here by mistake and you will get a current in exactly the wrong direction.

Why an emf appears at all is worth one line of mechanism. The rod is full of free electrons, and each of them is moving with the rod through the field, so each feels a magnetic force \(qvB\) that pushes it towards one end. Charge piles up there until the electric field it creates pushes back just as hard, and at that point the potential difference across the rod is steady. Setting the two forces equal for an electron in the middle gives \( qvB = qE = qV/L \), so \( V = BvL \) — the same answer the flux argument gave, from the other direction.

For a coil of \(N\) turns whose side is in the field, each turn generates \(BvL\) and they add in series:

\[ \varepsilon = BvLN \]

with one condition worth noticing: only the part of the coil inside the field generates anything. If the whole coil sits inside a uniform field, both sides generate equal emfs that oppose each other round the loop, and no current flows at all.

✏️Worked example 3 — the emf across an aircraft

An aircraft with a wingspan of 30 m flies at 200 m s\(^{-1}\). The vertical component of the Earth’s magnetic field is \( 5.0\times10^{-5} \) T. Estimate the potential difference between its wingtips.
\[ \varepsilon = BvL = 5.0\times10^{-5} \times 200 \times 30 = 0.30\ \text{V} \]
Why the vertical component, and not the whole field? Because \(B\), \(v\) and \(L\) must be mutually perpendicular. The wings are horizontal and across the motion, so the only part of the field that counts is the part perpendicular to both — the vertical component. Using the full field strength is the standard error here. And 0.30 V is real but useless: it drives no current, because there is no circuit to complete through the air.

🔄The AC generator

Rotate a coil steadily in a uniform field and the angle changes continuously, so \( \Phi = BA\cos\omega t \) and the flux linkage varies sinusoidally. Faraday’s law turns that cosine into a sine:

\[ \varepsilon = NBA\omega\sin\omega t \qquad\text{with peak value}\qquad \varepsilon_0 = NBA\omega \]
On the left, a rectangular coil rotating in a uniform horizontal magnetic field between two poles, with an arrow showing the angular velocity. On the right, two graphs against time with the same time axis. The upper graph is the flux linkage, a cosine starting at its maximum. The lower graph is the induced emf, a sine that is zero when the flux linkage is at its peak and greatest when the flux linkage is passing through zero. Markers connect the two, showing that the emf is largest exactly when the flux is changing fastest, which is when the coil lies along the field, and zero when the flux is momentarily stationary at its maximum.
The emf is zero when the flux is greatest and greatest when the flux is zero — because it is the gradient of the flux curve, not the flux. That quarter-cycle offset is the thing most often drawn wrong.
Four graphs of induced emf against time, each compared with the same original output drawn faintly behind it. The first is the original. The second doubles the magnetic field, which doubles the peak emf while leaving the period unchanged. The third doubles the angular velocity, which both doubles the peak emf and halves the period, so twice as many cycles fit in the same time and each is twice as tall. The fourth halves the angular velocity, which halves the peak emf and doubles the period. A note states that changing the field alters only the height, whereas changing the speed alters the height and the period together, because omega appears in both.
Changing \(B\) changes only the height. Changing \(\omega\) changes the height and the period together, because \(\omega\) appears in \( \varepsilon_0 = NBA\omega \) as well as in the timing. Sketching one without the other is the standard lost mark.

✏️Worked example 4 — a generator’s peak output

A coil of 50 turns and area \( 2.0\times10^{-2} \) m\(^{2}\) rotates at 50 revolutions per second in a field of 0.30 T. Find the peak emf.

Convert to angular velocity first: \( \omega = 2\pi f = 2\pi \times 50 = 314 \) rad s\(^{-1}\).

\[ \varepsilon_0 = NBA\omega = 50 \times 0.30 \times 2.0\times10^{-2} \times 314 = 94\ \text{V} \]
Revolutions per second is not \(\omega\). Forgetting the \(2\pi\) gives 15 V instead of 94 V — a factor of \(2\pi\) out, and the most common single error in generator questions. As a check, doubling the rotation rate would double this to 188 V and halve the period; changing the field would move only the height.

🔋rms values: what “230 volts” actually means

The generator’s output swings from \( +\varepsilon_0 \) to \( -\varepsilon_0 \) and averages zero, so quoting an average voltage would be useless. What matters is the power it can deliver, and power depends on \( I^{2} \), which is never negative.

Two graphs sharing a time axis. The upper one shows a sinusoidal current alternating equally above and below zero, with its peak value marked and its average clearly zero. The lower one shows the power, which is proportional to the current squared and is therefore always positive, rising and falling at twice the frequency between zero and a maximum. The average power is drawn as a horizontal line at exactly half the peak power. Beside them, the resulting relations: the root mean square current is the peak divided by the square root of two, the same for voltage, and the average power equals the rms voltage times the rms current, which is half the peak power.
The power curve is a \( \sin^{2} \) curve, not a sine: always positive, and averaging exactly half its peak. That factor of a half is where the \( \sqrt{2} \) comes from.

The root mean square value is defined so that it behaves like a steady direct current would: it is the square root of the mean of the square. For a sinusoid that works out to

Current\( I_{\text{rms}} = \dfrac{I_0}{\sqrt{2}} \)
Voltage\( V_{\text{rms}} = \dfrac{V_0}{\sqrt{2}} \)
Average power\( \bar{P} = V_{\text{rms}}I_{\text{rms}} = \tfrac{1}{2}I_0V_0 \)

Quoted AC values are always rms values. European mains at “230 V” is 230 V rms — its peak is \( 230\sqrt{2} = 325 \) V, which is what the insulation has to withstand. In the USA “120 V” means a peak of 170 V.

Resistance behaves itself throughout: \( R = V_{\text{rms}}/I_{\text{rms}} = V_0/I_0 \), so it does not matter which pair you use as long as you do not mix them.

✏️Worked example 5 — a lamp on the mains

A 60 W lamp runs from a 230 V (rms) mains supply. Find the rms current, the peak current, the peak voltage and the peak power.

rms current. Average power is \( V_{\text{rms}}I_{\text{rms}} \), so \( I_{\text{rms}} = 60/230 = 0.26 \) A.

Peaks. \( I_0 = \sqrt{2} \times 0.261 = 0.37 \) A and \( V_0 = \sqrt{2} \times 230 = 325 \) V.

Peak power. \( P_{\text{max}} = I_0V_0 = 0.369 \times 325 = 120 \) W.

Exactly twice the 60 W it is sold as — and it has to be, since the average of the \( \sin^{2} \) power curve is half its peak. So the lamp really does draw 120 W a hundred times a second and nothing in between; “60 W” is the steady power that would do the same job. Note also that using the peak voltage with the rms current, or the other way round, would give an answer \(\sqrt{2}\) out — the commonest slip here.

🌱Eddy currents

Everything so far assumed a wire. Change the flux through a solid lump of metal and the induced currents still flow — in closed loops within the metal itself. These are eddy currents, and they do exactly what Lenz’s law says: oppose the relative motion, while dissipating energy as heat.

Three consequences of eddy currents. First, a magnet dropped down a vertical copper pipe falls slowly, because the currents induced in the pipe wall oppose its motion; the pipe is not magnetic, and a non-magnetic object of the same shape falls normally. Second, an induction hob, where a rapidly changing field induces currents directly in the base of the pan and heats it, leaving the hob surface itself cool. Third, magnetic braking, where a metal disc spinning near an electromagnet is slowed without anything touching it, so there is nothing to wear out. A note states that in a transformer the same effect is a nuisance, which is why transformer cores are built from thin insulated laminations that break up the current loops.
The same physics is useful in a brake, essential in an induction hob and a nuisance in a transformer core — where it is suppressed by building the core from thin insulated laminations that break up the loops.
A magnet falls slowly down a copper pipe, and copper is not magnetic. That is the point of the demonstration. Nothing is attracting the magnet; the currents induced in the pipe wall create their own field that opposes the change, and the retarding force grows with speed until it nearly balances the weight. Drop an identical non-magnetic slug down the same pipe and it falls normally — which is the control that shows the effect is induction and not friction.

🔌Self-induction: a coil opposes its own current

Connect a coil across a battery and the current does not jump instantly to \( V/R \). It takes time to get there, and the reason is Faraday and Lenz applied to the coil by itself.

As the current grows, so does the coil’s own magnetic field, so the flux through its own turns is changing — and a changing flux induces an emf in that same coil. This is self-induction, and by Lenz’s law the induced emf opposes the change that caused it. While the current is building up, the induced emf acts against the build-up, which is why it is called a back emf.

On the left, a graph of current against time after a coil is connected to a supply: the current rises steeply at first and then more gradually, approaching its final steady value of V over R asymptotically rather than jumping to it immediately, an exponential growth. On the right, what happens at switch-off: the magnetic field collapses very rapidly, so the rate of change of flux is enormous and the induced emf is correspondingly large, drawn as a tall narrow spike that can far exceed the original supply voltage. A note explains that this is what fires the spark plug in a petrol engine, the spark needing a field of about ten to the four volts per centimetre.
At any instant the current is set by what is left over: \( I = (\text{supply pd} - \text{back emf})/R \). The back emf is largest at the start, when the current is changing fastest, and dies away as the current levels off.

Switching off is the dramatic half. The field collapses far faster than it was built, so \( \Delta\Phi/\Delta t \) is enormous and the induced emf can be very much larger than the supply that created it — large enough to jump a spark gap. That is exactly how the spark plugs in a petrol engine are fired: a coil’s magnetic field is built up and then collapsed, tens of times a second, and each collapse produces the several kilovolts needed to ignite the fuel.

Energy is stored in the magnetic field, and it has to go somewhere. Building the field took energy from the supply; collapsing it gives that energy back, and if you open a switch the only path available may be through the air. This is why switches in inductive circuits arc, why you should never break the circuit of a large electromagnet casually, and why the spark-plug trick works at all. The IB course does not ask for the mathematics of the build-up — it is another exponential, like radioactive decay — but it does ask for this reasoning.

🔭See it happen

PhET, Faraday’s Law and Generator. Move the magnet through the coil at different speeds and watch the meter: the emf tracks the speed, not the position. Then hold the magnet perfectly still inside the coil, where the flux is largest of all, and watch the meter read exactly zero — which is the whole of Faraday’s law in one observation.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. A single loop of area \( 2.5\times10^{-2} \) m\(^{2}\) lies in a field of 0.40 T, at 30° to the normal. Find the flux.
\( \Phi = BA\cos\theta = 0.40 \times 2.5\times10^{-2} \times \cos 30^\circ = 0.40 \times 2.5\times10^{-2} \times 0.866 = 8.7\times10^{-3} \) Wb. Had the question said 30° to the plane of the loop, the angle to the normal would be 60° and the answer would be \( 5.0\times10^{-3} \) Wb instead — always check which the question means.
2. State the flux linkage for that loop if it is replaced by a 300-turn coil of the same area in the same position.
\( N\Phi = 300 \times 8.7\times10^{-3} = 2.6 \) Wb. Each turn links the same flux, so the linkage is simply \(N\) times greater — which is why coils rather than single loops are used everywhere that a usable emf is wanted.
3. The flux through each turn of a 150-turn coil changes by \( 8.0\times10^{-4} \) Wb in 0.020 s. Find the average emf induced.
\( \varepsilon = N\dfrac{\Delta\Phi}{\Delta t} = 150 \times \dfrac{8.0\times10^{-4}}{0.020} = 6.0 \) V. Note that the flux itself is tiny; what produces a usable 6 V is the number of turns and the shortness of the time.
4. A bar magnet is pushed north-pole-first into a coil. State the polarity of the coil’s near face, and hence the direction of the induced current as seen from the magnet.
The near face becomes a north pole, so that it repels the approaching magnet and opposes the change. Seen from the magnet, the induced current therefore flows anticlockwise (by the right-hand grip rule applied to the coil: curl the fingers with the current and the thumb gives the north pole, which must point back at the magnet). Withdraw the magnet and everything reverses: the near face becomes a south pole and the current runs clockwise, now attracting the magnet and again opposing the change.
5. Explain why Lenz’s law must take the form it does, in terms of energy.
Suppose the induced current helped the motion instead of opposing it. The magnet would then accelerate; the flux would change faster; a larger current would be induced; the magnet would be pushed harder still. Its kinetic energy and the electrical energy in the coil would both grow without limit and without anything being supplied — energy from nothing, which is impossible. So the current must oppose the change. The work done pushing the magnet against that opposition is exactly the electrical energy that appears: Lenz’s law is conservation of energy in the language of induction.
6. A rod of length 0.25 m slides along rails at 3.0 m s\(^{-1}\) perpendicular to a 0.60 T field. Find the emf, and the current and retarding force if the circuit has resistance 1.5 Ω.
\( \varepsilon = BvL = 0.60 \times 3.0 \times 0.25 = 0.45 \) V. Then \( I = \varepsilon/R = 0.45/1.5 = 0.30 \) A, and the force on the current-carrying rod is \( F = BIL = 0.60 \times 0.30 \times 0.25 = 4.5\times10^{-2} \) N, directed opposite to the rod’s motion — Lenz’s law. Check the energy: mechanical power \( Fv = 0.135 \) W, electrical power \( \varepsilon I = 0.135 \) W. They match exactly, as they must.
7. An aircraft of wingspan 25 m flies horizontally at 250 m s\(^{-1}\) where the vertical component of the Earth’s field is \( 4.0\times10^{-5} \) T. Find the pd between the wingtips, and explain why no current flows.
\( \varepsilon = BvL = 4.0\times10^{-5} \times 250 \times 25 = 0.25 \) V. Only the vertical component counts, because \(B\), \(v\) and \(L\) must be mutually perpendicular and the wings are horizontal and across the motion. No current flows because there is no complete circuit — the emf exists across the wingtips but there is no conducting path through the air to return the charge.
8. A coil of 80 turns and area \( 1.5\times10^{-2} \) m\(^{2}\) rotates at 25 revolutions per second in a 0.25 T field. Find the peak emf.
\( \omega = 2\pi f = 2\pi \times 25 = 157 \) rad s\(^{-1}\), so \( \varepsilon_0 = NBA\omega = 80 \times 0.25 \times 1.5\times10^{-2} \times 157 = 47 \) V. Using 25 in place of \(\omega\) would give 7.5 V — wrong by exactly \(2\pi\), which is the classic slip in this calculation.
9. Sketch how the output of that generator changes if (a) the field is doubled, (b) the rotation rate is doubled.
(a) Doubling \(B\): the peak emf doubles, to 94 V. The period is unchanged, so the graph has the same shape stretched vertically. (b) Doubling \(\omega\): the peak emf doubles as well, since \( \varepsilon_0 = NBA\omega \), and the period halves — so twice as many cycles fit into the same time and each one is twice as tall. The distinction is the point of the question: \(B\) changes only the height, \(\omega\) changes the height and the timing together.
10. A small bar magnet is dropped through a vertical coil connected to a datalogger. Sketch the emf against time, and explain why the second peak is taller and narrower than the first.
Two pulses of opposite sign, with a moment of near-zero between them while the magnet is fully inside and the flux is momentarily not changing much. The second pulse is taller and narrower because the magnet has accelerated under gravity: it leaves faster than it entered, so the flux changes more quickly (a bigger emf) over a shorter time (a narrower pulse). The two pulses have opposite signs because the flux is increasing on the way in and decreasing on the way out, so the induced current opposes each in turn. The areas under the two pulses are equal, since each equals the same total flux change.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • HyperPhysics — Faraday’s law and Lenz’s law
  • The Physics Hypertextbook — electromagnetic induction