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12Electromagnetic induction and Faraday's law: what a changing flux does
Lay a loop of wire flat on the desk and clip a millivoltmeter across it. Wave a fridge magnet over the loop and the needle kicks; pull the magnet back and it kicks the other way. Hold the magnet still a centimetre above the wire, as close as you like, and the reading falls back to zero.
By the end of this section you can work out how much magnetic field passes through a loop, turn the rate at which that changes into volts, say which way round the current runs without guessing, and check the answer by matching electrical power out to mechanical power in.
In 60 seconds
A voltage appears round a circuit when the amount of magnetic field through it changes, its size is the rate of that change, and its direction always fights the change that produced it.
through a flat loop in a uniform field
$$\Phi_B = \vec B\cdot\vec A = BA\cos\theta$$
the field is the same over the whole loop; theta is measured from the normal to the loop, not from its plane; the unit is the
Faraday's law of induction
$$\varepsilon = -N\,\frac{d\Phi_B}{dt}$$
always; N is the number of turns, and in practice you compute the size from the rate of change and fix the direction separately
Charge pushed round by a flux change
$$q = \frac{N\,|\Delta\Phi_B|}{R}$$
you want total charge rather than current; no time appears, so a slow change and a fast one move the same charge
of a rod of length L
$$\varepsilon = BLv$$
rod, field and velocity mutually perpendicular; the retarding force is then $B^{2}L^{2}v/R$, always against the motion
both windings share the same changing flux and nothing is lost, so voltage and current move opposite ways
$$\oint\vec E\cdot d\vec l = -\frac{d\Phi_B}{dt}$$
a changing flux with no moving conductor; this field has closed loops and no potential, so points carry no voltage
Three most common mistakes
Using the angle between the field and the plane of the loop in $\Phi_B = BA\cos\theta$. The angle is measured from the normal. Face on gives $\theta = 0$ and the largest flux; edge on gives $\theta = 90^{\circ}$ and none, which is the reverse of what the word plane suggests.
Reading the minus sign as opposing the field. What is opposed is the change. In a loop whose flux is dropping, the makes flux the same way as the field that is disappearing; in a loop whose flux is steady, nothing flows at all.
Believing that a big flux gives a big emf. Only the rate of change appears. A coil sitting in one tesla generates nothing; a coil in the field of the Earth generates a usable voltage if you spin it fast enough.
The published weights are two midterms at twenty per cent each, quizzes at ten, homework at five, the final at twenty five and laboratory work at twenty. Nothing published says how many questions come from this week, so treat it as the weights do. What transfers beyond it is the habit of writing the flux down first and finishing with an energy check.
How much time do you have?
10 minutes
You leave able to do the step every question begins with: write down the flux, turn its rate of change into a voltage, and say which way the current goes.
The 60 second card · How much field goes through a loop · Faraday's law, and what the minus sign is for · Formula card · Mistake ledger
45 minutes
You add the three arrangements almost every exam question is built from: a conductor sliding on rails, a coil spinning in a field, and that same coil with a load on it.
The 60 second card · How much field goes through a loop · Faraday's law, and what the minus sign is for · The sliding rod: emf from motion · The rotating coil: where mains electricity comes from · Method box: which induction question is this · Method box: fixing the direction in four moves · Exam level example · Practice set C · Mistake ledger
full read
You can also explain why a motor draws a huge current at switch on, why a transformer refuses to work on a battery, why power lines run at hundreds of kilovolts, and what the electric field inside a ramped coil is doing with no charge to make it.
The opening pages · Prerequisites and pretest · Notation · Conventions · All seven concept blocks · Method boxes · Contrast pair · Scaffolding comes off · Exam level example · Practice sets A to D · Mistake ledger · Formula card · Glossary · Check yourself
By the end of this section
Compute the flux through a flat loop at any orientation, and set up the integral when the field varies across the loop.
Apply Faraday's law to get an , current and total charge, and use Lenz's law to state the direction with a viewpoint.
Analyse a conductor moving in a field: the motional emf, the retarding force, the , and the match between mechanical and electrical power.
Derive the emf of a coil rotating in a uniform field, find its peak and instantaneous values, and say where in the cycle the flux and the emf reach their extremes.
Explain a motor's back emf quantitatively, find the current at starting and at any speed, and account for the drag on a conductor crossing a field boundary.
Calculate transformer voltages and currents from the , and use it to find the heating loss on a transmission line.
Determine the induced electric field inside and outside a solenoid, and say why it cannot be described by a potential.
Syllabus coverage
Electromagnetic Induction and Faraday’s Law
Magnetic flux and the weber; Faraday's law in average and instantaneous form, and the charge a flux change moves; Lenz's law and the energy argument behind the minus sign; motional emf, the retarding force on a moving conductor and the power balance; the rotating coil generator; back emf, , and ; transformers, the turns ratio and power transmission; and the electric field a changing flux produces in empty space.
The week line is the phrase Electromagnetic Induction and Faraday's Law and carries no chapter number, so no section number is quoted anywhere on this page.
covered
The emf a circuit induces in itself
What happens when the changing flux through a coil is made by the coil's own current, and the energy in a coil's field.
Deferred to the later week line on coils and alternating currents. Every flux change here is driven from outside the circuit, so no result on this page needs the idea that a circuit reacts to its own current.
deferred
Alternating current circuits
How a resistor, a capacitor and a coil respond to a sinusoidal supply, and the averages that go with it.
Deferred to the later week line on alternating currents. The generator here produces a sinusoidal emf quoted as a peak value at a stated instant, never as an average or effective value, since those belong with the machinery for alternating circuits.
deferred
The complete set of field equations
The correction to the circulation law for the magnetic field when the electric field changes, and the wave the four equations give together.
Deferred to the last week line of the course. The circulation law for the induced electric field appears here in its own right, as the local form of Faraday's law, not as part of a set.
deferred
Induction cooktops, metal detectors and card readers
Everyday devices whose whole operating principle is a changing flux in a nearby conductor.
Not named on the week line and not examinable. They appear as one sentence apiece, to give the eddy current argument something recognisable to point at, and no number is quoted for any of them.
off_syllabus
Recall first
The field of a long straight current
A long straight wire carrying a steady current $I$ makes a field of size $B = \mu_0 I/(2\pi r)$ at perpendicular distance $r$, circling the wire in the sense given by the right hand grip rule. The grouped constant $\mu_0/2\pi = 2\times10^{-7}\ \mathrm{T\,m/A}$.
The flux through a loop beside a straight wire is not $BA$, because the field differs at every distance. That calculation needs this formula inside an integral, and it is the standard way an exam makes a flux question hard.
The field inside a long solenoid
Well inside a long solenoid with $n$ turns per metre carrying $I$, the field is uniform, along the axis, of size $B = \mu_0 n I$, and independent of the radius.
Half the induction problems here are built out of a solenoid, because it is the easiest way to make a known uniform field you can change at will.
The force a field exerts on a current carrying wire
A straight wire of length $L$ carrying current $I$ through a uniform field feels $\vec F = I\vec L\times\vec B$, of size $F = BIL\sin\theta$, where $\vec L$ points along the current.
The drag on a sliding rod, the counter torque of a generator and the braking of a metal plate are all this formula applied to a current the motion itself created.
The magnetic force on a moving charge
A charge $q$ moving with velocity $\vec v$ through a field feels $\vec F = q\vec v\times\vec B$, of size $qvB\sin\theta$, at right angles to both the velocity and the field.
This pushes the free charges to the ends of a moving rod, giving a second and independent derivation of the motional emf.
The torque on a current loop in a field
A flat coil of $N$ turns and area $A$ carrying $I$ in a uniform field feels a torque $\tau = NIAB\sin\theta$, with $\theta$ between the field and the coil's normal. Its magnetic dipole moment is $\mu = NIA$.
It is the counter torque of a generator: draw current from a spinning coil and the field pushes back, and this says how hard.
A source with internal resistance
A source of emf $\varepsilon$ with internal resistance $r$ driving an external resistance $R$ carries current $I = \varepsilon/(R+r)$, and its terminal voltage is $\varepsilon - Ir$.
An induced emf appears in wire that has resistance of its own. Coil resistance in a generator and armature resistance in a motor are internal resistances, handled exactly as a battery's is.
Power in a resistor
A resistance $R$ carrying current $I$ with voltage $V$ across it turns electrical energy into heat at a rate $P = IV = I^{2}R = V^{2}/R$.
Every energy check in this section ends here. Whatever mechanical power goes into a generator has to come out of this formula, and if it does not, something is wrong upstream.
The right hand grip rule for the field of a current
Point the right thumb along a current and the fingers curl the way its field circles it. For a current running anticlockwise round a loop, the field inside the loop points out of the page towards you.
The second sentence is the whole of the direction work here. Lenz's law says which way the induced field inside the loop must point, and this rule turns that into a current going one way round rather than the other.
Try it yourself first (3 questions)
1§12.0 — a magnet held still inside a coil●●○○○
Nobody expects you to know this yet; watch your own first instinct and then watch it break. A strong bar magnet is pushed into a coil of 400 turns whose ends are joined through a sensitive galvanometer, and held there, motionless, for a full minute.
Given
a coil of 400 turns with its ends joined through a galvanometer
a strong bar magnet resting inside the coil
nothing is moving and there is no battery anywhere in the circuit
Find
(a) What does the galvanometer read during that minute?
Hint 1/4
The question is not how strong the magnet is. It is whether anything about the situation is different at one moment from the next.
Hint 2/4
A current needs something to drive it. With no source in the circuit, the only candidate is whatever the magnet is doing.
Hint 3/4
Here the magnet is strong, it is well inside the 400 turn coil, and it is completely stationary for the whole minute.
Hint 4/4
The needle sits on zero. A magnet that is not moving drives nothing, however strong it is.
Show solutionAsk what could drive the current
$$\text{no battery} \Rightarrow \text{no chemical source of emf}$$
bare wire and a meter, so anything pushing charge round must come from outside
$$\text{nothing in the arrangement depends on } t$$
magnet, coil and field pattern are all frozen, so nothing has a rate of change to offer
$$I = 0$$
with nothing driving it, a resistive loop carries no current
Answer $$\boxed{\,I = 0\ \text{for as long as nothing moves}\,}$$
Check
Independent check by changing one thing. Pull the magnet out and the needle kicks; push it back and it kicks the other way; stop and it returns to zero. A reading that appears only during motion cannot be a property of the magnet's strength.
2§12.0 — the field inside a solenoid, recalled●●○○○
Pure recall from the previous week, and it will be used within the hour. A long solenoid is wound with 1200 turns per metre and carries $2.50\ \mathrm{A}$.
Given
$n = 1200\ \mathrm{m^{-1}}$
$I = 2.50\ \mathrm{A}$
$\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}$
Find
(a) Find the field well inside the solenoid.
(b) State how the answer would change at a point on the axis but half way to the wall, at half the radius.
Hint 1/4
You are being asked for a number from a formula you already have, and then for what that formula does not contain.
Hint 2/4
Well inside a long solenoid, $B = \mu_0 n I$, directed along the axis.
Hint 3/4
Here $n = 1200\ \mathrm{m^{-1}}$ and $I = 2.50\ \mathrm{A}$, with $\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}$.
Hint 4/4
The field is $3.77\times10^{-3}\ \mathrm{T}$, and it is the same at half the radius because the formula contains no radius at all.
Show solutionSubstitute
$$B = \mu_0 n I = (4\pi\times10^{-7})(1200)(2.50)$$
n is already turns per metre; a total and a length would need that division first
$$B = 3.77\times10^{-3}\ \mathrm{T}$$
three significant figures, matching the least precise datum in the question
Answer $$\boxed{\,B = 3.77\times10^{-3}\ \mathrm{T}\ \text{everywhere well inside}\,}$$
Check
Size check against the ruler used all course: $3.77\times10^{-3}\ \mathrm{T}$ is about seventy five times the field of the Earth, the right order for an air cored coil.
3§12.0 — force on a wire and the heat it makes●●●○○
The last piece of recall, with a trap in the second half. A straight wire of length $0.400\ \mathrm{m}$ lies at right angles to a uniform field of $0.250\ \mathrm{T}$ and carries $3.00\ \mathrm{A}$ through a total circuit resistance of $4.00\ \Omega$.
Given
$L = 0.400\ \mathrm{m}$, at right angles to the field
$B = 0.250\ \mathrm{T}$
$I = 3.00\ \mathrm{A}$
total circuit resistance $4.00\ \Omega$
Find
(a) Find the force on the wire.
(b) Find the rate at which the circuit turns electrical energy into heat.
(c) The wire is clamped and cannot move. At what rate is the magnetic force doing work on it?
Hint 1/4
Three questions that look like one. The third asks about work, and work needs a displacement.
Hint 2/4
$F = BIL$ for a wire at right angles to the field, $P = I^{2}R$ for the heating, and $P = Fv$ for the rate at which a force does work.
Hint 3/4
Here $B = 0.250\ \mathrm{T}$, $I = 3.00\ \mathrm{A}$, $L = 0.400\ \mathrm{m}$, $R = 4.00\ \Omega$, and the wire is clamped so that $v = 0$.
Hint 4/4
The force is $0.300\ \mathrm{N}$, the heating is $36.0\ \mathrm{W}$, and the magnetic force does no work at all because the wire does not move.
Show solutionThe force and the heating
$$F = BIL = (0.250)(3.00)(0.400) = 0.300\ \mathrm{N}$$
the sine is one for a wire perpendicular to the field, the only case where it may be left out silently
Independent check by energy bookkeeping. The battery is the only source, and with the wire clamped there is no mechanical output, so all $36.0\ \mathrm{W}$ must appear as heat, which part (b) found by a different formula.
Notation
symbol
reads as
means
watch out
$\Phi_B$
phi B, or the magnetic flux
how much magnetic field passes through a surface, $\Phi_B = \int\vec B\cdot d\vec A$, in webers
it belongs to a surface and not to a point, so asking for the flux at a place is meaningless; and the subscript matters, since the same letter without it was the electric flux in the Gauss's law week
$\varepsilon$
epsilon, or the emf
the electromotive force driven round a circuit, in volts: the work per unit charge in going once round
not to be confused with $\varepsilon_0$, the permittivity of free space; and despite the name it is not a force
$N$
N, the number of turns
how many times the wire goes round, so the same flux is cut N times and the emf is N times larger
distinct from lower case $n$, turns per metre in the solenoid formula; 500 turns over $0.250\ \mathrm{m}$ means $n = 2000\ \mathrm{m^{-1}}$
$\hat n$
n hat, the normal
the unit vector chosen to stick out of one face of the loop, fixing which side counts as the positive side
you choose it, the problem does not; once chosen it fixes the positive sense of circulation and both stay fixed for the rest of the problem
$\frac{d\Phi_B}{dt}$
d phi B by dt, the rate of change of the flux
how fast the flux through the loop is changing, in webers per second, which is the same as volts
the flux itself never appears in Faraday's law, only its rate of change; a large steady flux gives exactly nothing
$\omega$
omega, the angular speed
the rate at which a coil turns, in radians per second, related to the rotation rate by $\omega = 2\pi f$
questions almost always give the rate in revolutions per second or minute, and the factor of $2\pi$ is the most commonly dropped item here
$\varepsilon_0$
epsilon nought, the
in the generator formulas only, the largest value the alternating emf reaches, $\varepsilon_0 = NBA\omega$
written the same way as the permittivity of free space, so here the symbol is used only inside a generator formula and the words peak emf are attached whenever it appears in prose
$\varepsilon_b$
epsilon b, the back emf
the emf a motor generates because it is spinning, which acts against the supply that is driving it
it grows with speed and is zero at switch on, which is why the starting current is the largest a motor ever draws
$\Phi_B^{\rm link} = N\Phi_B$
the flux linkage
the flux through one turn times the number of turns, whose rate of change is the emf of the whole coil
meaningful only when every turn encloses the same flux, which is what a tightly wound coil provides
$\vec E_{\rm ind}$
E induced
the electric field produced by a changing magnetic flux, in volts per metre
its field lines close on themselves and its circulation is not zero, so no potential exists and points carry no voltage
Conventions used here
Which way a loop faces, and where the angle in the flux formula is measured from
Every loop here is given a normal direction first, and the angle $\theta$ in $\Phi_B = BA\cos\theta$ is always between the field and that normal, never between the field and the plane; those differ by ninety degrees and swapping them turns a maximum into a zero. Normal and circulation sense are tied by the right hand, and neither may change part way through a problem.
What the minus sign in Faraday's law means in practice on this page
Every answer here comes in two movements. First the size, from the rate of change of the flux, with every quantity positive. Then the direction, from Lenz's law, with a viewpoint attached: not anticlockwise but anticlockwise seen from above. A signed emf is quoted only where a sign convention has been fixed and stated, because marks are lost carrying a minus sign into arithmetic that has no use for it.
How the axes and the page are set up for the induction diagrams
As in the earlier magnetic sections, $x$ runs right across the page, $y$ up it and $z$ out of it towards you. A field or current into the page is a cross, out of the page a dot. Clockwise and anticlockwise mean as seen by the reader unless another viewpoint is named.
Which constants and how many digits this page uses
Answers are quoted to three significant figures, with $\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}$ and $g = 9.80\ \mathrm{m/s^{2}}$. The field of the Earth is taken as about $5\times10^{-5}\ \mathrm{T}$ and used as a ruler for plausibility, never as data unless a question supplies it.
What ideal is allowed to mean in this section
An ideal transformer loses no power and both windings share the same flux. An ideal motor or generator has all its resistance lumped in the armature and no friction. Voltmeters draw no current, ammeters have no resistance, and rails and wires have none unless a value is given. Departures from ideal are stated as an efficiency and applied at the end.
Which earlier results survive here and which do not
Results for the field made by a current, from the earlier week on the sources of the magnetic field, are used freely, because every current here changes slowly enough that its field at any instant is the field a steady current of that size would make. What does not survive is the habit of assuming every electric field has a potential, and the last concept block is about where that breaks.
How an emf and an induced current are quoted in an answer
An emf is quoted as a magnitude in volts together with the sense of the current it drives round a named loop, seen from a named side, and a current the same way. A bare number for an induced current is an incomplete answer here.
12.1How much field goes through a loop
Flux is field times the area facing it, and it is the one quantity a circuit actually responds to.
Everything so far has asked what a field does to a charge or a wire placed in it. The hook needs something else first: a number saying how much field a particular loop has caught.
Solvable with what we have
Find the field a current makes at a point, and the field inside a solenoid.
Find the force on a wire or a moving charge in a field you were handed.
Find the torque on a coil that is already carrying a current.
Not solvable yet
Say where the current in the hook came from, when no battery is in the loop.
Say why the meter reads only while the magnet is moving.
Say why the reading reverses when the motion reverses.
Treat the magnet as a source: it has a strong field, the loop is a conductor, so the field should push the electrons round it and a stronger magnet should push harder.
Why it fails
The bench says no. Hold the magnet still, as close as you like, and the reading is exactly zero. Move it, and a current appears whose size depends on how fast you move rather than on how strong the magnet is.
DefinitionDefinition 12.1: magnetic flux
Conditions
the surface is any surface bounded by the loop, and for a flat loop in a uniform field the flat one is obvious
$\theta$ is the angle between $\vec B$ and the normal $\hat n$ of the surface, and a normal direction must be chosen before the calculation starts
when the field varies across the surface, or the surface is not flat, the integral form is the one that applies
the unit is the weber, and $1\ \mathrm{Wb} = 1\ \mathrm{T\,m^{2}}$
Look at the loop from the direction the field comes from and see how big it looks. That apparent area times the field strength is the flux. Face on, the loop looks its full size and catches everything; edge on it looks like a line, has no apparent area, and catches nothing in the strongest field you can buy.
The same loop in the same $\textcolor{#1f6feb}{\text{field}}$, twice. On the left the $\textcolor{#d1690a}{\text{loop}}$ faces the field squarely and its $\textcolor{#8250df}{\text{shadow}}$ is its full height; on the right it is tilted, the shadow is shorter by $\cos\theta$, and the flux is smaller by that factor. Both loops are drawn edge on, so each appears as one thick line.
Looks like this, but is not
A bigger loop always catches more flux, and a stronger field always gives more flux. Both sound like arithmetic rather than physics, and both are written every year.
Neither survives a tilt. A square metre of loop edge on in one tesla has zero flux, while a postage stamp face on in the field of the Earth has a small but non zero flux. Flux belongs to the pair, not to the field or the loop, and the orientation can kill it outright.
angle between field and normal
flux through one turn
fraction of the maximum
$0^{\circ}$
$8.00\times10^{-3}\ \mathrm{Wb}$
1.000, the loop is face on
$30^{\circ}$
$6.93\times10^{-3}\ \mathrm{Wb}$
0.866, a large tilt costs little
$45^{\circ}$
$5.66\times10^{-3}\ \mathrm{Wb}$
0.707, half way in angle is not half way in flux
$60^{\circ}$
$4.00\times10^{-3}\ \mathrm{Wb}$
0.500, two thirds of the way in angle costs half the flux
$90^{\circ}$
$0$
0.000, edge on, and the field strength no longer matters
The column falls slowly at first and then fast. A thirty degree tilt loses only thirteen per cent of the flux; the last thirty degrees loses everything left. That shape matters later: a generator coil spends most of its cycle near the flat part, where the emf is small, and races through the steep part, where it peaks.
Flux through a small coil tilted inside a solenoid
A of 200 turns and radius $2.00\ \mathrm{cm}$ is held well inside a long solenoid wound with 1500 turns per metre and carrying $3.00\ \mathrm{A}$. The axis of the small coil is tilted at $30.0^{\circ}$ to the solenoid axis. Find the field inside the solenoid, the flux through one turn of the search coil, and its flux linkage.
Bracket the answer instead of repeating the arithmetic. The flux must lie between zero, for an edge on coil, and $BA = 7.11\times10^{-6}\ \mathrm{Wb}$, for a face on one. The answer is $87\%$ of that ceiling, and the sine by mistake would have given exactly half, which no thirty degree tilt can produce.
Three lines and one trigonometric factor. The difficulty is deciding that the angle handed to you is already the one the formula wants.
Notice the division of labour, which repeats for the rest of the section: the solenoid formula supplies a field, the flux formula supplies a number for the loop.
Flux through a rectangle lying beside a long straight wire
A long straight wire carries $15.0\ \mathrm{A}$. A rectangular loop of length $12.0\ \mathrm{cm}$ lies in the same plane as the wire, its long sides parallel to it, its near side $2.00\ \mathrm{cm}$ away and its far side $7.00\ \mathrm{cm}$ away. Find the flux through the loop.
Given
$I = 15.0\ \mathrm{A}$ in a long straight wire
loop length parallel to the wire, $L = 0.120\ \mathrm{m}$
near side at $a = 0.0200\ \mathrm{m}$, far side at $b = 0.0700\ \mathrm{m}$
the loop and the wire are coplanar
$\mu_0/2\pi = 2\times10^{-7}\ \mathrm{T\,m/A}$
Find
the magnetic flux through the loop
SolutionNotice that BA is not available here
$$B(r) = \frac{\mu_0 I}{2\pi r}$$
the field differs at every distance, so there is no single B to multiply by the area
$$dA = L\,dr$$
strips parallel to the wire, because B is constant along such a strip
Do the integral
$$\Phi_B = \int_a^b \frac{\mu_0 I}{2\pi r}L\,dr = \frac{\mu_0 I L}{2\pi}\int_a^b\frac{dr}{r}$$
everything but the one over r comes out, since none of it depends on r
$$\Phi_B = \frac{\mu_0 I L}{2\pi}\ln\!\frac{b}{a} = (2\times10^{-7})(15.0)(0.120)\ln\!\frac{7.00}{2.00}$$
the logarithm takes a ratio, so those distances may stay in centimetres
Independent bracketing with no integral. The near edge sits in $1.50\times10^{-4}\ \mathrm{T}$ and the far edge in $4.29\times10^{-5}\ \mathrm{T}$; times the area $6.00\times10^{-3}\ \mathrm{m^{2}}$ that brackets the flux between $2.57\times10^{-7}$ and $9.00\times10^{-7}\ \mathrm{Wb}$. The answer lies inside, nearer the lower end, because most of the area is far out.
One integral, and it is the only integral in this concept. Every later flux on this page is a product.
The logarithm is worth remembering as a shape. Doubling the far distance adds only what the logarithm of two is worth, which is why a rectangle a metre wide beside a mains cable picks up almost nothing.
Checkpoint
§12.1 — the angle that is easy to take from the wrong place●●○○○
Thirty seconds, no calculator. A flat circular coil of area $A$ sits in a uniform field $B$, tilted so that the plane of the coil makes $30^{\circ}$ with the field direction.
Given
a flat coil of area $A$ in a uniform field $B$
the plane of the coil is at $30^{\circ}$ to the field direction
Find
(a) What is the flux through the coil?
Hint 1/4
The number you have been given is not the number the formula wants. Find the relation between them before anything else.
Hint 2/4
$\Phi_B = BA\cos\theta$ where $\theta$ is measured from the normal to the loop, and the normal is at right angles to the plane.
Hint 3/4
Here the plane is at $30^{\circ}$ to the field, so the normal is at $90^{\circ}-30^{\circ} = 60^{\circ}$ to it.
Hint 4/4
$\Phi_B = BA\cos 60^{\circ} = 0.500\,BA$.
Show solutionConvert to the angle the formula uses
the normal is perpendicular to the plane, so the two angles add to a right angle
$$\Phi_B = BA\cos 60^{\circ} = 0.500\,BA$$
safe now that symbol and quantity mean the same thing
Answer $$\boxed{\,\Phi_B = 0.500\,BA\,}$$
Check
Check the two ends of the range. A plane angle of zero, the loop lying along the field, gives $BA\cos 90^{\circ} = 0$, right for an edge on loop; a plane angle of ninety gives $BA$, the maximum. The alternative reading of the angle fails both limits.
⚠ Taking the angle from the plane of the loop instead of from its normal
questions describe orientations however sounds natural in English, and a coil's plane is easier to picture than a normal sticking out of it
wrong$$\Phi_B = BA\cos(\text{angle to the plane})$$
right$$\Phi_B = BA\cos(\text{angle to the normal})$$
⚠ Multiplying by the number of turns inside the flux
linkage and flux are one letter apart and both quoted in webers, so the N gets absorbed early and applied again when the emf is computed
⚠ Using the area of the solenoid rather than of the coil inside it
two areas are on the page and the larger one is described first, so it is the one still in mind when the formula is written
wrong$$\Phi_B = B\,\pi R_{\rm solenoid}^{2}$$
right$$\Phi_B = B\,\pi r_{\rm coil}^{2}$$
⚠ Writing BA when the field varies across the loop
the product is quick and is the version that was practised, so it gets used before checking that the field is uniform over the surface
wrong$$\Phi_B = B(a)\cdot L(b-a)$$
right$$\Phi_B = \int_a^b B(r)L\,dr$$
12.2Faraday's law, and what the minus sign is for
The emf round a loop is the rate at which its flux linkage changes, and the current always fights that change.
We now have a number for a loop, and the bench says the meter reads only while that number is moving. The law is the sentence joining those two facts, and the hook falls out of it in one line.
TheoremLaw 12.2: Faraday's law of induction, with Lenz's law in the sign
Conditions
N turns, each enclosing the same flux, which is what a tightly wound coil provides
the flux may change because the field, the area or the orientation changes, and the law does not care which
the minus sign is with respect to a chosen normal and the circulation sense that the right hand ties to it
for an average emf over a finite interval, replace the derivative by $\Delta\Phi_B/\Delta t$
Count how many webers thread the coil, watch how fast that count changes, multiply by the number of turns, and you have volts. The minus sign says the loop is a bad loser: whatever you did to its flux, the current it makes tries to undo it. Increase the flux and it pushes back; decrease it and it tries to hold on.
Proof
The size of the law is experimental and the bench gives it. The sign can be derived, and it is worth doing because the sign is where the marks go.
Suppose the sign were the other way, so that the induced current helped the change. Push a magnet a little towards a loop and the induced current would pull the magnet in further.
That pull increases the flux faster, which increases the current, which increases the pull. Nothing limits it: the magnet accelerates in on its own while the loop heats up.
Energy would appear from nowhere, at a growing rate, out of a lump of iron and a ring of copper. So the assumption is wrong and the induced effect must oppose the change.
The working rule takes two movements. Ask which way the flux through the loop points now and whether it is growing or shrinking. Then ask which way the induced field inside the loop must point to resist that: opposite if growing, along it if shrinking.
Then use the grip rule backwards: an anticlockwise current makes a field out of the page inside the loop, a clockwise one makes a field into the page. That turns the required field direction into a current direction.
The whole of the direction work, in two pictures. The $\textcolor{#1f6feb}{\text{field}}$ points out of the page through the $\textcolor{#d1690a}{\text{loop}}$ in both panels; the only difference is whether it is growing or shrinking, and the $\textcolor{#8250df}{\text{induced current}}$ reverses accordingly. Nothing depends on the field being out of the page rather than into it.
Looks like this, but is not
The induced current opposes the field. The short version everybody remembers, right about half the time, which is worse than always being wrong.
Put a loop in a steady field of two tesla: if the rule were about the field there would be a large current, and there is none. Now let that field fall. The induced current makes flux in the same direction as the field that is disappearing, so it runs with the field. What is opposed is never the field, only the change.
time taken
average emf
average current
total charge moved
$1.00\ \mathrm{s}$
$0.400\ \mathrm{V}$
$0.0400\ \mathrm{A}$
$0.0400\ \mathrm{C}$
$0.100\ \mathrm{s}$
$4.00\ \mathrm{V}$
$0.400\ \mathrm{A}$
$0.0400\ \mathrm{C}$
$0.0100\ \mathrm{s}$
$40.0\ \mathrm{V}$
$4.00\ \mathrm{A}$
$0.0400\ \mathrm{C}$
$0.00100\ \mathrm{s}$
$400\ \mathrm{V}$
$40.0\ \mathrm{A}$
$0.0400\ \mathrm{C}$
Three columns change by a factor of a thousand down the table and the fourth does not move. That last column is why a flux change measures a field well: the charge depends on the size of the change and the resistance, not on how fast you made it, so a hand movement nobody can time still gives an accurate answer.
A coil pulled out of a field in a fifth of a second
A square coil of 150 turns and side $5.00\ \mathrm{cm}$ lies with its plane perpendicular to a uniform field of $0.400\ \mathrm{T}$, which points into the page. The coil is pulled completely out of the field region to the right in $0.200\ \mathrm{s}$. Coil and leads together have resistance $8.00\ \Omega$. Find the average emf, the average current, and the total charge past any point of the wire.
Given
$N = 150$ turns, side $0.0500\ \mathrm{m}$, so $A = 2.50\times10^{-3}\ \mathrm{m^{2}}$
$B = 0.400\ \mathrm{T}$, perpendicular to the coil, into the page
$\Delta t = 0.200\ \mathrm{s}$ to leave the field completely
$R = 8.00\ \Omega$
Find
the average emf, the average current, the total charge, and the direction of the current
SolutionThe flux before and after
$$\Phi_i = BA = (0.400)(2.50\times10^{-3}) = 1.00\times10^{-3}\ \mathrm{Wb}$$
the cosine is one, and this is one turn, so the 150 is deliberately not used yet
Independent check on the charge by changing the experiment. Pull the coil out over $2.00\ \mathrm{s}$ instead: the emf and current fall tenfold but the charge is still $1.88\times10^{-2}\ \mathrm{C}$. A quantity that survives a tenfold change of timing was computed correctly.
Four short stages, only the last needing a right hand. The time appeared in two of them and cancelled out of the third.
Keep the order: flux, change in flux, divide by the time, multiply by the turns, and only then argue the direction. Mixing direction work into the arithmetic is where the minus sign starts multiplying things it should not.
A coil that never moves, in a field that is being ramped up
A circular coil of 40 turns and radius $3.00\ \mathrm{cm}$ lies flat with its plane perpendicular to a field pointing out of the page as drawn. The field is increased steadily from $0.100\ \mathrm{T}$ to $0.700\ \mathrm{T}$ in $0.0500\ \mathrm{s}$. The coil has resistance $2.50\ \Omega$. Find the emf, the current and its direction, and the charge that flows.
the same charge formula, needing the change in B and not its final value
Answer $$\boxed{\,\varepsilon = 1.36\ \mathrm{V},\quad I = 0.543\ \mathrm{A}\ \text{clockwise from above},\quad q = 2.71\times10^{-2}\ \mathrm{C}\,}$$
Check
Cross check through a formula not used to get the current. The charge came from $N\Delta\Phi_B/R$, and multiplying the current by the duration gives $2.71\times10^{-2}\ \mathrm{C}$, the same number by a route sharing no arithmetic.
Nothing moved in this problem, and that is the point: the coil is bolted down and the emf is as real as if it had been yanked out.
There the field was fixed and the area changed; here the area is fixed and the field changes. Faraday's law does not distinguish them: find which factor of $BA\cos\theta$ is moving and differentiate that one.
Checkpoint
§12.2 — which way, when the field is dying away●●○○○
Thirty seconds, no calculator. A single circular loop of copper lies flat on the page, with a field pointing out of the page through it that is being reduced steadily towards zero.
Given
a closed loop of copper lying in the plane of the page
the field through it points out of the page
the field is decreasing steadily
Find
(a) Which way does the induced current run, as seen by the reader?
Hint 1/4
Ask what the loop is losing, not what it is sitting in.
Hint 2/4
The induced current makes a field that opposes the change. Flux falling means the induced field must point the same way as the flux that is disappearing.
Hint 3/4
Here the existing flux points out of the page and it is falling, so the induced field inside the loop must also point out of the page.
Hint 4/4
A field out of the page inside a loop needs an anticlockwise current, by the grip rule.
Show solutionName the change, then oppose the change
$$\frac{d\Phi_B}{dt} < 0\ \text{with}\ \vec B\ \text{out of the page}$$
writing the sign explicitly stops the argument drifting into a claim about the field itself
$$\vec B_{\rm induced}\ \text{out of the page inside the loop}$$
to resist a loss you supply more of what is going, the whole content of the minus sign
$$\Rightarrow\ I\ \text{anticlockwise as the reader sees it}$$
grip rule read backwards: fingers curling anticlockwise put the thumb out of the page
Answer $$\boxed{\,I\ \text{anticlockwise, seen by the reader}\,}$$
Check
Independent check by energy. An anticlockwise current makes the loop a small magnet whose north face points at the reader, the face the vanishing field pointed towards, so you must do work to remove the source. Clockwise would have pushed it away for free while the loop heated up.
⚠ Opposing the field instead of opposing the change
the slogan is four words shorter and works whenever the flux is increasing, which is most textbook figures
A conductor of length L moving at speed v across a field B carries an emf BLv, and the price of the current is drag.
So far the flux changed because a field grew or a coil was yanked away. Now let the area change instead, in the simplest arrangement there is, and a mechanical price tag appears.
TheoremResult 12.3: motional emf and the drag that comes with it
Conditions
the rod, its velocity and the field are mutually perpendicular; otherwise only the perpendicular components count
the rod slides on rails completing a circuit of total resistance R, the rails having no resistance unless a value is given
the field is uniform over the whole circuit and steady in time
the drag expression assumes the rod's own resistance is already included in R
$$\boxed{\,\varepsilon = BLv\,,\qquad I = \frac{BLv}{R}\,,\qquad F_{\rm drag} = \frac{B^{2}L^{2}v}{R}\ \text{against the motion}\,}$$
Sweeping a rod of length L sideways at speed v wipes out an area Lv every second, and each square metre carries B webers, so the flux changes at BLv webers per second, which is BLv volts. The force term says it from the other side: the faster you go the more current you make, and the more current, the harder the field pulls back.
Proof
Two independent routes give the same answer, worth seeing once because in this section they are the only two mechanisms there are.
By the flux rule. Let the rod sit a distance $x$ from the closed end. The circuit encloses $Lx$, so $\Phi_B = BLx$, and only $x$ depends on time, giving $d\Phi_B/dt = BLv$.
By the force on the carriers, with no flux at all. Every free charge in the rod is carried sideways at $v$ through the field, so each feels $qvB$ along the rod, pushing charge to one end until the electric field it builds up stops the next charge arriving.
Balance those: $qE = qvB$, so $E = vB$ inside the rod. That field is uniform along the rod, so the ends differ by $EL = BLv$, which is what the flux rule gave.
Now the drag. The current $I = BLv/R$ runs along a rod of length $L$ in the field, and $F = BIL$ gives $F = B^{2}L^{2}v/R$. By Lenz it can only oppose the motion, since a force helping it would be the runaway of the previous concept.
Finally the energy account, the check to run on every problem of this type. Mechanical power in is $Fv = B^{2}L^{2}v^{2}/R$; electrical power out is $I^{2}R = B^{2}L^{2}v^{2}/R$. Equal, with nothing left over.
The same physics twice, once with a hand supplying the energy and once with gravity. The $\textcolor{#1f6feb}{\text{field}}$ goes into the page, the $\textcolor{#d1690a}{\text{rod and rails}}$ form the circuit, and the $\textcolor{#8250df}{\text{induced current}}$ and its drag share a colour because they are the same event. The drag points against the velocity in both panels, and it has to.
Looks like this, but is not
If no current flows, there is no emf. An open circuit, a rod with nothing joined to its ends, nothing to measure.
Put a voltmeter across the ends of that isolated rod while it moves and it reads $BLv$. The magnetic force has already pushed charge to the ends and it stays there as long as the motion lasts. An emf is a push per unit charge, and a push exists whether or not anything is free to move under it.
speed
emf
current
force you must apply
power you must supply
$1.00\ \mathrm{m/s}$
$0.228\ \mathrm{V}$
$0.0379\ \mathrm{A}$
$8.63\times10^{-3}\ \mathrm{N}$
$8.63\times10^{-3}\ \mathrm{W}$
$2.00\ \mathrm{m/s}$
$0.455\ \mathrm{V}$
$0.0758\ \mathrm{A}$
$1.73\times10^{-2}\ \mathrm{N}$
$3.45\times10^{-2}\ \mathrm{W}$
$4.00\ \mathrm{m/s}$
$0.910\ \mathrm{V}$
$0.152\ \mathrm{A}$
$3.45\times10^{-2}\ \mathrm{N}$
$0.138\ \mathrm{W}$
$8.00\ \mathrm{m/s}$
$1.82\ \mathrm{V}$
$0.303\ \mathrm{A}$
$6.90\times10^{-2}\ \mathrm{N}$
$0.552\ \mathrm{W}$
The first four columns double when the speed doubles, and the last one quadruples. The force grows with speed and the power with its square, exactly as air resistance does, so a conductor moving in a field behaves like something moving through treacle rather than something with friction. It is also why the falling rod reaches a terminal speed at all.
A rod pushed along rails: emf, current, force and the energy account
A metal rod of length $0.350\ \mathrm{m}$ slides without friction along horizontal rails in a uniform field of $0.650\ \mathrm{T}$ into the page, perpendicular to the plane of the rails. The rails are closed at one end by a $6.00\ \Omega$ resistor and have no resistance of their own. The rod is kept at a steady $2.40\ \mathrm{m/s}$. Find the emf, the current, the applied force, and check the mechanical power against the electrical power.
the rate the resistor heats, with no reference to the mechanics
Answer $$\boxed{\,\varepsilon = 0.546\ \mathrm{V},\quad I = 0.0910\ \mathrm{A},\quad F = 2.07\times10^{-2}\ \mathrm{N},\quad P = 4.97\times10^{-2}\ \mathrm{W}\,}$$
Check
The two powers are the verification, reached along separate paths, one from force and velocity and the other from current and resistance, with no line shared. Size check too: fifty milliwatts keeps an indicator lamp dim and two hundredths of a newton is the weight of two grams.
Four short lines. The energy check was free, because both powers were already available from numbers computed for other reasons.
Run that check on every problem here. It costs one line and catches the errors that matter, since a wrong current or force will almost never give two matching powers.
A rod released on vertical rails: the speed at which it stops accelerating
A rod of mass $0.0500\ \mathrm{kg}$ and length $0.400\ \mathrm{m}$ slides without friction down vertical rails closed at the top by a $1.50\ \Omega$ resistor, in a uniform field of $1.20\ \mathrm{T}$ perpendicular to the plane of the rails. It is released from rest. Find the terminal speed, the current there, and confirm the energy account. Take $g = 9.80\ \mathrm{m/s^{2}}$.
into the resistor: the rod is a generator fuelled by its own height
Answer $$\boxed{\,v_t = 3.19\ \mathrm{m/s},\quad I = 1.02\ \mathrm{A},\quad P = 1.56\ \mathrm{W}\,}$$
Check
Independent check on the force balance rather than the power. At the answer speed the drag is $BIL = 0.490\ \mathrm{N}$ and the weight is $mg = 0.490\ \mathrm{N}$, from different formulas and different data. That is what terminal speed means.
One equation set up and rearranged, then two substitutions. The difficulty is entirely in the first line.
Notice the shape, $v_t = mgR/(B^{2}L^{2})$. Shorting the rails makes R almost zero and the rod hangs in the field. That is a magnetic brake, the next concept with the geometry changed.
Checkpoint
§12.3 — what happens to the drag when the resistance is halved●●●○○
Thirty seconds, no calculator. A rod is pulled along rails at a fixed speed in a fixed field. Someone swaps the resistor for one of half the resistance and keeps the speed exactly the same.
Given
the same rod, the same field and the same steady speed
the external resistance is halved
Find
(a) What happens to the force needed to keep the rod moving?
Hint 1/4
Two things did not change: the emf and the field. Work out which quantity in the chain the resistance actually controls.
Hint 2/4
$\varepsilon = BLv$ is untouched by the resistance, but $I = \varepsilon/R$ and $F = BIL$ both are.
Hint 3/4
Here $B$, $L$ and $v$ are all unchanged while $R$ becomes $R/2$, so the emf is the same and the current is twice what it was.
Hint 4/4
Twice the current through the same rod in the same field is twice the force.
Show solutionWhich link the resistance sits in
$$\varepsilon = BLv \quad \text{unchanged}$$
the emf comes from the motion and would be the same with the circuit open
the only place the resistance appears, so the only quantity that changes here
$$F = BIL \to B(2I)L = 2F$$
the force is proportional to the current and nothing else in it moved
Answer $$\boxed{\,F\ \text{doubles}\,}$$
Check
Independent check through the power. At the same speed the mechanical power is $Fv$, so doubling the force doubles it; electrically the emf is unchanged and the resistance halved, so $\varepsilon^{2}/R$ also doubles. No other option gives two sides that agree.
⚠ Putting the resistance into the emf
everything else in the problem depends on R, so it gets attached to the one quantity that does not
wrong$$\varepsilon = \frac{BLv}{R}$$
right$$\varepsilon = BLv,\qquad I = \frac{BLv}{R}$$
⚠ Using the length of the rails instead of the length of the rod
both are lengths in the same picture and the rails are the longer and more prominent of the two
wrong$$\varepsilon = B\,x\,v$$
right$$\varepsilon = B\,L\,v$$
⚠ Setting the applied force equal to zero because the speed is constant
constant speed is correctly linked to zero net force, and the word net then gets dropped
⚠ Equating the weight to a force per unit length or to a power
three quantities have to balance in terminal speed problems, and the units of what is being balanced get lost when the algebra comes before the sentence
wrong$$mg = \frac{B^{2}L^{2}v^{2}}{R}$$
right$$mg = \frac{B^{2}L^{2}v}{R}$$
12.4The rotating coil: where mains electricity comes from
Spinning a coil in a fixed field changes its flux smoothly and produces a sinusoidal emf of peak value NBA omega.
The sliding rod runs out of rail. An emf that lasts needs a flux that changes forever without anything running away, and rotation is the only motion that repeats. That one change turns the last concept into a power station.
TheoremResult 12.4: the emf of a coil rotating in a uniform field
Conditions
a flat coil of N turns and area A turning at constant $\omega$ about an axis perpendicular to a uniform field
time is measured from an instant when the coil is face on, so the flux starts at its maximum
$\omega$ is in radians per second, and a rotation rate f in revolutions per second means $\omega = 2\pi f$
the coil's own resistance is treated as an internal resistance in series with the load
The flux swings between plus and minus its largest value once per turn, and the emf is how fast it is swinging. So the emf is zero when the coil is face on, where the flux is at a maximum and momentarily flat, and largest when the coil is edge on, where the flux is racing through zero. The peak voltage is turns times field times area times angular speed.
Proof
Start from the flux. The coil is flat with area $A$, the field is uniform, and the angle from the coil's normal is $\omega t$ if it turns steadily from the face on position. So $\Phi_B = BA\cos\omega t$.
Nothing else in the flux depends on time: the field and area are fixed and only the angle turns, so all of it sits in the cosine.
Differentiate: $d\Phi_B/dt = -BA\omega\sin\omega t$, where the $\omega$ comes out by the chain rule and is the factor that people lose.
Faraday's law with N turns gives $\varepsilon = NBA\omega\sin\omega t$. The two minus signs cancelled, which is luck of this starting time and not a general rule.
The sine reaches one at most, so the peak emf is $\varepsilon_0 = NBA\omega$, and the emf alternates because the sine does: for half of every turn the current runs one way round and for the other half the other way.
It must alternate. The face that pointed along the field half a turn ago now points against it, so the rotation that was increasing the flux is now decreasing it, and by Lenz's law the current reverses with it.
One full turn of the coil. The $\textcolor{#1f6feb}{\text{flux}}$ swings as a cosine while the $\textcolor{#8250df}{\text{emf}}$ follows a sine, and the dashed line marks the quarter turn where the flux passes through zero and the emf peaks. The two are drawn on the same axes because the second is the slope of the first, not because they share units.
Looks like this, but is not
The emf is largest when the flux is largest, because that is when the coil has the most field through it. It is the most common wrong sentence in this material.
Look at the figure. At maximum flux the curve is flat: the coil is face on, and turning it either way changes the flux by almost nothing, so the emf there is zero. It peaks a quarter turn later, with the coil edge on and the flux racing through zero. Large and changing fast are different properties, and only the second appears in Faraday's law.
rotation rate
angular speed
peak emf
what that would run
$10.0\ \mathrm{rev/s}$
$62.8\ \mathrm{rad/s}$
$90.5\ \mathrm{V}$
a small tool, if you could turn it that fast by hand
$25.0\ \mathrm{rev/s}$
$157\ \mathrm{rad/s}$
$226\ \mathrm{V}$
close to a domestic supply, at half the usual frequency
$50.0\ \mathrm{rev/s}$
$314\ \mathrm{rad/s}$
$452\ \mathrm{V}$
the European mains frequency, and a peak well above its quoted value
$60.0\ \mathrm{rev/s}$
$377\ \mathrm{rad/s}$
$543\ \mathrm{V}$
the frequency used across the Americas
The last column is proportional to the first, with no curvature: doubling the speed doubles the voltage, because $\omega$ appears once. That is why every generator in a grid is locked to the same rotation rate, and why a turbine that slows delivers power at the wrong voltage rather than merely less of it.
Peak emf of a 250 turn coil, and its value an instant later
A flat coil of 250 turns, each of area $1.20\times10^{-2}\ \mathrm{m^{2}}$, rotates at $60.0$ revolutions per second about an axis perpendicular to a uniform field of $0.480\ \mathrm{T}$. Find the angular speed, the peak emf, and the emf $1/480\ \mathrm{s}$ after the coil is face on.
Independent check on the peak by a route with no sine in it. Over a quarter cycle the linkage falls from $1.44\ \mathrm{Wb}$ to zero in $1/240\ \mathrm{s}$, an average emf of $346\ \mathrm{V}$; for a sinusoid that average is $2/\pi$ of the peak, and $(543)(2/\pi) = 346\ \mathrm{V}$.
Three lines, one a unit conversion. Drop it and the answer is $2\pi$ times too small while still looking like a sensible voltage, which is why it gets its own line.
Every generator question here is this example with one of the four factors unknown. Asked for the turns or the speed, it is the same equation rearranged, with the conversion still on the first line.
A generator with a lamp on it: the torque you have to supply
A generator coil of 120 turns, each of area $2.50\times10^{-2}\ \mathrm{m^{2}}$, turns at $50.0$ revolutions per second in a uniform field of $0.350\ \mathrm{T}$. The coil has resistance $3.00\ \Omega$ and feeds a load of $150\ \Omega$. Find the peak emf, the peak current, and the torque needed at the instant the current peaks. Confirm it with an energy argument.
Independent check by power, using nothing from the torque line. At the peak instant the electrical power out is $\varepsilon_0 I_0 = 711\ \mathrm{W}$ and the mechanical power in is $\tau\omega = 711\ \mathrm{W}$. They could not match with the wrong angle or a dropped coil resistance.
Five lines, and the only new physics is the earlier torque formula applied to a current the generator made itself.
This is the sentence that makes a generator make sense: taking current out makes it harder to turn, by exactly the amount that pays for the current. Disconnect the load and the effort falls away, which is why a bicycle dynamo is easy to pedal with the lamp off.
Checkpoint
§12.4 — where in the turn the output vanishes●●●○○
Thirty seconds, no calculator. A flat coil spins steadily in a uniform field and its output is watched on an oscilloscope.
Given
a flat coil rotating at constant angular speed
a uniform field perpendicular to the axis of rotation
Find
(a) At which orientation of the coil is the output emf momentarily zero?
Hint 1/4
The emf is not the flux, it is how fast the flux is changing. Ask where the flux is momentarily not changing at all.
Hint 2/4
$\Phi_B = BA\cos\omega t$, so the emf vanishes wherever the cosine is at a turning point, that is where the flux is largest or smallest.
Hint 3/4
Here the flux is largest when the coil faces the field squarely, that is when the plane of the coil is perpendicular to the field direction.
Hint 4/4
That face on orientation is the one where the output momentarily vanishes.
Show solutionSet the emf to zero and read off the orientation
$$\varepsilon = NBA\omega\sin\omega t = 0 \Rightarrow \sin\omega t = 0$$
none of N, B, A or omega is zero in a working generator, so only the sine can vanish
$$\omega t = 0,\ \pi \Rightarrow \cos\omega t = \pm 1$$
the same instants at which the cosine, and therefore the flux, takes its extreme values
$$\Phi_B = \pm BA \Rightarrow \text{coil face on to the field}$$
full flux BA means the normal lies along the field, the plane perpendicular to it
Answer $$\boxed{\,\varepsilon = 0\ \text{when the coil is face on, twice per revolution}\,}$$
Check
Independent check by counting. A sinusoid crosses zero twice per cycle, and the coil passes the face on orientation twice per revolution and the edge on twice as well; the flux argument decides which pair. The alternative would put four zeros in a cycle, which no sinusoid has.
⚠ Leaving out the factor of two pi
machines are specified in revolutions per second or minute, and the formula wants radians per second
wrong$$\varepsilon_0 = NBA f$$
right$$\varepsilon_0 = NBA(2\pi f)$$
⚠ Putting the peak emf where the flux peaks
both are called maxima and the flux was computed first, so it is the one on the page when the sentence gets written
the coil's own resistance is quoted separately and reads like a property of the machine rather than part of the circuit
wrong$$I_0 = \frac{\varepsilon_0}{R}$$
right$$I_0 = \frac{\varepsilon_0}{R+r}$$
⚠ Using the coil's own area twice, once as area and once as turns
a coil of 120 turns each of area A invites reading the total area as 120A, and then N appears again in the formula
wrong$$\varepsilon_0 = B(NA)\omega N$$
right$$\varepsilon_0 = NBA\omega$$
12.5Motors that fight back, and metal that will not move freely
Anything turning or sliding in a field makes an emf opposing whatever drives it, and that opposition sets the current.
A generator turns motion into current and resists being turned. Run the machine the other way, feeding current in and taking motion out, and the resistance does not disappear: it becomes a voltage fighting the supply.
RuleResult 12.5: back emf, counter torque and eddy currents
Conditions
the armature is a source of emf $\varepsilon_b$ in series with its own resistance R, driven by a supply V
$\varepsilon_b$ is proportional to the rotation speed, since it is the generator emf of the same coil in the same field
the mechanical power expression assumes no friction, so every watt not lost as heat leaves as useful work
the eddy current statement is qualitative: the current pattern in a slab depends on its shape, and only the direction of the force is fixed
$$\boxed{\,I = \frac{V-\varepsilon_b}{R}\,,\qquad P_{\rm mech} = \varepsilon_b I\,,\qquad P_{\rm supply} = VI = \varepsilon_b I + I^{2}R\,}$$
A motor is a generator with the argument running backwards. The faster it spins, the more voltage it makes against the supply, and the less is left to drive current through the winding. At full speed that difference is small and the motor sips current; at switch on it makes no voltage at all and the supply sees bare copper.
Proof
There is nothing new in this concept, only a change of viewpoint, so the argument is short.
An armature spinning in a field is a coil whose flux is changing, so it generates an emf. That is the generator result, and it does not care what is turning the coil.
In a motor the supply is turning the coil. Faraday's law is indifferent and the emf appears anyway, and Lenz's law fixes its direction: it opposes the rotation producing it, and the only way to oppose a rotation driven by a supply is to oppose that supply.
Now go round the circuit. The supply raises the potential by V, the winding drops it by IR, and the back emf drops it by $\varepsilon_b$. Setting the total to zero gives $I = (V-\varepsilon_b)/R$.
Multiply through by I: $VI = I^{2}R + \varepsilon_b I$. The supply pays the left side; the first term on the right is heat in the winding, so the second must be the mechanical output.
Eddy currents are the same physics on a solid conductor. A plate entering a field region has a changing flux through any loop drawn in it, so currents circulate, lie in the field, feel a force, and by Lenz that force opposes the motion. The lost kinetic energy appears as heat spread through the metal.
Why a sheet of metal will not fall freely past a magnet. As the $\textcolor{#d1690a}{\text{plate}}$ moves into the $\textcolor{#1f6feb}{\text{field region}}$, the flux through any loop in the metal grows, so an $\textcolor{#8250df}{\text{eddy current}}$ circulates. Its right hand side lies in the field and feels a force, its left hand side does not, and the imbalance drags on the whole plate. The metal need not be magnetic.
Looks like this, but is not
A motor draws the most current when working hardest, at full speed under full load. More work out, more current in: it sounds like common sense.
The largest current a motor draws is at switch on, when it is doing no work at all. The speed is zero, so the back emf is zero and the supply sees only the winding resistance, small on purpose. That is why the lights dim when a fridge compressor starts and not while it runs, and why a jammed drill is in danger of burning out.
speed
back emf
current drawn
mechanical power out
heat in the winding
stalled, or the first instant
$0\ \mathrm{V}$
$100\ \mathrm{A}$
$0\ \mathrm{W}$
$12000\ \mathrm{W}$
a quarter of full speed
$28.8\ \mathrm{V}$
$76.0\ \mathrm{A}$
$2190\ \mathrm{W}$
$6930\ \mathrm{W}$
half of full speed
$57.6\ \mathrm{V}$
$52.0\ \mathrm{A}$
$3000\ \mathrm{W}$
$3240\ \mathrm{W}$
three quarters of full speed
$86.4\ \mathrm{V}$
$28.0\ \mathrm{A}$
$2420\ \mathrm{W}$
$941\ \mathrm{W}$
full speed
$115\ \mathrm{V}$
$4.00\ \mathrm{A}$
$461\ \mathrm{W}$
$19.2\ \mathrm{W}$
Two things here are worth more than the formula. The current column runs backwards from the intuition: largest when the machine is doing nothing, smallest when it runs properly. And the mechanical power humps near half speed, where the motor is most powerful and also most wasteful. A stalled motor turns twelve kilowatts into heat in a winding built to shed twenty watts.
Why the starting current of a motor is twenty five times its running current
A direct current motor runs from a $120\ \mathrm{V}$ supply with an armature resistance of $1.20\ \Omega$, and at normal running speed it draws $4.00\ \mathrm{A}$. Find the back emf at running speed, the current at the instant of switching on, the mechanical power at running speed, and the current if a load slows it to half speed.
Given
$V = 120\ \mathrm{V}$
$R = 1.20\ \Omega$ armature resistance
$I = 4.00\ \mathrm{A}$ at normal running speed
the back emf is proportional to the speed
Find
the back emf, the starting current, the mechanical power, and the current at half speed
SolutionBack emf from the loop equation
$$\varepsilon_b = V - IR = 120 - (4.00)(1.20) = 115\ \mathrm{V}$$
the back emf is not measured; it is what is left after the resistive drop
The instant of switching on
$$\omega = 0 \Rightarrow \varepsilon_b = 0$$
the back emf is a generator emf, and a generator not turning generates nothing
Independent energy audit at running speed. The supply delivers $480\ \mathrm{W}$ and the winding turns $19.2\ \mathrm{W}$ into heat, leaving $461\ \mathrm{W}$, which matches the mechanical power from $\varepsilon_b I$ by a different formula.
Four short parts sharing one equation, needing only the fact that the back emf is proportional to the speed.
The last part is the one to carry away. The current is set by $V - \varepsilon_b$, and when two large numbers nearly cancel, a small change in one changes the difference enormously.
Why the lights dim for a moment when the fridge starts
A $120\ \mathrm{V}$ supply reaches a room through wiring of total resistance $0.400\ \Omega$. A lamp of resistance $240\ \Omega$ is on. A motor of armature resistance $1.20\ \Omega$ is then switched on in parallel with it. Find the socket voltage before and at the instant of starting, and the factor by which the lamp dims.
Independent check by computing the lamp powers directly rather than through the ratio: $59.8\ \mathrm{W}$ before and $33.7\ \mathrm{W}$ during starting, whose quotient is $0.563$, the number the ratio gave without computing either power.
One parallel combination and two dividers. The only induction is the sentence setting the back emf to zero.
Notice what did the damage: not the motor, but the wiring. With zero resistance wiring nothing would have dimmed, which is why heavy appliances get their own circuit back to the board.
Checkpoint
§12.5 — the moment a motor draws the most current●●○○○
Thirty seconds, no calculator. An electric drill is switched on, run up to full speed with no load, then pressed hard into steel until it slows almost to a stop.
Given
the supply voltage is the same throughout
the armature resistance is the same throughout
the back emf is proportional to the speed
Find
(a) At which moment is the current drawn from the supply largest?
Hint 1/4
The current is set by a difference of two voltages. Ask which of the two moves, and which way.
Hint 2/4
$I = (V-\varepsilon_b)/R$, and $\varepsilon_b$ is proportional to the speed.
Hint 3/4
Here V and R never change, and the speed is highest when running free and nearly zero when the bit is jammed in the steel.
Hint 4/4
The smallest back emf gives the largest current, so nearly stalled is the worst case.
Show solutionTrack the one quantity that changes
$$I = \frac{V-\varepsilon_b}{R}$$
V and R are fixed, so all the behaviour of the current sits in the back emf
Independent check at the other end of the range. Spin a motor faster than the supply can drive it, so $\varepsilon_b > V$, and the formula gives a negative current: the machine has become a generator. One equation describing both machines, changing sign in the right place, is a strong sign it is being read correctly.
⚠ Using the supply voltage to find the mechanical output
the supply voltage is printed on the machine and comes to hand, while the back emf has to be computed first
wrong$$P_{\rm mech} = VI$$
right$$P_{\rm mech} = \varepsilon_b I$$
⚠ Assuming the running current also flows at startup
the question usually quotes one current, easily taken as a property of the motor rather than of one operating point
12.6Two coils on one core, and why the grid runs at hundreds of kilovolts
Two windings sharing one changing flux have voltages in the ratio of their turns, and currents in the inverse ratio.
Nothing so far has needed the flux through one circuit to be made by another. Put two coils on the same iron core and that is exactly what happens, and the arithmetic that follows is why a power station can be two hundred kilometres from a kettle.
TheoremResult 12.6: the ideal transformer
Conditions
both windings are threaded by the same flux, which is what the closed iron core is for
no power is lost, so all the input appears in the output; a real transformer carries an efficiency applied at the end
the supply must be changing with time; a steady current gives a steady flux and no output at all
resistance of the windings themselves is ignored unless a value is given
Both coils are wrapped round the same lump of iron, so the same flux changes at the same rate through every turn of both. Each turn is worth the same number of volts, so the coil with more turns has more volts across it, in the ratio of the turns. Nothing is created: ten times the voltage means a tenth of the current, their product being the power.
Proof
Let $\Phi$ be the flux through one turn of the core. Because the core is a closed loop of iron carrying both windings, the same $\Phi$ threads every turn of both; that is the whole content of the word ideal here.
Faraday's law on each winding: $V_p = N_p\,d\Phi/dt$ and $V_s = N_s\,d\Phi/dt$. The rate of change is the same quantity in both lines, which makes the next step possible.
Divide one by the other and $d\Phi/dt$ cancels: $V_s/V_p = N_s/N_p$. Notice how little was needed. No property of iron, no current, no frequency.
Now conservation of energy, a separate assumption rather than a consequence. If nothing is lost, $V_pI_p = V_sI_s$.
Combining gives $I_s/I_p = N_p/N_s$: the current ratio is the turns ratio upside down, so the higher voltage winding carries the smaller current and can use thinner wire.
Finally, why none of this works on a battery. Every line above contains $d\Phi/dt$. A steady current gives a steady flux and both voltages are zero, so a transformer on a battery does nothing but get warm, apart from one brief pulse at the moment of connection.
The whole device. The $\textcolor{#1f6feb}{\text{flux}}$ runs round the closed core, so both $\textcolor{#d1690a}{\text{windings}}$ see the same rate of change and each turn is worth the same volts. Drawn with twice as many turns on the left as on the right, which halves the voltage and doubles the current on the way through.
Looks like this, but is not
A gives you more power out than you put in. Ten volts in, a thousand out, from a lump of iron with no moving parts.
The current went down by a hundred at the same time, so the product is unchanged, and a real one gives slightly less out than in. A transformer changes how power is packaged, not how much there is. Otherwise you could feed the output back into the input and never buy electricity again, which is the runaway Lenz's law rules out.
line voltage
current in the line
power lost as heat
fraction of the power lost
$20.0\ \mathrm{kV}$
$1000\ \mathrm{A}$
$8.00\ \mathrm{MW}$
$40.0\%$
$50.0\ \mathrm{kV}$
$400\ \mathrm{A}$
$1.28\ \mathrm{MW}$
$6.40\%$
$150\ \mathrm{kV}$
$133\ \mathrm{A}$
$142\ \mathrm{kW}$
$0.711\%$
$400\ \mathrm{kV}$
$50.0\ \mathrm{A}$
$20.0\ \mathrm{kW}$
$0.100\%$
The last column falls by a factor of four hundred down the table, because the loss goes as the square of the current and the current as one over the voltage. At twenty kilovolts nearly half the generated power heats the countryside; at four hundred kilovolts a thousandth of it does. That column is why every high voltage line exists.
A doorbell transformer, ideal and then real
A transformer runs a doorbell. Its primary is on a $240\ \mathrm{V}$ supply and its secondary delivers $8.00\ \mathrm{V}$ at $3.00\ \mathrm{A}$. Find the turns ratio, the primary current if it is ideal, and the primary current if it is $92.0\%$ efficient.
Given
$V_p = 240\ \mathrm{V}$, $V_s = 8.00\ \mathrm{V}$
$I_s = 3.00\ \mathrm{A}$
efficiency $92.0\%$ in the second part
Find
the turns ratio and the primary current, ideal and real
Check the direction of the correction rather than repeating it. A real transformer must draw more current than an ideal one for the same output, and $0.109 > 0.100$. Multiplying by the efficiency instead would have given less than the ideal, which is impossible.
Three lines. The only decision is which way round the efficiency goes, and the plausibility check settles it.
A sanity habit for this concept: work out the power first and carry it, rather than juggling four quantities in two ratios.
Twenty megawatts down a line, at two different voltages
A power station delivers $20.0\ \mathrm{MW}$ along a transmission line of total resistance $8.00\ \Omega$. Find the line current, the power lost as heat and the fraction of the delivered power it represents, first at $20.0\ \mathrm{kV}$ and then at $400\ \mathrm{kV}$.
Given
power to be transmitted $P = 20.0\ \mathrm{MW}$
line resistance $R = 8.00\ \Omega$
case one: line voltage $20.0\ \mathrm{kV}$
case two: line voltage $400\ \mathrm{kV}$
Find
the current, the heating loss and the loss fraction in each case
Independent check by scaling instead of arithmetic. The loss goes as $1/V^{2}$, so twenty times the voltage divides it by four hundred, and $8.00\ \mathrm{MW}/400 = 20.0\ \mathrm{kW}$, the number computed from scratch in the second case.
Two divisions and two squarings. The entire engineering argument for the national grid fits into six lines of arithmetic.
This argument needs a transformer because no simple device changes a direct voltage twentyfold without wasting the difference, which is why the grid is alternating rather than steady.
Checkpoint
§12.6 — a transformer connected to a battery●●○○○
Thirty seconds, no calculator. A student wires the primary of a step up transformer to a car battery, leaves it connected, and puts a voltmeter across the secondary.
Given
the primary is across a battery giving a steady voltage
the transformer is a normal step up transformer with an iron core
the connection has been in place for several seconds
Find
(a) What does the voltmeter on the secondary read?
Hint 1/4
Ask what physically connects the two coils. It is not a wire.
Hint 2/4
$V_s = N_s\,d\Phi/dt$. The secondary responds to the rate of change of the flux, not to the flux.
Hint 3/4
Here the battery gives a steady current, so the flux in the core settles to a steady value and its rate of change is zero.
Hint 4/4
A steady flux induces nothing, so the reading is zero once the initial transient has died away.
Show solutionGo back past the ratio to the law it came from
$$V_s = N_s\frac{d\Phi}{dt}$$
the turns ratio came from this line, and a strange consequence is worth tracing back
the core flux is set by the primary current, and a constant current gives constant flux
$$V_s = 0$$
and the number of turns multiplies zero, which no ratio can rescue
Answer $$\boxed{\,V_s = 0\ \text{for a steady primary current}\,}$$
Check
Independent check by energy. A steady secondary voltage across a load would deliver power continuously, drawn from the battery through a magnetic link that never changes. Nothing would be changing while energy flowed, which cannot happen.
⚠ Turning the turns ratio upside down for the currents
one ratio is written down and reused for the other quantity without asking which way conservation of power sends it
wrong$$\frac{I_s}{I_p} = \frac{N_s}{N_p}$$
right$$\frac{I_s}{I_p} = \frac{N_p}{N_s}$$
⚠ Computing the transmission loss from the line voltage
the line voltage is the number the question emphasises, and the formula $V^{2}/R$ is available and gives an answer
⚠ Multiplying by the efficiency when the output is what is known
efficiency is habitually a factor below one that gets multiplied in, without checking which power is the larger
wrong$$P_{\rm in} = \eta P_{\rm out}$$
right$$P_{\rm in} = \frac{P_{\rm out}}{\eta}$$
⚠ Expecting an output from a steady supply
the turns ratio contains no time and looks like a property of the device, so it seems to apply to any voltage
wrong$$V_s = \frac{N_s}{N_p}V_{\rm battery}$$
right$$V_s = N_s\frac{d\Phi}{dt} = 0\ \text{for a steady current}$$
12.7A changing flux makes an electric field, with or without a wire
The emf round a loop is a line integral of an electric field that exists whether or not a conductor is there.
One question has been open since the second concept. When a coil sits still in a field that is being ramped, nothing pushes the charges sideways, so what drives them round? The answer costs one new statement and takes away one old habit.
TheoremResult 12.7: the induced electric field
Conditions
the loop is any closed path, real or imaginary, and the flux is through any surface bounded by it
the explicit results are for a long solenoid of radius R, uniform inside and negligible outside
the circular symmetry is what lets E come out of the integral; without it the law is still true but does not give E
this field comes from the changing flux alone, and any field of static charges would be added separately
$$\boxed{\,\oint\vec E\cdot d\vec l = -\frac{d\Phi_B}{dt}\,,\qquad E = \frac{r}{2}\frac{dB}{dt}\ (r\le R)\,,\qquad E = \frac{R^{2}}{2r}\frac{dB}{dt}\ (r\ge R)\,}$$
Walk once round a closed loop adding up the push the electric field gives you, and the total is the rate at which flux is pouring through the loop. Inside the solenoid the field grows with how far out you stand, because a bigger circle catches more flux while its own length grows more slowly. Outside, no new flux is caught, so the same total is shared round a longer path.
Proof
Start from what an emf is: the work done per unit charge in going once round the loop. From an electric field that work is $\oint\vec E\cdot d\vec l$, and setting it equal to $-d\Phi_B/dt$ is Faraday's law with the wire taken away.
Now specialise to a solenoid being ramped. Take a circular path of radius $r$ on the axis. By symmetry nothing distinguishes one point of it from another, so E has the same size everywhere on it and must lie along the circle rather than across it.
So the integral is trivial: $\oint\vec E\cdot d\vec l = E(2\pi r)$. This is the same manoeuvre as pulling B out of an Amperian loop, and it fails for the same reason when the symmetry is absent.
Inside, $r\le R$, the circle catches $\Phi_B = B\pi r^{2}$, so $E(2\pi r) = \pi r^{2}\,dB/dt$ and $E = (r/2)\,dB/dt$: zero on the axis, growing in a straight line.
Outside, $r\ge R$, the circle catches only the flux inside the solenoid, $\Phi_B = B\pi R^{2}$, so $E = R^{2}\,dB/dt/(2r)$, falling as one over the distance.
The two expressions agree at $r = R$, as they must, and the peak of the whole picture sits at the winding itself.
One consequence undoes a habit from the electric weeks. The circulation of this field round a closed loop is not zero, so a charge carried once round returns with more energy than it started with, no potential function exists, and asking for the voltage at a point is meaningless. That is a genuine difference from the field of static charges.
How strong the $\textcolor{#8250df}{\text{induced electric field}}$ is as a function of distance from the axis of a solenoid being ramped. It rises in a straight line inside, peaks at the winding, and falls as one over the distance outside. A place well outside the coil can sit in a stronger induced field than a place just inside it.
Looks like this, but is not
There must be charges somewhere making this field, since electric fields come from charges. The solenoid is neutral, the air is neutral, and nothing in the room is charged.
Any field made by charges has zero circulation round every closed loop, which was the whole content of the earlier work on potential. This one has a circulation equal to the rate of change of the flux, so no arrangement of charges can produce it. A changing magnetic field is a second, independent source of electric field.
distance from the axis
inside or outside
induced electric field
$0.500\ \mathrm{cm}$
inside
$1.27\times10^{-4}\ \mathrm{V/m}$
$1.00\ \mathrm{cm}$
inside
$2.54\times10^{-4}\ \mathrm{V/m}$
$2.50\ \mathrm{cm}$
at the winding
$6.36\times10^{-4}\ \mathrm{V/m}$
$5.00\ \mathrm{cm}$
outside
$3.18\times10^{-4}\ \mathrm{V/m}$
$10.0\ \mathrm{cm}$
outside
$1.59\times10^{-4}\ \mathrm{V/m}$
Compare two entries directly. Half a centimetre inside the coil, where the magnetic field is at full strength, the induced electric field is $1.27\times10^{-4}\ \mathrm{V/m}$; ten centimetres outside, where there is no magnetic field at all, it is larger. Asking where the induced electric field is strongest is a different question from asking where the magnetic field is.
The electric field inside and outside a solenoid with a rising current
A long solenoid of radius $2.50\ \mathrm{cm}$ is wound with 900 turns per metre and its current is increased steadily at $45.0\ \mathrm{A/s}$. Find the rate at which the field inside is rising, and the induced electric field at $1.00\ \mathrm{cm}$ and at $5.00\ \mathrm{cm}$ from the axis.
Continuity at the boundary, which neither answer used. At $r = R$ the inside formula gives $6.36\times10^{-4}\ \mathrm{V/m}$ and the outside one reduces to the same expression. Two formulas from different flux calculations meet exactly at the winding.
Three substitutions and one decision, the decision being which side of the winding each point is on.
The induced electric field spills out well beyond the magnetic field that made it, which is why a changing current in one place can drive a current some distance away with nothing in between.
A ring with no battery in it, and where its energy comes from
A large coil makes a uniform field over a circular region of radius $0.200\ \mathrm{m}$, increasing at $6.00\ \mathrm{T/s}$. A copper ring of radius $0.100\ \mathrm{m}$ and resistance $0.0500\ \Omega$ lies inside that region, concentric and perpendicular to the field. Find the induced electric field at the ring, the emf round it, the current and the power dissipated.
Given
uniform field over a region of radius $0.200\ \mathrm{m}$
$dB/dt = 6.00\ \mathrm{T/s}$
ring radius $r = 0.100\ \mathrm{m}$, resistance $0.0500\ \Omega$
the ring is concentric and perpendicular to the field
this energy comes from whatever drives the large coil, now slightly harder to ramp
Answer $$\boxed{\,E = 0.300\ \mathrm{V/m},\quad \varepsilon = 0.188\ \mathrm{V},\quad I = 3.77\ \mathrm{A},\quad P = 0.711\ \mathrm{W}\,}$$
Check
The two routes to the emf are the verification and were kept apart: one multiplied a field by a circumference, the other an area by a rate of change of field. As a size check, a fifth of a volt round a copper ring giving a few amperes is why a ring is a bad thing to wear near an induction heater.
Four lines, one of which was optional and was done anyway because it is the cheapest possible check.
Ask where you would put a voltmeter to measure that $0.188\ \mathrm{V}$. There is no answer: the emf is spread round the whole ring, and two voltmeters on the same points with leads passing opposite sides of the coil would read differently.
Checkpoint
§12.7 — where the induced electric field is strongest●●●○○
Thirty seconds, no calculator. A long solenoid of radius R has its current ramped steadily upwards, and a small charged bead is placed at various distances from the axis to see where the push is biggest.
Given
a long solenoid of radius R, with a steadily rising current
the field inside is uniform, and outside it is negligible
Find
(a) At what distance from the axis is the induced electric field largest?
Hint 1/4
Ask two separate questions for a circle of radius r: how much flux does it catch, and how long is it?
Hint 2/4
$E(2\pi r) = d\Phi_B/dt$. Inside, the flux caught grows as $r^{2}$; outside, it stops growing altogether.
Hint 3/4
So inside, $E = (r/2)\,dB/dt$, which increases with r; outside, $E = R^{2}\,dB/dt/(2r)$, which decreases with r.
Hint 4/4
One expression rises and the other falls, so the largest value is where they meet, at the winding.
Show solutionCompare the two regions
$$r\le R:\quad E = \frac{r}{2}\frac{dB}{dt}\ \text{increases with}\ r$$
the flux caught goes as the area and the path only as the circumference
$$r\ge R:\quad E = \frac{R^{2}}{2r}\frac{dB}{dt}\ \text{decreases with}\ r$$
the numerator is frozen at the solenoid's flux while the denominator keeps growing
$$\Rightarrow E_{\max}\ \text{at}\ r = R,\quad E_{\max} = \frac{R}{2}\frac{dB}{dt}$$
a function that rises then falls peaks at the turning point, and both expressions agree there
Answer $$\boxed{\,E_{\max} = \frac{R}{2}\frac{dB}{dt}\ \text{at}\ r = R\,}$$
Check
Independent check at the axis, where no formula is needed. A circle of vanishing radius catches vanishing flux, so the circulation of E round it is essentially zero, and by symmetry E itself must be zero there. The formula gives zero at $r = 0$ too.
⚠ Using the radius of the solenoid instead of the radius of the point inside it
two radii are given and the one belonging to the apparatus feels more important than the one belonging to the question
wrong$$E = \frac{R}{2}\frac{dB}{dt}\quad (r<R)$$
right$$E = \frac{r}{2}\frac{dB}{dt}\quad (r<R)$$
⚠ Using the inside formula at a point outside
the inside formula is simpler, it was derived first, and nothing in the arithmetic complains if the wrong one is used
right$$\oint\vec E\cdot d\vec l \neq 0 \Rightarrow \text{no potential exists}$$
⚠ Expecting the induced electric field to point away from the axis
radial is what electric fields look like in every picture from the first weeks of the course, and the axis looks like a centre to point away from
wrong$$\vec E_{\rm ind} \parallel \hat r$$
right$$\vec E_{\rm ind}\ \text{tangential, in closed circles round the axis}$$
Which induction question is this, decided before any arithmetic
Any question in which something is induced. Thirty seconds here decides which of four formulas you are about to use, and stops the commonest error, reaching for the rotating coil formula when nothing is rotating.
Write the flux down as a product and ask which factor is moving
Every problem starts with $\Phi_B = BA\cos\theta$ and usually one factor is changing. If B, the emf is $NA\,dB/dt$; if A, it is a sliding conductor and $\varepsilon = BLv$; if $\theta$, a rotating coil and a sinusoid.
Check whether the field is uniform across the loop
If it is, the flux is a product with no integral. If not, which in practice means a loop beside a straight wire, you need $\int B(r)L\,dr$ and a logarithm appears.
Ask whether the question wants a current or a charge
A current needs the time; a total charge does not, since $q = N|\Delta\Phi_B|/R$ has none in it. A vague or unmeasurable duration is the clue that charge is wanted.
Do the direction as a separate operation, in words
Never carry the minus sign through the arithmetic. Compute a size, then take the direction apart with the method box below.
Finish with an energy check and a size check
If anything moves, check that $Fv$ or $\tau\omega$ equals $I^{2}R$. If nothing moves, check the size: a hand movement near a magnet gives millivolts, a generator hundreds of volts.
Where it goes wrong
Using $\varepsilon = BLv$ on a problem where nothing is sliding, because it is the shortest formula on the page.
Using the rotating coil formula for a coil flipped once rather than spun; a single flip is a charge question, not a sinusoid.
Multiplying by the area when the field varies across the loop, which is the whole difficulty of the straight wire geometry.
Producing a signed emf and then not knowing what the sign refers to, because no normal was ever chosen.
Fixing the direction in four moves, every time, with no memorised cases
Every question asking which way the induced current flows, which is most of them. The four moves never need a special case for magnets, coils or sliding rods.
Which way does the flux through the loop point right now
Not which way the magnet points, and not the current in some other wire. Stand at the loop, look at the field through it, and say into the page or out of it.
Is that flux growing or shrinking
This is the step that gets skipped. A magnet approaching and one receding give opposite answers with everything else identical.
Which way must the induced field inside the loop point
Growing means oppose, so the induced field points the other way; shrinking means support, so it points the same way. Say it aloud before going on.
Convert that into a current with the right hand
Curl the right hand round the loop and the thumb gives the field inside it: out of the page needs an anticlockwise current, into the page a clockwise one. State the viewpoint.
Where it goes wrong
Skipping step two and answering the same way for a magnet arriving as for one leaving.
Opposing the field rather than opposing the change, which gives the wrong answer for every shrinking flux.
Giving a rotational sense with no viewpoint, so that the answer is only half stated.
Using the left hand out of habit from some other rule; every rule in this course is a right hand rule.
The energy audit that catches errors arithmetic cannot
At the end of every problem where something moves, turns or is driven. It costs one line, uses numbers you already have, and fails loudly when a current or force is wrong.
Name the mechanical side
A sliding rod: $P = Fv$. A rotating coil or armature: $P = \tau\omega$. A falling conductor at terminal speed: $P = mgv$. Compute it from mechanical quantities only, with no reference to the circuit.
Name the electrical side
$P = I^{2}R$ over the whole circuit, or $P = \varepsilon I$ at the source, or $VI$ minus $I^{2}R$ for a motor. Compute it from electrical quantities only.
Set them equal, and mean it
At constant speed they must be equal. If the body is accelerating, the difference is the rate of change of kinetic energy; if there is friction, it is the friction. Anything else is an error upstream.
If they disagree, look at the current first
A factor of two usually means a resistance left out or counted twice; a factor of N means it was applied to the wrong quantity; a factor of $2\pi$ means a rotation rate was not converted.
Where it goes wrong
Comparing a power with an energy, which have different units and will never agree.
Forgetting that at terminal speed the kinetic energy is not changing, so there is no third term to hide a discrepancy in.
Using the supply power rather than the back emf power for the mechanical output of a motor.
A loop dragged through a field that fills all space
A square loop of side $0.200\ \mathrm{m}$ and resistance $2.00\ \Omega$ is pulled at $3.00\ \mathrm{m/s}$ through a uniform field of $0.750\ \mathrm{T}$ that extends far beyond the loop in every direction. Find the induced emf and the current.
Given
square loop, side $L = 0.200\ \mathrm{m}$, resistance $2.00\ \Omega$
$B = 0.750\ \mathrm{T}$, uniform everywhere, perpendicular to the loop
$v = 3.00\ \mathrm{m/s}$, in the plane of the loop
Find
the emf and the current
SolutionWrite the flux, then differentiate it
$$\Phi_B = BA = (0.750)(0.0400) = 3.00\times10^{-2}\ \mathrm{Wb}$$
the loop is inside the field wherever it goes, so all of it contributes
$$\frac{d\Phi_B}{dt} = 0 \Rightarrow \varepsilon = 0,\ I = 0$$
nothing on the right depends on time; the loop moved, its flux did not
Check it with the two edges, since the leading edge certainly has an emf
the leading edge is a rod of length L at speed v, with a motional emf like any other
$$\varepsilon_{\rm trail} = 0.450\ \mathrm{V}\ \text{in the opposite sense round the loop}$$
the trailing edge does the same, and round the loop the two are traversed oppositely
Answer $$\boxed{\,\varepsilon = 0,\qquad I = 0\,}$$
Check
Independent check by energy. A current would heat the resistor and something would have to pay for it, but the drag on the leading edge and the push on the trailing edge cancel too, so the loop coasts.
The same loop leaving the same field at its edge
The same square loop of side $0.200\ \mathrm{m}$ and resistance $2.00\ \Omega$ is pulled at $3.00\ \mathrm{m/s}$ out of a region of uniform field $0.750\ \mathrm{T}$, at the moment when the leading edge is outside the region and the trailing edge is still inside. Find the emf and the current.
Given
the same loop and the same speed
$B = 0.750\ \mathrm{T}$ inside the region and zero outside
at this instant the trailing edge is in the field and the leading edge is not
Find
the emf and the current
SolutionOnly the part still inside counts
$$\Phi_B = BLx,\qquad x = \text{length still inside}$$
the field is zero over the part that has left, the only difference from the previous problem
$$\frac{d\Phi_B}{dt} = BL\frac{dx}{dt} = -BLv$$
x shrinks at the speed of the loop, which is where the single factor of v enters
the same arithmetic as before, with no second edge to cancel it this time
$$I = \frac{0.450}{2.00} = 0.225\ \mathrm{A}$$
and the direction keeps the flux the loop is losing: clockwise for a field into the page
Answer $$\boxed{\,\varepsilon = 0.450\ \mathrm{V},\qquad I = 0.225\ \mathrm{A}\,}$$
Check
Independent check by force. The trailing edge carries $0.225\ \mathrm{A}$ in the field, so it feels $BIL = 0.0338\ \mathrm{N}$ pulling the loop back. Mechanical power $0.101\ \mathrm{W}$, electrical power $0.101\ \mathrm{W}$: somebody is paying, which is what distinguishes this case.
Same loop, same field, same speed, same everything a careless reading would notice, and one gives nothing while the other gives nearly a quarter of an ampere; the only difference is how many edges are in the field.
How to tell them apart
Do not ask whether the conductor is moving. Ask how many of its edges lie in the field, and whether the flux through the whole loop is different a moment later. Motion inside a uniform field changes no flux and induces nothing; motion across the boundary changes flux and induces everything. Count edges and you can answer these without a formula.
Scaffolding comes off
The common skeleton
Draw the loop, choose a normal direction for it, and write down the field passing through it.
Write the flux as $\Phi_B = BA\cos\theta$ and identify which of the three factors depends on time.
Differentiate that factor alone, leaving the others outside as constants, to get $d\Phi_B/dt$.
Multiply by the number of turns to get the size of the emf, keeping every quantity positive.
Divide by the resistance for the current, and if a total charge is wanted use $N|\Delta\Phi_B|/R$ instead.
Argue the direction separately: which way the flux points, whether it grows or shrinks, which way the induced field must point, then the right hand.
Close with an energy or size check: mechanical power against electrical power, or a comparison with a known voltage.
1 · fully worked
A coil in a field that is switched off in a fortieth of a second
A circular coil of 80 turns and radius $4.00\ \mathrm{cm}$ lies in a uniform field of $0.550\ \mathrm{T}$ perpendicular to its plane, out of the page. The field is reduced steadily to zero in $0.0250\ \mathrm{s}$. The coil has resistance $12.0\ \Omega$. Find the average emf, the current and its direction, the charge that flows, and the heat generated.
Given
$N = 80$, radius $4.00\times10^{-2}\ \mathrm{m}$
$B$ falls from $0.550\ \mathrm{T}$ to zero, out of the page
$\Delta t = 0.0250\ \mathrm{s}$
$R = 12.0\ \Omega$
Find
the emf, the current with its direction, the charge and the heat
Two cross checks, neither repeating a line above. Charge from current and duration is $1.84\times10^{-2}\ \mathrm{C}$, matching the charge from flux and resistance; heat from $\varepsilon I\,\Delta t = 0.163\ \mathrm{J}$ matches the value from $I^{2}R\,\Delta t$.
Seven lines for four answers, and three of the seven were reused by more than one part.
Nine volts out of a coil the size of a coaster. Induced voltages are not small; what is small is the energy behind them, and $0.163\ \mathrm{J}$ would not warm a teaspoon of water measurably.
2 · you write the reasoning
Easier than the last one: one turn, no charge, no heat, three lines. A single square loop of side $0.100\ \mathrm{m}$ lies flat in a field out of the page increasing steadily at $2.00\ \mathrm{T/s}$. All three lines are correct. Before opening the model reasons, say in your own words why each is allowed, paying particular attention to what the third line is really claiming.
reasoning
The first line is geometry with no physics in it, and it is on its own line for a reason: the area is fixed for the whole problem, so it sits outside the derivative in the next line rather than being differentiated.
reasoning
The second line has no N in it because there is one turn, and the rate of change of the field is used directly rather than a change over an interval, because steadily says the two are the same. Note what is absent: the value of the field never appears. A loop in half a tesla rising at this rate and one in fifty tesla give the same twenty millivolts.
reasoning
The third line is the only one doing physics. The flux out of the page is growing, so the loop opposes the growth by making flux into the page inside itself, and by the grip rule that needs a clockwise current. Reverse the field direction or the growth and the answer reverses; reverse both and it comes back.
3 · find the buried error
Harder than the last one: a rotating coil, so a rate of rotation has to be handled, and a question about where in the cycle the output sits. A flat coil of 200 turns, each of area $3.00\times10^{-2}\ \mathrm{m^{2}}$, rotates at $25.0$ revolutions per second in a field of $0.150\ \mathrm{T}$. A student writes the four lines below and concludes that the peak emf is $22.5\ \mathrm{V}$ and the output is zero when the plane of the coil contains the field. Exactly two lines are faulty. Find them.
the two buried errors (2)
⚠ step 1
Revolutions per second is not radians per second. One revolution is $2\pi$ radians, so $\omega = 2\pi(25.0) = 157\ \mathrm{rad/s}$, and every later number using it is too small by $2\pi$. The correct peak emf is $141\ \mathrm{V}$, not $22.5\ \mathrm{V}$.
Machines are specified in revolutions and formulas written in radians, so the conversion sits between two habits. It is also invisible: twenty five is a reasonable number to see in a formula, the units are the ones the question used, and twenty two volts is a plausible output for a small generator.
right
Convert on its own line before anything else, and write the units on the result. A generator turning twenty five times a second should produce something in the range of a mains supply rather than a torch battery, and $141\ \mathrm{V}$ fits while $22.5\ \mathrm{V}$ does not.
⚠ step 3
A coil whose plane contains the field is edge on, not face on. Its normal is then perpendicular to the field, so $\cos\theta = 0$ and the flux is zero rather than maximum. The flux is largest when the plane is perpendicular to the field.
The plane of the coil is along the field sounds like the coil is lined up with it, and lined up sounds like maximum. The formula is stated in terms of the normal, an invisible line nobody draws, so the sentence gets translated using the visible object instead. It is the flux confusion from earlier, arriving where it costs a whole answer.
right
Sketch the loop edge on and ask how many field lines get through. Better, translate every orientation into an angle from the normal before writing trigonometry: plane along the field means normal across it means $\theta = 90^{\circ}$ means zero flux, and therefore maximum emf rather than zero.
4 · the bare problem
§12.3 — a rod on rails, with its own resistance this time●●●○○
No scaffolding now. A rod of length $0.500\ \mathrm{m}$ and resistance $1.00\ \Omega$ slides at a steady $4.00\ \mathrm{m/s}$ along frictionless rails of no resistance, closed at one end by a $5.00\ \Omega$ resistor, in a uniform field of $0.800\ \mathrm{T}$ perpendicular to the plane of the rails.
Given
$L = 0.500\ \mathrm{m}$, rod resistance $1.00\ \Omega$
external resistor $5.00\ \Omega$, rails of no resistance
$B = 0.800\ \mathrm{T}$, perpendicular
$v = 4.00\ \mathrm{m/s}$, constant
Find
(a) Find the emf and the current.
(b) Find the force that must be applied to keep the speed constant.
(c) Show that the mechanical power supplied equals the total electrical power dissipated.
(d) What fraction of the power is wasted inside the rod itself rather than delivered to the resistor?
Hint 1/4
The rod is a source with internal resistance and the resistor is the load. Once the emf is known, every part is a standard circuit question.
Here $B = 0.800\ \mathrm{T}$, $L = 0.500\ \mathrm{m}$, $v = 4.00\ \mathrm{m/s}$, $R = 5.00\ \Omega$ and the rod's own $r = 1.00\ \Omega$.
Hint 4/4
The emf is $1.60\ \mathrm{V}$, the current $0.267\ \mathrm{A}$, the force $0.107\ \mathrm{N}$, both powers $0.427\ \mathrm{W}$, and one sixth is wasted in the rod.
the current cancels, so only the ratio of resistances matters, at any speed
Answer $$\boxed{\,\varepsilon = 1.60\ \mathrm{V},\ I = 0.267\ \mathrm{A},\ F = 0.107\ \mathrm{N},\ P = 0.427\ \mathrm{W},\ 16.7\%\ \text{wasted}\,}$$
Check
The power agreement in part (c) is the main check, from two disjoint calculations. A cheaper second one: the rod's terminal voltage $\varepsilon - Ir = 1.33\ \mathrm{V}$ equals the load voltage $IR = 1.33\ \mathrm{V}$, as a series circuit requires.
Full exam-style question
A loop crossing a strip of field: emf, current, force and total heatexam format
A square loop of side $0.150\ \mathrm{m}$ and resistance $0.0250\ \Omega$ moves at a constant $2.00\ \mathrm{m/s}$ to the right, in the plane of the page, completely through a strip of uniform field of $0.350\ \mathrm{T}$ into the page. The strip is $0.400\ \mathrm{m}$ wide along the direction of motion. Find, with reasons: (a) the emf while entering, (b) the current and its direction while entering, (c) the emf while completely inside, (d) the force needed while entering, and (e) the total heat over the whole passage.
$B = 0.350\ \mathrm{T}$ into the page, in a strip $0.400\ \mathrm{m}$ wide
$v = 2.00\ \mathrm{m/s}$, constant, perpendicular to the strip boundaries
the strip is wider than the loop, so the loop is fully inside it for part of the journey
Find
four quantities during entry, one while fully inside, and the total heat
Solution
This is the shape most examiners choose for a full question here, because it tests three things in one diagram: computing a motional emf, knowing that a loop moving inside a uniform field produces nothing, and closing the energy account. A mistake in the first part does not destroy the rest, which is why part marks are usually generous.
Independent check on the heat by mechanical work, with no electrical quantity. A force of $0.221\ \mathrm{N}$ acts over $0.150\ \mathrm{m}$ of entry and again of exit, so the work is $6.62\times10^{-2}\ \mathrm{J}$, matching the heat from $I^{2}R$. Including the middle stretch would break it at once.
Five parts, one formula each, two of them answered with the word zero once the middle stage is recognised.
Sketch the current against time and you get a positive pulse, a flat stretch of nothing, and a negative pulse of the same size. A metal detector, a card reader and a bicycle speed sensor all produce that shape.
Practice
A · concept 4 questions
1§12.3 — motion without a boundary●●○○○
One mark, and the claim sounds safe because every word in it is true separately. A closed rectangular loop of copper is carried at a steady speed across a laboratory in which a uniform field fills the entire room, perpendicular to the loop. Someone says a current flows because a conductor is moving through a magnetic field.
Given
a closed loop of copper, moving at constant velocity
a uniform field filling the whole region, perpendicular to the loop
the loop never leaves the field
Find
(a) True or false: a current flows. Give your reason in one sentence.
Hint 1/4
Do not ask whether the conductor moves. Ask whether the number the law responds to is different a moment later.
Hint 2/4
$\varepsilon = -N\,d\Phi_B/dt$, and $\Phi_B = BA\cos\theta$ for this loop.
Hint 3/4
Here B is the same everywhere, the loop's area is fixed and its orientation never changes, so all three factors are constant.
Hint 4/4
False: the flux never changes, so there is no emf and no current, however fast the loop is carried.
Show solutionThe flux rule
$$\Phi_B = BA = \text{constant}$$
B is the same everywhere, so moving the loop takes it from field B to field B
$$\varepsilon = -\frac{d\Phi_B}{dt} = 0$$
the derivative of a constant is zero, and no amount of speed changes that
equal lengths at equal speed in the same field, so equal emfs
$$\oint: \ BLv - BLv = 0$$
round the loop one is traversed upwards and the other downwards, so they subtract
Answer $$\boxed{\,\varepsilon = 0,\ I = 0\,}$$
Check
Independent check by energy. A current would heat the loop and someone would have to pay for it, but the forces on the two edges are also equal and opposite, so no work is done and the loop coasts.
2§12.2 — a south pole coming down towards a loop●●●○○
A flat circular loop of wire lies on a bench. A bar magnet is held above it with its south pole pointing down at the loop, and is lowered towards it. You are looking down from above.
Given
a horizontal closed loop lying on a bench
a bar magnet above it, south pole downwards, moving down
the viewpoint is from above, looking down
Find
(a) Which way does the induced current run in the loop?
Hint 1/4
Two separate questions hide in this one: which way does the field through the loop point, and is it growing or shrinking.
Hint 2/4
Field lines run out of a north pole and into a south pole, so just below a south pole they point up into the magnet. Then Lenz's law and the grip rule.
Hint 3/4
Here the field through the loop points upwards, towards the viewer, and it is growing because the magnet is coming closer.
Hint 4/4
The induced field inside the loop must point downwards, away from the viewer, which needs a clockwise current seen from above.
Show solutionWhich way and whether growing
$$\vec B\ \text{through the loop points upwards}$$
lines enter a south pole, so beneath it they are heading upwards into the magnet
$$\frac{d\Phi_B}{dt} > 0$$
the magnet is closer each moment, so more field passes through
Oppose, then convert
$$\vec B_{\rm induced}\ \text{downwards inside the loop}$$
growing flux is opposed, so the induced field points the other way
$$\Rightarrow I\ \text{clockwise seen from above}$$
grip rule read backwards: fingers curling clockwise from above put the thumb downwards
Answer $$\boxed{\,I\ \text{clockwise, seen from above}\,}$$
Check
Independent check by force rather than field. The loop must repel the magnet, since attracting it would let the magnet accelerate in on its own while the loop heated up, and a clockwise current from above presents a south face upwards.
3§12.1 — the orientation that gives the most flux●●○○○
Half a mark, and it is the half that decides several other answers. A flat coil in a uniform field can be turned to any orientation. Someone claims the flux is largest when the plane of the coil is parallel to the field, because that is when it is lined up with it.
Given
a flat coil of fixed area in a uniform field
the coil may be turned to any orientation
Find
(a) True or false, and state the orientation that actually gives the largest flux.
Hint 1/4
Picture the coil edge on to the field and ask how many field lines get through it.
Hint 2/4
$\Phi_B = BA\cos\theta$, with $\theta$ measured between the field and the normal to the coil, not between the field and the plane.
Hint 3/4
Here the plane being parallel to the field means the normal is perpendicular to it, so $\theta = 90^{\circ}$.
Hint 4/4
False: that orientation gives zero flux, and the largest flux is when the plane is perpendicular to the field.
Show solutionTranslate the description into the angle the formula uses
$$\text{plane}\ \parallel\ \vec B \Rightarrow \hat n \perp \vec B \Rightarrow \theta = 90^{\circ}$$
the normal is perpendicular to the plane, so a statement about the plane must be turned round
Independent check by counting lines. Draw the field as parallel lines and the coil as a hoop: along its rim no lines pass through, whatever the field strength; face on, every line in its cross section does.
4§12.2 — what more turns really buys you●●●●○
One mark, and it separates two things that usually move together. A coil is wound from thin wire and placed in a field changing at a steady rate. Someone rewinds it with twice as many turns of the same wire, each of the same area, and claims this doubles the emf and therefore the current.
Given
the same kind of wire, so the resistance per turn is the same
twice as many turns, each of the same area
the same rate of change of field
the circuit is the coil on its own, with nothing else in it
Find
(a) True or false, and say what happens to the current.
Hint 1/4
Two quantities change when you add turns, not one. Find the second one before answering.
Hint 2/4
$\varepsilon = NA\,dB/dt$ and $I = \varepsilon/R$, and R is the resistance of the wire that makes up the coil.
Hint 3/4
Here doubling the turns doubles the length of wire in the coil, so it doubles R as well as doubling the emf.
Hint 4/4
The first half is true and the second is false: the emf doubles, the resistance doubles with it, and the current is unchanged.
Show solutionWhat doubling the turns does to each quantity
Independent check by power. The emf doubles and the current is unchanged, so $\varepsilon I$ doubles; on the other side $I^{2}R$ has the same current and twice the resistance, which also doubles.
B · computation 8 questions
1§12.1 — flux and flux linkage of a tilted rectangular coil●●●○○
A rectangular coil of 60 turns measures $8.00\ \mathrm{cm}$ by $12.0\ \mathrm{cm}$ and sits in a uniform field of $0.250\ \mathrm{T}$, its normal at $55.0^{\circ}$ to the field.
Given
$N = 60$ turns, sides $8.00\times10^{-2}\ \mathrm{m}$ and $0.120\ \mathrm{m}$
$B = 0.250\ \mathrm{T}$, uniform
the angle between the normal and the field is $55.0^{\circ}$
Find
(a) Find the flux through one turn.
(b) Find the flux linkage of the whole coil.
(c) By what angle would the coil have to be turned, from where it is, to double the flux through it?
Hint 1/4
Part (c) is a question about the cosine, not about the coil. Work out what value the cosine would need to take.
Hint 2/4
$\Phi_B = BA\cos\theta$ with $\theta$ from the normal, and the linkage is $N\Phi_B$.
Hint 3/4
Here $A = (8.00\times10^{-2})(0.120)$, $B = 0.250\ \mathrm{T}$ and $\theta = 55.0^{\circ}$, whose cosine is $0.574$.
Hint 4/4
The flux is $1.38\times10^{-3}\ \mathrm{Wb}$, the linkage $8.26\times10^{-2}\ \mathrm{Wb}$, and doubling needs a cosine of $1.147$, which is impossible.
Size check by bracketing. The flux must lie between zero and $BA = 2.40\times10^{-3}\ \mathrm{Wb}$, and $1.38\times10^{-3}$ is $57\%$ of that, a little above the half a sixty degree tilt would cost.
2§12.2 — a flux that reverses rather than merely falling●●●○○
A coil of 250 turns sits in a field being changed by an electromagnet. The flux through one turn goes from $+3.50\times10^{-4}\ \mathrm{Wb}$ to $-1.50\times10^{-4}\ \mathrm{Wb}$, reversing direction, in $0.0200\ \mathrm{s}$.
Given
$N = 250$ turns
flux per turn goes from $+3.50\times10^{-4}\ \mathrm{Wb}$ to $-1.50\times10^{-4}\ \mathrm{Wb}$
$\Delta t = 0.0200\ \mathrm{s}$
Find
(a) Find the size of the change in flux through one turn.
(b) Find the average emf induced in the coil.
Hint 1/4
The two flux values have opposite signs. Decide what that means for the size of the change before computing anything.
Hint 2/4
$|\bar\varepsilon| = N|\Delta\Phi_B|/\Delta t$, and $\Delta\Phi_B$ is the final value minus the initial one, signs included.
Hint 3/4
Here the initial value is $+3.50\times10^{-4}\ \mathrm{Wb}$ and the final one is $-1.50\times10^{-4}\ \mathrm{Wb}$, over $0.0200\ \mathrm{s}$.
Hint 4/4
The change is $5.00\times10^{-4}\ \mathrm{Wb}$ in size, and the average emf is $6.25\ \mathrm{V}$.
Show solutionKeep the signs until the very last step
Independent check by splitting the interval. The flux falls from $3.50\times10^{-4}$ to zero and then goes to $1.50\times10^{-4}$ the other way, and the stages add to $5.00\times10^{-4}\ \mathrm{Wb}$, which is what the signed subtraction gave.
3§12.2 — measuring a field with a flip coil●●●●○
A search coil of 500 turns, each of area $4.00\ \mathrm{cm^{2}}$, lies face on in a uniform field of unknown strength. It is flipped through $180^{\circ}$ and the charge that flows is $6.40\times10^{-5}\ \mathrm{C}$. Coil and meter together have resistance $250\ \Omega$.
(a) Find the size of the change in flux through one turn during the flip.
(b) Find the field.
(c) Explain in one sentence why the answer does not depend on how quickly the flip was made.
Hint 1/4
Ask what the flux is before the flip and what it is after, being careful about the sign of the second one.
Hint 2/4
$q = N|\Delta\Phi_B|/R$, and turning a coil through half a turn reverses the sign of its flux.
Hint 3/4
Here the flux goes from $+BA$ to $-BA$, so the size of the change is $2BA$, with $N = 500$, $A = 4.00\times10^{-4}\ \mathrm{m^{2}}$, $q = 6.40\times10^{-5}\ \mathrm{C}$ and $R = 250\ \Omega$.
Hint 4/4
The flux change is $6.40\times10^{-5}/500 \times 250 = 3.20\times10^{-5}\ \mathrm{Wb}$, giving $B = 0.0400\ \mathrm{T}$.
three significant figures, and the units come out as tesla
Answer $$\boxed{\,|\Delta\Phi_B| = 3.20\times10^{-5}\ \mathrm{Wb},\qquad B = 0.0400\ \mathrm{T}\,}$$
Check
Size check against the course ruler. Forty millitesla is about eight hundred times the field of the Earth, the range of a small laboratory electromagnet. Dropping the factor of two would give $0.0800\ \mathrm{T}$, also plausible, so the real check is on the physics.
4§12.3 — the voltage across the wings of an aircraft●●●○○
An aircraft of wingspan $35.0\ \mathrm{m}$ flies horizontally at $250\ \mathrm{m/s}$ where the vertical component of the Earth's field is $5.00\times10^{-5}\ \mathrm{T}$.
Given
wingspan $L = 35.0\ \mathrm{m}$
speed $v = 250\ \mathrm{m/s}$, horizontal
vertical component of the field $5.00\times10^{-5}\ \mathrm{T}$
Find
(a) Find the potential difference between the wingtips.
(b) State whether a current flows through the aircraft as a result, and why.
(c) Explain why the horizontal component of the Earth's field was not used.
Hint 1/4
The wing is a rod moving through a field. The only question is which component of the field counts.
Hint 2/4
$\varepsilon = BLv$ needs the field, the rod and the velocity to be mutually perpendicular; only the component of the field perpendicular to both contributes.
Hint 3/4
Here the wing is horizontal and across the motion and the velocity is horizontal, so the perpendicular component is the vertical one, $5.00\times10^{-5}\ \mathrm{T}$.
Hint 4/4
The voltage is $(5.00\times10^{-5})(35.0)(250) = 0.438\ \mathrm{V}$.
Show solutionPick the component, then substitute
$$\vec F = q\vec v\times\vec B \Rightarrow \text{only}\ B_{\perp}\ \text{pushes charge along the wing}$$
the force must lie along the wing, and only the vertical field crossed into the velocity does that
Size check. Half a volt from the Earth's field seems large until you notice thirty five metres of wing and a quarter of a kilometre per second. Cross check by the flux rule: the wing sweeps $8750\ \mathrm{m^{2}}$ per second, which at $5.00\times10^{-5}\ \mathrm{T}$ is $0.438\ \mathrm{Wb/s}$.
5§12.3 — a rod pivoted at one end and swept round●●●●○
A straight rod of length $0.600\ \mathrm{m}$ rotates in a horizontal plane about a vertical axis through one end, at $120$ revolutions per minute. A uniform field of $0.400\ \mathrm{T}$ is parallel to the axis, perpendicular to the plane the rod sweeps.
Given
rod length $L = 0.600\ \mathrm{m}$, pivoted at one end
rotation rate $120\ \mathrm{rev/min}$
$B = 0.400\ \mathrm{T}$, perpendicular to the plane of rotation
Find
(a) Find the angular speed in radians per second.
(b) Find the potential difference between the ends of the rod.
(c) Say which end is at the higher potential if the field points upwards and the rod turns anticlockwise seen from above.
Hint 1/4
Different parts of the rod move at different speeds, so $BLv$ has no single v. Ask what v is at a distance r from the pivot.
Hint 2/4
A short piece $dr$ at radius r moves at $v = \omega r$ and contributes $dV = B\omega r\,dr$; integrating from zero to L gives $\tfrac12 B\omega L^{2}$.
Hint 3/4
Here $\omega = 2\pi(120/60) = 12.6\ \mathrm{rad/s}$, $B = 0.400\ \mathrm{T}$ and $L = 0.600\ \mathrm{m}$.
Hint 4/4
The voltage is $\tfrac12(0.400)(12.6)(0.360) = 0.905\ \mathrm{V}$, with the far end positive.
Show solutionWhy the plain formula fails, and what replaces it
$$v(r) = \omega r$$
the pivot is stationary and the tip fastest, so no single speed describes the rod
$$d\varepsilon = B\,v(r)\,dr = B\omega r\,dr$$
each piece is a tiny rod at its own speed, and the pieces are in series
Independent check by the flux rule, which never mentions the integral. In one turn the rod sweeps a disc of area $1.131\ \mathrm{m^{2}}$ in $0.500\ \mathrm{s}$, so the flux swept per second is $0.905\ \mathrm{Wb/s}$.
6§12.4 — designing a coil for a wanted peak voltage●●●○○
A generator coil of area $2.00\times10^{-2}\ \mathrm{m^{2}}$ is to turn at $50.0$ revolutions per second in a field of $0.500\ \mathrm{T}$ and must give a peak emf of $170\ \mathrm{V}$.
Given
$A = 2.00\times10^{-2}\ \mathrm{m^{2}}$
$f = 50.0\ \mathrm{rev/s}$
$B = 0.500\ \mathrm{T}$
required peak emf $170\ \mathrm{V}$
Find
(a) Find the number of turns needed.
(b) State how many whole turns would actually be wound, and what peak emf that gives.
(c) Find the peak emf if the same coil were turned at $60.0$ revolutions per second instead.
Hint 1/4
This is the peak emf formula with a different unknown. Rearrange it before putting any numbers in.
Hint 2/4
$\varepsilon_0 = NBA\omega$ with $\omega = 2\pi f$, so $N = \varepsilon_0/(BA\omega)$.
Hint 3/4
Here $\varepsilon_0 = 170\ \mathrm{V}$, $B = 0.500\ \mathrm{T}$, $A = 2.00\times10^{-2}\ \mathrm{m^{2}}$ and $\omega = 2\pi(50.0) = 314\ \mathrm{rad/s}$.
Hint 4/4
The answer is $54.1$ turns, so 55 whole turns would be wound, giving $173\ \mathrm{V}$.
Independent check on part (c) by full substitution rather than scaling: $(55)(0.500)(2.00\times10^{-2})(2\pi)(60.0) = 207\ \mathrm{V}$, confirming both the proportionality and that no $2\pi$ went missing.
7§12.6 — sizing the secondary of a small transformer●●○○○
A transformer with 1200 turns on its primary is connected to a $240\ \mathrm{V}$ supply. Its secondary must deliver $12.0\ \mathrm{V}$ at $2.50\ \mathrm{A}$. Treat it as ideal.
Independent check through the current ratio, deliberately not used above. $I_s/I_p$ should equal $N_p/N_s = 20.0$, and $2.50/0.125 = 20.0$. A slip either way would show as a factor of four hundred.
8§12.7 — two places with the same induced electric field●●●●○
A long solenoid of radius $3.00\ \mathrm{cm}$ is wound with 1200 turns per metre. Its current is increased steadily at $30.0\ \mathrm{A/s}$.
(a) Find the rate at which the field inside is increasing.
(b) Find the induced electric field at $1.50\ \mathrm{cm}$ from the axis.
(c) Find the induced electric field at $6.00\ \mathrm{cm}$ from the axis, and comment on the two answers.
Hint 1/4
Two of the three parts need a decision before a formula: is the point inside the solenoid or outside it.
Hint 2/4
$dB/dt = \mu_0 n\,dI/dt$; then $E = (r/2)\,dB/dt$ inside and $E = R^{2}\,dB/dt/(2r)$ outside.
Hint 3/4
Here $R = 3.00\ \mathrm{cm}$, so $1.50\ \mathrm{cm}$ is inside and $6.00\ \mathrm{cm}$ is outside, with $n = 1200\ \mathrm{m^{-1}}$ and $dI/dt = 30.0\ \mathrm{A/s}$.
Hint 4/4
Both answers come out as $3.39\times10^{-4}\ \mathrm{V/m}$, which is not a coincidence.
only the current depends on time, so the geometry comes out as constants
Inside and outside, with the test made explicit each time
$$1.50\ \mathrm{cm} < R \Rightarrow E = \frac{r}{2}\frac{dB}{dt} = (7.50\times10^{-3})(4.52\times10^{-2}) = 3.39\times10^{-4}\ \mathrm{V/m}$$
writing the comparison first makes the branch a decision rather than an accident
$$6.00\ \mathrm{cm} > R \Rightarrow E = \frac{R^{2}}{2r}\frac{dB}{dt} = (7.50\times10^{-3})(4.52\times10^{-2}) = 3.39\times10^{-4}\ \mathrm{V/m}$$
the coefficient $R^{2}/2r$ comes out to $7.50\times10^{-3}$ as well, which is why the two answers coincide
Answer $$\boxed{\,\frac{dB}{dt} = 4.52\times10^{-2}\ \mathrm{T/s},\qquad E = 3.39\times10^{-4}\ \mathrm{V/m}\ \text{at both radii}\,}$$
Check
Independent check by algebra rather than arithmetic. The two branches give equal fields when $r_1r_2 = R^{2}$, and $(1.50)(6.00) = 9.00\ \mathrm{cm^{2}} = R^{2}$ exactly, confirming both numbers without recomputing either.
C · exam level 4 questions
1§12.4 — the shape of the output as a loop crosses a strip●●●○○
A square loop is pulled at constant speed straight through a strip of uniform field much wider than itself, so there is a stretch during which it is completely inside. The emf is recorded from before it reaches the strip until after it has left.
Given
a square loop, constant speed, moving perpendicular to the edges of the strip
the strip of field is wider than the loop
the field inside the strip is uniform and zero outside it
Find
(a) Which description matches the graph of induced emf against time?
Hint 1/4
Split the journey into three stages and answer the same question in each: is the flux through the loop growing, steady, or shrinking?
Hint 2/4
$\varepsilon = -d\Phi_B/dt$, so a growing flux and a shrinking flux give emfs of opposite sign and a steady flux gives none.
Hint 3/4
Here the flux grows while the loop enters, stays at $BL^{2}$ while it is fully inside, and falls while it leaves.
Hint 4/4
So the graph is a pulse one way, a flat stretch at zero, and a pulse the other way.
Show solutionEntering
$$\Phi_B = BLx,\qquad \varepsilon = BLv$$
x grows at the speed of the loop, so the emf is constant while entry lasts
the area gained at the front is the area lost at the back, so nothing changes
Leaving
$$\Phi_B = BL(L-y),\qquad \varepsilon = -BLv$$
the same size as at entry, the same edge crossing the same boundary, with the opposite sign
Answer $$\boxed{\,+BLv,\quad 0,\quad -BLv\,}$$
Check
Independent check by the areas under the graph. The total change in flux linkage is zero, since the loop starts and ends outside the field, and the area under an emf against time graph is exactly that change, so the two pulses must cancel.
2§12.3 — a loop falling into a field and slowing down●●●●○
A square loop of side $0.120\ \mathrm{m}$, mass $0.0150\ \mathrm{kg}$ and resistance $0.0400\ \Omega$ falls from rest with its plane vertical, entering bottom edge first a horizontal boundary below which there is a uniform field of $0.680\ \mathrm{T}$ perpendicular to the loop and above which there is none. Take $g = 9.80\ \mathrm{m/s^{2}}$.
Given
square loop, side $L = 0.120\ \mathrm{m}$, mass $0.0150\ \mathrm{kg}$, resistance $0.0400\ \Omega$
$B = 0.680\ \mathrm{T}$ below the boundary, zero above it
the loop falls with its plane vertical, bottom edge first
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the terminal speed the loop would reach while entering the field.
(b) Find the current at that speed.
(c) Find the rate at which the loop is generating heat there.
(d) State what happens to the motion once the loop is completely below the boundary, and why.
Hint 1/4
Terminal speed means the acceleration has become zero. Write the force balance before writing any induction formula.
Hint 2/4
$F_{\rm drag} = B^{2}L^{2}v/R$ and at terminal speed it equals $mg$, so $v_t = mgR/(B^{2}L^{2})$.
Hint 3/4
Here $m = 0.0150\ \mathrm{kg}$, $R = 0.0400\ \Omega$, $B = 0.680\ \mathrm{T}$ and $L = 0.120\ \mathrm{m}$.
Hint 4/4
The terminal speed is $0.883\ \mathrm{m/s}$, the current $1.80\ \mathrm{A}$, the heating $0.130\ \mathrm{W}$, and once fully inside the loop falls freely again.
Show solutionForce balance first, formula second
$$mg = \frac{B^{2}L^{2}v_t}{R}$$
zero acceleration, so weight and magnetic drag are equal; induction supplies only the drag
the rate gravitational energy turns into heat rather than speed
After the boundary
$$\text{both edges in the field} \Rightarrow \frac{d\Phi_B}{dt} = 0 \Rightarrow I = 0$$
the two horizontal edges give equal and cancelling emfs: the contrast pair again
$$\Rightarrow a = g\ \text{once more}$$
no current means no drag, so only gravity acts from that speed onwards
Answer $$\boxed{\,v_t = 0.883\ \mathrm{m/s},\quad I = 1.80\ \mathrm{A},\quad P = 0.130\ \mathrm{W},\quad \text{free fall afterwards}\,}$$
Check
Independent check on the force balance rather than the power. At the answer speed the drag is $BIL = 0.147\ \mathrm{N}$ and the weight is $mg = 0.147\ \mathrm{N}$, from different formulas and different data.
3§12.5 — a motor under load, from full speed down to a jam●●●●○
A direct current motor runs from a $240\ \mathrm{V}$ supply. Its armature resistance is $0.800\ \Omega$ and at full speed it draws $12.0\ \mathrm{A}$.
Given
$V = 240\ \mathrm{V}$
armature resistance $R = 0.800\ \Omega$
current at full speed $12.0\ \mathrm{A}$
the back emf is proportional to the speed
Find
(a) Find the back emf at full speed.
(b) Find the mechanical power delivered at full speed.
(c) Find the efficiency at full speed.
(d) A load drags the motor down to sixty per cent of full speed. Find the new current.
Hint 1/4
Everything here comes out of one loop equation. Write it first and identify which symbol each part is asking about.
Hint 2/4
$V = \varepsilon_b + IR$, mechanical power is $\varepsilon_b I$, and the supplied power is $VI$.
Hint 3/4
Here $V = 240\ \mathrm{V}$, $R = 0.800\ \Omega$ and $I = 12.0\ \mathrm{A}$ at full speed, with the back emf proportional to speed.
Hint 4/4
The back emf is $230\ \mathrm{V}$, the output $2760\ \mathrm{W}$, the efficiency $96.0\%$, and at sixty per cent of speed the current rises to $127\ \mathrm{A}$.
Independent check on part (d) by the power account. The supply then delivers $30500\ \mathrm{W}$ and the winding turns $12900\ \mathrm{W}$ into heat, leaving $17600\ \mathrm{W}$, which matches $\varepsilon_b'I'$.
4§12.6 — a generator, a line and two transformers●●●●○
A generating station produces $25.0\ \mathrm{MW}$ at $22.0\ \mathrm{kV}$. A transformer raises this to $275\ \mathrm{kV}$ for a line of total resistance $6.00\ \Omega$, and a second transformer at the far end brings it down. Treat both as ideal.
Given
$P = 25.0\ \mathrm{MW}$ generated at $22.0\ \mathrm{kV}$
line voltage $275\ \mathrm{kV}$, line resistance $6.00\ \Omega$
ideal transformers
Find
(a) Find the turns ratio of the step up transformer.
(b) Find the current in the transmission line.
(c) Find the power lost as heat in the line, and what percentage of the generated power that is.
(d) Find the loss if the same power were sent along the same line at the generating voltage of $22.0\ \mathrm{kV}$ instead.
Hint 1/4
The line resistance never has the line voltage across it. Decide which quantity the loss is computed from before starting.
Hint 2/4
$N_s/N_p = V_s/V_p$, then $I = P/V$ for the line, and $P_{\rm loss} = I^{2}R$.
Hint 3/4
Here $P = 25.0\ \mathrm{MW}$, the step up goes from $22.0\ \mathrm{kV}$ to $275\ \mathrm{kV}$, and $R = 6.00\ \Omega$.
Hint 4/4
The ratio is $12.5$ to one, the line current $90.9\ \mathrm{A}$, the loss $49.6\ \mathrm{kW}$ or $0.198\%$, and at the low voltage it would be $7.75\ \mathrm{MW}$.
Show solutionThe transformer and the line current
$$\frac{N_s}{N_p} = \frac{275}{22.0} = 12.5$$
voltages and turns are in the same ratio, and the power does not enter
Independent check on part (d) by scaling rather than squaring a large number. The loss goes as $1/V^{2}$, so raising the voltage by $12.5$ divides it by $156$, and $7.75\ \mathrm{MW}/156 = 49.7\ \mathrm{kW}$, matching part (c).
D · interleaved 4 questions
1§12.2 — a rheostat, a solenoid and a coil wound round it●●●●○
A long solenoid of radius $2.00\ \mathrm{cm}$ with 800 turns per metre and resistance $1.50\ \Omega$ is connected to a $12.0\ \mathrm{V}$ battery of internal resistance $0.500\ \Omega$ through a rheostat. A search coil of 50 turns and radius $3.00\ \mathrm{cm}$ is wound round the outside of the solenoid near its middle, with resistance $8.00\ \Omega$. The rheostat is adjusted so that the solenoid current rises steadily from $1.00\ \mathrm{A}$ to $4.00\ \mathrm{A}$ over $0.250\ \mathrm{s}$.
solenoid current rises from $1.00\ \mathrm{A}$ to $4.00\ \mathrm{A}$ in $0.250\ \mathrm{s}$
Find
(a) Find the resistance of the rheostat at the start, when the current is $1.00\ \mathrm{A}$.
(b) Find the change in the field inside the solenoid.
(c) Find the emf induced in the search coil.
(d) Find the current in the search coil.
Hint 1/4
Two radii are given and only one of them belongs in the flux. Decide which before doing any arithmetic.
Hint 2/4
For the circuit, $I = \mathcal{E}/(R_{\rm total})$; for the field, $B = \mu_0 n I$; for the emf, $\varepsilon = N A\,dB/dt$ with A the area through which the field actually passes.
Hint 3/4
Here the solenoid has radius $2.00\ \mathrm{cm}$ and the search coil $3.00\ \mathrm{cm}$, and there is no field outside the solenoid.
Hint 4/4
The rheostat is $10.0\ \Omega$, the field changes by $3.02\times10^{-3}\ \mathrm{T}$, the emf is $7.58\times10^{-4}\ \mathrm{V}$ and the current $9.47\times10^{-5}\ \mathrm{A}$.
a hundred microamperes, which is why a search coil needs a sensitive meter
Answer $$\boxed{\,R = 10.0\ \Omega,\quad \Delta B = 3.02\times10^{-3}\ \mathrm{T},\quad \varepsilon = 7.58\times10^{-4}\ \mathrm{V},\quad I = 9.47\times10^{-5}\ \mathrm{A}\,}$$
Check
Independent check by a limiting case. Wind the search coil with a radius of a metre and nothing physical changes, since all the field is inside the solenoid, so the emf must be unchanged. Only the formula using the solenoid's area has that property.
2§12.3 — a sliding rod and the field its own current makes●●●●○
A rod of length $0.250\ \mathrm{m}$ slides at $6.00\ \mathrm{m/s}$ along frictionless rails in a uniform field of $0.400\ \mathrm{T}$ perpendicular to their plane. The circuit has total resistance $0.500\ \Omega$. One of the leads carrying the current away is long and straight.
$\mu_0/2\pi = 2\times10^{-7}\ \mathrm{T\,m/A}$; the field of the Earth is about $5\times10^{-5}\ \mathrm{T}$
Find
(a) Find the current in the circuit.
(b) Find the magnetic field $4.00\ \mathrm{cm}$ from that long straight lead, at a place far from everything else.
(c) Compare that field with the field of the Earth.
(d) Find the mechanical power being supplied to the rod, and check it against the electrical power.
Hint 1/4
Nothing about the field of the lead can be started until the current is known, and the current is an induction question.
Hint 2/4
$\varepsilon = BLv$ and $I = \varepsilon/R$ for the first part, then $B = \mu_0 I/(2\pi r)$ for the second, and $P = Fv$ against $P = I^{2}R$ for the last.
Hint 3/4
Here $B = 0.400\ \mathrm{T}$, $L = 0.250\ \mathrm{m}$, $v = 6.00\ \mathrm{m/s}$, $R = 0.500\ \Omega$ and the field point is $4.00\times10^{-2}\ \mathrm{m}$ from the lead.
Hint 4/4
The current is $1.20\ \mathrm{A}$, the field $6.00\times10^{-6}\ \mathrm{T}$, about an eighth of the Earth's, and both powers are $0.720\ \mathrm{W}$.
The power agreement is the main check and the two sides share no line. A cheaper second one on part (b): doubling the distance should halve the field, and at $8.00\ \mathrm{cm}$ the formula gives $3.00\times10^{-6}\ \mathrm{T}$.
3§12.2 — an induced emf charging a capacitor●●●●○
A coil of 300 turns, each of area $5.00\times10^{-3}\ \mathrm{m^{2}}$, lies with its plane perpendicular to a field increasing at a steady $0.800\ \mathrm{T/s}$. Its ends are joined through a $500\ \Omega$ resistor to an uncharged $25.0\ \mu\mathrm{F}$ capacitor. The coil's own resistance is negligible.
Given
$N = 300$ turns, $A = 5.00\times10^{-3}\ \mathrm{m^{2}}$, plane perpendicular to the field
$dB/dt = 0.800\ \mathrm{T/s}$, steady
$R = 500\ \Omega$ in series with $C = 25.0\times10^{-6}\ \mathrm{F}$
the capacitor starts uncharged
Find
(a) Find the emf of the coil.
(b) Find the current at the instant the connection is made.
(c) Find the final charge on the capacitor and the energy stored in it.
(d) Find the time constant of the charging.
Hint 1/4
The coil is a battery of fixed voltage, because the field rises at a steady rate. After that it is an ordinary charging problem.
Hint 2/4
$\varepsilon = NA\,dB/dt$; then $I_0 = \varepsilon/R$, $Q = C\varepsilon$, $U = \tfrac12 C\varepsilon^{2}$ and $\tau = RC$.
The emf is $1.20\ \mathrm{V}$, the initial current $2.40\ \mathrm{mA}$, the final charge $3.00\times10^{-5}\ \mathrm{C}$ storing $1.80\times10^{-5}\ \mathrm{J}$, and the time constant $1.25\times10^{-2}\ \mathrm{s}$.
Independent check on the energy through the source rather than the capacitor formula. The source pushes $3.00\times10^{-5}\ \mathrm{C}$ round at $1.20\ \mathrm{V}$, doing $3.60\times10^{-5}\ \mathrm{J}$; exactly half ends up stored, the signature of charging through a resistance.
4§12.7 — a field with no charges behind it●●●●●
A large coil makes a uniform field over a circular region of radius $0.150\ \mathrm{m}$, increasing at a steady $2.40\ \mathrm{T/s}$. A free electron sits at rest $0.100\ \mathrm{m}$ from the axis of the region.
Given
field region of radius $0.150\ \mathrm{m}$, $dB/dt = 2.40\ \mathrm{T/s}$
the electron is at $r = 0.100\ \mathrm{m}$ from the axis, initially at rest
(a) Find the induced electric field at the electron.
(b) Find the electron's initial acceleration, and say in which direction it starts to move.
(c) Find the point charge that would have to sit $1.00\ \mathrm{m}$ away to produce a field of the same size.
(d) Explain in one sentence why no arrangement of charges could reproduce the field of part (a) everywhere.
Hint 1/4
Parts (a) and (c) use two different laws for the same quantity, and part (d) asks what stops them being the same thing.
Hint 2/4
$E = (r/2)\,dB/dt$ inside the region; $F = eE$ and $a = F/m$; and $E = kQ/d^{2}$ for a point charge.
Hint 3/4
Here $r = 0.100\ \mathrm{m}$ is inside the region of radius $0.150\ \mathrm{m}$, $dB/dt = 2.40\ \mathrm{T/s}$, and the comparison distance is $1.00\ \mathrm{m}$.
Hint 4/4
The field is $0.120\ \mathrm{V/m}$, the acceleration $2.11\times10^{10}\ \mathrm{m/s^{2}}$ along the circle through the electron, and the equivalent point charge is $1.33\times10^{-11}\ \mathrm{C}$.
Independent check on part (a) through the emf rather than the field. The emf round the circle is $7.54\times10^{-2}\ \mathrm{V}$, and dividing by the circumference $0.628\ \mathrm{m}$ gives $0.120\ \mathrm{V/m}$.
Mistake ledger (32 entries)
⚠ Taking the angle from the plane of the loop instead of from its normal
questions describe orientations however sounds natural in English, and a coil's plane is easier to picture than a normal sticking out of it
wrong$$\Phi_B = BA\cos(\text{angle to the plane})$$
right$$\Phi_B = BA\cos(\text{angle to the normal})$$
⚠ Multiplying by the number of turns inside the flux
linkage and flux are one letter apart and both quoted in webers, so the N gets absorbed early and applied again when the emf is computed
⚠ Equating the weight to a force per unit length or to a power
three quantities have to balance in terminal speed problems, and the units of what is being balanced get lost when the algebra comes before the sentence
wrong$$mg = \frac{B^{2}L^{2}v^{2}}{R}$$
right$$mg = \frac{B^{2}L^{2}v}{R}$$
⚠ Leaving out the factor of two pi
machines are specified in revolutions per second or minute, and the formula wants radians per second
wrong$$\varepsilon_0 = NBA f$$
right$$\varepsilon_0 = NBA(2\pi f)$$
⚠ Putting the peak emf where the flux peaks
both are called maxima and the flux was computed first, so it is the one on the page when the sentence gets written
right$$\oint\vec E\cdot d\vec l \neq 0 \Rightarrow \text{no potential exists}$$
⚠ Expecting the induced electric field to point away from the axis
radial is what electric fields look like in every picture from the first weeks of the course, and the axis looks like a centre to point away from
wrong$$\vec E_{\rm ind} \parallel \hat r$$
right$$\vec E_{\rm ind}\ \text{tangential, in closed circles round the axis}$$
⚠ Adding the magnitudes of two flux values instead of subtracting them, or the reverse
when the flux reverses the two magnitudes have to add, and a habit of subtracting whatever is on the page halves the answer silently
wrong$$|\Delta\Phi_B| = |\Phi_f| - |\Phi_i|$$
right$$\Delta\Phi_B = \Phi_f - \Phi_i\ \text{with signs, then take the size}$$
⚠ Forgetting the factor of two when a coil is flipped through half a turn
the coil looks the same after a flip and the word does not sound like a reversal, but the flux has gone from plus BA to minus BA
wrong$$|\Delta\Phi_B| = BA$$
right$$|\Delta\Phi_B| = 2BA$$
⚠ Using the tip speed for a rod that is pivoted at one end
the plain formula assumes every part of the rod moves at the same speed, and on a pivoted rod the inner parts are slower, which the integral accounts for and the shortcut does not
both windings threaded by the same changing flux; no losses; the supply must be changing with time
Heating loss on a transmission line
$$P_{\rm loss} = I^{2}R = \frac{P^{2}R}{V^{2}}$$
R is the resistance of the line itself; V is the line voltage and P the power being delivered
Induced electric field
$$\oint\vec E\cdot d\vec l = -\frac{d\Phi_B}{dt}$$
any closed path, with or without a conductor on it; E can only be extracted when the symmetry is circular
Induced electric field of a solenoid being ramped
$$E = \frac{r}{2}\frac{dB}{dt}\ (r\le R),\qquad E = \frac{R^{2}}{2r}\frac{dB}{dt}\ (r\ge R)$$
long solenoid of radius R, uniform field inside and negligible outside
Check yourself
Close the page and write out from memory: the definition of magnetic flux and which angle goes into it; Faraday's law in both forms and the charge a flux change moves; the four moves that fix the direction of an induced current; the motional emf of a straight rod and of a pivoted one, with the drag and the terminal speed; the emf of a rotating coil and where in the cycle it peaks; the loop equation for a motor and what it says about starting current; the two transformer ratios and which way up each goes; and the induced electric field inside and outside a ramped solenoid.
Compute the flux when the question describes the orientation by the plane rather than the normal, and set up the integral for a loop beside a straight wire?
c-flux
Get an emf, a current and a total charge from a flux change, handle a flux that reverses sign, and state the direction with a viewpoint?
c-faraday-lenz
Find the emf, current, drag and terminal speed for a conductor on rails, and show in one line that the two powers match?
c-motional
Convert a rotation rate into radians per second unprompted, compute a peak emf, and say without hesitating where in the cycle the emf is largest?
c-generator
Explain why a motor draws its largest current at startup, compute it, and split the supplied power into output and heat?
c-backemf
Get transformer turns, voltages and currents the right way up, and compute a line loss from the current rather than the line voltage?
c-transformer
Compute the induced electric field inside and outside a solenoid, say where it is largest, and explain why a voltmeter reading depends on where its leads pass?
c-induced-e
Glossary (23 terms)
magnetic fluxmanyetik akı
The amount of magnetic field passing through a surface, $\Phi_B = \int\vec B\cdot d\vec A$, which for a flat loop in a uniform field is $BA\cos\theta$ with $\theta$ measured from the normal. It belongs to a surface, not to a point.
weberweber
The SI unit of magnetic flux, one tesla times one square metre. One weber per second through a single turn is one volt.
flux linkageakı halkalanması
The flux through one turn multiplied by the number of turns, $N\Phi_B$, in webers. Its rate of change is the emf of the whole coil, and it is meaningful only when every turn encloses the same flux.
yüzey normali
The direction perpendicular to a surface, chosen by whoever is doing the problem, which fixes the positive side of the loop and, through the right hand, the positive sense of circulation round it.
Faraday's law of induction
The statement that the emf round a circuit is minus the rate of change of the magnetic flux linkage through it. It does not care why the flux is changing, and the flux itself never appears, only its rate of change.
Lenz's law
The rule contained in the minus sign of Faraday's law: an induced current opposes the change in flux that produced it. It follows from energy conservation, since the opposite convention would let energy appear from nothing.
induced emfindüklenen emk
The work done per unit charge in taking a charge once round a circuit, produced here by a changing magnetic flux rather than by a chemical cell. Measured in volts, and present whether or not a current can flow.
induced currentindüklenen akım
The current an induced emf drives round a closed circuit, the emf divided by the total resistance. Quoting one without saying which way round is an incomplete answer.
motional emf
The emf produced when a conductor moves through a magnetic field, of size $BLv$ for a rod moving perpendicular to both. It follows either from the flux swept out or from the magnetic force on the free charges.
magnetic brakingmanyetik frenleme
The slowing of a conductor moving in a magnetic field by the force acting on the current the motion itself induces. The force grows with speed, which is why it gives a terminal speed rather than a stop.
terminal speedsınır hız
The steady speed reached when a speed dependent retarding force has grown to balance the driving force. For a conductor falling on rails it is $mgR/(B^{2}L^{2})$.
eddy currentgirdap akımı
A closed loop of current circulating inside a solid conductor, induced by a changing flux through it. It needs a good conductor rather than a magnetic material, and it is behind magnetic damping and induction heating.
generatorjeneratör
A machine producing an emf by turning a coil in a magnetic field, giving $NBA\omega\sin\omega t$, which reverses once per half turn. Drawing current from it makes it harder to turn, by the amount that pays for the current.
peak emftepe emk
The largest value an alternating emf reaches in a cycle, $NBA\omega$ for a rotating coil. It is not the value quoted for a domestic supply.
back emfzıt emk
The emf a motor generates because its armature is spinning, opposing the supply that drives it. It is proportional to the speed, so it is zero at switch on, which is when the current is largest.
counter torque
The torque a field exerts on a generator coil because of the current it is delivering, always opposing the rotation. It is why a dynamo becomes hard to turn the moment its lamp is switched on.
armatureendüvi
The rotating winding of a motor or generator, where the emf is induced or the driving torque acts. Its resistance is treated as a battery's internal resistance is.
transformertransformatör
Two windings on a shared core, so that the same changing flux threads both. It changes voltage and current in inverse proportion, leaves the power unchanged, and does nothing with a steady supply.
turns ratiosarım oranı
The ratio of secondary turns to primary turns in a transformer, equal to the ratio of the voltages and the inverse of the ratio of the currents.
step up transformeryükseltici transformatör
A transformer with more turns on the secondary than the primary, giving higher voltage and smaller current. It is what puts power onto a transmission line.
induced electric field
The electric field produced by a changing magnetic flux, with field lines that close on themselves and a circulation round a closed loop that is not zero. No potential function exists for it.
search coilarama bobini
A small coil of known area and turns, used to measure a field by flipping or removing it and measuring the charge that flows. It works because that charge does not depend on how fast the movement was.
A field in which the work done on a charge taken round a closed path is not zero, so no potential energy function can be defined. The induced electric field is the first example in this course.
What comes next
§13 · Inductance, Electromagnetic Oscillations, and AC Circuits
Every changing flux here was driven from outside: a magnet moved, a rod slid, another coil was ramped. The next section asks what happens when the flux through a circuit is made by that circuit's own current, so that it reacts to itself, and what that does to a coil and a capacitor left alone together.
Sources
D. Giancoli, Physics for Scientists and Engineers with Modern Physics, 5th edition The set textbook. Its treatment of electromagnetic induction covers the same ground, and its end of chapter problems on loops entering field regions and on loaded transformers run a level harder than these, which is the right next step.
Course syllabus, week 12 line The scope of this section comes from the week line, Electromagnetic Induction and Faraday's Law, and nowhere else. The line carries no chapter number, so no chapter or section number is quoted anywhere on this page, and later week lines have been left out rather than previewed.
Course syllabus, catalogue description and assessment weights The catalogue description lists Faraday's law among the course topics, which is what places this material inside it. The exam note on the card uses only the published weights: two midterms at twenty per cent each, quizzes ten, homework five, the final twenty five, laboratory twenty.
SI units and constants The weber is used as the SI unit of magnetic flux throughout, one weber being one tesla square metre. The permeability of free space is $4\pi\times10^{-7}\ \mathrm{T\,m/A}$, and $g$, the elementary charge and the electron mass take the same values as in earlier sections.