7 concepts20 worked examples30 exercises4 exam-level7 figures
What are you here for?
10Review and Consolidation II: capacitance, current, DC circuits, and magnetism
One figure, four weeks of material. A battery drives a small network; somewhere in it a capacitor is filling up, and one of the branches is a straight wire lying across the gap of a magnet. The last line asks for a single number, the force on that wire a long time after the switch closed, and nothing in the figure tells you where to start.
By the end of this section you can take any question built out of weeks six to nine, say in one sentence which quantity it is really after and which route is cheapest, carry it through with units attached at every step, and finish with a check that catches a dropped factor of two before the paper leaves your hand.
In 60 seconds
Four weeks reduce to one habit: work out what the charge is doing, then reach for the formula family that answers the question actually asked.
Capacitance, and the charge it parks
$$C = \frac{Q}{V},\qquad Q = CV$$
any pair of conductors with equal and opposite charge on them; C depends on the geometry and the filling, never on how much charge you put there
Current, , and power
$$I = \frac{\Delta Q}{\Delta t},\qquad V = IR,\qquad R = \frac{\rho L}{A},\qquad P = IV = I^{2}R = \frac{V^{2}}{R}$$
charge is on the move: any question naming a current, a voltage drop, a heating rate or an energy bill
a switch is thrown and the question names a time, or asks for the value at the instant of closing or a long time later
The magnetic force, and the circle it makes
$$\vec F = q\,\vec v\times\vec B,\qquad \vec F = I\,\vec L\times\vec B,\qquad r = \frac{mv_\perp}{\vert q\vert B},\qquad T = \frac{2\pi m}{\vert q\vert B}$$
anything moving, or any current, sitting in a magnetic field; the force is always at right angles to both the motion and the field
Loops, crossed fields, and the Hall bar
$$\tau = NIAB\sin\theta,\qquad v = \frac{E}{B},\qquad V_H = \frac{IB}{nqt}$$
a coil free to turn, a beam that goes straight through both fields, or a strip with a sideways voltage across it
Three most common mistakes
Combining capacitors with the resistor rule. Two 6 microfarad capacitors in series make 3 microfarads, not 12. Nothing in the arithmetic complains, the units still come out right, and the rest of the question is then quietly wrong.
Treating a capacitor as if it had one behaviour. It has three: at the instant an empty one is switched on it behaves as a plain wire, a long time later it behaves as a break in the circuit, and in between it is neither of those.
Feeding a magnetic formula a current that was never checked against the circuit. The force formula is usually the easy half; the marks are lost upstream, working out how much current is really in that branch.
The published weights for this course are Midterm 1 twenty per cent, Midterm 2 twenty per cent, quizzes ten per cent, homework five per cent, the final twenty five per cent and the laboratory twenty per cent. Everything on this page is revision of material already covered; no new result is introduced here.
How much time do you have?
10 minutes
You walk away with the map, the route rule and the formula card, which between them decide the opening two lines of nearly every question in this block.
The 60 second card · Formula card · Five quantities, and the arrows that turn one into another · Choosing the route in the first thirty seconds · Mistake ledger
45 minutes
You add the two places where marks actually leak: the mirrored combination rules, and knowing which instant of a timed circuit the question is standing at. Then the ladder tells you whether you can do it without scaffolding.
The 60 second card · Five quantities, and the arrows that turn one into another · Choosing the route in the first thirty seconds · Series and parallel, mirrored · Networks, and the clock inside them · The magnetic force: square to the motion, and useless for speed · Fading ladder · Mistake ledger
full read
Everything above plus the two energy accounts, the four magnetic templates, the exam length worked example and the four practice tiers. The interleaved set at the end is the one that predicts your exam mark, because there nobody tells you which week the question came from.
Hook and card · Conventions · Pre test · All seven concepts · Method boxes · Contrast pairs · Fading ladder · Exam length example · Practice A to D · Check yourself
By the end of this section
Translate between charge, current, resistance and energy for any element in a circuit, and name which definition or law you are using at each step.
Choose, before computing anything, which of the four routes a question wants and which instant of the circuit it is standing at.
Combine resistors and capacitors in series and in parallel without mixing the two sets of rules, and find the voltage on any single element afterwards.
Account for energy in a circuit, separating what is stored in a capacitor from what is dissipated in a resistor, and quote each with the right unit.
Solve a network that will not reduce, using the junction and loop rules, and read a timed circuit correctly at the instant of switching and in the steady state.
Apply the magnetic force to a moving charge and to a current carrying wire, including its direction, and state correctly what it does and does not change.
Recognise which magnetic template a question belongs to, from the helix and the velocity selector to the turning loop and the Hall strip, and carry it through.
Syllabus coverage
Catch up and Review
The four weeks behind us: capacitance and dielectrics, and resistance, direct current circuits with Kirchhoff's rules and the RC circuit, and magnetism, meaning the force on moving charge, on current carrying wires and on current loops.
The syllabus line for this week carries no chapter numbers, so no chapter number is quoted anywhere on this page. Everything reviewed here has already been introduced; nothing beyond that material is used.
covered
Recall first
Capacitance of a parallel plate capacitor
$C = \dfrac{\varepsilon_0 A}{d}$ for two plates of area $A$ a distance $d$ apart in vacuum. The field between them is uniform and equal to $V/d$.
One interleaved question builds a capacitor out of a plate area and a gap and asks for the capacitance, the charge and then the force on an electron inside it; without this the geometry cannot be turned into a number.
's law and
$V = IR$ for an ohmic conductor, with $R = \dfrac{\rho L}{A}$: the resistance grows with length, falls with cross sectional area, and $\rho$ is a property of the material alone.
Two practice questions build a resistor out of a described piece of wire rather than handing you an ohm value, and the resistivity route is the only way in.
Terminal voltage of a real source
$V_{ab} = \varepsilon - Ir$ while the source delivers current, and $V_{ab} = \varepsilon + Ir$ while it is being charged. The terminal voltage equals the emf only when no current flows.
The exam length example and one of the ladder rungs both quote an internal resistance, and both would be wrong by several volts without it.
The uniform field between charged plates
Between two large parallel plates a distance $d$ apart with a potential difference $V$ across them, the field is uniform with magnitude $E = V/d$, pointing from the positive plate to the negative one.
The crossed field template needs $E$, and questions usually give a plate voltage and a gap rather than a field strength.
Work and kinetic energy from a potential difference
A charge $q$ moving through a potential difference $\Delta V$ has its kinetic energy changed by $\Delta K = -q\,\Delta V$, so a charge accelerated from rest through $V$ arrives with $\tfrac12 mv^{2} = \vert q\vert V$.
One interleaved question accelerates a particle first and then bends it, and the speed handed to the magnetic stage comes from this relation.
The dot product and the cross product
$\vec a\cdot\vec b = ab\cos\theta$ is a number, largest when the two are parallel and zero when they are perpendicular. $\vec a\times\vec b$ is a vector of magnitude $ab\sin\theta$, perpendicular to both, zero when they are parallel.
The statement that a magnetic force does no work is a dot product being zero, and every direction in this week's material is a cross product.
Try it yourself first (3 questions)
1§10.3 — two capacitors joined end to end●●○○○
Before starting, a check on which rules are where. Two capacitors are joined in series, and you want the single capacitance that could replace the pair. Getting this wrong is not a problem; it is the single most common slip in this material and the reason for the third concept below.
in a series chain each capacitor carries the same charge, and the charge sits underneath in $V = Q/C$, so it is $1/C$ that adds
$$C = \frac{6.00}{5} = 1.20\ \mathrm{\mu F}$$
the reciprocal at the end is the step most often forgotten
Answer $$\boxed{\;C = 1.20\ \mathrm{\mu F}\;}$$
Check
Bound check: a series combination must be smaller than every member, and $1.20 < 2.00$. The parallel answer $5.00\ \mathrm{\mu F}$ fails that test instantly.
2§10.5 — reading a circuit a long time after switching●●○○○
A second check, this time on instants. A battery, a resistor and an initially empty capacitor are wired in series and the switch is closed.
Given
$\varepsilon = 10.0\ \mathrm{V}$, ideal
$R = 5.00\ \mathrm{k\Omega}$
$C = 4.00\ \mathrm{\mu F}$, uncharged at $t = 0$
the reading is taken a long time after closing
Find
(a) What is the current in the loop then?
(b) What is the voltage across the capacitor then?
Hint 1/4
Ask what the phrase a long time is doing to the circuit, rather than reaching for the exponential.
Hint 2/4
In the steady state the charge on the capacitor has stopped changing, so $I = dQ/dt = 0$, and the loop rule then puts the whole emf across the capacitor.
Hint 3/4
With $I = 0$ there is no drop across the $5.00\ \mathrm{k\Omega}$ resistor, so $V_C = \varepsilon = 10.0\ \mathrm{V}$.
Hint 4/4
$I = 0$ and $V_C = 10.0\ \mathrm{V}$.
Show solutionSteady state means nothing is changing
$$I = \frac{dQ}{dt} = 0$$
the charge on the capacitor has reached its final value, so no more charge is crossing
$$V_R = IR = 0$$
a resistor with no current in it drops no voltage, whatever its resistance
Check the stored charge for consistency: $Q = CV_C = 40.0\ \mathrm{\mu C}$, and the time constant $\tau = RC = 0.0200\ \mathrm{s}$ tells you a long time here means a tenth of a second.
3§10.6 — what a magnetic field does to a speed●●○○○
The trap in this one is deliberate and most people fall into it the first time. A charged particle is fired into a region of strong uniform magnetic field and follows a curved path.
Given
a uniform magnetic field, constant in time
a charged particle, no other force present
the path through the region is curved
Find
(a) True or false: the particle leaves the region moving faster than it entered.
Hint 1/4
The path curves, so something is certainly accelerating the particle. Ask whether that acceleration has any component along the direction of travel.
Hint 2/4
The magnetic force is $q\vec v\times\vec B$, which is perpendicular to $\vec v$, so the power it delivers is $\vec F\cdot\vec v = 0$.
Hint 3/4
Zero power for the whole journey means the kinetic energy at the exit equals the kinetic energy at the entry, however curved the path was in between.
Hint 4/4
So it leaves at exactly the speed it entered with, and the statement is false.
Show solutionSeparate turning from speeding up
$$\vec F = q\,\vec v\times\vec B \perp \vec v$$
a cross product is perpendicular to both factors, so the force has no component along the motion
$$P = \vec F\cdot\vec v = 0 \ \Rightarrow\ \Delta K = 0$$
no power delivered over any interval, so the kinetic energy is the same at exit as at entry
Answer $$\boxed{\;\text{False: the speed is unchanged}\;}$$
Check
Consistency with the geometry: the radius $r = mv/(\vert q\vert B)$ is constant only while $v$ is, and the paths observed in these fields are circles, not spirals.
Notation
symbol
reads as
means
watch out
$C$
C
capacitance, in farads; a property of two conductors and whatever fills the space between them
not the coulomb, which is the unit $\mathrm{C}$ in upright type; a sentence with both in it is worth reading twice
$Q$
Q
the magnitude of the charge on one plate of a capacitor, or the charge that has passed a point in a circuit
in a series chain of capacitors every capacitor carries the same $Q$, which is the fact that makes the series rule what it is
$I$
I
the current in a branch, in , taken as the flow of positive charge
the same symbol is used for the steady value and for the value at one instant; where both matter the time dependent one is written $I(t)$
$\varepsilon$
epsilon
the electromotive force of a source, in volts: the energy it gives each coulomb before any internal loss
not the same as the terminal voltage $V_{ab}$, which is smaller whenever the source is delivering current
$\rho$
rho
resistivity, in ohm metres, a property of the material alone
the same Greek letter was used for volume charge density earlier in the course; here it always carries units of $\Omega\,\mathrm{m}$
$\tau$
tau
in a circuit, the time constant $RC$ in seconds; in a magnetic problem, the torque on a loop in newton metres
the two meanings never appear in the same equation, and the units tell them apart instantly
$\vec B$
B vector
the magnetic field at a stated point, a vector measured in teslas
$B$ without the arrow is its magnitude and is never negative; a minus sign in front of it is a statement about direction
$\vec\mu$
mu vector
the magnetic dipole moment of a current loop, $NI\vec A$, pointing along the normal given by curling the fingers with the current
its direction is perpendicular to the plane of the loop, not in it, which is where most torque sign errors are born
$n$
n
the number of mobile charge carriers per cubic metre in a conductor
distinct from $N$, the number of turns in a coil; the two appear within a page of each other in this material
$v_d$
v sub d
drift speed, the slow average crawl of the carriers, typically a fraction of a millimetre per second
not the speed at which the current starts flowing round the circuit, which is close to the speed of light
Conventions used here
Which quantities this review holds fixed
Wires have no resistance, ammeters none and voltmeters infinite resistance, unless the question says otherwise. A source is ideal unless an internal resistance is quoted next to it. Capacitors have no leakage. When a real component is meant, the question says so and the number is given.
Current direction and loop signs, restated for this review
Every current arrow is a conventional current: the direction positive charge would drift. Assumed directions may be wrong, and a negative answer simply means the arrow was drawn backwards, which is not an error to be corrected. Walking a loop, a resistor traversed with the arrow contributes $-IR$ and against it $+IR$; a source traversed from its short plate to its long plate contributes $+\varepsilon$ and the other way $-\varepsilon$.
Q and V for a capacitor, restated for this review
$Q$ always means the magnitude of the charge on one plate, not the sum of the two, which is zero. $V$ always means the magnitude of the potential difference between the plates. Both are quoted as positive numbers, and the sign work is done separately when a capacitor sits inside a loop.
Axes, the page, and the angle in magnetic formulas here
The page is the $xy$ plane, $x$ to the right and $y$ upwards, so $z$ points out of the page towards the reader. Dots mark a field out of the page and crosses a field into it. The angle $\theta$ in $F = \vert q\vert vB\sin\theta$ and $F = BIL\sin\theta$ is measured between the velocity, or the wire, and the field; in $\tau = \mu B\sin\theta$ it is measured between the loop's dipole moment and the field, which is ninety degrees away from the angle between the loop's plane and the field.
Constants and digits used on this review page
$e = 1.60\times10^{-19}\ \mathrm{C}$, $m_p = 1.67\times10^{-27}\ \mathrm{kg}$, $m_e = 9.11\times10^{-31}\ \mathrm{kg}$, $m_\alpha = 6.64\times10^{-27}\ \mathrm{kg}$, $1\ \mathrm{u} = 1.66\times10^{-27}\ \mathrm{kg}$, $\varepsilon_0 = 8.85\times10^{-12}\ \mathrm{C^{2}/(N\,m^{2})}$. Answers are quoted to three significant figures throughout, and intermediate values are carried at full precision.
What counts as ideal on this page
A capacitor labelled with one number is a single capacitance with no series resistance of its own. A resistor is ohmic and its value does not drift with temperature unless a temperature coefficient is given. A magnetic field described as uniform has the same magnitude and direction everywhere in the stated region and is exactly zero outside it.
Time in a circuit: which instant a symbol refers to
Where a circuit is timed, $t = 0$ is the instant the switch changes state. A symbol with no time attached means the steady state, that is, the situation a long time after the last switching, when nothing is changing any more. Where both are wanted the question asks for them separately, and this page keeps them separate too.
10.1Five quantities, and the arrows that turn one into another
Charge either sits on plates or moves along a wire, and every formula belongs to one state or the other.
Four weeks have each handed over their own formulas, and nothing has yet said how the sets fit together.
Solvable with what we have
the capacitance of two conductors and the energy in the gap
the current a battery pushes through a resistor chain
every branch current in a two loop network
the force on a wire, once told its current
Not solvable yet
the force on a wire whose current the circuit must supply
how long after switching the stored energy reaches half
how much of a battery's energy was stored, how much burned
which route a question wants, when it will not say
Try the first. A $12.0\ \mathrm{V}$ battery drives $R_1 = 3.00\ \mathrm{\Omega}$ in series with $R_2 = 5.00\ \mathrm{\Omega}$, and the second is a wire of length $0.250\ \mathrm{m}$ lying square across a field of $0.400\ \mathrm{T}$. Start with the part we know: the wire has $5.00\ \mathrm{\Omega}$ across $12.0\ \mathrm{V}$, so $I = 2.40\ \mathrm{A}$ and $F = BIL = 0.240\ \mathrm{N}$.
Why it fails
The current in a series element is set by the whole loop, not by that element. The loop here has $8.00\ \mathrm{\Omega}$ in it, so the real current is $1.50\ \mathrm{A}$ and the real force is $0.150\ \mathrm{N}$. The magnetic formula was never in trouble; the number fed into it came from a circuit that does not exist.
RuleRule 10.1: the two definitions the rest of the block leans on
Conditions
the current is the same at every cross section of a single unbranched element, because charge does not pile up anywhere in a wire once things are steady
the capacitance belongs to the two conductors and to whatever fills the gap; putting more charge on them does not change it
both statements are definitions rather than laws, so no experiment can violate them; what an experiment can do is tell you whether $V$ and $I$ are proportional, which is a separate claim
The current is how many coulombs cross a chosen surface each second. The capacitance is how many coulombs a pair of conductors takes per volt across them, so a large capacitance swallows a lot of charge for very little voltage.
Charge sitting still is on the left, charge on the move is on the right, and each arrow carries the formula that makes that particular trip. A question is a request to walk one of these arrows.
Looks like this, but is not
The resistor uses up some of the current, so less comes back to the battery than left it. It matches the way people talk about a bulb consuming current, and the arithmetic never contradicts it out loud.
Put an ammeter each side of the $5.00\ \mathrm{\Omega}$ resistor and both read $1.50\ \mathrm{A}$. What it takes is energy, not charge: it turns $11.25\ \mathrm{W}$ into heat while every coulomb that goes in comes out again. Charge is conserved, energy is not free, and those two sentences are the junction rule and the loop rule.
Current first, then force, on a 25.0 cm wire in a 0.400 T field
A $12.0\ \mathrm{V}$ battery of negligible internal resistance drives $R_1 = 3.00\ \mathrm{\Omega}$ in series with $R_2 = 5.00\ \mathrm{\Omega}$. The second resistor is a straight wire $0.250\ \mathrm{m}$ long lying at right angles to a uniform magnetic field of $0.400\ \mathrm{T}$. Find the magnetic force on it.
series, so the same current passes through both and the resistances add; the wire cannot be treated on its own because the current has to get past $R_1$ first
the wire is square to the field, so $\sin\theta = 1$ and the force takes its largest possible value for this current
$$F = 0.150\ \mathrm{N}$$
three significant figures, matching the least precise datum given
Answer $$\boxed{\;F = 0.150\ \mathrm{N}\;}$$
Check
Independent check on the current: the two resistors must share the $12.0\ \mathrm{V}$ in the ratio of their resistances, giving $4.50\ \mathrm{V}$ and $7.50\ \mathrm{V}$, which sum to $12.0\ \mathrm{V}$ and give $1.50\ \mathrm{A}$ in each. On size: $0.150\ \mathrm{N}$ is roughly the weight of a fifteen gram object, which is the order of magnitude you expect from an ampere or so in a field of a few tenths of a tesla.
Two formulas, one from week eight and one from week nine, and the entire difficulty was in which order to use them.
The magnetic half of a question like this is rarely where marks go missing. Whenever a magnetic quantity depends on a current, treat the circuit as the real question and the field as the last line.
The same 1.50 A counted in electrons, and the charge parked on a 220 microfarad capacitor
The same steady current of $1.50\ \mathrm{A}$ flows in the circuit above. Separately, a $220\ \mathrm{\mu F}$ capacitor is charged to $12.0\ \mathrm{V}$. (a) How many electrons pass a point in the wire each second? (b) How much charge sits on one plate of the capacitor? (c) How long would that current take to deliver that much charge?
Sanity on size: a millicoulomb is a tiny amount of charge and an ampere is a large current, so a millisecond is exactly the timescale you should expect. If the answer had come out in seconds, one of the two prefixes was dropped.
This is why a capacitor discharging through a low resistance can deliver a huge current for a very short time: the stored charge is small, but nothing stops it leaving quickly.
Checkpoint
§10.1 — charge delivered by a steady current●○○○○
Thirty seconds, one multiplication. A steady current runs in a wire while a stopwatch runs beside it.
Given
$I = 0.250\ \mathrm{A}$, steady
the current runs for $4.00$ minutes
the wire is unbranched
Find
(a) How much charge passes a given cross section of the wire in that time?
Hint 1/4
You are asked for a charge, and you have been given a rate and a duration, so this is one multiplication and a unit conversion.
Hint 2/4
For a steady current, $\Delta Q = I\,\Delta t$, with the time in seconds because an ampere is a coulomb per second.
Hint 3/4
Here $I = 0.250\ \mathrm{A}$ and $\Delta t = 4.00\ \mathrm{min} = 240\ \mathrm{s}$, so $\Delta Q = (0.250)(240)$.
Hint 4/4
$\Delta Q = 60.0\ \mathrm{C}$.
Show solutionOne multiplication, in SI units
$$\Delta Q = I\,\Delta t = (0.250)(240) = 60.0\ \mathrm{C}$$
the current is steady, so no integral is needed; the minutes have to become seconds first because the ampere is defined per second
Count it in electrons instead: $60.0/1.60\times10^{-19} = 3.75\times10^{20}$ of them, which is the same physical statement reached without using the coulomb at all.
⚠ Letting one element set the current in a series loop
the element in question is the one the sentence is about, so the eye goes to it and the rest of the loop stops being visible
wrong$$I = \frac{\varepsilon}{R_2}$$
right$$I = \frac{\varepsilon}{R_1 + R_2}$$
⚠ Feeding microfarads straight into a formula
the prefix is printed on the component and gets copied along with the number, and the answer still looks like a number
⚠ Believing the current is smaller after a resistor than before it
everyday language says a device consumes current, and the word consume implies something is used up
wrong$$I_{\rm after} < I_{\rm before}$$
right$$I_{\rm after} = I_{\rm before},\quad \text{what is lost is energy}$$
10.2Choosing the route in the first thirty seconds
The last line of a question names the quantity, and the quantity names the route you should be on.
We have the map; what we do not yet have is a rule for reading a question and knowing which arrow on that map it is asking us to walk.
MethodRule 10.2: the three questions that pick the route
Conditions
answer them in this order; the second and third are only worth asking once the first has narrowed things down
the answers are read off the question's own wording, not off which formula you happen to remember best
if the first question has two answers, the problem has two parts, and they are almost always solved in the order circuit first, field second
$$\boxed{\begin{aligned}&\textbf{1.}\ \ \text{What quantity does the last line name?}\\[2pt]&\textbf{2.}\ \ \text{Is there a switch, a clock, or the words a long time?}\\[2pt]&\textbf{3.}\ \ \text{Is a charge or a current sitting in a magnetic field?}\end{aligned}}$$
First ask what you are being marked on: a force or a torque sends you to the magnetic formulas, a current or a power sends you to the circuit, a stored charge or a stored energy sends you to the capacitor. Then ask whether time appears at all: if it does not, the circuit is in its steady state and a capacitor is simply a break. Then ask whether anything is moving inside a field, because that step always comes last and always needs a current that the earlier steps supplied.
Read it top down, starting from the quantity the question ends with. The left and right branches are usually one line of work; the middle branch is where the second question has to be asked before anything else happens.
Looks like this, but is not
Whatever the question hands you, use it: if a capacitance is printed on the figure, the capacitance belongs in the answer. Data given and data needed usually do coincide, which is exactly what makes this habit survive so long.
In a steady state circuit the capacitance never enters the current calculation at all. It sets how much charge ends up parked, and nothing else. A question can print $C = 5.00\ \mathrm{\mu F}$ on the figure, ask for the current in one branch, and the correct working never mentions that number once. Its job in that question is to tell you the branch carries no current.
Energy stored in a 5.00 microfarad capacitor a long time after the switch closes
A $6.00\ \mathrm{V}$ battery is in series with $R_1 = 2.00\ \mathrm{k\Omega}$ and $R_2 = 3.00\ \mathrm{k\Omega}$. A $5.00\ \mathrm{\mu F}$ capacitor is connected across $R_2$. A long time after the switch closes, how much energy is stored in the capacitor?
Given
$\varepsilon = 6.00\ \mathrm{V}$, ideal
$R_1 = 2.00\ \mathrm{k\Omega}$ in series with $R_2 = 3.00\ \mathrm{k\Omega}$
$C = 5.00\ \mathrm{\mu F}$, connected across $R_2$
the reading is taken a long time after switching
Find
the energy stored in the capacitor in the steady state
SolutionDecide what a long time means for the capacitor branch
$$I_C = 0\ \text{in the steady state}$$
nothing is changing any more, so the charge on the capacitor is constant and no charge is crossing into it; the branch is a break and can be ignored while the current is worked out
the voltage across the capacitor is the one that goes in the formula; using the $6.00\ \mathrm{V}$ of the source would be answering a different question
Cross check by charge: $Q = CV_2 = 18.0\ \mathrm{\mu C}$, and $U = \tfrac12 QV_2 = \tfrac12(18.0\times10^{-6})(3.60) = 3.24\times10^{-5}\ \mathrm{J}$, reached without ever squaring a voltage. On size, tens of microjoules is what a small capacitor at a few volts holds; a joule would have meant a prefix went missing.
Notice the order the routes came in: question two first, because the answer to it deleted a whole branch, and only then question one.
Radius and period of a proton in a 0.150 T field, and why only one of them needs the speed
A proton enters a uniform magnetic field of $0.150\ \mathrm{T}$ at $2.40\times10^{5}\ \mathrm{m/s}$, moving at right angles to the field. (a) What is the radius of its path? (b) How long does one full circuit take?
Given
$B = 0.150\ \mathrm{T}$, uniform
$v = 2.40\times10^{5}\ \mathrm{m/s}$, perpendicular to $B$
the radius of the circular path and the period of the motion
SolutionRecognise the template before computing
$$\vec F \perp \vec v\ \Rightarrow\ \text{uniform circular motion}$$
the force is square to the velocity at every instant, so the speed is fixed and only the direction turns; that is the definition of the circular template
the speed has cancelled out of this one; a faster proton runs a bigger circle in exactly the same time, which is the fact the cyclotron is built on
Answer $$\boxed{\;r = 1.67\ \mathrm{cm},\qquad T = 4.37\times10^{-7}\ \mathrm{s}\;}$$
Check
Independent route to the period: $T = 2\pi r/v = 2\pi(1.67\times10^{-2})/(2.40\times10^{5}) = 4.37\times10^{-7}\ \mathrm{s}$, computed from the radius rather than from the formula, and the two agree.
Two questions, one field, and only one of them cared how fast the proton was going. Deciding that in advance is worth more than remembering either formula.
Checkpoint
§10.2 — reading the route off the last line●●○○○
A question ends with the sentence: a long time after the switch closes, what is the current through the $8.00\ \mathrm{\Omega}$ resistor? The figure shows a battery, three resistors and one capacitor.
Given
a battery of known emf
three resistors, one of them $8.00\ \mathrm{\Omega}$
one capacitor of known capacitance, in a branch of its own
the phrase a long time after the switch closes
Find
(a) What is the first thing to do with the capacitor?
Hint 1/4
Do not compute anything yet. Ask only what the words a long time do to the picture.
Hint 2/4
In the steady state nothing is changing, so no charge is crossing into or out of the capacitor and the current in its branch is zero.
Hint 3/4
A branch carrying no current can be erased while the other currents are found; here that leaves a battery and three resistors. The capacitance value is not needed for this part at all.
Hint 4/4
Treat the capacitor branch as a break in the circuit and solve what is left.
Show solutionRead the steady state condition off the definition of current
steady state means nothing changes with time, and a current into the capacitor would change its charge
$$\Rightarrow\ \text{the branch behaves as an open circuit}$$
zero current through a branch is exactly what a break in the wire would give, so the two are interchangeable for this part of the calculation
Answer $$\boxed{\;\text{erase the capacitor branch, then solve the resistor network}\;}$$
Check
Test the rule at the other extreme: an empty capacitor at the instant of switching has zero voltage across it, which is what a plain wire would give. The two limits are opposite, which is why naming the instant matters before anything is computed.
⚠ Starting from the data instead of from the question
the given numbers are visible on the page and the required quantity is one sentence at the end, so the eye starts in the wrong place
right$$q = C\varepsilon\ \text{, the } t\to\infty \text{ limit}$$
10.3Series and parallel, mirrored: why capacitors are the odd ones out
Resistors add where they share a current, capacitors where they share a voltage, and one upside down fraction explains the swap.
Both weeks handed over a pair of combination rules, and the pairs are the wrong way round with respect to each other, which is where most of the arithmetic in this block goes wrong.
TheoremTheorem 10.3: the four combination rules, and the one line that decides them
Conditions
the elements are ideal and the wires joining them have no resistance of their own
series means the same current passes through each element, which for capacitors means the same charge ends up on each
parallel means the same potential difference sits across each element, which requires both ends to be genuinely joined by wire
a network that is neither, such as a bridge, needs the loop rules instead and no amount of staring will reduce it
Line elements up end to end so the same thing passes through all of them, and the resistances add while the capacitances add as reciprocals. Put them side by side across the same two points and it is the other way round. There is one reason and it is one line long: a resistor's defining ratio has the voltage on top and a capacitor's has it underneath.
Proof
Two elements in series carry the same charge through them, and the voltages across them add: $V = V_1 + V_2$.
For resistors, $V_i = IR_i$ with the same $I$ in both, so $V = I(R_1+R_2)$ and the pair behaves as a single resistance $R_1+R_2$.
For capacitors, $V_i = Q/C_i$ with the same $Q$ on both, so $V = Q\left(\frac{1}{C_1}+\frac{1}{C_2}\right)$ and the pair behaves as a single capacitance obeying $\frac{1}{C} = \frac{1}{C_1}+\frac{1}{C_2}$.
The two arguments are word for word the same except that the voltage sits on top in one definition and underneath in the other. Run them again for elements side by side, where the voltage is shared and the currents or charges add, and the rules swap back.
The top row shares a current, the bottom row shares a voltage, and each cell states which side of the fraction its sum lands on. Reading the row label first, rather than the component, is what keeps the four apart.
Looks like this, but is not
Adding a second capacitor always gives you more capacitance. It is true side by side, and it feels right for the same reason two buckets hold more than one.
In series, two $10.0\ \mathrm{\mu F}$ capacitors give $5.00\ \mathrm{\mu F}$, less than either of them alone. The two inner plates are joined and carry equal and opposite charge, so the pair behaves like one capacitor with twice the gap, and doubling a gap halves the capacitance. Add a third in series and it falls again, to $3.33\ \mathrm{\mu F}$.
Three capacitors across 24.0 V, and the voltage each one settles at
A $6.00\ \mathrm{\mu F}$ capacitor and a $3.00\ \mathrm{\mu F}$ capacitor are joined in series, and that pair is joined in parallel with a $4.00\ \mathrm{\mu F}$ capacitor. The whole arrangement is placed across a $24.0\ \mathrm{V}$ supply. Find the equivalent capacitance, the total charge drawn from the supply, and the voltage across the $6.00\ \mathrm{\mu F}$ capacitor.
Given
$C_1 = 6.00\ \mathrm{\mu F}$ in series with $C_2 = 3.00\ \mathrm{\mu F}$
that pair in parallel with $C_3 = 4.00\ \mathrm{\mu F}$
the supply provides $V = 24.0\ \mathrm{V}$ across the whole arrangement
Find
the equivalent capacitance, the total charge, and the voltage across the 6.00 microfarad capacitor
the innermost combination is done first, exactly as with brackets in algebra; working microfarads throughout is safe here because every capacitance carries the same prefix
$$C_{12} = 2.00\ \mathrm{\mu F}$$
smaller than either member, which is the signature of a series pair and a useful sign that the right rule was used
Check the series branch adds up: $V_2 = Q_{12}/C_2 = 48.0/3.00 = 16.0\ \mathrm{V}$, and $8.00 + 16.0 = 24.0\ \mathrm{V}$, which is the supply voltage. If those two had not summed to the supply, the charge would have been wrong.
Two combination rules, in the right order, and one division per capacitor after that.
The ratio $8.00$ to $16.0$ is the reciprocal of the ratio $6.00$ to $3.00$. In a series chain of capacitors, voltage divides in inverse proportion to capacitance, which is the exact opposite of how it divides among series resistors.
The same three numbers read as resistances, and how far the answers move
Replace the capacitances of the previous example by resistances of the same numbers: $R_1 = 6.00\ \mathrm{\Omega}$ in series with $R_2 = 3.00\ \mathrm{\Omega}$, that pair in parallel with $R_3 = 4.00\ \mathrm{\Omega}$, all across $24.0\ \mathrm{V}$. Find the equivalent resistance, the total current, and the voltage across the $6.00\ \mathrm{\Omega}$ resistor.
Given
$R_1 = 6.00\ \mathrm{\Omega}$ in series with $R_2 = 3.00\ \mathrm{\Omega}$
that pair in parallel with $R_3 = 4.00\ \mathrm{\Omega}$
$V = 24.0\ \mathrm{V}$ across the whole arrangement
Find
the equivalent resistance, the total current, and the voltage across the 6.00 ohm resistor
SolutionCollapse the series pair, with the other rule
Junction check: the other branch carries $24.0/4.00 = 6.00\ \mathrm{A}$, and $2.67 + 6.00 = 8.67\ \mathrm{A}$, which is the total. Any error in the equivalent resistance would break this sum.
Same three numbers, same wiring, and the $6.00$ element ends up with $8.00\ \mathrm{V}$ in one case and $16.0\ \mathrm{V}$ in the other. Nothing about the figure warns you which; only the word capacitor or resistor does.
Checkpoint
§10.3 — two equal capacitors joined in series●●○○○
Ten seconds, no calculator. Two identical capacitors are joined end to end and the pair is placed across a supply.
Given
$C_1 = C_2 = 4.00\ \mathrm{\mu F}$
they are joined in series
no other component is present
Find
(a) What single capacitance behaves the same way as the pair?
Hint 1/4
You are asked for one number that could replace two, so this is a combination rule; the only decision is which of the four applies.
Hint 2/4
For capacitors in series the reciprocals add: $\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$.
Hint 3/4
Here $C_1 = C_2 = 4.00\ \mathrm{\mu F}$, so $\frac{1}{C} = \frac{1}{4.00} + \frac{1}{4.00} = \frac{1}{2.00}\ \mathrm{\mu F^{-1}}$.
Hint 4/4
$C = 2.00\ \mathrm{\mu F}$, half of either one.
Show solutionReciprocals, because it is a series pair of capacitors
the shared quantity in a series chain is the charge, and the charge sits underneath in the capacitor definition, so it is the reciprocals that add
$$C = 2.00\ \mathrm{\mu F}$$
half of either member, which is the check that the series rule and not the parallel rule was used
Answer $$\boxed{\;C = 2.00\ \mathrm{\mu F}\;}$$
Check
Physical check: joining two identical capacitors in series is the same as doubling the plate separation of one of them, and capacitance falls as the gap grows, so the answer has to be smaller than $4.00\ \mathrm{\mu F}$.
⚠ Adding capacitances in series
the resistor rules are met first and practised more, so the hand writes the sum before the eye has registered which component it is looking at
10.4Stored versus burned: two energy accounts that never mix
A capacitor's joules can be handed back in full; a resistor's joules have already left the room as heat.
Both weeks produced an energy formula, one of them a quantity in joules and the other a rate in watts, and questions routinely ask for both in the same breath.
TheoremTheorem 10.4: the two energy accounts of this block
Conditions
the first formula assumes the capacitor was empty to begin with, so the whole charge had to be carried across the growing voltage
the three forms of each are algebraically identical; pick the one whose two quantities you already know rather than the one you remember first
the second is a rate, so getting a quantity of heat out of it needs a multiplication by time, and a varying current needs the average of $I^{2}$ rather than the square of the average
The first is a stock: joules parked in the gap of a capacitor, recoverable in full if you let it empty into something useful. The second is a rate: joules per second walking out of a resistor as heat, and no rewiring will bring them back.
Proof
Charge the capacitor in slices. When $q$ has already been carried across, the voltage between the plates is $q/C$, so carrying the next $dq$ across costs $dW = (q/C)\,dq$.
Add the slices from empty to full: $W = \int_0^{Q}\frac{q}{C}\,dq = \frac{Q^{2}}{2C}$.
The very first slice was free and the very last one cost the full $Q/C$, so the average cost was half of the final voltage. That average is the entire origin of the factor of one half, and it is why $U = QV$ without the half is the commonest wrong answer in this material.
Voltage against charge for a capacitor is a straight line through the origin, so the work done filling it is the area of a triangle rather than a rectangle. The factor of one half is that triangle and nothing more mysterious.
Looks like this, but is not
Charging a capacitor from a battery through a resistor stores the battery's energy in the capacitor. Energy is conserved and the capacitor is the only thing left holding any at the end, so where else would it go.
Half of it, at most, and never more. Moving a charge $Q = C\varepsilon$ out of a battery of emf $\varepsilon$ costs the battery $C\varepsilon^{2}$ joules, while the capacitor ends up holding $\tfrac12 C\varepsilon^{2}$. The other half was turned into heat in the resistor on the way, and the size of that resistor makes no difference at all: a small one burns it quickly and a large one slowly, and both burn exactly half.
voltage V (V)
charge Q (mC)
stored energy U (mJ)
3.00
1.41
2.12
6.00
2.82
8.46
9.00
4.23
19.0
12.0
5.64
33.8
18.0
8.46
76.1
Doubling the voltage from 3.00 V to 6.00 V doubles the charge and quadruples the energy, and every further doubling does the same again. If your energy column ever grows at the same rate as your charge column, the square has been left out.
Where a 9.00 V battery's energy goes while a 470 microfarad capacitor fills
A $470\ \mathrm{\mu F}$ capacitor, initially empty, is charged through a $2.20\ \mathrm{k\Omega}$ resistor by a $9.00\ \mathrm{V}$ battery. When the charging is over, how much energy has the battery supplied, how much is stored, and how much was turned into heat?
Given
$C = 470\ \mathrm{\mu F}$, initially uncharged
$R = 2.20\ \mathrm{k\Omega}$
$\varepsilon = 9.00\ \mathrm{V}$, ideal battery
charging is allowed to finish
Find
the energy supplied, the energy stored, and the energy dissipated
every coulomb leaves the battery at the full $9.00\ \mathrm{V}$, because a battery holds its terminals at a fixed emf regardless of how full the capacitor already is
Check the split algebraically rather than numerically: $U_{\rm battery} = C\varepsilon^{2}$ and $U_C = \tfrac12 C\varepsilon^{2}$, so the ratio is exactly two to one no matter what $C$, $\varepsilon$ or $R$ are. Getting anything other than a clean half means an arithmetic slip, not a physical effect.
The time constant here is $\tau = RC = 1.03\ \mathrm{s}$, so all of this happens in about five seconds.
Efficiency fifty per cent, always, for charging an empty capacitor from a fixed voltage through any resistance. It is one of the few results in this course that does not depend on a single given number.
Which of two series resistors gets hotter, and how the answer flips in parallel
A $6.00\ \mathrm{\Omega}$ and an $18.0\ \mathrm{\Omega}$ resistor are placed in series across a $24.0\ \mathrm{V}$ supply. (a) Find the power in each and the total. (b) How much heat comes out in one hour? (c) Rewire them in parallel across the same supply and find the two powers again.
with a shared voltage the power is inversely proportional to the resistance, so now the smaller resistor is the hotter one and the ranking has reversed
Independent check on the series total: $P = IV = (1.00)(24.0) = 24.0\ \mathrm{W}$, computed without splitting the circuit at all, and it matches the sum of the parts.
There is no such thing as the hotter resistor on its own. Ask first what the pair shares, and the ranking follows; get that backwards and the answer is not slightly wrong but reversed.
Checkpoint
§10.4 — energy stored in a charged capacitor●○○○○
Thirty seconds. A capacitor has been charged and then disconnected, and it sits there holding its voltage.
Given
$C = 220\ \mathrm{\mu F}$
$V = 50.0\ \mathrm{V}$ across its plates
no other component is connected
Find
(a) How much energy is stored in it?
Hint 1/4
You have a capacitance and a voltage and you want an energy, so one of the three stored energy forms fits directly.
Hint 2/4
Use $U = \tfrac12 CV^{2}$, the form whose two ingredients you already have.
Hint 3/4
Here $C = 220\times10^{-6}\ \mathrm{F}$ and $V = 50.0\ \mathrm{V}$, so $U = \tfrac12(220\times10^{-6})(50.0)^{2}$.
Hint 4/4
$U = 0.275\ \mathrm{J}$.
Show solutionOne substitution into the stored energy formula
the capacitance is converted to farads before substituting, and the voltage is squared before the half is applied
Answer $$\boxed{\;U = 0.275\ \mathrm{J}\;}$$
Check
Reach it another way: $Q = CV = 11.0\ \mathrm{mC}$ and $U = \tfrac12 QV = \tfrac12(11.0\times10^{-3})(50.0) = 0.275\ \mathrm{J}$, with no square anywhere in the calculation.
⚠ Dropping the factor of one half in the stored energy
the charge really did leave the battery at the full voltage, so $QV$ feels like the honest answer and it is, for the battery, not for the capacitor
wrong$$U_C = QV$$
right$$U_C = \tfrac12 QV$$
⚠ Squaring the wrong thing when the current varies
the average of the current is the quantity that is easy to write down, and squaring it afterwards looks harmless
⚠ Reporting a power where a quantity of energy was asked for
watts and joules both feel like energy words, and the numbers are often of similar size
wrong$$E = I^{2}R$$
right$$E = I^{2}Rt$$
10.5Networks, and the clock inside them
Reduce the network if it will reduce, use the two loop rules if it will not, and always name the instant first.
Series and parallel handle a network only as long as it can be peeled apart; the moment a second source appears in a branch of its own, it cannot be, and a different tool is needed.
RuleRule 10.5: the two circuit rules, and the capacitor's three faces
Conditions
the junction rule is charge conservation and holds at every instant, steady or not
the loop rule is energy conservation per unit charge and needs a consistent sign convention chosen before the first term is written
the exponential forms assume one resistance and one capacitance in a single loop; a network has to be reduced to that shape first, and then $\tau = R_{\rm eq}C_{\rm eq}$
an assumed current direction that turns out negative is not an error and must not be flipped halfway through
Whatever flows into a junction flows out of it again, because charge does not collect at a point of wire. Walk any closed loop, adding the rises and subtracting the drops, and you come back to the voltage you started from. And when a capacitor is filling, the charge climbs towards its final value along an exponential whose only timescale is the resistance times the capacitance.
Two sources, three branches, and no pair of resistors that is either purely in series or purely in parallel. This is the shape that forces the loop rules, and it is also the shape used in both worked examples below.
Looks like this, but is not
A bigger charging resistor means less charge on the capacitor at the end. It certainly means less current at every instant, and every other feature of the circuit does get weaker when $R$ grows.
In a single loop the final charge is $C\varepsilon$ and the resistance is nowhere in that expression. Double $R$ and you double the time constant and halve the initial current, but the capacitor still ends up at the full emf with the same charge on it, just later. What the resistor controls is the schedule, not the destination.
time t (s)
charge q (mC)
capacitor voltage (V)
current (mA)
0
0
0
4.09
0.717
2.12
4.50
2.05
1.03
2.67
5.69
1.51
2.07
3.66
7.78
0.554
3.10
4.02
8.55
0.204
The second row is the half way point, reached at 0.693 times the time constant rather than at the time constant itself. Notice that the charge column and the current column always add up in the same way: whatever fraction of the charge is in place, that same fraction of the current has gone.
Three branch currents in a network with two batteries
In the network of the figure, $\varepsilon_1 = 12.0\ \mathrm{V}$ with $R_1 = 4.00\ \mathrm{\Omega}$ in the left branch, $\varepsilon_2 = 6.00\ \mathrm{V}$ with $R_2 = 3.00\ \mathrm{\Omega}$ in the right branch, and $R_3 = 6.00\ \mathrm{\Omega}$ in the middle branch between $a$ and $b$. Both sources drive current towards node $a$. Find the current in each branch.
Given
$\varepsilon_1 = 12.0\ \mathrm{V}$, $R_1 = 4.00\ \mathrm{\Omega}$, left branch, from $b$ to $a$
$\varepsilon_2 = 6.00\ \mathrm{V}$, $R_2 = 3.00\ \mathrm{\Omega}$, right branch, from $b$ to $a$
$R_3 = 6.00\ \mathrm{\Omega}$, middle branch, from $a$ to $b$
all sources ideal apart from the resistances shown
Find
the current in all three branches
SolutionPick one unknown instead of three
$$V \equiv V_a - V_b$$
the two outer branches and the middle branch all join the same two nodes, so one node voltage fixes all three currents and the junction rule then gives a single equation instead of a system
negative, which means the second battery is being driven backwards and is charging rather than delivering; the arrow was drawn the wrong way and that is allowed
$$I_3 = \frac{6.667}{6.00} = 1.11\ \mathrm{A}$$
positive, flowing from $a$ down to $b$ through the middle resistor as assumed
Power balance, which uses none of the algebra above: the first source delivers $\varepsilon_1I_1 = 16.0\ \mathrm{W}$ and the second absorbs $\varepsilon_2\vert I_2\vert = 1.33\ \mathrm{W}$, a net $14.7\ \mathrm{W}$; the three resistors dissipate $7.11 + 0.148 + 7.41 = 14.7\ \mathrm{W}$. The two totals agree, which they could not do if any branch current were wrong.
One unknown instead of three, at the cost of noticing that all three branches share the same two nodes.
The negative sign is the most informative number in the answer. It says the twelve volt source is charging the six volt one, which is exactly what happens when a car with a good battery jump starts one with a flat battery.
The same network with a capacitor in the middle branch, a long time later
Take the same two outer branches, $\varepsilon_1 = 12.0\ \mathrm{V}$ with $R_1 = 4.00\ \mathrm{\Omega}$ and $\varepsilon_2 = 6.00\ \mathrm{V}$ with $R_2 = 3.00\ \mathrm{\Omega}$, but replace the middle resistor by a $100\ \mathrm{\mu F}$ capacitor. A long time after the switch closes, find the current, the voltage across the capacitor and the charge on it.
$C = 100\ \mathrm{\mu F}$ in the middle branch, between nodes $a$ and $b$
steady state reached
Find
the steady current, the capacitor voltage and its charge
SolutionDelete the branch that carries no current
$$I_C = 0 \ \Rightarrow\ \text{one loop is left}$$
in the steady state the charge on the capacitor is constant, so nothing crosses into it and the middle branch is a break; what remains is a single loop through both outer branches
Reach the node voltage from the other side: $V_{ab} = \varepsilon_2 + IR_2 = 6.00 + (0.857)(3.00) = 8.57\ \mathrm{V}$, adding rather than subtracting because the current runs into that source. Two independent paths, same number.
Same picture, one component changed, and the current went from three branch currents to one. The phrase a long time did more work here than any formula did.
Checkpoint
§10.5 — current at the instant an empty capacitor is switched on●●○○○
A single loop is assembled and the switch is closed at $t=0$. The capacitor has no charge on it at all beforehand.
Given
$\varepsilon = 9.00\ \mathrm{V}$, ideal
$R = 3.00\ \mathrm{k\Omega}$
$C = 10.0\ \mathrm{\mu F}$, uncharged at $t=0$
all three in series with the switch
Find
(a) What is the current in the loop immediately after the switch closes?
Hint 1/4
Ask what the voltage across the capacitor is at that first instant, and let the loop rule do the rest.
Hint 2/4
An uncharged capacitor has $V_C = Q/C = 0$, so at $t=0$ the whole emf appears across the resistor: $I_0 = \varepsilon/R$.
Hint 3/4
Here $\varepsilon = 9.00\ \mathrm{V}$ and $R = 3.00\ \mathrm{k\Omega}$, so $I_0 = 9.00/3.00\times10^{3}$.
Hint 4/4
$I_0 = 3.00\ \mathrm{mA}$.
Show solutionUse the state of the capacitor, not its capacitance
$$V_C(0) = \frac{Q(0)}{C} = 0$$
no charge means no potential difference, whatever the capacitance happens to be
the loop rule with one term equal to zero; the capacitance never enters, which is why it is not needed in the answer
Answer $$\boxed{\;I_0 = 3.00\ \mathrm{mA}\;}$$
Check
Check the other end of the story: as $t\to\infty$ the capacitor reaches $9.00\ \mathrm{V}$, the loop rule leaves nothing across the resistor and the current is zero. The two limits are a wire and a break, and the exponential joins them.
⚠ Flipping an assumed current direction because the answer came out negative
a negative current looks like an error message, and the instinct is to correct it rather than to read it
wrong$$I_2 = -0.222\ \mathrm{A}\ \Rightarrow\ \text{redo with the arrow reversed}$$
right$$I_2 = -0.222\ \mathrm{A}\ \Rightarrow\ \text{the arrow was backwards, the number is right}$$
⚠ Using the source emf as the capacitor voltage in a network
in the single loop case those two really are equal at the end, and the habit survives into circuits where they are not
wrong$$Q = C\varepsilon$$
right$$Q = CV_{ab},\quad V_{ab} \ne \varepsilon \text{ in general}$$
⚠ Putting only the charging resistor into the time constant of a network
the formula is written $\tau = RC$ with a single letter, which quietly suggests there is only one resistance to consider
wrong$$\tau = R_1C$$
right$$\tau = R_{\rm eq}C \ \text{seen from the capacitor's terminals}$$
The charge climbs fast and then slowly, and the only thing setting the horizontal scale is $\tau = RC$. After one time constant it is 63 per cent of the way there, after three about 95 per cent, and it never formally arrives.
10.6The magnetic force: square to the motion, and useless for changing a speed
It bends the path and never alters the speed, which is why circles and helices keep appearing in this material.
Everything so far has been charge in wires; the last week put charge in open space and gave it a force that behaves like nothing else in the course.
TheoremTheorem 10.6: the magnetic force and the three things that follow from it
Conditions
the field must be uniform over the region the particle explores, otherwise the path is not a circle
only the component of the velocity perpendicular to the field bends; the parallel component is untouched, which is what turns a circle into a helix
gravity is neglected throughout, which is safe for charged particles here: the magnetic force in these problems is many orders of magnitude larger
the same formula written with $I\vec L$ in place of $q\vec v$ gives the force on a straight wire, because a current is a stream of moving charge
$$\boxed{\;\textcolor{#1a7f37}{\vec F = q\,\vec v\times\vec B},\quad F = \vert q\vert vB\sin\theta,\quad \textcolor{#1a7f37}{r = \frac{mv_\perp}{\vert q\vert B}},\quad T = \frac{2\pi m}{\vert q\vert B}\;}$$
The force is at right angles to both the velocity and the field, so it can steer but never push along the direction of travel. A charge fired square into a uniform field therefore runs round a circle whose radius grows with momentum and shrinks with field, and it completes that circle in a time that does not depend on how fast it was going.
Proof
The force is perpendicular to the velocity at every instant, so the rate at which it does work is $P = \vec F\cdot\vec v = 0$.
Zero power means the kinetic energy never changes, so the speed is constant and only the direction of travel turns.
A constant speed with a constant force of constant magnitude always aimed square to the motion is uniform circular motion. Setting $\vert q\vert vB = mv^{2}/r$ and cancelling one factor of $v$ gives $r = mv/(\vert q\vert B)$.
The period follows from $T = 2\pi r/v = 2\pi m/(\vert q\vert B)$, and the speed has cancelled out entirely. A faster particle runs a proportionally bigger circle in exactly the same time, which is the fact a cyclotron is built on.
The force is redrawn at every point along the path and always points at the same centre, which is the geometric definition of a circle. The velocity arrow never gets longer.
Looks like this, but is not
A stronger magnet gives the particle more energy. Everything about magnets in ordinary life is about pushing and pulling harder, and a bigger field does give a bigger force.
Double the field and the force doubles, the radius halves and the particle goes round twice as often, and it still arrives at every point of that tighter circle with precisely the speed it entered with. Changing a speed needs a force with a component along the velocity, and $q\vec v\times\vec B$ never has one. If a question reports a change in kinetic energy, something other than the magnetic field did it.
An electron at 3.60 million metres per second in a 4.50 millitesla field
An electron travelling at $3.60\times10^{6}\ \mathrm{m/s}$ enters a uniform magnetic field of $4.50\ \mathrm{mT}$ at right angles to it. (a) What is the radius of its path? (b) How long does one revolution take? (c) How many revolutions per second is that?
Given
$v = 3.60\times10^{6}\ \mathrm{m/s}$, perpendicular to the field
one over the period, and the answer lands in the radio band, which is exactly where the drive electronics of such machines operate
Answer $$\boxed{\;r = 4.56\ \mathrm{mm},\qquad T = 7.95\ \mathrm{ns},\qquad f = 126\ \mathrm{MHz}\;}$$
Check
Independent route to the period, using the radius just found rather than the formula: $T = 2\pi r/v = 2\pi(4.56\times10^{-3})/(3.60\times10^{6}) = 7.95\times10^{-9}\ \mathrm{s}$. The two agree, which they would not if the mass or the charge had been misread.
If the same electron had entered at half the speed, the circle would have been half as wide and the period would have been identical. Whenever a question changes the speed and asks about timing, that is the fact being tested.
Force on a 18.0 cm wire carrying 2.50 A at 35 degrees to a 0.320 T field
A straight wire of length $0.180\ \mathrm{m}$ carries a current of $2.50\ \mathrm{A}$ and lies in a uniform field of $0.320\ \mathrm{T}$, making an angle of $35.0^{\circ}$ with the field direction. Find the magnitude of the force on it, and say how much larger the force would be if the wire were turned square to the field.
Given
$L = 0.180\ \mathrm{m}$, straight
$I = 2.50\ \mathrm{A}$
$B = 0.320\ \mathrm{T}$, uniform
the angle between the wire and the field is $35.0^{\circ}$
Find
the force on the wire, and the maximum force available at this current
SolutionSubstitute into the wire form of the magnetic force
Limiting case check: setting $\theta = 0$ in the formula gives zero, and a wire lying along the field really should feel nothing, because then every charge in it is moving parallel to the field. Setting $\theta = 90^{\circ}$ recovers $BIL$, the value computed in the second part.
A useful habit for any magnetic question: compute $BIL$ or $\vert q\vert vB$ first, then apply the sine. It gives you an upper bound to sanity check against, and it separates the physics from the trigonometry.
Checkpoint
§10.6 — can a magnetic field speed a particle up●●○○○
One line, no numbers. A free proton is fired into a region of uniform magnetic field and follows a curved path through it.
Given
the field is uniform and constant in time
no electric field and no other force acts
the proton enters at some angle to the field and leaves later
Find
(a) True or false: the uniform magnetic field can increase the kinetic energy of the proton.
Hint 1/4
Ask what determines whether a force changes a speed, rather than what determines whether it changes a path.
Hint 2/4
The rate at which a force does work is $P = \vec F\cdot\vec v$, and the magnetic force is perpendicular to $\vec v$ at every instant, so that dot product is zero.
Hint 3/4
Zero power for the whole journey means zero change in kinetic energy, no matter how strong the field is or how long the proton stays in it.
Hint 4/4
So the statement is false: the speed on the way out equals the speed on the way in.
Show solutionCompute the power delivered by the force
$$P = \vec F\cdot\vec v = (q\,\vec v\times\vec B)\cdot\vec v = 0$$
a cross product is perpendicular to both of its factors, so its dot product with either of them vanishes identically, whatever the angle between $\vec v$ and $\vec B$
$$\frac{dK}{dt} = P = 0 \ \Rightarrow\ v = \text{constant}$$
the work energy theorem with zero power on the right; the direction of $\vec v$ is free to change and does
Answer $$\boxed{\;\text{False: } \Delta K = 0 \text{ for any uniform magnetic field}\;}$$
Check
Consistency check against the radius formula: $r = mv/(\vert q\vert B)$ is a constant only if $v$ is, and the circular path observed in every one of these experiments would be a spiral if the speed drifted. The geometry and the energy statement agree.
⚠ Letting the magnetic force change the speed
it is a force, and forces accelerate things, so the exception feels like a special rule rather than a consequence
wrong$$\Delta K = F\,d = \vert q\vert vB\,d$$
right$$\Delta K = 0,\qquad \vec F\perp\vec v$$
⚠ Using the full speed in the radius when the entry is at an angle
the speed is the number printed in the question and the perpendicular component has to be worked out
wrong$$r = \frac{mv}{\vert q\vert B}\ \text{for any entry angle}$$
diagrams often show the wire against a plate or a page edge, and the eye takes the angle from whatever is nearest
wrong$$F = BIL\sin(\text{angle to the page})$$
right$$F = BIL\sin(\text{angle between the wire and } \vec B)$$
10.7Four magnetic templates, and the question each one answers
Helix, crossed fields, turning loop, Hall strip: almost every magnetic exam question is one of these four wearing a costume.
The circle is the core case, but the questions that actually get set are four variations on it, and each is recognisable from one phrase in the stem.
MethodMethod 10.7: matching a magnetic question to its template
Conditions
the helix appears when the entry is at an angle other than ninety degrees, so that part of the velocity survives untouched
the crossed field result assumes the electric and magnetic forces are exactly opposite, which is a statement about geometry before it is a formula
the torque expression assumes the field is uniform across the whole loop, so that the net force on it is zero and only a couple remains
the Hall expression assumes a steady state has been reached, in which the sideways electric field exactly balances the sideways magnetic push on the carriers
A particle entering at an angle keeps its along field speed and spirals, advancing one pitch per revolution. Two fields at right angles let exactly one speed through undeflected, the ratio of the two field strengths. A loop of current in a field feels no net push but does feel a twist, largest when the loop faces edge on to the field. And a current carrying strip in a field develops a small voltage across its width, whose size counts the carriers inside it.
Only one speed survives the trip: too slow and the electric force wins and the particle is pushed down, too fast and the magnetic force wins and it is pushed up. Neither the charge nor the mass appears in the surviving condition.
Looks like this, but is not
A velocity selector picks out particles of a given energy. It does sort a mixed beam, and energy is the usual thing one sorts beams by, so the word fits the setting.
It selects a speed and nothing else. Two ions of different mass with the same speed both go straight through, and they carry very different kinetic energies. That is precisely why a mass spectrometer needs a second stage: the selector guarantees a known speed, and only then does the radius in the analyser field become a clean measurement of the mass.
Identifying an ion from a 8.84 cm radius in a mass spectrometer
Ions pass through a velocity selector with $E = 3.20\times10^{4}\ \mathrm{V/m}$ and $B = 0.180\ \mathrm{T}$, then enter an analysing field of $B' = 0.480\ \mathrm{T}$ where they follow a semicircle of radius $8.84\ \mathrm{cm}$. The ions carry a single elementary charge. Find the mass of the ion in kilograms and in atomic mass units.
Plausibility rather than repetition: the answer had to be a whole number of atomic mass units to within measurement error, and it came out as 23.0 without being forced. Had the number landed on 23.7 the calculation, not the ion, would be under suspicion. The circle diameter of 17.7 cm is also the right size for an instrument that fits on a bench.
Two stages, one formula each, and the only link between them is the speed.
Whenever a magnetic question hands you two field regions, expect the first to fix a speed and the second to measure something with it. Doing them in the other order leaves two unknowns in one equation.
Torque on a 40 turn coil at 55 degrees, and where the maximum lies
A rectangular coil of $40$ turns, each of area $0.0120\ \mathrm{m^{2}}$, carries a current of $0.750\ \mathrm{A}$ in a uniform field of $0.220\ \mathrm{T}$. The coil's magnetic moment makes an angle of $55.0^{\circ}$ with the field. Find the magnetic moment, the torque, and the largest torque the coil could feel in this field.
Given
$N = 40$ turns, $A = 0.0120\ \mathrm{m^{2}}$ each
$I = 0.750\ \mathrm{A}$
$B = 0.220\ \mathrm{T}$, uniform
the angle between $\vec\mu$ and $\vec B$ is $55.0^{\circ}$
Find
the magnetic moment, the torque at this angle, and the maximum torque
SolutionMagnetic moment first, because it collapses three numbers into one
Ratio check: $6.49\times10^{-2}/0.0792 = 0.819$, and $\sin 55.0^{\circ} = 0.819$. The two parts of the answer are consistent with each other by construction, so any mismatch would have been an arithmetic slip rather than a physics error.
The torque vanishes at $0^{\circ}$ and at $180^{\circ}$, but only the first of those is stable. Nudge the coil away from $0^{\circ}$ and the torque pushes it back; nudge it away from $180^{\circ}$ and the torque carries it further, all the way round. That is why a compass needle has a preferred end.
Checkpoint
§10.7 — which template a stem is describing●●●○○
A question reads: a beam containing several isotopes passes undeflected through a region where an electric field and a magnetic field are at right angles to each other and to the beam. It then asks what the beam's speed is.
Given
an electric field and a magnetic field, mutually perpendicular
the beam passes through undeflected
the beam contains particles of several different masses
the beam's speed is asked for
Find
(a) Which quantities does the answer depend on?
Hint 1/4
Ask which forces are present and what undeflected tells you about their sum, before reaching for any formula.
Hint 2/4
Undeflected means the net force is zero, so $qE = qvB$, and the charge cancels from both sides.
Hint 3/4
Here the two fields are the only data that survive that cancellation, giving $v = E/B$ with no mass and no charge left in it.
Hint 4/4
So the speed depends on the two field strengths and on nothing else.
Show solutionBalance the two forces and see what cancels
$$qE = qvB$$
undeflected means the electric and magnetic forces are equal and opposite, which is a statement about magnitudes once the directions are checked
$$v = \frac{E}{B}$$
the charge divides out of both sides, and the mass was never in either force; only the two field strengths remain
Answer $$\boxed{\;v = \frac{E}{B},\ \text{independent of } q \text{ and } m\;}$$
Check
Check the units: volts per metre divided by teslas is $\mathrm{(V/m)/(N\,s\,C^{-1}m^{-1})} = \mathrm{m/s}$, a speed, as required.
⚠ Taking the torque angle from the plane of the loop
the loop is the visible object and its plane is what a diagram shows, while the moment is an arrow you have to add yourself
wrong$$\tau = NIAB\sin(\text{angle between the plane and } \vec B)$$
right$$\tau = NIAB\sin(\text{angle between } \vec\mu \text{ and } \vec B)$$
⚠ Putting the charge into the velocity selector condition
every other magnetic formula in the week contains the charge, so leaving it out feels like an omission
wrong$$v = \frac{qE}{B}$$
right$$v = \frac{E}{B}$$
⚠ Using the selector field in the analyser radius
both stages are magnetic and the two field symbols look alike on a hurried page
wrong$$m = \frac{qBr}{v}$$
right$$m = \frac{qB'r}{v}$$
Torque against the angle between the loop's moment and the field. It is zero when the two are lined up, largest when they are square, and zero again when the loop has been turned right over.
Deciding what to compute, in the first thirty seconds
Every question in this block, before the pen touches the paper. It is worth doing even when the answer feels obvious, because the questions that feel obvious are the ones that punish the wrong route hardest.
Read the last line first
The final sentence names the quantity you are marked on. A force or a torque means the magnetic formulas. A current, a voltage or a power means the circuit. A stored charge or a stored energy means a capacitor. If two of these appear, the circuit part always comes first.
Ask whether time is in the question at all
If there is no switch, no clock and no phrase like a long time, the circuit is in its steady state and every capacitor is a break. If a switch has just been thrown, an initially empty capacitor is a plain wire. Only if an actual value of $t$ is named do you need the exponential.
Reduce the network if it will reduce
Look for pairs that genuinely share a current or genuinely share a voltage, and collapse them one pair at a time, writing each intermediate value down. If two sources sit in different branches, stop: the network will not reduce and the loop rules are the tool.
Choose signs and directions once, in writing
Draw an arrow on every branch and say which way you are walking each loop. Do not change either one later. A current that comes out negative is a correct answer with a backwards arrow, not a mistake.
Work symbolically until the last line
Put the numbers in only when the expression is complete. It makes the units check possible, it makes the limiting case check possible, and it stops a rounded intermediate value from moving the last digit of the answer.
Check before moving on
Sum the branch currents at a junction, sum the voltages around a loop, compare total power in with total power out, and ask whether the order of magnitude is one a laboratory could produce. Any one of these catches most errors; all four together catch nearly all of them.
Where it goes wrong
Starting the magnetic calculation before the branch current is known, so a correct formula is fed a wrong number.
Applying series and parallel rules to a network containing two sources in different branches, which produces a clean looking number that is simply wrong.
Using the source emf where the element's own voltage belongs, which is the single most expensive substitution error in this block.
Never writing down which instant the question is asking about, and then mixing a $t=0$ value with a steady state one in the same line.
Reading a timed circuit at its three instants
Any question containing a switch, a capacitor and the words immediately, a long time, or a numerical time.
At t = 0, replace an empty capacitor by a wire
An uncharged capacitor has $V = Q/C = 0$ across it, and a component with zero voltage across it behaves exactly as a piece of wire. Redraw the circuit with that substitution and solve it as a plain resistor network.
At t = 0, replace a charged capacitor by a battery
If the capacitor already carries charge $Q_0$, it holds $V_0 = Q_0/C$ across its terminals at the instant of switching, and behaves as a source of that voltage. Its polarity is set by which plate carries the positive charge.
For the steady state, replace every capacitor by a break
Erase the capacitor branch entirely and solve the resistor network that is left. Then read off the potential difference between the two points where the capacitor used to be; that is the voltage it has settled at.
For anything in between, find the two ends first
The quantity you want always moves exponentially from its $t=0$ value to its steady state value: $X(t) = X_\infty + (X_0 - X_\infty)e^{-t/\tau}$. Getting the two ends right is most of the work, and the exponential is then bookkeeping.
Get the time constant from the capacitor's own point of view
Look out from the capacitor's two terminals with every source switched off, and work out the resistance you see. That is the $R$ in $\tau = RC$, and in a network it is rarely the value of any single printed resistor.
Where it goes wrong
Using the exponential when the question only asked about one of the two ends, which turns a one line answer into five lines of unnecessary work.
Putting the charging resistor alone into the time constant when other resistors also sit in the capacitor's path.
Treating an already charged capacitor as a wire at $t=0$, which is only correct if it started empty.
Mixing the charging formula and the discharging formula, which differ by whether the exponential is subtracted from one or standing alone.
Getting a magnetic direction right, every time
Whenever the answer is a direction rather than a number, and whenever a sign in a later step depends on which way something pushes.
Write down the two input directions in words
Say out loud which way the velocity or the current points and which way the field points, using the page convention: $x$ to the right, $y$ up, $z$ out of the page. Dots mean out of the page, crosses mean into it.
Apply the right hand rule to the vectors, not to the particle
Point the fingers along $\vec v$ or along $\vec L$, curl them towards $\vec B$, and the thumb gives $\vec v\times\vec B$. This is a statement about two arrows and has nothing yet to do with the sign of the charge.
Apply the sign of the charge afterwards, as a separate decision
For a positive charge the force is along $\vec v\times\vec B$; for a negative one it is exactly opposite. Doing this as its own step, after the geometry, is what stops the two reversals from being applied at once or not at all.
Sanity check with a perpendicular test
The answer must be perpendicular to both inputs. If your force points even partly along the velocity or along the field, the rule has been misapplied, and it is worth restarting rather than patching.
Where it goes wrong
Reversing twice for an electron, once in the rule and once for the sign, and arriving back where you started.
Curling the fingers from $\vec B$ towards $\vec v$ rather than the other way, which reverses the answer because the cross product is not commutative.
Taking the angle from the page edge or from a plate rather than from the field direction.
Forgetting that a force perpendicular to the velocity changes nothing about the speed, and then reporting an energy change.
Two resistors in series across 18.0 V: which one gets more voltage
A $6.00\ \mathrm{\Omega}$ and a $3.00\ \mathrm{\Omega}$ resistor are joined in series across an ideal $18.0\ \mathrm{V}$ supply. Find the current and the voltage across each.
Given
$R_1 = 6.00\ \mathrm{\Omega}$, $R_2 = 3.00\ \mathrm{\Omega}$, in series
Sum check: $12.0 + 6.00 = 18.0\ \mathrm{V}$, the supply voltage, as the loop rule requires.
Two capacitors in series across 18.0 V: which one gets more voltage
A $6.00\ \mathrm{\mu F}$ and a $3.00\ \mathrm{\mu F}$ capacitor are joined in series across an ideal $18.0\ \mathrm{V}$ supply. Find the charge on each and the voltage across each.
Given
$C_1 = 6.00\ \mathrm{\mu F}$, $C_2 = 3.00\ \mathrm{\mu F}$, in series
Sum check: $6.00 + 12.0 = 18.0\ \mathrm{V}$. Independently, the energy stored is $\tfrac12(2.00\times10^{-6})(18.0)^{2} = 324\ \mathrm{\mu J}$, which also equals the sum of $\tfrac12 QV$ for the two capacitors separately.
The same two numbers wired the same way: the larger resistor takes $12.0\ \mathrm{V}$ and the larger capacitor takes $6.00\ \mathrm{V}$, so the voltage divides in opposite senses even though nothing about the picture has changed.
How to tell them apart
Ask what the series connection forces the pair to share. Resistors share a current and the voltage sits on top of the defining ratio, so voltage follows resistance. Capacitors share a charge and the voltage sits underneath, so voltage runs against capacitance. One glance at whether the shared quantity is above or below the line settles every one of these.
An electron pushed through 5.00 kV per metre for 2.00 cm
An electron starts from rest and is pushed by a uniform electric field of $5.00\times10^{3}\ \mathrm{V/m}$ through a distance of $2.00\ \mathrm{cm}$ along the field direction. How fast is it moving at the end?
Given
$E = 5.00\times10^{3}\ \mathrm{V/m}$, uniform
$d = 2.00\times10^{-2}\ \mathrm{m}$, along the force
electron from rest: $\vert q\vert = 1.60\times10^{-19}\ \mathrm{C}$, $m = 9.11\times10^{-31}\ \mathrm{kg}$
Find
the final speed
SolutionEnergy gained from the field
$$\Delta K = \vert q\vert Ed = (1.60\times10^{-19})(5.00\times10^{3})(2.00\times10^{-2})$$
the electric force has a component along the motion, so it does work; this is the only route that avoids computing an acceleration
$$\Delta K = 1.60\times10^{-17}\ \mathrm{J}$$
equivalently $100\ \mathrm{eV}$, since the electron crossed a potential difference of $100\ \mathrm{V}$
Cross check by potential difference: $Ed = 100\ \mathrm{V}$, and an electron accelerated through $100\ \mathrm{V}$ has $100\ \mathrm{eV}$ of energy, which is $1.60\times10^{-17}\ \mathrm{J}$; the same number reached without multiplying three quantities together.
The same electron sent into a 5.00 millitesla field instead
The same electron, now travelling at $5.93\times10^{6}\ \mathrm{m/s}$, enters a uniform magnetic field of $5.00\ \mathrm{mT}$ at right angles to it. What is the radius of its path, and how fast is it moving after a quarter turn?
Given
$v = 5.93\times10^{6}\ \mathrm{m/s}$, perpendicular to the field
a few millimetres, which is why these deflections are done in small vacuum tubes
Speed after a quarter turn
$$P = \vec F\cdot\vec v = 0 \ \Rightarrow\ v = 5.93\times10^{6}\ \mathrm{m/s}$$
the magnetic force is square to the motion at every instant, so it does no work and the speed is untouched by the turn
Answer $$\boxed{\;r = 6.75\ \mathrm{mm},\qquad v = 5.93\times10^{6}\ \mathrm{m/s}\ \text{unchanged}\;}$$
Check
Order of magnitude check on the period: $T = 2\pi m/(\vert q\vert B) = 7.16\ \mathrm{ns}$, so a quarter turn takes under two nanoseconds. That is short enough that nothing else has time to act, which is why the idealisation holds.
The same electron in two uniform fields of comparable strength: the electric one changed its speed by six million metres per second and did not turn it, while the magnetic one turned it through ninety degrees and did not change its speed at all.
How to tell them apart
Ask whether the force has a component along the velocity. The electric force does not care which way the particle is moving, so it generally does work and changes the energy. The magnetic force is built from a cross product with the velocity, so it never has a component along the motion and never changes the energy. Any question mentioning speed, work or kinetic energy is asking about an electric field somewhere, even if a magnet dominates the picture.
Scaffolding comes off
The common skeleton
Name the quantity the last line asks for, and let it choose the route.
Name the instant: the moment of switching, the steady state, or a stated time.
Reduce the network as far as it will go, writing every intermediate value down.
Work back outwards to the element the question is about, and get its own current or its own voltage.
Feed that number into the formula for the quantity asked for, with units carried through.
Check: junction sums, loop sums, power in against power out, and whether the size is one a bench could produce.
1 · fully worked
Force on the 12.0 ohm branch wire, a long time after the switch closes
A $24.0\ \mathrm{V}$ ideal battery drives $R_1 = 8.00\ \mathrm{\Omega}$ in series with a parallel pair, $R_2 = 12.0\ \mathrm{\Omega}$ and $R_3 = 6.00\ \mathrm{\Omega}$. The $R_2$ element is a straight wire of length $0.200\ \mathrm{m}$ lying at right angles to a uniform field of $0.350\ \mathrm{T}$. Find the force on it.
Given
$\varepsilon = 24.0\ \mathrm{V}$, ideal
$R_1 = 8.00\ \mathrm{\Omega}$ in series with the parallel pair
$R_2 = 12.0\ \mathrm{\Omega}$ and $R_3 = 6.00\ \mathrm{\Omega}$ in parallel
the $R_2$ element is a wire, $L = 0.200\ \mathrm{m}$, square to $B = 0.350\ \mathrm{T}$
Junction check: the other branch carries $8.00/6.00 = 1.33\ \mathrm{A}$, and $0.667 + 1.33 = 2.00\ \mathrm{A}$, the total. Power check: the battery supplies $(24.0)(2.00) = 48.0\ \mathrm{W}$, and $I^{2}R_1 = 32.0\ \mathrm{W}$ plus $I^{2}R_{23} = 16.0\ \mathrm{W}$ is the same $48.0\ \mathrm{W}$.
Four circuit lines to get one current, then one line of magnetism. That ratio is typical.
Using the total $2.00\ \mathrm{A}$ instead of the branch $0.667\ \mathrm{A}$ would have trebled the answer, and nothing about the units or the size would have warned you.
2 · you write the reasoning
Easier than the rung above: one loop, no parallel section and no magnetic step. A battery of emf $9.00\ \mathrm{V}$ and internal resistance $0.500\ \mathrm{\Omega}$ is connected to a $4.00\ \mathrm{\Omega}$ resistor. Find the current, the terminal voltage and the power delivered to the external resistor. The steps are given; write the reason for each one before opening the model answers.
reasoning
The internal resistance is in the same loop as the external one and carries the same current, so the two add exactly as any pair of series resistances would. Treating the battery as ideal here would give $2.25\ \mathrm{A}$, twelve per cent too high.
reasoning
The terminal voltage is what the outside world actually receives, and it falls below the emf by the drop inside the source. A voltmeter across the battery reads this, not the $9.00\ \mathrm{V}$ printed on its label, and the gap grows with the current drawn.
reasoning
The current is the same in both resistors, so the form with $I^{2}$ in it needs no extra work. Cross checking with $P = I V_{ab} = (2.00)(8.00) = 16.0\ \mathrm{W}$ gives the same number, and the difference from the total $\varepsilon I = 18.0\ \mathrm{W}$ is the $2.00\ \mathrm{W}$ wasted inside the battery.
3 · find the buried error
Harder than the rung above: two capacitors, a resistor and a steady state, and the two combination rules are both in play. A $20.0\ \mathrm{V}$ battery, a $2.00\ \mathrm{k\Omega}$ resistor and two capacitors of $3.00\ \mathrm{\mu F}$ and $6.00\ \mathrm{\mu F}$ in series are all wired in one loop. A long time after the switch closes, how much energy is stored in the $3.00\ \mathrm{\mu F}$ capacitor? Below is a student's four step solution. Exactly two of the steps are wrong. Find them.
the two buried errors (2)
⚠ step 2
The two capacitors are in series, so it is the reciprocals that add: $\frac{1}{C_{\rm eq}} = \frac{1}{3.00}+\frac{1}{6.00}$, giving $2.00\ \mathrm{\mu F}$, not $9.00\ \mathrm{\mu F}$.
The resistor rules are learned first and used far more often, and nothing about the arithmetic or the units complains when they are applied to a capacitor. The answer even looks reasonable: bigger than either member, which is what adding two of anything usually gives.
right
$C_{\rm eq} = 2.00\ \mathrm{\mu F}$, which is smaller than either capacitor, as every series combination must be.
⚠ step 4
The voltage used is the battery's $20.0\ \mathrm{V}$, but the $3.00\ \mathrm{\mu F}$ capacitor only has its own share of that across it. With the corrected charge $Q = 40.0\ \mathrm{\mu C}$, its voltage is $Q/C_1 = 13.3\ \mathrm{V}$.
The source voltage is the number printed on the figure and the element voltage has to be worked out, so the eye reaches for the one that is already there. The step also ignores the charge computed in the line immediately above it, which is the clue that something has gone loose.
right
$U_1 = \tfrac12 C_1V_1^{2} = \tfrac12(3.00\times10^{-6})(13.33)^{2} = 2.67\times10^{-4}\ \mathrm{J}$, or equivalently $U_1 = Q^{2}/(2C_1)$ with $Q = 40.0\ \mathrm{\mu C}$.
4 · the bare problem
§10.5 — stored energy in a parallel pair, steady state●●●○○
No scaffolding this time. A $15.0\ \mathrm{V}$ ideal battery, a $1.00\ \mathrm{k\Omega}$ resistor and two capacitors of $2.00\ \mathrm{\mu F}$ and $4.00\ \mathrm{\mu F}$ in parallel with each other are all wired in a single loop with a switch.
Given
$\varepsilon = 15.0\ \mathrm{V}$, ideal
$R = 1.00\ \mathrm{k\Omega}$
$C_1 = 2.00\ \mathrm{\mu F}$ in parallel with $C_2 = 4.00\ \mathrm{\mu F}$
the resistor, the switch and the capacitor pair are all in one loop
the reading is taken a long time after the switch closes
Find
(a) How much charge sits on the $4.00\ \mathrm{\mu F}$ capacitor?
(b) How much energy is stored in it?
Hint 1/4
Name the instant first, and let that decide what the resistor is doing.
Hint 2/4
In the steady state no current flows, so the resistor drops nothing and the full emf sits across the capacitor pair. Capacitors in parallel share that same voltage: $V_1 = V_2 = \varepsilon$.
Hint 3/4
With $\varepsilon = 15.0\ \mathrm{V}$ and $C_2 = 4.00\ \mathrm{\mu F}$: $Q_2 = C_2\varepsilon$ and $U_2 = \tfrac12 C_2\varepsilon^{2}$.
Hint 4/4
$Q_2 = 60.0\ \mathrm{\mu C}$ and $U_2 = 4.50\times10^{-4}\ \mathrm{J}$.
Show solutionName the instant and delete the resistor's effect
$$I = 0 \ \Rightarrow\ V_R = IR = 0$$
steady state means the charges have stopped moving, and a resistor carrying no current drops no voltage regardless of its value
$$V_1 = V_2 = \varepsilon = 15.0\ \mathrm{V}$$
the pair is in parallel, so both capacitors have the same two nodes and therefore the same voltage, which the loop rule fixes at the full emf
Independent form of the energy: $U_2 = \tfrac12 Q_2V = \tfrac12(6.00\times10^{-5})(15.0) = 4.50\times10^{-4}\ \mathrm{J}$, with no square in it. Consistency with the pair: $C_{\rm eq} = 6.00\ \mathrm{\mu F}$ stores $\tfrac12(6.00\times10^{-6})(15.0)^{2} = 675\ \mathrm{\mu J}$ in total, and $450 + 225 = 675\ \mathrm{\mu J}$.
Full exam-style question
Exam length: a real battery, a mixed network, a capacitor and a wire in a fieldexam format
A battery of emf $36.0\ \mathrm{V}$ and internal resistance $1.00\ \mathrm{\Omega}$ is connected to $R_1 = 5.00\ \mathrm{\Omega}$ in series with a parallel pair, $R_2 = 20.0\ \mathrm{\Omega}$ and $R_3 = 30.0\ \mathrm{\Omega}$. A $50.0\ \mathrm{\mu F}$ capacitor is connected across $R_3$. The $R_2$ element is a straight wire of length $0.150\ \mathrm{m}$ lying at right angles to a uniform magnetic field of $0.250\ \mathrm{T}$. A long time after the switch closes, find (a) the current supplied by the battery, (b) the terminal voltage, (c) the current in each of the two parallel branches, (d) the charge and energy stored in the capacitor, and (e) the magnetic force on the $R_2$ wire.
Two independent checks. Junction: $1.20 + 0.800 = 2.00\ \mathrm{A}$, the supply current. Power: the emf delivers $(36.0)(2.00) = 72.0\ \mathrm{W}$, while $I^{2}r = 4.00\ \mathrm{W}$, $I^{2}R_1 = 20.0\ \mathrm{W}$ and $I^{2}R_{23} = 48.0\ \mathrm{W}$ sum to the same $72.0\ \mathrm{W}$. A single wrong branch current would break both.
Five parts, and only the last one is magnetism. Four of the five marks live in the circuit.
Notice how late the capacitance was used. Its only role in parts (a) to (c) was to tell you that one branch carried no current, and its numerical value did not appear until part (d).
Practice
A · concept 4 questions
1§10.1 — is current used up on the way round●●○○○
One mark, and it separates the people who have read the definition from the people who have collected the formulas. A single loop contains a battery and one resistor, and two identical ammeters are placed in the loop, one on each side of the resistor.
Given
a single unbranched loop
one resistor, with an ammeter immediately before it and another immediately after
both ammeters are ideal and identical
Find
(a) True or false: the second ammeter reads less than the first.
Hint 1/4
Ask what a resistor takes from the circuit, and whether that thing is the same as what an ammeter measures.
Hint 2/4
An ammeter measures charge per second. Charge is conserved and cannot accumulate anywhere in a wire, so the junction rule applies at every point of an unbranched loop: whatever goes in comes out.
Hint 3/4
The two readings are therefore equal. What the resistor removes is energy per coulomb, which a voltmeter across it would show as a drop in potential, not a drop in current.
Hint 4/4
So the statement is false: both ammeters read the same current.
Show solutionApply charge conservation to the wire itself
$$\sum I_{\rm in} = \sum I_{\rm out}\ \text{at every point}$$
charge cannot pile up in a wire; if it did, an electric field would build there and immediately push it on
$$I_{\rm before} = I_{\rm after}$$
an unbranched loop has one current, and it is the same everywhere in the loop
Name the thing that does fall
$$V_{\rm before} - V_{\rm after} = IR$$
the potential drops across the resistor by exactly $IR$, and that product times the current is the power turned into heat
Energy check: the battery supplies $\varepsilon I$ watts and the resistor removes $I^{2}R$ watts, and for a single loop $\varepsilon = IR$ makes these equal. The books balance in energy while the charge simply circulates.
2§10.4 — rewiring a capacitor pair from parallel to series●●●○○
A $2.00\ \mathrm{\mu F}$ and a $6.00\ \mathrm{\mu F}$ capacitor are connected in parallel across a $12.0\ \mathrm{V}$ supply and allowed to settle. They are then disconnected, rewired in series, and reconnected across the same $12.0\ \mathrm{V}$ supply.
Check by parts: in series the charge is $Q = (1.50\times10^{-6})(12.0) = 18.0\ \mathrm{\mu C}$ on each, giving $V_1 = 9.00\ \mathrm{V}$ and $V_2 = 3.00\ \mathrm{V}$, and $\tfrac12 QV_1 + \tfrac12 QV_2 = 81.0 + 27.0 = 108\ \mathrm{\mu J}$.
3§10.7 — net force and net torque on a current loop●●●○○
A closed rectangular loop of wire carries a steady current and sits in a uniform magnetic field, tilted so that its plane is neither along the field nor square to it. Nothing is holding it in place.
Given
a closed rectangular loop carrying a steady current
a uniform magnetic field, the same everywhere across the loop
the loop is tilted with respect to the field and free to move
Find
(a) True or false: the net magnetic force on the loop is zero, yet the loop can still start rotating.
Hint 1/4
Separate two questions that sound like one: does the loop get pushed anywhere, and does it get twisted.
Hint 2/4
For a closed loop in a uniform field the forces on opposite sides are equal and opposite, so they add to zero; but they act along different lines, which is the definition of a couple.
Hint 3/4
A couple has zero resultant force and a nonzero moment, $\tau = NIAB\sin\theta$, which vanishes only when the loop's moment is along the field.
Hint 4/4
So both halves of the statement hold, and the answer is true.
the vector sum of the side elements around any closed loop is zero, and the field is the same for all of them because it is uniform
$$\vec\tau = \vec\mu\times\vec B \ne \vec 0\ \text{unless}\ \vec\mu\parallel\vec B$$
the opposing forces act along different lines, so their moments add rather than cancel
Answer $$\boxed{\;\text{True: zero net force, nonzero torque}\;}$$
Check
Limiting case: turn the loop until its moment lines up with the field and the torque does vanish, which is the resting position of a compass needle. Both statements agree there, as they must.
4§10.1 — a wire drawn out to twice its length●●●○○
A piece of copper wire is drawn through a die so that it becomes twice as long. Nothing is added or removed, so the volume of copper is unchanged and the wire becomes correspondingly thinner.
Given
the same piece of copper before and after, so the volume is unchanged
the final length is twice the initial length
the resistivity of copper is unchanged by the drawing
the original resistance is $R_0$
Find
(a) What is the new resistance in terms of the old one?
Hint 1/4
Two things about the wire changed, not one. Write down what happened to each of the two geometric quantities in the resistance formula.
Hint 2/4
$R = \rho L/A$. The length doubled; since the volume $LA$ is fixed, doubling $L$ must halve $A$.
Hint 3/4
So $R_{\rm new} = \rho(2L)/(A/2) = 4\rho L/A$, with the resistivity unchanged because it is still the same copper.
Hint 4/4
$R_{\rm new} = 4R_0$.
Show solutionTrack both geometric factors, not just the obvious one
$$L' = 2L,\qquad L'A' = LA \ \Rightarrow\ A' = \frac{A}{2}$$
the volume of metal is fixed by the amount of copper present, so lengthening it must thin it in exact proportion
the resistivity is a property of copper and is untouched by drawing
Answer $$\boxed{\;R' = 4R_0\;}$$
Check
General form as a check: for a fixed volume, $R \propto L^{2}$, so tripling the length would give nine times the resistance. Doubling gives four, which is what was found.
B · computation 7 questions
1§10.3 — a mixed capacitor network across 30.0 V●●●○○
Three capacitors are arranged and a supply is connected across the whole arrangement. Every capacitor starts empty and the reading is taken once everything has settled.
Given
$C_1 = 12.0\ \mathrm{\mu F}$ in series with $C_2 = 4.00\ \mathrm{\mu F}$
that series pair in parallel with $C_3 = 5.00\ \mathrm{\mu F}$
the supply provides $30.0\ \mathrm{V}$ across the whole arrangement
Find
(a) Find the equivalent capacitance.
(b) Find the voltage across $C_1$.
(c) Find the total energy stored.
Hint 1/4
Work from the inside out: collapse whichever pair is genuinely in series or genuinely in parallel first, then treat the result as one element.
Hint 2/4
Series capacitors: $\frac{1}{C_{12}} = \frac{1}{C_1}+\frac{1}{C_2}$. Parallel capacitors: $C_{\rm eq} = C_{12} + C_3$. Then $Q = CV$ and $U = \tfrac12 C_{\rm eq}V^{2}$.
Hint 3/4
With $C_1 = 12.0\ \mathrm{\mu F}$, $C_2 = 4.00\ \mathrm{\mu F}$, $C_3 = 5.00\ \mathrm{\mu F}$ and $V = 30.0\ \mathrm{V}$: $C_{12} = (12.0)(4.00)/16.0$, then add $5.00$, then $Q_{12} = C_{12}V$ and divide by $C_1$.
Power audit: the emf produces $(18.0)(3.00) = 54.0\ \mathrm{W}$; the internal resistance takes $4.50\ \mathrm{W}$, $R_1$ takes $13.5\ \mathrm{W}$ and the parallel pair takes $36.0\ \mathrm{W}$, summing to $54.0\ \mathrm{W}$. Also $I_2 = 2.00\ \mathrm{A}$ and $I_3 = 1.00\ \mathrm{A}$ add to the $3.00\ \mathrm{A}$ supplied.
3§10.5 — timing a capacitor as it fills●●●●○
A switch is closed at $t = 0$ on a single loop containing a battery, a resistor and an empty capacitor.
Given
$\varepsilon = 12.0\ \mathrm{V}$, ideal
$R = 33.0\ \mathrm{k\Omega}$
$C = 47.0\ \mathrm{\mu F}$, uncharged at $t = 0$
the switch closes at $t = 0$
Find
(a) Find the time constant and the final charge.
(b) Find the charge at $t = 2.00\ \mathrm{s}$.
(c) How long until the charge reaches ninety per cent of its final value?
Hint 1/4
Two of the three parts are about the ends of the story and one is about the middle; get the ends first because the middle is written in terms of them.
Hint 2/4
$\tau = RC$, $Q_\infty = C\varepsilon$, and $q(t) = Q_\infty(1 - e^{-t/\tau})$. Inverting the last of these gives $t = -\tau\ln(1 - q/Q_\infty)$.
Hint 3/4
With $R = 33.0\times10^{3}\ \mathrm{\Omega}$, $C = 47.0\times10^{-6}\ \mathrm{F}$ and $\varepsilon = 12.0\ \mathrm{V}$: $\tau = (33.0\times10^{3})(47.0\times10^{-6})$, $Q_\infty = (47.0\times10^{-6})(12.0)$, and for ninety per cent, $t = \tau\ln 10$.
Consistency between parts: $2.00\ \mathrm{s}$ is $1.29\,\tau$ and $3.57\ \mathrm{s}$ is $2.30\,\tau$, so the second time is later and the charge at it must be higher, which it is. Current check: $I(0) = \varepsilon/R = 0.364\ \mathrm{mA}$ falling to $0.100\ \mathrm{mA}$ at $t = 2.00\ \mathrm{s}$, and the fraction of charge in place plus the fraction of current remaining equals one, as it always does.
4§10.1 — resistance and heating of a length of copper wire●●●○○
A cable is specified by its material and its dimensions rather than by an ohm value, which is how a real extension lead is sold.
the current is the same all along the wire, so this form is the natural one
Answer $$\boxed{\;R = 0.514\ \mathrm{\Omega},\quad V = 6.17\ \mathrm{V},\quad P = 74.0\ \mathrm{W}\;}$$
Check
Cross check on the power without using the resistance twice: $P = IV = (12.0)(6.17) = 74.0\ \mathrm{W}$. On size, seventy four watts is roughly an old filament lamp, which matches the everyday observation that a fully loaded thin cable becomes noticeably warm.
5§10.6 — an alpha particle circling in a 0.850 T field●●●○○
An alpha particle, which is a helium nucleus carrying two elementary charges, is injected square into a uniform magnetic field.
$v = 1.50\times10^{6}\ \mathrm{m/s}$, perpendicular to the field
$B = 0.850\ \mathrm{T}$, uniform
Find
(a) Find the radius of its path.
(b) Find the time for one revolution.
(c) Find its kinetic energy.
Hint 1/4
Two of these are about the geometry of the circle and one is about the particle alone; only one of the three needs the field at all.
Hint 2/4
$r = mv/(\vert q\vert B)$, $T = 2\pi m/(\vert q\vert B)$, and $K = \tfrac12 mv^{2}$, which has nothing to do with the magnet.
Hint 3/4
With $m = 6.64\times10^{-27}\ \mathrm{kg}$, $\vert q\vert = 3.20\times10^{-19}\ \mathrm{C}$, $v = 1.50\times10^{6}\ \mathrm{m/s}$ and $B = 0.850\ \mathrm{T}$: the denominator $\vert q\vert B = 2.72\times10^{-19}$ appears in both of the first two.
the magnetic force does no work, so this value is fixed by how the particle was produced, not by the field it is in
Answer $$\boxed{\;r = 3.66\ \mathrm{cm},\quad T = 1.53\times10^{-7}\ \mathrm{s},\quad K = 7.47\times10^{-15}\ \mathrm{J}\;}$$
Check
Independent route to the period from the radius: $T = 2\pi r/v = 2\pi(3.662\times10^{-2})/(1.50\times10^{6}) = 1.53\times10^{-7}\ \mathrm{s}$. In electronvolts the energy is $7.47\times10^{-15}/1.60\times10^{-19} = 4.66\times10^{4}\ \mathrm{eV}$, a plausible figure for a laboratory alpha source.
6§10.7 — torque on a 25 turn coil at two angles●●●○○
A rectangular coil carries a steady current in a uniform field and is free to turn about an axis through its centre.
Given
$N = 25$ turns, area $A = 8.00\times10^{-3}\ \mathrm{m^{2}}$ each
$I = 2.00\ \mathrm{A}$
$B = 0.400\ \mathrm{T}$, uniform
Find
(a) Find the magnetic moment of the coil.
(b) Find the largest torque it can experience in this field.
(c) Find the torque when its moment is at $30.0^{\circ}$ to the field.
Hint 1/4
Everything about the coil apart from its orientation collapses into a single quantity; compute that first and the rest is one multiplication each.
Hint 2/4
$\mu = NIA$, then $\tau = \mu B\sin\theta$ with the angle measured between the moment and the field, so the maximum is $\mu B$ at $90^{\circ}$.
Hint 3/4
With $N = 25$, $I = 2.00\ \mathrm{A}$, $A = 8.00\times10^{-3}\ \mathrm{m^{2}}$ and $B = 0.400\ \mathrm{T}$: $\mu = (25)(2.00)(8.00\times10^{-3})$, then multiply by $B$, then by $\sin 30.0^{\circ}$.
Ratio check: the third answer is exactly half the second, and $\sin 30.0^{\circ}$ is exactly one half, so the two parts are consistent. Units check: $\mathrm{A\,m^{2}\cdot T} = \mathrm{A\,m^{2}\cdot N\,A^{-1}m^{-1}} = \mathrm{N\,m}$, a torque.
7§10.7 — counting carriers with a Hall measurement●●●●○
A flat metal strip carries a current along its length while a magnetic field is applied through its face, and a small voltage appears across its width.
Plausibility rather than repetition: a metal has of order $10^{29}$ atoms per cubic metre, so about one free electron per atom is exactly what a good conductor should show. An answer of $10^{22}$ would describe a semiconductor and an answer of $10^{35}$ nothing physical at all.
C · exam level 4 questions
1§10.5 — charge on a capacitor wired across one of two series resistors●●●●○
Exam length, one mark for the number and the rest for knowing which voltage belongs in it. A capacitor is connected across the second of two series resistors and the circuit is left until nothing is changing.
Given
$\varepsilon = 12.0\ \mathrm{V}$, ideal battery
$R_1 = 3.00\ \mathrm{\Omega}$ in series with $R_2 = 6.00\ \mathrm{\Omega}$
$C = 20.0\ \mathrm{\mu F}$, connected across $R_2$
steady state reached
Find
(a) How much charge is stored on the capacitor?
Hint 1/4
Decide first what the capacitor branch is doing to the current, then decide which voltage it has settled at.
Hint 2/4
In the steady state the capacitor branch carries no current, so the two resistors form a plain series loop and the capacitor sits at whatever $R_2$ has across it: $Q = CV_2$ with $V_2 = IR_2$.
Hint 3/4
With $\varepsilon = 12.0\ \mathrm{V}$, $R_1 = 3.00\ \mathrm{\Omega}$ and $R_2 = 6.00\ \mathrm{\Omega}$: $I = 12.0/9.00 = 1.33\ \mathrm{A}$ and $V_2 = (1.333)(6.00) = 8.00\ \mathrm{V}$, then $Q = (20.0\times10^{-6})(8.00)$.
Hint 4/4
$Q = 1.60\times10^{-4}\ \mathrm{C}$.
Show solutionErase the branch that carries nothing
$$I_C = 0\ \Rightarrow\ \text{plain series loop}$$
no current crosses into a capacitor once the charge has stopped changing, so the resistors can be solved on their own
Check by the other resistor: $V_1 = (1.333)(3.00) = 4.00\ \mathrm{V}$, and $4.00 + 8.00 = 12.0\ \mathrm{V}$, the emf. The stored energy is $\tfrac12(20.0\times10^{-6})(8.00)^{2} = 640\ \mathrm{\mu J}$, a plausible size for a small capacitor at a few volts.
2§10.5 — a student's solution to a loop with two opposing batteries●●●●○
A single loop contains two batteries whose positive terminals face each other, so that they push current in opposite senses, together with two resistors. A student's four step solution is shown; exactly one step is wrong.
Given
$\varepsilon_1 = 15.0\ \mathrm{V}$ and $\varepsilon_2 = 6.00\ \mathrm{V}$, opposing each other in the same loop
$R_1 = 4.00\ \mathrm{\Omega}$ and $R_2 = 5.00\ \mathrm{\Omega}$, both in that loop
Step 1. Both sources drive the loop, so the emfs add: $\varepsilon_{\rm tot} = 15.0 + 6.00 = 21.0\ \mathrm{V}$.
Step 2. The resistors are in the same unbranched loop, so $R_{\rm tot} = 4.00 + 5.00 = 9.00\ \mathrm{\Omega}$.
(a) Which step is wrong, and what should the current and the power in $R_2$ actually be?
Hint 1/4
Walk the loop once in your head and ask what each source does to the potential as you pass through it.
Hint 2/4
The loop rule adds a rise of $+\varepsilon$ when a source is traversed from its negative plate to its positive one and a drop of $-\varepsilon$ the other way. Two sources facing each other cannot both be rises around the same loop.
Hint 3/4
With $\varepsilon_1 = 15.0\ \mathrm{V}$ opposing $\varepsilon_2 = 6.00\ \mathrm{V}$ and $R_{\rm tot} = 9.00\ \mathrm{\Omega}$, the driving voltage is the difference: $I = (15.0 - 6.00)/9.00$.
Hint 4/4
Step 1 is wrong; $I = 1.00\ \mathrm{A}$ and $P_2 = 5.00\ \mathrm{W}$.
Show solutionWalk the loop and write each term with its own sign
the same formula the student used, now with a current that comes from a correct loop equation
Answer $$\boxed{\;\text{Step 1 is wrong};\quad I = 1.00\ \mathrm{A},\quad P_2 = 5.00\ \mathrm{W}\;}$$
Check
Power audit as an independent check: the fifteen volt source delivers $(15.0)(1.00) = 15.0\ \mathrm{W}$, the six volt source absorbs $(6.00)(1.00) = 6.00\ \mathrm{W}$, and the resistors take $4.00 + 5.00 = 9.00\ \mathrm{W}$. The books balance at $15.0 = 6.00 + 9.00$, which the student's $2.33\ \mathrm{A}$ could not do.
3§10.6 — force on a branch wire in a resistor network●●●●○
A network is built and one of its branches happens to be a straight wire lying in a magnetic field. The question asks about the force on that wire, but the wire's current is not given.
Given
$\varepsilon = 30.0\ \mathrm{V}$, ideal battery
$R_1 = 6.00\ \mathrm{\Omega}$ in series with the parallel pair
$R_2 = 20.0\ \mathrm{\Omega}$ and $R_3 = 5.00\ \mathrm{\Omega}$ in parallel
the $R_2$ element is a straight wire, $L = 0.250\ \mathrm{m}$, square to $B = 0.500\ \mathrm{T}$
Find
(a) What is the magnitude of the magnetic force on the $R_2$ wire?
Hint 1/4
The magnetic formula is one line; everything before it is circuit work. Ask which current actually passes through that particular wire.
Hint 2/4
Reduce the network to find the total current, work back out to the voltage across the parallel section, then $I_2 = V_{23}/R_2$ and $F = BI_2L$.
Junction check: $I_3 = 12.0/5.00 = 2.40\ \mathrm{A}$, and $0.600 + 2.40 = 3.00\ \mathrm{A}$, the supply current. On size, a fraction of an ampere in half a tesla over a quarter of a metre giving tens of millinewtons is the right order for a bench demonstration.
4§10.4 — heat produced while a capacitor fills●●●●○
A capacitor is charged from empty through a resistor and the process is allowed to finish. The question asks about the heat, not about the store.
Given
$\varepsilon = 12.0\ \mathrm{V}$, ideal battery
$R = 1.50\ \mathrm{k\Omega}$
$C = 220\ \mathrm{\mu F}$, empty at the start
charging is allowed to run to completion
Find
(a) How much energy is turned into heat in the resistor over the whole charging process?
Hint 1/4
Rather than integrating the current, account for the energy: the battery pays, the capacitor keeps some, and the resistor gets the rest.
Hint 2/4
The battery supplies $Q\varepsilon = C\varepsilon^{2}$ while the capacitor keeps $\tfrac12 C\varepsilon^{2}$, so the resistor receives $\tfrac12 C\varepsilon^{2}$, exactly the same as the capacitor and independent of $R$.
Hint 3/4
With $C = 220\times10^{-6}\ \mathrm{F}$ and $\varepsilon = 12.0\ \mathrm{V}$: $U_R = \tfrac12(220\times10^{-6})(12.0)^{2}$.
Hint 4/4
$U_R = 1.58\times10^{-2}\ \mathrm{J}$.
Show solutionAccount for the energy instead of integrating
Check the ratio rather than the arithmetic: the answer must be exactly half of $C\varepsilon^{2} = 3.17\times10^{-2}\ \mathrm{J}$, and it is. Any answer that depends on $R$ contradicts the algebra, since $R$ cancels out of the integral of $I^{2}R\,dt$ completely.
D · interleaved 4 questions
1§10.D — a parallel plate capacitor, from geometry to the force on an electron●●●○○
A pair of parallel plates is built to a stated size and a voltage is placed across it. A stray electron finds itself in the gap.
Given
plate area $A = 1.20\times10^{-2}\ \mathrm{m^{2}}$
separation $d = 1.20\ \mathrm{mm} = 1.20\times10^{-3}\ \mathrm{m}$, vacuum between
the field between large parallel plates is uniform, so the potential difference is simply the field times the gap
Force on one electron
$$F = eE = (1.60\times10^{-19})(2.00\times10^{5}) = 3.20\times10^{-14}\ \mathrm{N}$$
the definition of the electric field as force per unit charge, run backwards
Answer $$\boxed{\;C = 88.5\ \mathrm{pF},\quad Q = 21.2\ \mathrm{nC},\quad E = 2.00\times10^{5}\ \mathrm{V/m},\quad F = 3.20\times10^{-14}\ \mathrm{N}\;}$$
Check
Independent route to the field, through the surface charge density: $\sigma = Q/A = 2.124\times10^{-8}/1.20\times10^{-2} = 1.77\times10^{-6}\ \mathrm{C/m^{2}}$, and $E = \sigma/\varepsilon_0 = 2.00\times10^{5}\ \mathrm{V/m}$, which is the same number reached from the charge instead of from the voltage.
2§10.D — accelerate an electron, then bend it●●●●○
An electron gun and a small magnet in sequence. The electron is produced at rest, pushed across a gap, and then enters a field region at right angles to the field.
Given
the electron starts from rest and crosses a potential difference of $250\ \mathrm{V}$
it then enters a uniform field of $B = 1.20\times10^{-2}\ \mathrm{T}$, perpendicular to its motion
(a) Find the speed at which it enters the magnetic field.
(b) Find the radius of its path in that field.
(c) State what its speed is after half a revolution.
Hint 1/4
Two different fields do two different jobs here; decide which of them can change a speed and which cannot before computing anything.
Hint 2/4
The electric stage: $\tfrac12 mv^{2} = \vert q\vert V$. The magnetic stage: $r = mv/(\vert q\vert B)$, and $\Delta K = 0$ because the magnetic force does no work.
Hint 3/4
With $V = 250\ \mathrm{V}$, $m = 9.11\times10^{-31}\ \mathrm{kg}$, $\vert q\vert = 1.60\times10^{-19}\ \mathrm{C}$ and $B = 1.20\times10^{-2}\ \mathrm{T}$: $v = \sqrt{2(1.60\times10^{-19})(250)/(9.11\times10^{-31})}$, then divide $mv$ by $\vert q\vert B$.
Hint 4/4
$v = 9.37\times10^{6}\ \mathrm{m/s}$, $r = 4.45\ \mathrm{mm}$, and the speed after half a turn is still $9.37\times10^{6}\ \mathrm{m/s}$.
Show solutionThe electric stage, which sets the speed
$$\tfrac12 mv^{2} = \vert q\vert V$$
the work done by the accelerating field goes entirely into kinetic energy because the electron started from rest
Timing check: the period is $T = 2\pi m/(\vert q\vert B) = 2.98\times10^{-9}\ \mathrm{s}$, so half a turn takes about $1.5\ \mathrm{ns}$; and $\pi r/v = \pi(4.446\times10^{-3})/(9.371\times10^{6}) = 1.49\times10^{-9}\ \mathrm{s}$ agrees, reached from the geometry rather than the formula.
3§10.D — how much charge is actually in a charged capacitor●●●○○
A capacitor is charged and then disconnected. The question is about the sheer number of electrons involved, and about what would happen if that much charge were ever separated in open space.
the same charge treated as two isolated point charges, which is a thought experiment rather than a description of the capacitor
$$F = 3.60\times10^{6}\ \mathrm{N}$$
millions of newtons, which is the size of the force that makes large charge separations impossible to maintain in air
Answer $$\boxed{\;Q = 2.00\times10^{-2}\ \mathrm{C},\quad N = 1.25\times10^{17},\quad F = 3.60\times10^{6}\ \mathrm{N}\;}$$
Check
Consistency with the stored energy: $U = \tfrac12 QV = \tfrac12(2.00\times10^{-2})(200) = 2.00\ \mathrm{J}$, which is a joule or two, the sort of energy that gives a real jolt if the capacitor is discharged through a finger. The Coulomb figure and the energy figure describe the same charge in two very different geometries.
A capacitor works precisely because the two opposite charges are close together. Pull them a metre apart and the arrangement stops being possible; keep them a millimetre apart with an insulator between and it becomes an ordinary component.
4§10.D — the potential difference between two points that share no element●●●●○
Two chains of resistors are connected side by side across the same battery, and the question asks about two points in the middle of the chains, with nothing wired between them.
Given
$\varepsilon = 24.0\ \mathrm{V}$, ideal battery
left chain: $4.00\ \mathrm{\Omega}$ then $8.00\ \mathrm{\Omega}$, with the midpoint called $P$
right chain: $6.00\ \mathrm{\Omega}$ then $6.00\ \mathrm{\Omega}$, with the midpoint called $S$
both chains are connected across the same battery terminals
nothing is connected between $P$ and $S$
Find
(a) Find $V_P - V_S$.
(b) If a $3.00\ \mathrm{\mu F}$ capacitor is now connected between $P$ and $S$, find the charge it settles at.
Hint 1/4
The two points are not the ends of any single component, so no one formula gives the answer directly; find each point's potential separately and subtract.
Hint 2/4
Each chain is an independent series loop across the same $24.0\ \mathrm{V}$, so $I = \varepsilon/R_{\rm chain}$ in each, and the midpoint potential is the emf minus the drop across the first resistor. Then $Q = C(V_P - V_S)$.
Hint 3/4
With $\varepsilon = 24.0\ \mathrm{V}$: the left chain has $12.0\ \mathrm{\Omega}$ and so carries $2.00\ \mathrm{A}$, giving $V_P = 24.0 - (2.00)(4.00)$; the right chain also has $12.0\ \mathrm{\Omega}$ and carries $2.00\ \mathrm{A}$, giving $V_S = 24.0 - (2.00)(6.00)$.
Check from the other terminal: measuring up from the negative terminal instead, $V_P = (2.00)(8.00) = 16.0\ \mathrm{V}$ and $V_S = (2.00)(6.00) = 12.0\ \mathrm{V}$, the same two numbers and the same difference. A bridge like this is balanced only when the two chains split their voltage in the same ratio, which these do not.
Mistake ledger (21 entries)
⚠ Letting one element set the current in a series loop
the element in question is the one the sentence is about, so the eye goes to it and the rest of the loop stops being visible
wrong$$I = \frac{\varepsilon}{R_2}$$
right$$I = \frac{\varepsilon}{R_1 + R_2}$$
⚠ Feeding microfarads straight into a formula
the prefix is printed on the component and gets copied along with the number, and the answer still looks like a number
diagrams often show the wire against a plate or a page edge, and the eye takes the angle from whatever is nearest
wrong$$F = BIL\sin(\text{angle to the page})$$
right$$F = BIL\sin(\text{angle between the wire and } \vec B)$$
⚠ Taking the torque angle from the plane of the loop
the loop is the visible object and its plane is what a diagram shows, while the moment is an arrow you have to add yourself
wrong$$\tau = NIAB\sin(\text{angle between the plane and } \vec B)$$
right$$\tau = NIAB\sin(\text{angle between } \vec\mu \text{ and } \vec B)$$
⚠ Putting the charge into the velocity selector condition
every other magnetic formula in the week contains the charge, so leaving it out feels like an omission
wrong$$v = \frac{qE}{B}$$
right$$v = \frac{E}{B}$$
⚠ Using the selector field in the analyser radius
both stages are magnetic and the two field symbols look alike on a hurried page
wrong$$m = \frac{qBr}{v}$$
right$$m = \frac{qB'r}{v}$$
Formula card
Current and capacitance, the two definitions
$$I = \frac{\Delta Q}{\Delta t},\qquad C = \frac{Q}{V}$$
the current is the same at every cross section of an unbranched element; the capacitance depends on geometry and filling, never on how much charge is present
Ohm's law and the resistance of a shaped conductor
$$V = IR,\qquad R = \frac{\rho L}{A}$$
at fixed temperature; the second form needs a uniform cross section along the whole length
one equivalent resistance and one equivalent capacitance in a single loop; the resistance is the one seen from the capacitor's own terminals with the sources switched off
The capacitor at its two extreme instants
$$t = 0\ \text{(empty)}:\ \text{behaves as a wire};\qquad t\to\infty:\ \text{behaves as a break}$$
the first applies only if the capacitor started with no charge on it; a precharged capacitor behaves instead as a source of $Q_0/C$
The magnetic force and its circle
$$\vec F = q\,\vec v\times\vec B,\quad \vec F = I\,\vec L\times\vec B,\quad r = \frac{mv_\perp}{\vert q\vert B},\quad T = \frac{2\pi m}{\vert q\vert B}$$
uniform field; only the perpendicular component of the velocity bends, and the force never does work
the helix needs an entry at an angle; the selector needs the two forces exactly opposed; the torque needs a field uniform across the whole loop; the Hall result needs a steady state
Check yourself
Close the page and write out from memory: the two definitions that everything else in this block rests on; the four combination rules and the one sentence that decides which is which; the two energy formulas and which one is a rate; the two circuit rules and the sign convention that goes with them; what an empty capacitor looks like at the instant of switching and a long time later; the magnetic force in both of its forms and the two things it can never do; and the four magnetic templates with the phrase in a question that gives each one away. Then compare with the formula card and mark only what was missing.
Say what a resistor takes from a circuit and what it does not, and build a resistance from a material and a shape?
c-map
Read a question's last line and say in one sentence which route it wants and which instant it is standing at?
c-choose
Write all four combination rules down without hesitating, and say why the capacitor pair is the mirror image of the resistor pair?
c-combinations
Explain where the factor of one half in the stored energy comes from, and say what fraction of a battery's energy survives a charging run?
c-energy-power
Set up a two loop network with signs chosen once and read the answer correctly when a current comes out negative?
c-network-time
State what a uniform magnetic field can and cannot change about a free particle, and produce the radius and the period without looking?
c-magnetic-force
Name the four magnetic templates and the phrase in a question stem that identifies each one?
c-magnetic-templates
Glossary (18 terms)
electric currentelektrik akımı
The rate at which charge crosses a chosen surface, measured in amperes. By convention its direction is the direction positive carriers would move, which in a metal is opposite to the way the electrons actually drift.
ampereamper
The SI unit of current, equal to one coulomb per second. It is one of the base units of the SI system, which is why the coulomb is defined from it rather than the other way round.
akım yoğunluğu
Current per unit cross sectional area, in amperes per square metre. Two wires can carry the same current at very different current densities, and it is the density rather than the current that decides how hot a conductor gets.
sürüklenme hızı
The slow average velocity that the mobile carriers acquire on top of their fast random motion, typically well under a millimetre per second. It is not the speed at which the current appears to start when a switch is closed.
resistivityözdirenç
A property of a material alone, in ohm metres, measuring how strongly it opposes the flow of charge. Multiplying it by length and dividing by cross sectional area turns it into the resistance of a particular piece.
iletkenlik
The reciprocal of resistivity, in siemens per metre. A good conductor has a high conductivity and a low resistivity, and the two words carry exactly the same information.
resistancedirenç
The ratio of the potential difference across a conductor to the current through it, in ohms. For an ohmic material it is a constant at fixed temperature; for others it is still defined at each point but is no longer a single number.
ohmohm
The SI unit of resistance, equal to one volt per ampere. An ohm multiplied by a farad gives a second, which is the basis of every time constant in this material.
Ohm's law
The empirical statement that the current through certain conductors is proportional to the potential difference across them. It is not a law of nature in the way charge conservation is; plenty of ordinary components disobey it.
ohmic material
A material whose resistance stays constant as the voltage across it is varied, so that a graph of current against voltage is a straight line through the origin. Metals at fixed temperature are close to ohmic; diodes and gases are not.
The fractional change in resistivity per degree of temperature change. It is positive for metals, whose resistance rises as they warm, and negative for many semiconductors.
The conversion of electrical energy into heat in a resistance, at a rate of $I^{2}R$ watts. It is the reason a cable warms up under load and the reason a fuse works at all.
elektriksel güç
The rate at which electrical energy is converted, in watts. For any two terminal element it is the current through it multiplied by the potential difference across it, whatever the element happens to be.
kilowatt hourkilovat saat
A unit of energy equal to one kilowatt sustained for one hour, that is $3.60\times10^{6}$ joules. It is the unit an electricity meter counts in, which is why it appears in questions about running costs.
devre elemanı
Any two terminal component treated as a single object with one defining relation between its voltage and its current, such as a resistor, a capacitor or a source. Combination rules are statements about replacing several elements by one.
geçici rejim
The part of a circuit's behaviour that dies away after a switch is thrown, as opposed to the steady state that survives. In an RC circuit the transient decays exponentially with the time constant, and after about five of them it is undetectable.
nodedüğüm
A point where three or more branches meet, at which the junction rule applies. Two components joined end to end with nothing else attached do not form a node in the useful sense, because no current can divide there.
gerilim bölücü
A chain of series elements used to produce a fraction of an applied voltage. With resistors the fraction is proportional to resistance; with capacitors it is inversely proportional to capacitance, which is why the same circuit shape behaves oppositely in the two cases.
What comes next
§11 · Sources of Magnetic Field
Everything about magnetism so far has started from a field that was simply handed to you, with no account of where it came from. That question is next: currents do not only feel magnetic fields, they make them, and once that is quantified the whole subject closes on itself.
Sources
D. Giancoli, Physics for Scientists and Engineers with Modern Physics, 5th edition The set textbook for the course. Everything reviewed on this page comes from its treatment of capacitance and dielectrics, electric current and resistance, direct current circuits, and magnetism.
SI values of the physical constants used here $e = 1.60\times10^{-19}\ \mathrm{C}$, $m_p = 1.67\times10^{-27}\ \mathrm{kg}$, $m_e = 9.11\times10^{-31}\ \mathrm{kg}$, $m_\alpha = 6.64\times10^{-27}\ \mathrm{kg}$, $1\ \mathrm{u} = 1.66\times10^{-27}\ \mathrm{kg}$, $\varepsilon_0 = 8.85\times10^{-12}\ \mathrm{C^{2}/(N\,m^{2})}$, $k = 8.99\times10^{9}\ \mathrm{N\,m^{2}/C^{2}}$, and the resistivity of copper $1.68\times10^{-8}\ \mathrm{\Omega\,m}$.
The course syllabus line for this week The line reads catch up and review and carries no chapter numbers, so no chapter number is quoted anywhere on this page and no new material is introduced.