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09Magnetism: the force on moving charge, on currents, and on loops
A loudspeaker cone moves in and out about forty times a second for the lowest note you can hear, and about twenty thousand times a second for the highest. Nothing touches it. The only thing that changes is the current in a coil of wire sitting in the gap of a small permanent magnet. Eight sections of this course can tell you exactly what that current is and how much heat it makes in the wire, and not one of them can tell you why the coil moves at all.
By the end of this section you can take any charge or any current, put it into a field that has been handed to you, and produce the force on it, the direction of that force, the path the charge then follows, the torque on a loop, and the voltage that appears sideways across a strip of metal, all with signs that you can defend.
In 60 seconds
A magnetic field pushes on moving charge and on nothing else, the push is sideways to both the velocity and the field, and because it is always sideways it bends paths and twists loops without ever changing a speed or adding an energy.
Force on a moving charge
$$\vec F = q\,\vec v \times \vec B,\qquad F = |q|vB\sin\theta$$
always; theta is the angle between the velocity and the field, and the direction comes from the with the sign of q applied afterwards
The magnetic force does no work
$$\vec F\cdot\vec v = 0 \;\Rightarrow\; \frac{dK}{dt}=0$$
any time a question asks what happened to the speed, the kinetic energy or the magnitude of the velocity in a magnetic field: nothing did
a closed loop or coil in a uniform field; the net force is zero and only the twist survives
Energy of a dipole in a field
$$U = -\vec\mu\cdot\vec B = -\mu B\cos\theta$$
you are asked for the work needed to turn a coil, or which orientation is stable
Crossed fields and the
$$v=\frac{E}{B},\qquad V_H=\frac{IB}{nqt}$$
a velocity selector picks one speed out of a beam; the Hall formula gives the sideways voltage across a strip of thickness t along the field
Three most common mistakes
Using the speed instead of the perpendicular component in the radius, or putting the speed into the period, where it does not belong. Only $v_\perp$ sets the radius, and the period contains no speed at all.
Letting the magnetic force change the kinetic energy. It cannot: it is perpendicular to the velocity at every instant, so it does zero work and the speed never changes.
Reading the right hand rule off for a negative charge without flipping the answer. The rule gives the direction of $\vec v\times\vec B$; the force on an electron points the opposite way.
The two midterms and the final together carry $65\%$ of the course and the quizzes another $10\%$, so this is written exam material; a further $20\%$ of the course is laboratory work, which the syllabus lists separately and which this page does not try to cover.
How much time do you have?
10 minutes
You leave able to do the two things that start almost every question on this material: get the size and the direction of the force on a moving charge or a current, and turn that force into the radius and the period of a circular path.
The 60 second card · Formula card · The magnetic field, and the one experiment that defines it · The force on a moving charge, and why it never speeds anything up · Going round in a circle: radius, period, and what the speed does not affect · Mistake ledger
45 minutes
You add the parts that separate a set up mark from a full mark: the force on a wire, the torque on a coil and the energy that goes with it, and the two crossed field devices that turn up in exams every year.
The 60 second card · The magnetic field, and the one experiment that defines it · The force on a moving charge, and why it never speeds anything up · Going round in a circle: radius, period, and what the speed does not affect · The force on a current carrying wire · Torque on a loop, and the magnetic dipole moment · Crossed fields: the velocity selector and the mass spectrometer · Method box: a charge is fired into a field · Method box: getting the direction right every time · Exam level example · Practice set C · Mistake ledger
full read
You can also handle , bent wires and closed loops, the stable and unstable orientations of a coil, the Hall voltage and what its sign tells you about the carriers, and every sign convention that these carry with them.
The opening pages · Prerequisites and pretest · Notation · Conventions · All seven concept blocks · Method boxes · Contrast pair · Scaffolding comes off · Exam level example · Practice sets A to D · Mistake ledger · Formula card · Glossary · Check yourself
By the end of this section
State what a magnetic field is by the force it produces, quote the in base units, and say which of a stationary charge, a charge moving along the field and a charge moving across it feels a force.
Compute the magnitude and the direction of the force on a charge moving in a uniform field, including for a negative charge, and show that the force does no work.
Predict the radius, the period and the frequency of the path a charged particle follows in a uniform field, and separate the velocity into the part that circles and the part that drifts along the field.
Calculate the force a uniform field exerts on a straight current carrying wire at any angle, extend it to a bent wire and to a closed loop, and use it in a balance of forces.
Determine the magnetic dipole moment of a coil, the torque a uniform field exerts on it, its potential energy, and the work needed to turn it from one orientation to another.
Analyse a velocity selector and a mass spectrometer, obtaining the selected speed from the two field strengths and the mass of an ion from the radius of its path.
Interpret a Hall measurement: obtain the number density of the carriers from the sideways voltage, and read the sign of the carriers off the polarity.
Syllabus coverage
Magnetism
Magnets, poles and field lines; the magnetic field defined by the force on a moving charge; the tesla; the force on a moving charge and the fact that it does no work; circular and helical motion in a uniform field, with the radius, the period and the frequency; the force on a straight current carrying wire, on a bent wire and on a closed loop; the torque on a current loop, the magnetic dipole moment and its potential energy; crossed fields, the velocity selector, the mass spectrometer and the of the electron; and the Hall effect.
The week line is the single word Magnetism and carries no chapter number, so no chapter or section number of the textbook is quoted anywhere on this page.
covered
How a current produces a magnetic field
The field of a straight wire, of a loop and of a solenoid, and the law that produces them from the current.
Deferred to the section on sources of the magnetic field. On this page every field is handed to you as data, and no field is ever calculated from a current. That a current does produce a field is used only as the reason why two magnets and two wires interact at all, never as a formula.
deferred
Fields that change with time
The voltage a changing field drives round a loop, and everything built on it.
Deferred to the section on electromagnetic induction. Every field on this page is steady, and every loop that appears is either held still or turned slowly enough that the question only asks for the torque on it.
deferred
Relativistic correction to the radius
What happens to the radius formula when the speed becomes comparable with the speed of light.
Not named on the week line and not examinable here. It is mentioned in one line only because every worked speed on this page is checked against the speed of light so that you can see the formula being used inside the range where it is honest; the fastest particle here reaches under seven per cent of the speed of light.
off_syllabus
Recall first
Force, mass and circular motion
A particle moving in a circle of radius $r$ at constant speed $v$ has an acceleration $v^{2}/r$ pointing at the centre, so the net force on it is $mv^{2}/r$ towards the centre. Nothing about the source of that force enters the statement.
The whole of the circular motion section is this identity with the magnetic force written on the left hand side. If you cannot write $F=mv^{2}/r$ from memory, the radius formula will look like something to be memorised rather than something that follows in one line.
Work done by a force perpendicular to the motion
Work is $\vec F\cdot d\vec s$. If the force stays at right angles to the displacement then the dot product is zero at every instant, so the work is zero and the kinetic energy is unchanged, no matter how long the force acts.
It is the single most useful fact in this section. It is why a magnetic field never changes a speed, and it is what lets you write down half the answers in a question before doing any arithmetic.
Electric field and the force on a charge
A charge $q$ in an electric field feels $\vec F=q\vec E$, in the direction of $\vec E$ if $q$ is positive and against it if $q$ is negative. This force acts whether the charge is moving or not.
The velocity selector puts an electric force and a magnetic force on the same particle at the same time. Telling the two apart in one picture is the whole trick, and the contrast pair on this page is built on it.
Energy gained crossing a potential difference
A charge $q$ released from rest and taken across a potential difference $V$ arrives with kinetic energy $qV$, so $\tfrac12 mv^{2}=qV$ and $v=\sqrt{2qV/m}$.
Almost every exam question in this section starts by accelerating a particle electrically and only then lets it into the magnetic field. That first stage is entirely the earlier material and this is the one line it needs.
Current, drift velocity and carrier density
The current in a conductor of cross sectional area $A$ is $I=nqv_dA$, where $n$ is the number of mobile carriers per cubic metre, $q$ the charge on each and $v_d$ the average drift speed. Drift speeds in a metal are tiny, of order $10^{-4}\ \mathrm{m/s}$.
The Hall effect is this relation and the magnetic force put together. Without $I=nqv_dA$ the Hall voltage formula has to be memorised; with it, the formula is three lines of algebra.
Torque about an axis
The torque of a force about an axis is the force times the perpendicular distance from the axis to the line of the force. Two equal and opposite forces whose lines are a distance $d$ apart make a of torque $Fd$ about every axis, so no axis needs to be chosen.
The torque on a current loop is exactly such a couple. Knowing that the answer is the same about every axis is what makes the derivation short and what lets you check it from a different pivot.
Vector components and the right handed set of axes
With $\hat x$ to the right, $\hat y$ up and $\hat z$ out of the page, the cross products of the unit vectors are $\hat x\times\hat y=\hat z$, $\hat y\times\hat z=\hat x$ and $\hat z\times\hat x=\hat y$, and reversing the order of any of them reverses the sign.
Every direction on this page can be checked with these three lines, which is worth having when the right hand rule and a negative charge are in the same problem and your hand has run out of fingers.
Try it yourself first (3 questions)
1§09.0 — a force that stays sideways●○○○○
Before anything magnetic appears, one question about forces in general. A ball on the end of a string is whirled in a horizontal circle at constant speed. The tension in the string is the only horizontal force on it. Getting this wrong is not a problem; it tells you which line of the circular motion block to slow down at.
Given
the speed is constant
the string is inextensible, so the radius is constant
the tension points from the ball towards the centre at every instant
Find
(a) How much work does the tension do on the ball during one complete revolution?
Hint 1/4
You are being asked for work, so the only thing that matters is the angle between the force and the direction of travel at each instant.
Hint 2/4
Work is $\vec F\cdot d\vec s = F\,ds\cos\alpha$, where $\alpha$ is the angle between the force and the small step of the path.
Hint 3/4
Here the tension points at the centre and the ball moves along the circle, so at every instant $\alpha = 90^{\circ}$ and $\cos\alpha = 0$.
Hint 4/4
The work is zero, and it is zero instant by instant rather than only over a whole loop.
Show solution
Dot product, not round trip; the closed path is right for wrong reasons.
Write work as a dot product
$$dW = \vec F\cdot d\vec s = F\,ds\cos\alpha$$
work only counts the component of the force along the step, which is what the cosine picks out
$$\alpha = 90^{\circ}\ \text{at every instant}$$
the tension points at the centre while the step is along the tangent, and a radius meets a circle at right angles
Add up the steps
$$W=\oint F\cos(90^{\circ})\,ds = 0$$
every term in the sum is zero, so no cancellation between positive and negative parts is being relied on
Answer $$\boxed{\,W = 0\ \mathrm{J}\,,\ \text{and }\tfrac12 mv^{2}\text{ is unchanged}\,}$$
Check
Independent check from the other end: the speed is stated to be constant, so the kinetic energy is constant, so by the work energy theorem the net work must be zero. Two routes, same answer, and neither used the mass.
2§09.0 — the speed an accelerating voltage buys●●○○○
The second thing this page assumes is the first stage of almost every magnetic problem in an exam: a particle is dropped from rest across a potential difference and arrives somewhere with a speed. A proton starts at rest and is accelerated through a potential difference of $500\ \mathrm{V}$.
Given
$q = +1.60\times10^{-19}\ \mathrm{C}$
$m_p = 1.67\times10^{-27}\ \mathrm{kg}$
$V = 500\ \mathrm{V}$, starting from rest
Find
(a) Find the kinetic energy it arrives with, in joules, and its speed.
Hint 1/4
This is an energy question, not a force question: nothing needs to be known about the shape of the field or the distance travelled, only about the two ends of the trip.
Hint 2/4
A charge released from rest across a potential difference $V$ arrives with kinetic energy $qV$, so $\tfrac12 mv^{2}=qV$.
Hint 3/4
Here $q = 1.60\times10^{-19}\ \mathrm{C}$, $V = 500\ \mathrm{V}$ and $m_p = 1.67\times10^{-27}\ \mathrm{kg}$.
Hint 4/4
The proton arrives with $8.00\times10^{-17}\ \mathrm{J}$ and a speed of $3.10\times10^{5}\ \mathrm{m/s}$.
Show solution
Energy first, speed second; the joules are reused all section.
The energy, straight from the potential difference
going through the energy avoids ever needing the accelerating field or the gap between the plates
$$v = 3.10\times10^{5}\ \mathrm{m/s}$$
three figures, matching the least precise datum
Answer $$\boxed{\,K = 8.00\times10^{-17}\ \mathrm{J},\qquad v = 3.10\times10^{5}\ \mathrm{m/s}\,}$$
Check
Order of magnitude check: $v/c = 3.10\times10^{5}/3.00\times10^{8} = 0.00103$, one tenth of one per cent of the speed of light, so the non relativistic formula was allowed. A useful rule of thumb falls out: a proton needs about half a kilovolt to reach a few hundred kilometres per second.
3§09.0 — the trap this section is built to spring●●○○○
This one is designed to catch a guess that most people make before they meet any magnetism, and it is worth making the guess out loud so that you notice later when the page contradicts it. A charged particle is fired into a region where there is a strong, steady magnetic field and nothing else.
Given
the field is uniform and does not change with time
the particle enters with some speed at some angle to the field
gravity and every other force are negligible
Find
(a) True or false: a strong enough field can be used to speed the particle up. Answer before reading on, and give your reason in one sentence.
Hint 1/4
The question is really about the direction of the magnetic force relative to the direction of travel, not about how strong the field is.
Hint 2/4
The magnetic force on a moving charge is $\vec F = q\vec v\times\vec B$, and a cross product is perpendicular to both of the vectors that went into it.
Hint 3/4
So $\vec F$ is perpendicular to $\vec v$ here, whatever the size of $B$ and whatever the angle of entry.
Hint 4/4
False: the field can change the direction of travel as much as you like but never the speed.
Show solution
Argued from direction, not size, ruling out every field strength.
Where the force points
$$\vec F = q\,\vec v\times\vec B \perp \vec v$$
a cross product is perpendicular to both factors, which is a statement about the operation and not about the physics
What that does to the energy
$$\frac{dK}{dt} = \vec F\cdot\vec v = 0$$
the rate of change of kinetic energy is the power delivered, and the power is a dot product of two perpendicular vectors
$$|\vec v| = \text{constant}$$
constant kinetic energy with constant mass means constant speed, though the direction is free to change
Answer $$\boxed{\,\text{False: } |\vec v| \text{ cannot change; only its direction can}\,}$$
Check
Independent check by a limiting case: if a magnetic field could add energy you could build a device that gains energy from a permanent magnet indefinitely, since a permanent magnet needs no supply. No such device has ever worked, which is the experimental face of the same statement.
Notation
symbol
reads as
means
watch out
$\vec B$
B vector
the magnetic field at a point, a vector, in teslas
it is not a force and not a force per unit charge; only when it is crossed with a velocity does anything with the units of force appear
$B$
B
the magnitude of the magnetic field, a positive number
the same letter without the arrow is always a magnitude here, so a minus sign in front of it is a statement about direction that belongs in the picture instead
$\mathrm{T}$
tesla
the SI unit of magnetic field, one newton per ampere metre
it is a large unit; a fridge magnet is a few hundredths of a tesla and the Earth's field is a few times $10^{-5}\ \mathrm{T}$, so a problem quoting several tesla is describing a laboratory electromagnet
$\theta$
theta
the angle between the two vectors named in whichever formula it appears in
between $\vec v$ and $\vec B$ in the force law but between $\vec\mu$ and $\vec B$ in the torque law, and since $\vec\mu$ is perpendicular to the loop those two conventions differ by ninety degrees
$v_\perp,\ v_\parallel$
v perpendicular and v parallel
the components of the velocity across the field and along it
only $v_\perp$ appears in the radius; $v_\parallel$ is untouched by the field and simply carries the particle along, which is what turns a circle into a helix
$\vec L$
L vector
a vector whose length is the length of wire in the field and whose direction is the direction of the conventional current in it
for a wire that bends inside the field it is the straight vector from where the current enters the field to where it leaves, not the length of wire you would measure with a tape
$\vec\mu$
mu vector
the magnetic dipole moment of a loop or coil, $NI\vec A$, in ampere square metres
its direction is perpendicular to the plane of the loop, given by curling the right hand fingers along the current, and it is a property of the loop alone with no field in it
$n$
n
the number of mobile charge carriers per cubic metre in a conductor
in the Hall formula it sits in the denominator, so the small $n$ of a semiconductor is exactly what makes its Hall voltage large enough to measure comfortably
$V_H$
V Hall
the potential difference that appears across the width of a current carrying strip placed in a field
it is measured across the strip, at right angles to the current, and its polarity rather than its size is what identifies the sign of the carriers
Conventions used here
How the axes and the page are set up in this section
Throughout this section $x$ points to the right across the page, $y$ points up the page, and $z$ points out of the page towards you. A field into the page is therefore the $-z$ direction and is drawn as a grid of small crosses; a field out of the page is $+z$ and would be drawn as dots. Every cross product is taken in this right handed order, so that $\hat x\times\hat y=\hat z$.
Where the angle in the magnetic force formulas is measured from
In $F=|q|vB\sin\theta$ the angle $\theta$ is the one between $\vec v$ and $\vec B$ themselves, taken between $0$ and $180$ degrees, never the angle to a surface or to a normal. In $F=BIL\sin\theta$ it is the angle between the wire and the field. In the torque formulas $\theta$ changes meaning to the angle between $\vec\mu$ and $\vec B$, and $\vec\mu$ is perpendicular to the plane of the loop, so a loop lying along the field has $\theta=90$ degrees. That switch is deliberate and is flagged again where it happens.
What the sign of the charge is allowed to do
The letter $q$ carries its own sign. Magnitudes are written with $|q|$, so $F=|q|vB\sin\theta$ is never negative. The direction is always found in two moves: first point $\vec v\times\vec B$ with the right hand as though the charge were positive, then reverse the answer if the charge is negative. A minus sign is never allowed to sit in a picture as an arrow drawn backwards and also in the arithmetic as a negative number, because that counts it twice.
Which constants this page uses and to how many digits
Answers are quoted to three significant figures. The constants used are the elementary charge $e=1.60\times10^{-19}\ \mathrm{C}$, the electron mass $m_e=9.11\times10^{-31}\ \mathrm{kg}$, the proton mass $m_p=1.67\times10^{-27}\ \mathrm{kg}$, the unified $u=1.66\times10^{-27}\ \mathrm{kg}$, and $g=9.80\ \mathrm{m/s^{2}}$. The tesla is the SI unit of magnetic field, and the older appears only in the one place where a field is quoted in it, with $1\ \mathrm{T}=10^{4}\ \mathrm{G}$.
What a uniform field is allowed to mean here
Every field in this section is given, steady and uniform over the region where the charge or the wire sits, unless a problem says otherwise. Uniform means the same magnitude and the same direction everywhere in that region, so a field can be uniform across a small coil while varying over the size of the room. Where a field is stated as coming from a magnet or from a current, that origin is background information only and no formula on this page produces a field from its source.
How a magnetic field direction is stated in words
A field is described either by a direction in the plane of the page, or by the phrases into the page and out of the page. Field lines point in the direction a compass needle's north end would turn to, and outside a magnet they run from the north pole to the south pole. A north pole is a pole that seeks geographic north, which is why the buried under the Earth's northern hemisphere behaves as a south magnetic pole.
9.1The magnetic field, and the one experiment that defines it
A magnetic field is whatever pushes on moving charge, and the push comes out sideways.
Everything so far has been charges pushing on charges. Now a force that ignores a charge until it moves.
Solvable with what we have
Find the force between two charges and the field around them.
Find the charge a pair of conductors takes at a given voltage.
Find the current in every branch of a circuit.
Find the speed a charge gains across a potential difference.
Not solvable yet
Say why a compass swings when a nearby current is switched on.
Say why a loudspeaker coil moves with nothing touching it.
Say why an electron beam bends between magnet poles while a resting one does not.
Say why breaking a magnet never yields a single loose pole.
The obvious move is to treat the two ends of a magnet as two charges and reach for Coulomb's law: like ends repel, unlike ends attract, and the force falls off fast.
Why it fails
One measurement kills it. A charge at rest next to a strong magnet feels no force at all, which no arrangement of fixed charges could produce. Set the same charge moving and a force appears, sideways to its motion. Whatever this field is, it is not electric.
DefinitionDefinition 9.1: the magnetic field, and the tesla
Conditions
A test charge $q$ moving with velocity $\vec v$ through the point of interest
The force measured is the one that vanishes when the charge is brought to rest, so any electric force has been subtracted first
$\theta$ is the angle between $\vec v$ and $\vec B$, taken between $0$ and $180$ degrees
$$\boxed{\,\vec F = q\,\vec v\times\vec B\,,\qquad F = |q|\,v\,B\sin\theta\,,\qquad 1\ \mathrm{T} = 1\ \frac{\mathrm{N}}{\mathrm{A\cdot m}}\,}$$
The field is defined backwards, by the force it produces rather than by what it is made of. Turn a test charge until the force vanishes: that line is the field. Turn it ninety degrees away and the force is largest. Divide that largest force by the charge and the speed, so one tesla is one newton on one coulomb passing at one metre per second.
The field of a bar magnet, drawn as $\textcolor{#1f6feb}{\text{lines}}$ that never begin and never end. Outside the magnet they run from the north end to the south end; inside the metal they carry straight on from south to north and join up. Cut the magnet anywhere across and you have cut some of these loops, which is why the two halves come out as two complete magnets rather than as a loose north pole and a loose south pole.
Looks like this, but is not
A magnet picks up a paper clip and a rubbed comb picks up scraps of paper, so magnetism must be electricity wearing a different hat.
The comb works on anything light and neutral, moving or not. The magnet ignores the paper, works on iron charged or not, and its grip on a passing electron depends on how fast that electron goes and which way. A force that switches on only when its target moves is not anything Coulomb's law produces.
The same charge at three angles to the same field
A small bead carrying $q = +3.00\ \mathrm{\mu C}$ is fired through a uniform field of magnitude $B = 0.600\ \mathrm{T}$ at a speed of $250\ \mathrm{m/s}$. We fire it three times: once along the field, once straight across it, and once at $40.0^{\circ}$ to it. This is the defining experiment of the box, done with numbers.
Given
$q = +3.00\times10^{-6}\ \mathrm{C}$
$v = 250\ \mathrm{m/s}$
$B = 0.600\ \mathrm{T}$
three runs, at $\theta = 0^{\circ}$, $90^{\circ}$ and $40.0^{\circ}$ to the field
Find
the magnitude of the magnetic force in each of the three runs
Solution
Perpendicular case first, then a sine; the product repeats otherwise.
The run along the field
$$F = |q|vB\sin 0^{\circ} = 0\ \mathrm{N}$$
the sine kills it; travelling along a field line is indistinguishable, as far as this force goes, from sitting still
The run across the field
$$F = |q|vB = (3.00\times10^{-6})(250)(0.600)$$
at ninety degrees the sine is one, which is where the force is largest and where the definition of B is anchored
$$F = 4.50\times10^{-4}\ \mathrm{N}$$
three figures, and the direction is perpendicular to both the velocity and the field
The run at forty degrees
$$F = (4.50\times10^{-4})\sin 40.0^{\circ}$$
only the size changes with angle; the direction of the force is fixed by the plane containing v and B, not by the angle inside it
$$F = 2.89\times10^{-4}\ \mathrm{N}$$
about sixty four per cent of the maximum, which is what a sine of forty degrees is
Independent check on the middle answer by units alone: $\mathrm{C}\cdot(\mathrm{m/s})\cdot\mathrm{T} = \mathrm{C\,m\,s^{-1}}\cdot\mathrm{N\,A^{-1}m^{-1}} = \mathrm{N}$, since a coulomb per second is an ampere. The tesla was defined to make exactly this cancellation work.
The angle enters only through a sine, so a charge fired along a field line feels nothing at all. That single zero is worth more than the other two answers: whenever a question says a charged particle went through a magnetic field in a straight line, this is the reason.
How big is one tesla? The Earth's field against a laboratory magnet
Fields in problems arrive as bare numbers, and a bare number is only useful once you know what counts as big. The Earth's field where you are standing is roughly $5\times10^{-5}\ \mathrm{T}$, which in the older unit is about half a gauss. Take an electron moving at $1.00\times10^{6}\ \mathrm{m/s}$ across it.
Plausibility from a different direction: this force bends the electron onto a circle of radius $m v/(eB) = 0.114\ \mathrm{m}$, about the width of a hand. That is exactly the scale on which the Earth's field visibly disturbs an old cathode ray tube, which is why such tubes had to be shielded.
Two multiplications and one division. The comparison with the weight is the part worth keeping: for electrons and protons in any field bigger than the Earth's, gravity is not a small correction, it is nothing.
Checkpoint
§09.1 — when is there no magnetic force at all●○○○○
Thirty seconds, and it is a check on the definition rather than on any arithmetic. A charged particle is placed in a region where a strong uniform magnetic field is switched on, and it feels no magnetic force whatsoever.
Given
the field is uniform and definitely not zero
the particle definitely carries charge
the only force in question is the magnetic one
Find
(a) Which single conclusion is forced by that observation?
Hint 1/4
You are being handed a zero and asked which factor in a product produced it, so start by writing the product down.
Hint 2/4
The magnitude is $F=|q|vB\sin\theta$, a product of four things: the size of the charge, the speed, the field and the sine of the angle between velocity and field.
Hint 3/4
Here $|q|\neq0$ and $B\neq0$ are both given, so the zero has to come from $v$ or from $\sin\theta$.
Hint 4/4
Either the particle is not moving, or it is moving along the field line so that the sine vanishes.
Show solution
Written as a product; ask which factor vanished, not four arguments.
Write the magnitude as a product
$$F=|q|\,v\,B\sin\theta = 0$$
a product vanishes only when one of its factors does, so the question becomes a list of four candidates
$$|q|\neq 0,\quad B\neq 0$$
both are excluded by the wording of the problem, leaving two
antiparallel counts as well as parallel, since the sine of one hundred and eighty degrees is also zero
Answer $$\boxed{\,v=0\ \text{ or }\ \vec v\parallel\vec B\,}$$
Check
Check by the contrapositive: if the particle were moving at any angle other than zero or one hundred and eighty degrees, the sine would be nonzero and so would the force. No third possibility survives.
⚠ Expecting a stationary charge to feel a magnetic force
electric fields push charges whether they move or not, and the magnetic case looks like more of the same until you notice the velocity in the formula
wrong$$\vec F_{\rm mag} = q\vec B$$
right$$\vec F_{\rm mag} = q\,\vec v\times\vec B \;\Rightarrow\; \vec F_{\rm mag}=0 \text{ when } \vec v=0$$
⚠ Treating the tesla as a small unit because the numbers in problems are small
fields of a few tenths of a tesla appear in every exercise, so the tesla starts to feel like a modest everyday amount
9.2The force on a moving charge, and why it never speeds anything up
Point the cross product with your right hand, then flip the answer if the charge is negative.
The definition gives the size of the force and one sentence about its direction. Directions are where marks are lost, so it is worth being slow here.
RuleRule 9.2: direction of the magnetic force, and the work it does
Conditions
$\vec v$ and $\vec B$ are the velocity and the field at the same instant and the same point
The sign of $q$ is applied after the cross product has been pointed, never inside it
The statement about work holds instant by instant, not merely over a closed path
$$\boxed{\,\vec F = q\,\vec v\times\vec B\ \perp\ \vec v\,,\qquad P=\vec F\cdot\vec v = 0\,,\qquad \frac{dK}{dt}=0\,}$$
Lay the fingers of your right hand along the velocity and curl them into the field; your thumb then points along $\vec v\times\vec B$, which is the force if the charge is positive and the opposite of the force if it is negative. Whichever it turns out to be, that direction is at right angles to the velocity, so the force can steer the particle but can never push it forwards or hold it back. The speed, and with it the kinetic energy, is untouchable by a magnetic field.
Proof
The claim to prove is the second one: that a magnetic field cannot change a kinetic energy. It is one line, and it is worth seeing because it is the reason so many answers in this section can be written down before any arithmetic.
The rate at which any force feeds energy into a particle is the power $P=\vec F\cdot\vec v$.
Here $\vec F = q\,\vec v\times\vec B$, so $P = q\,(\vec v\times\vec B)\cdot\vec v$.
A cross product is perpendicular to both of the vectors that made it, so $(\vec v\times\vec B)$ is perpendicular to $\vec v$, and the dot product of perpendicular vectors is zero. Hence $P=0$ at every instant, for every $q$, every $\vec v$ and every $\vec B$.
Since $P = dK/dt$, the kinetic energy is constant, and with a constant mass so is the speed. Notice what was not assumed: the field did not have to be uniform, steady, or weak.
The defining picture. Both charges move to the right with the same $\textcolor{#d1690a}{\text{velocity}}$ through the same $\textcolor{#1f6feb}{\text{field into the page}}$, and both $\textcolor{#8250df}{\text{forces}}$ come out at right angles to the page directions of travel and of the field. The two differ only in sign, and that alone turns the force upside down: point $\vec v\times\vec B$ with the right hand, then reverse it if the charge is negative.
Looks like this, but is not
The force is perpendicular to the velocity, so the particle must go round in a circle. That is what the next block is about, so it sounds safe enough here.
It is only true if the velocity is entirely across the field. A particle fired at some other angle keeps its along-the-field component forever, because the force has no component along $\vec B$ to change it, and the path becomes a helix that drifts steadily along the field line while circling. Perpendicular force guarantees constant speed; it does not guarantee a closed path.
A proton crossing a 0.750 T field at 30 degrees
A proton moves at $4.00\times10^{5}\ \mathrm{m/s}$ through a uniform field of $0.750\ \mathrm{T}$, its velocity making an angle of $30.0^{\circ}$ with the field. Find the force on it, and say where that force points.
Given
$q = +1.60\times10^{-19}\ \mathrm{C}$
$v = 4.00\times10^{5}\ \mathrm{m/s}$
$B = 0.750\ \mathrm{T}$
$\theta = 30.0^{\circ}$ between $\vec v$ and $\vec B$
$m_p = 1.67\times10^{-27}\ \mathrm{kg}$
Find
the magnitude of the magnetic force and its direction
Solution
Size and direction apart; components repeat what the sine does.
grouping the first three factors first keeps the powers of ten in one place and the trigonometry in another
Direction, by the right hand
$$\hat F = \widehat{\vec v\times\vec B}$$
the charge is positive, so the force is along the cross product itself with no flip
$$\vec F \perp \vec v \ \text{and}\ \vec F\perp\vec B$$
it therefore sticks out of the plane that contains the velocity and the field, and no component of it lies in that plane
Answer $$\boxed{\,F = 2.40\times10^{-14}\ \mathrm{N}\,,\ \text{perpendicular to the plane of } \vec v \text{ and } \vec B\,}$$
Check
Plausibility: the proton's weight is $m_pg = 1.64\times10^{-26}\ \mathrm{N}$, so this force is about $1.5\times10^{12}$ times larger. The acceleration it produces is $F/m_p = 1.44\times10^{13}\ \mathrm{m/s^{2}}$, which for a particle this light is entirely ordinary.
Two thirds of the arithmetic here was the product $|q|vB$, and only the last factor knew anything about geometry. Getting into the habit of computing the maximum force first and then multiplying by a sine keeps the two kinds of error apart.
An electron heading left through a field out of the page
An electron moves in the $-x$ direction at $2.00\times10^{6}\ \mathrm{m/s}$ through a uniform field $\vec B = 0.500\ \mathrm{T}$ pointing in the $+z$ direction, that is, out of the page. Find the force on it as a vector. This is the case that catches people out, because two reversals are in play at once.
Given
$\vec v = (-2.00\times10^{6},\,0,\,0)\ \mathrm{m/s}$
$\vec B = (0,\,0,\,0.500)\ \mathrm{T}$
$q = -e = -1.60\times10^{-19}\ \mathrm{C}$
axes: $x$ right, $y$ up, $z$ out of the page
Find
the force vector, with its direction stated in words
Solution
Cross product first, charge sign last, so both reversals stay checkable.
Cross the vectors first, with the sign of the charge left outside
$$\vec v\times\vec B = (v_yB_z-v_zB_y,\ v_zB_x-v_xB_z,\ v_xB_y-v_yB_x)$$
writing the components out is slower than the hand rule but it cannot be got backwards, which makes it the check rather than the guess
only the middle component survives, so the product points along $+y$, up the page
Now apply the sign of the charge
$$\vec F = q(\vec v\times\vec B) = (-1.60\times10^{-19})(0,\ 1.00\times10^{6},\ 0)$$
the electron's negative charge reverses the direction the right hand gave and does nothing else
$$\vec F = (0,\ -1.60\times10^{-13},\ 0)\ \mathrm{N}$$
so the force is down the page, opposite to what the bare cross product said
Answer $$\boxed{\,\vec F = -1.60\times10^{-13}\,\hat y\ \mathrm{N}\ \ (\text{magnitude } 1.60\times10^{-13}\ \mathrm{N}, \text{ down the page})\,}$$
Check
Independent check of the size only: $|q|vB = (1.60\times10^{-19})(2.00\times10^{6})(0.500) = 1.60\times10^{-13}\ \mathrm{N}$, with $\sin\theta=1$ because the velocity lies in the page and the field comes out of it. The magnitude route knows nothing about direction, so it is a genuinely separate calculation.
Two reversals were applied: one from the velocity pointing along $-x$, one from the charge being negative. Applying only one of them is the single most common way to lose the mark, and doing the components explicitly is how to avoid it.
Checkpoint
§09.2 — an electron in a field into the page●●○○○
Thirty seconds with your right hand. An electron is moving to the right across this page, and there is a uniform magnetic field pointing into the page.
Given
the electron moves in the $+x$ direction, to the right
$\vec B$ points into the page, the $-z$ direction
the charge on the electron is negative
Find
(a) Which way does the magnetic force on the electron point?
Hint 1/4
Do this in two separate moves, and do not try to fold the sign of the charge into the hand rule itself.
Hint 2/4
First point $\vec v\times\vec B$ with the right hand as though the charge were positive, then reverse the result because it is not.
Hint 3/4
Here $\vec v = v\hat x$ to the right and $\vec B = -B\hat z$ into the page, and the unit vector identity you need is $\hat x\times\hat z = -\hat y$.
Hint 4/4
The cross product points up the page, so the force on a negative charge points down the page.
Show solution
Unit vectors rather than a hand, charge sign kept separate.
The cross product on its own
$$\hat x\times\hat z = -\hat y \;\Rightarrow\; \hat x\times(-\hat z) = +\hat y$$
using the fixed right handed set of unit vectors rather than a hand, so the answer can be checked on paper
$$\vec v\times\vec B = vB\,\hat y$$
up the page, for a positive charge
Apply the sign
$$\vec F = q(\vec v\times\vec B) = (-e)(vB\hat y) = -evB\,\hat y$$
the negative charge reverses it, giving a force down the page
Answer $$\boxed{\,\vec F = -evB\,\hat y\ :\ \text{down the page}\,}$$
Check
Independent check by symmetry: a proton moving the same way would be pushed up, and an electron is the same magnitude of charge with the opposite sign, so it must be pushed exactly the other way. Two particles, opposite answers, one calculation.
⚠ Forgetting to reverse the hand rule for a negative charge
the right hand rule is drilled until it feels like the whole answer, and the sign of the charge sits outside the geometry where it is easy to leave behind
9.3Going round in a circle: radius, period, and what the speed does not affect
Across the field the path curls into a circle; along the field nothing happens, so in general it is a helix.
We now have a force of fixed size that always points at right angles to the motion. That is the exact recipe for circular motion, so the path follows without any new physics.
TheoremResult 9.3: radius, period and frequency in a uniform field
Conditions
The field is uniform over the whole of the path, and steady
No other force acts, which for a subatomic particle in any laboratory field includes gravity
The speed is small compared with the speed of light, so that the momentum is $mv$
$v_\perp$ is the component of the velocity across the field; the component along it is untouched
The radius is the momentum across the field divided by the charge times the field, so a faster particle simply swings wider. The period is the mass divided by the charge times the field, multiplied by two pi, and it contains no speed at all: double the speed and the particle goes round a circle of twice the radius in exactly the same time. Everything that identifies the particle enters only as the ratio of its mass to its charge, which is why this motion is a measuring instrument for that ratio and for nothing else.
Proof
Take the velocity to be entirely across the field first, so $v_\perp = v$. The magnitude of the magnetic force is then $|q|vB$ and it points at the centre of the turn.
For any particle moving on a circle of radius $r$ at speed $v$, the net force towards the centre is $mv^{2}/r$. That is kinematics and knows nothing about magnetism.
Setting the two equal: $|q|vB = \dfrac{mv^{2}}{r}$.
One power of $v$ cancels from both sides, which is the whole trick, and rearranging gives $r=\dfrac{mv}{|q|B}$.
The period is the circumference over the speed: $T=\dfrac{2\pi r}{v}=\dfrac{2\pi}{v}\cdot\dfrac{mv}{|q|B}=\dfrac{2\pi m}{|q|B}$, and the second $v$ cancels too, which is why the period does not depend on how fast the particle is going.
If part of the velocity lies along the field, that part feels no force at all, since $\vec v_\parallel\times\vec B=0$. It carries on unchanged while the perpendicular part circles, so the path is a helix: replace $v$ by $v_\perp$ in the radius and leave the period alone.
A charge that enters a uniform field at right angles goes round a circle, and the reason is entirely in the geometry: the $\textcolor{#8250df}{\text{force}}$ is always at ninety degrees to the $\textcolor{#d1690a}{\text{velocity}}$, so it can only turn it. A force of fixed size that is always perpendicular to the motion is exactly the definition of uniform circular motion, and setting $|q|vB$ equal to $mv^{2}/r$ gives the radius in one line.
Looks like this, but is not
A faster particle has further to travel, so it must take longer to get round. Every other circular motion problem you have met behaves that way, so the instinct is well earned.
It is wrong here because the radius is not fixed in advance. Doubling the speed doubles the circumference, but it doubles the speed as well, and the two changes cancel exactly. On a string of fixed length the radius cannot respond, so a faster ball really does come round sooner; in a magnetic field the geometry adjusts itself and the clock does not notice.
kinetic energy (keV)
speed (m/s)
radius (cm)
period (ns)
0.075
$5.13\times10^{6}$
1.17
14.3
0.300
$1.03\times10^{7}$
2.34
14.3
1.20
$2.05\times10^{7}$
4.68
14.3
Each row has four times the energy of the one above it, so twice the speed, and the radius doubles with it: $1.17$, $2.34$, $4.68$ centimetres. The period sits at $14.3\ \mathrm{ns}$ in all three rows and would sit there for any energy in this range, because $T=2\pi m/(|q|B)$ has nowhere to put a speed. The fastest of the three is travelling at under seven per cent of the speed of light, which is why the non relativistic formula can be trusted across the whole table.
An electron of 1.20 keV in a field of 2.50 mT
An electron is accelerated to a kinetic energy of $1.20\ \mathrm{keV}$ and then enters a uniform field of $2.50\times10^{-3}\ \mathrm{T}$ at right angles to it. Find the radius of its path, the time it takes to go round once, and how many times a second it goes round.
a frequency in the radio band, which is what makes this motion easy to drive electrically
Answer $$\boxed{\,r = 4.68\ \mathrm{cm}\,,\qquad T = 1.43\times10^{-8}\ \mathrm{s}\,,\qquad f = 69.9\ \mathrm{MHz}\,}$$
Check
Independent check of the period by the other route: $T = 2\pi r/v = 2\pi(4.68\times10^{-2})/(2.05\times10^{7}) = 1.43\times10^{-8}\ \mathrm{s}$. The two calculations share no algebra, and they agree. Speed check: $v/c = 0.068$, so the non relativistic formulas were legitimate.
Notice the shape of the work: energy, then speed, then radius. Almost every exam question in this section has that same spine, and the only thing that changes from question to question is how the first line hands you the energy.
A proton entering at 60 degrees: the helix and its pitch
A proton enters a uniform field of $0.300\ \mathrm{T}$ with a speed of $4.00\times10^{5}\ \mathrm{m/s}$, its velocity making $60.0^{\circ}$ with the field. Find the radius of the helix it follows and its , meaning the distance it advances along the field in one full turn.
Given
$v = 4.00\times10^{5}\ \mathrm{m/s}$ at $60.0^{\circ}$ to $\vec B$
Independent check by a ratio that needs neither the mass nor the field: the pitch divided by the circumference must equal $v_\parallel/v_\perp = \cot 60.0^{\circ} = 0.577$. Here $4.37/(2\pi\times1.21) = 0.575$, agreeing to the rounding.
The velocity was resolved once, and everything afterwards used one component or the other and never the full speed. If an answer in a helix problem contains $v$ rather than $v_\perp$ or $v_\parallel$, it is wrong.
A helix is not a new phenomenon; it is a circle plus a constant drift, computed separately and then put back together. That is worth remembering because the two pieces never interfere with each other.
Checkpoint
§09.3 — doubling the speed●○○○○
Thirty seconds, no calculator. A proton is going round a circle of radius $8.00\ \mathrm{cm}$ in a uniform magnetic field, taking $2.40\ \mathrm{\mu s}$ per revolution. A second, identical proton is injected into the same field with twice the speed.
Given
first proton: $r_1 = 8.00\ \mathrm{cm}$ and $T_1 = 2.40\ \mathrm{\mu s}$
second proton: same mass, same charge, same field, twice the speed
both enter perpendicular to the field
Find
(a) State the radius and the period of the second proton.
Hint 1/4
Do not compute the field. Ask instead which of the two quantities contains the speed and which does not.
Hint 2/4
$r = mv/(|q|B)$ is proportional to the speed, while $T = 2\pi m/(|q|B)$ has no speed in it at all.
Hint 3/4
Here the mass, the charge and the field are all unchanged, and only $v$ has doubled, starting from $r_1 = 8.00\ \mathrm{cm}$ and $T_1 = 2.40\ \mathrm{\mu s}$.
Hint 4/4
The radius doubles to $16.0\ \mathrm{cm}$ and the period stays at $2.40\ \mathrm{\mu s}$.
Show solution
Scaled from the case given; $B$ was withheld as unnecessary.
Independent consistency check: $T = 2\pi r/v$ must hold for both. Doubling both $r$ and $v$ leaves that ratio alone, so the two statements agree rather than merely coexisting.
⚠ Putting the full speed into the radius when the entry is at an angle
the formula is usually first met for perpendicular entry, where $v_\perp$ and $v$ happen to be the same number, so the subscript looks decorative
wrong$$r=\frac{mv}{|q|B}\ \text{ for entry at } 60^{\circ}$$
⚠ Using the charge with its sign inside the radius
the sign is genuinely part of $q$ everywhere else on this page, so leaving it in feels consistent
wrong$$r=\frac{m v_\perp}{qB} = \text{negative for an electron}$$
right$$r=\frac{m v_\perp}{|q|B}>0,\quad \text{the sign decides the sense of rotation, not the size}$$
9.4The force on a current carrying wire
A current is a queue of moving charges, so the same law applies with I L replacing q v.
Nothing about the force law asked for a lone particle. A wire carrying a current is a great many charges all moving the same way at once, and adding them up costs three lines.
TheoremResult 9.4: the force a uniform field puts on a current
Conditions
The field is uniform over the whole length of wire that lies inside it
$\vec L$ points along the conventional current and its length is the length of wire in the field
For a wire that bends inside the field, $\vec L$ is the straight vector from where the current enters to where it leaves
$\theta$ is the angle between the wire and the field
$$\boxed{\,\vec F = I\,\vec L\times\vec B\,,\qquad F = BIL\sin\theta\,,\qquad \vec F_{\rm closed\ loop} = \vec 0 \ \text{ in a uniform field}\,}$$
The force on a straight wire is the field times the current times the length in the field, cut down by the sine of the angle between wire and field, and it points at right angles to both. The direction comes from the right hand exactly as it did for a single charge, with the current playing the part of the velocity, and no sign has to be flipped because conventional current already points the way positive charge would go. If the wire returns to where it started, the vector from entry to exit is zero and so is the total force, however complicated the shape in between.
Proof
Take a straight piece of wire of length $L$ and cross section $A$, carrying carriers of charge $q$ with number density $n$, drifting at speed $v_d$ along the wire.
The force on one carrier is $q\,\vec v_d\times\vec B$, and there are $nAL$ of them in the piece, all with the same drift velocity, so the total is $\vec F = (nAL)\,q\,\vec v_d\times\vec B$.
Group the factors as $\vec F = (nqv_dA)\,(L\hat v_d)\times\vec B$, where $\hat v_d$ is the unit vector along the drift.
The bracket $nqv_dA$ is exactly the current $I$, from the earlier section on current; and $L\hat v_d$ is what we are calling $\vec L$. Hence $\vec F = I\,\vec L\times\vec B$.
For a wire that bends, cut it into short straight pieces $d\vec L$ and add: $\vec F=I\left(\int d\vec L\right)\times\vec B$ when $\vec B$ is the same everywhere, and $\int d\vec L$ between two points is just the straight vector joining them, whatever route the wire took.
Around a closed loop that vector is zero, so the net force on any closed loop in a uniform field is zero. The loop can still be twisted, which is the next block.
The free body diagram for a current carrying rod held up by a magnetic field. Two forces act on it and no more: the $\textcolor{#8250df}{\text{magnetic force}}$ $BIL$ up the page, because the $\textcolor{#d1690a}{\text{current}}$ runs to the right through a $\textcolor{#1f6feb}{\text{field into the page}}$, and the weight $mg$ down. Setting the two equal is the whole of the levitation problem, and reversing the current turns the support into a push in the other direction.
Looks like this, but is not
More wire in the field means more force, so a long tangled wire feels more push than a short straight one. It sounds like the safest statement on the page.
Only the straight vector from entry to exit counts. Bend the wire back on itself and the forces on the two halves point opposite ways and cancel; run it round a closed loop and the total is exactly zero no matter how many metres of copper are in there. What matters is displacement, not length of wire, and the two are the same thing only for a straight run.
angle to the field
$\sin\theta$
force (N)
$0^{\circ}$
0.000
0.000
$30.0^{\circ}$
0.500
0.648
$45.0^{\circ}$
0.707
0.916
$60.0^{\circ}$
0.866
1.122
$90.0^{\circ}$
1.000
1.296
The force is zero for a wire laid along the field and largest for one laid across it, and at forty five degrees it is already seventy per cent of the way to its maximum rather than half of it, because a sine is not a straight line. The last row is the number to carry in your head: $BIL$ is the most a given wire can be pushed with, and every other orientation is a fraction of it.
A wire crossing the poles of a magnet at 55 degrees
A long straight wire carries $12.0\ \mathrm{A}$. A section of it $0.450\ \mathrm{m}$ long lies inside the uniform $0.240\ \mathrm{T}$ field between the poles of a magnet, making an angle of $55.0^{\circ}$ with the field. Find the force on that section.
Given
$I = 12.0\ \mathrm{A}$
$L = 0.450\ \mathrm{m}$ inside the field
$B = 0.240\ \mathrm{T}$
$\theta = 55.0^{\circ}$ between the wire and the field
Find
the magnitude of the force on the section inside the field
Solution
$BIL$ first, sine second, so the bound stays visible.
$\theta$ is the angle between the wire and the field, which is what the problem stated
$$F = 1.06\ \mathrm{N}$$
three figures; the direction is perpendicular to both the wire and the field
Answer $$\boxed{\,F = 1.06\ \mathrm{N}\,}$$
Check
Plausibility: a newton is about the weight of a small apple, and this is a fairly stiff wire in a fairly strong laboratory field, so a force you could feel with a fingertip is the right order. Bounds check: the answer must lie between $0$ and $BIL = 1.30\ \mathrm{N}$, and $1.06$ does.
Only the length actually inside the field entered the calculation. The rest of the wire may be a hundred metres long; with no field on it, it feels nothing.
What current makes a rod float?
A straight rod of mass $18.0\ \mathrm{g}$ and length $0.220\ \mathrm{m}$ rests across two horizontal rails in a uniform horizontal field of $0.420\ \mathrm{T}$ that runs perpendicular to the rod. Find the current that would make the rod hover, and say which way it has to flow.
SI units throughout: the mass had to come out of grams first
$$I = 1.91\ \mathrm{A}$$
and the current must run in the direction that makes $I\vec L\times\vec B$ point upwards; reversing it doubles the downward load instead
Answer $$\boxed{\,I = 1.91\ \mathrm{A}\,,\ \text{directed so that } I\vec L\times\vec B \text{ points up}\,}$$
Check
Independent check by working backwards through a different quantity: at $1.91\ \mathrm{A}$ the magnetic force is $BIL = (0.420)(1.91)(0.220) = 0.177\ \mathrm{N}$, and the weight is $mg = (1.80\times10^{-2})(9.80) = 0.176\ \mathrm{N}$. Order of magnitude: about two amperes is an ordinary bench current, and eighteen grams is a light aluminium rod, so the numbers describe a demonstration that can actually be set up.
Two forces, one equation. The only place this goes wrong is the direction: half of all attempts pick the current that presses the rod down, and the picture is the only defence.
A semicircular wire: the length that counts is the chord
A wire carrying $8.00\ \mathrm{A}$ is bent into a semicircle of radius $0.120\ \mathrm{m}$ and placed in a uniform field of $0.350\ \mathrm{T}$ perpendicular to the plane of the semicircle. The current enters at one end of the diameter and leaves at the other. Find the total force on the bent wire.
Given
$I = 8.00\ \mathrm{A}$
$R = 0.120\ \mathrm{m}$, semicircular arc
$B = 0.350\ \mathrm{T}$, perpendicular to the plane of the arc
current enters at one end of the diameter and leaves at the other
Find
the total magnetic force on the semicircular wire
Solution
Chord, not integral: in a uniform field only endpoints survive.
Replace the arc by the straight vector across it
$$\vec L = \int d\vec L = \text{(exit point)} - \text{(entry point)}$$
the field is uniform, so it comes outside the integral and only the sum of the little displacement vectors is left
$$|\vec L| = 2R = 0.240\ \mathrm{m}$$
the diameter, not the arc length $\pi R = 0.377\ \mathrm{m}$; the extra copper contributes nothing to the total
Apply the straight wire result to that chord
$$F = BI|\vec L| = (0.350)(8.00)(0.240)$$
the chord is perpendicular to the field, since the field is perpendicular to the whole plane, so the sine is one
$$F = 0.672\ \mathrm{N}$$
directed along the perpendicular bisector of the diameter, in the plane of the arc
Answer $$\boxed{\,F = 0.672\ \mathrm{N}\,,\ \text{along the perpendicular bisector of the diameter}\,}$$
Check
Independent check by direct integration, which uses none of the shortcut: an element at angle $\phi$ contributes $BIR\,d\phi$, whose component along the bisector is $BIR\sin\phi\,d\phi$. Integrating from $0$ to $\pi$ gives $BIR\int_0^{\pi}\sin\phi\,d\phi = 2BIR = BI(2R)$, the same answer, and the sideways components cancel in pairs.
If the wire had been closed into a full circle the entry and exit points would coincide, $\vec L$ would be zero, and so would the force. That is the statement in the box, and it is what makes the next block about torque necessary: a loop that cannot be pushed can still be turned.
Checkpoint
§09.4 — a closed loop in a uniform field●●○○○
Thirty seconds. A closed loop of wire of some awkward shape carries a steady current and sits entirely inside a uniform magnetic field, at some arbitrary orientation.
Given
the loop is closed, so the current returns to where it started
the field is uniform over the whole loop
the shape of the loop is not stated and does not matter
Find
(a) True or false: the total magnetic force on the loop is zero. Give the reason in one sentence.
Hint 1/4
The question is about the sum of the forces on all the little pieces of the wire, so ask what that sum depends on.
Hint 2/4
For a uniform field, $\vec F = I\left(\int d\vec L\right)\times\vec B$, and $\int d\vec L$ between two points is the straight vector joining them.
Hint 3/4
Here the two points are the same point, because the loop is closed, so that straight vector has length zero.
Hint 4/4
True: the net force is zero, though this says nothing at all about the torque.
Show solution
Field pulled out of the integral; a rectangle would not generalise.
Pull the field out of the sum
$$\vec F=\oint I\,d\vec L\times\vec B = I\left(\oint d\vec L\right)\times\vec B$$
this step is legitimate only because $\vec B$ is the same at every point of the loop, which is the one condition being used
Evaluate the remaining integral geometrically
$$\oint d\vec L = \vec 0$$
adding up displacement vectors round a closed path returns you to the start, whatever the route
$$\vec F = \vec 0$$
zero crossed with anything is zero
Answer $$\boxed{\,\vec F_{\rm net} = \vec 0\ \text{ for any closed loop in a uniform field}\,}$$
Check
Independent check on a case where the answer can be seen: a square loop with its plane containing the field. The two sides perpendicular to the field feel equal and opposite forces, and the two sides along it feel nothing at all, so the total is zero without any integration.
⚠ Using the length of wire instead of the straight vector across the field
the formula is written with an $L$ in it and the wire has a length, so the tape measure answer looks like the right one
wrong$$F = BI(\pi R)\ \text{ for a semicircular arc}$$
right$$F = BI(2R),\qquad \vec L = \text{entry to exit, in a straight line}$$
⚠ Concluding that a loop with zero net force does nothing
zero force is the usual test for equilibrium, and the torque is easy to forget when the forces have already cancelled
9.5Torque on a loop, and the magnetic dipole moment
A uniform field cannot push a loop anywhere, but it can twist it, and the twist is what motors are built on.
The last block ended with a loop that feels no net force. That is not the same as a loop that does nothing, and the difference is the whole of this one.
TheoremResult 9.5: magnetic dipole moment, torque and energy
Conditions
A flat closed loop of $N$ turns, each of area $A$, carrying current $I$, in a uniform field
$\vec A$ is normal to the plane of the loop, its direction given by curling the right hand fingers along the current
Here $\theta$ is the angle between $\vec\mu$ and $\vec B$, which is ninety degrees away from the angle between the plane of the loop and the field
The zero of the energy is taken with $\vec\mu$ at right angles to $\vec B$
Everything the field needs to know about the loop is packed into one vector: the number of turns times the current times the area, pointing out of the face that the current goes anticlockwise around. The field then twists that vector towards itself, hardest when the two are at right angles and not at all when they are lined up, and the energy is lowest when they point the same way. A compass needle does exactly this, which is why a current loop and a small magnet are interchangeable as far as a uniform field is concerned.
Proof
Take a rectangular loop with sides $a$ and $b$, one turn, current $I$, in a uniform field $\vec B$, and let the rotation axis run through the middle parallel to the two sides of length $a$.
The two sides of length $b$ are at some angle to the field and feel forces, but those forces are directed along the axis or cancel across it and produce no twist about it.
Each of the two sides of length $a$ is perpendicular to $\vec B$, so each feels a force of size $F=BIa$, and the two are opposite in direction: a couple.
The two forces act along lines separated by $b\sin\theta$, where $\theta$ is the angle between the normal of the loop and the field, so the torque is $\tau = F\,b\sin\theta = BIab\sin\theta = IAB\sin\theta$ with $A=ab$.
For $N$ turns wound on the same frame this happens $N$ times over, giving $\tau = NIAB\sin\theta = \mu B\sin\theta$ with $\mu = NIA$. Writing it as $\vec\tau=\vec\mu\times\vec B$ adds the direction: the twist swings $\vec\mu$ towards $\vec B$.
The energy follows by integrating the work done against that torque: $U(\theta)=\int_{90^{\circ}}^{\theta}\mu B\sin\theta'\,d\theta' = -\mu B\cos\theta$, which is $-\vec\mu\cdot\vec B$. It is lowest at $\theta=0$, so lining up is the stable orientation, and highest at $\theta=180^{\circ}$, which is an equilibrium that falls over at the slightest push.
The derivation used a rectangle, but any flat loop can be built out of small rectangles whose internal edges carry equal and opposite currents and cancel, so the result holds for any shape with $A$ its area.
A rectangular loop seen edge on, so that its plane runs into the page. The current comes out of the page on the upper side, marked with a dot, and goes into the page on the lower side, marked with a cross. The two $\textcolor{#8250df}{\text{forces}}$ are equal and opposite, so the loop as a whole is not pushed anywhere by the $\textcolor{#1f6feb}{\text{field}}$, but their lines of action are a distance apart and the pair twists the loop until $\vec\mu$ swings into line with $\vec B$.
Looks like this, but is not
The torque must be biggest when the loop faces the field square on, the way a sail catches the most wind when it faces the breeze. The sail picture is the reason this is such a durable error.
A loop facing the field square on has its normal along the field, so $\theta=0$ and the torque is zero: that is the orientation it settles into, not the one it is driven hardest from. The torque is largest when the plane of the loop contains the field, so the loop is edge on to it, because that is where the normal is at ninety degrees to the field. A sail transfers momentum, a loop transfers only twist, and the two peak at opposite orientations.
Torque on a 25 turn coil at 30 degrees
A rectangular coil of $25$ turns measures $0.120\ \mathrm{m}$ by $0.080\ \mathrm{m}$ and carries a current of $3.00\ \mathrm{A}$. It sits in a uniform field of $0.450\ \mathrm{T}$, with its normal making an angle of $30.0^{\circ}$ with the field. Find its magnetic dipole moment and the torque on it.
Independent check from the forces rather than from the formula: each of the two $0.120\ \mathrm{m}$ sides carries $25$ turns of $3.00\ \mathrm{A}$ across the field and feels $F = NBIa = (25)(0.450)(3.00)(0.120) = 4.05\ \mathrm{N}$. Their lines of action are separated by $b\sin\theta = (0.080)(0.500) = 0.0400\ \mathrm{m}$, so the couple is $\tau = F\times$ separation $= (4.05)(0.0400) = 0.162\ \mathrm{N\,m}$. Same number, no formula.
Splitting the work into a coil part and a field part pays off whenever the question then changes the field, the angle or both: only the second half has to be redone.
The work needed to turn that coil, and which way it would rather sit
Take the same coil, with $\mu = 0.720\ \mathrm{A\,m^{2}}$, in the same $0.450\ \mathrm{T}$ field. How much work must be done against the field to turn it from lined up with the field ($\theta=0$) to at right angles to it ($\theta = 90^{\circ}$), and how much to turn it all the way round to $\theta=180^{\circ}$?
Given
$\mu = 0.720\ \mathrm{A\,m^{2}}$
$B = 0.450\ \mathrm{T}$
start at $\theta = 0$
finish at $\theta = 90.0^{\circ}$, then at $\theta = 180^{\circ}$
Find
the external work needed for each of the two turns
Solution
Energy at both ends, not an integral; the path cannot matter.
Independent check by integrating the torque instead of using the energy: $W=\int_0^{\pi/2}\mu B\sin\theta\,d\theta = \mu B[-\cos\theta]_0^{\pi/2} = \mu B = 0.324\ \mathrm{J}$. The two routes share only the value of $\mu B$. Plausibility: $0.324\ \mathrm{J}$ is roughly the work of lifting a $100\ \mathrm{g}$ apple by a third of a metre, which is the right feel for turning a small coil against a laboratory magnet.
Three energies computed, two subtractions. Nothing here needed the shape of the coil, only the single number $\mu$.
Notice which orientation is stable. At $\theta=0$ the energy is at its minimum and a small nudge brings the coil back; at $\theta = 180^{\circ}$ it is at a maximum, so the coil balances there and falls away from the smallest disturbance. A compass needle at rest is sitting in the first of those two.
Checkpoint
§09.5 — which orientation is twisted hardest●●○○○
Thirty seconds, and it is the angle convention that is being tested rather than any arithmetic. A single current carrying loop sits in a uniform magnetic field and can be held at any orientation.
Given
the loop is flat and closed, and carries a steady current
the field is uniform and does not change
$\vec\mu$ is perpendicular to the plane of the loop
Find
(a) At which orientation is the torque on the loop largest?
Hint 1/4
The answer is decided entirely by which pair of directions the angle in the formula is measured between, so write the formula down before picturing anything.
Hint 2/4
$\tau = \mu B\sin\theta$ with $\theta$ the angle between $\vec\mu$ and $\vec B$, and $\vec\mu$ is perpendicular to the plane of the loop.
Hint 3/4
So the torque is largest when $\sin\theta = 1$, meaning $\vec\mu$ is at ninety degrees to $\vec B$, which for the loop itself means the plane of the loop is edge on to the field.
Hint 4/4
The torque peaks when the field lies in the plane of the loop, and it vanishes when the field is perpendicular to that plane.
Show solution
Formula maximised first; the picture that springs to mind is wrong.
Maximise the formula
$$\tau = \mu B\sin\theta \ \text{ is largest at } \sin\theta = 1 \Rightarrow \theta = 90^{\circ}$$
the only variable is the angle, so this is a one line maximisation
Translate the angle back into a picture of the loop
$$\theta = \angle(\vec\mu,\vec B) = 90^{\circ} \iff \vec B \ \text{lies in the plane of the loop}$$
because $\vec\mu$ is the normal, a right angle between $\vec\mu$ and $\vec B$ means the field lies flat in the loop's own plane
Answer $$\boxed{\,\tau_{\max} = \mu B\ \text{ when the field lies in the plane of the loop}\,}$$
Check
Independent check with the couple picture instead of the formula: the forces on the two sides of a rectangular loop are separated by $b\sin\theta$, and that separation is greatest when the loop is edge on, which is the same configuration by a different argument.
⚠ Measuring the torque angle from the plane of the loop instead of from its normal
the loop is the visible object and its plane is what you can see, while the normal is an arrow that only exists on paper
wrong$$\tau = \mu B\sin(\text{angle between the loop's plane and } \vec B)$$
⚠ Leaving the number of turns out of the dipole moment
the area and the current are given as single numbers and $N$ arrives separately, so it is easy to treat it as a description of the coil rather than a factor
wrong$$\mu = IA$$
right$$\mu = NIA$$
⚠ Expecting a net force on a loop in a uniform field
a torque is a mechanical effect and it feels as though something must be pushing the loop for it to move at all
9.6Crossed fields: the velocity selector and the mass spectrometer
Set an electric push against a magnetic one and only a single speed gets through undeflected.
So far the magnetic force has been alone in the problem. Put an electric force beside it and the two can be made to cancel, which turns out to be the most useful thing either of them does.
MethodResult 9.6: crossed fields, and what each stage measures
Conditions
$\vec E$ and $\vec B$ are perpendicular to each other and both perpendicular to the beam
The particle is undeflected, which is the condition being imposed and not something to be derived
In the analyser stage the field $B'$ is perpendicular to the velocity, so the whole speed goes into the radius
$$\boxed{\,qE = qvB \ \Longrightarrow\ v = \frac{E}{B}\,,\qquad\text{then}\qquad r=\frac{mv}{qB'} \ \Longrightarrow\ m = \frac{qB'r}{v}\,}$$
In the first stage the electric force and the magnetic force are pointed at each other and the particle goes straight through only if they are equal in size. Both of them carry a factor of the charge, so the charge cancels and the selected speed is the ratio of the two field strengths, the same for a proton, an electron or a uranium ion. In the second stage the beam of known speed is bent by a magnetic field alone, and since the radius is proportional to the mass, where a particle lands on the detector is a direct reading of its mass.
Proof
In the selector, take the beam along $+x$, the electric field along $-y$ so that it pushes a positive charge downwards, and the magnetic field into the page along $-z$.
The electric force is $\vec F_E = q\vec E$, of size $qE$, downwards.
The magnetic force is $q\vec v\times\vec B = q(v\hat x)\times(-B\hat z) = qvB\,\hat y$, of size $qvB$, upwards. The two are opposite, which is the point of crossing the fields.
Undeflected means the two cancel: $qE = qvB$. The charge cancels from both sides, so $v = E/B$ for every particle whatever its charge and mass, and even the sign of the charge drops out because reversing it reverses both forces together.
A particle slower than $E/B$ has too little magnetic force and is pushed down by the electric field; a faster one is pushed up. Only the exact speed survives the slit, which is why the arrangement selects rather than merely deflects.
In the analyser the electric field is switched off and only $B'$ acts, so the beam follows a circle of radius $r=mv/(qB')$. With $v$ now known from the first stage, measuring $r$ gives $m$, and the detector is effectively a ruler calibrated in mass.
The velocity selector. The $\textcolor{#1a7f37}{\text{electric field}}$ between the plates points down and pushes a positive charge down; the $\textcolor{#1f6feb}{\text{field into the page}}$ pushes the same charge up, by an amount that grows with its speed. Only the speed at which the two $\textcolor{#8250df}{\text{forces}}$ are equal survives the trip, and since both forces carry the same factor $q$, that speed is the same for every particle whatever its charge and whatever its mass.
Looks like this, but is not
Since heavier ions are harder to push about, a velocity selector must let the heavy ones through preferentially. Mass is the thing that resists being pushed, so it ought to matter.
Neither force in the balance contains the mass. The condition $qE = qvB$ is about forces, not accelerations, and it is satisfied or not before the mass gets any say. Mass enters only in the second stage, where a magnetic field alone bends the selected beam and heavier ions swing wider. Keeping the two stages apart is what makes the instrument work: one measures speed, the other measures mass.
Setting a velocity selector to 4.00 x 10^5 m/s
A velocity selector uses a field of $3.20\times10^{4}\ \mathrm{V/m}$ between its plates and a magnetic field of $0.0800\ \mathrm{T}$ across the gap, the two at right angles to each other and to the beam. Which speed passes through undeflected, and does the answer change if singly charged ions are replaced by doubly charged ones?
Given
$E = 3.20\times10^{4}\ \mathrm{V/m}$
$B = 0.0800\ \mathrm{T}$
the two fields are perpendicular to each other and to the beam
Find
the selected speed, and whether it depends on the charge
Solution
Balanced symbolically, so the charge cancels and part two is free.
Balance the two forces
$$qE = qvB$$
the undeflected condition is a statement about forces, so it is written before any numbers appear
$$v = \frac{E}{B}$$
the charge cancels, which is the single most important feature of this device
volts per metre divided by teslas gives metres per second, which is worth checking once and then trusting
Answer the second question without any new calculation
$$v = E/B \ \text{contains no } q \ \text{and no } m$$
doubling the charge doubles both forces at once, so the balance point does not move
Answer $$\boxed{\,v = 4.00\times10^{5}\ \mathrm{m/s}\,,\ \text{the same for every charge and every mass}\,}$$
Check
Unit check as an independent test: $\mathrm{V/m}$ divided by $\mathrm{T}$ is $(\mathrm{J\,C^{-1}m^{-1}})/(\mathrm{N\,A^{-1}m^{-1}}) = \mathrm{m/s}$, since a joule is a newton metre and an ampere is a coulomb per second. Plausibility: $4\times10^{5}\ \mathrm{m/s}$ is what a singly charged ion picks up across a few hundred volts, which is the sort of voltage such an instrument runs at.
A device whose answer contains neither the charge nor the mass is unusual and worth noticing. It is exactly that blindness which makes it a useful first stage: it hands the next stage a beam whose speed is known without having been told anything about what is in the beam.
Separating neon 20 from neon 22
Singly charged neon ions leave the selector above at $4.00\times10^{5}\ \mathrm{m/s}$ and enter a region of uniform field $0.450\ \mathrm{T}$ perpendicular to their velocity, where they follow semicircles and strike a detector back on the entry line. Neon has two common , of mass $20u$ and $22u$. Find where each lands and how far apart the two marks are.
Given
$v = 4.00\times10^{5}\ \mathrm{m/s}$ entering perpendicular to the field
$B' = 0.450\ \mathrm{T}$
$q = +1.60\times10^{-19}\ \mathrm{C}$ for both isotopes
the half turn doubles the difference in radius, which is why the detector is placed there rather than a quarter turn round
Answer $$\boxed{\,r_{20}=0.184\ \mathrm{m},\quad r_{22}=0.203\ \mathrm{m},\quad \Delta d = 3.69\ \mathrm{cm}\,}$$
Check
Independent check on the separation without using either radius: $\Delta d = 2\Delta r$ and $\Delta r/r_{20} = \Delta m/m_{20} = 2/20 = 10.0\%$, so $\Delta d$ should be ten per cent of $d_{20} = 0.369\ \mathrm{m}$, which is $0.0369\ \mathrm{m}$. Plausibility: a few centimetres is comfortably resolvable on a detector, which is why mass spectrometers can tell isotopes apart at all.
One radius computed in full, the second obtained by a ratio. Doing the second from scratch is not wrong, but it is where rounding errors creep in and where a mistyped mass goes unnoticed.
Checkpoint
§09.6 — reading a selector setting●○○○○
Thirty seconds. A velocity selector is built with a magnetic field of $0.250\ \mathrm{T}$, and it is required to pass particles moving at $2.00\times10^{6}\ \mathrm{m/s}$.
metres per second times teslas gives volts per metre
$$E \ \text{contains no } q$$
so doubling the charge doubles both forces and leaves the balance untouched
Answer $$\boxed{\,E = 5.00\times10^{5}\ \mathrm{V/m},\ \text{independent of } q\,}$$
Check
Check by scaling rather than by running the division backwards: the balance fixes only the ratio $E/B$, so at $0.500\ \mathrm{T}$ the plates would need $1.00\times10^{6}\ \mathrm{V/m}$ for the same particles. Units agree too: $\mathrm{T}\cdot\mathrm{m/s} = \mathrm{N/C} = \mathrm{V/m}$.
⚠ Thinking the selected speed depends on the charge or the mass
every other formula on this page contains one or the other, so a formula containing neither looks incomplete
wrong$$v = \frac{qE}{mB}$$
right$$qE = qvB \Rightarrow v = \frac{E}{B}$$
⚠ Using the selector field in the analyser stage
both stages have a magnetic field and problems often quote the two numbers close together
wrong$$r = \frac{mv}{qB_{\rm selector}}$$
right$$r = \frac{mv}{qB'_{\rm analyser}},\qquad B' \ \text{is the field in the bending region}$$
⚠ Confusing the radius with the distance to the detector
the radius is what the formula gives and the detector distance is what the ruler measures, and the problem asks for whichever one you are not thinking about
wrong$$d_{\rm detector} = r$$
right$$d_{\rm detector} = 2r \ \text{ after a half turn back to the entry line}$$
9.7The Hall effect: a sideways voltage that counts the carriers
Push the carriers sideways until the voltage they pile up pushes back exactly as hard.
One last use of the same force, and this time it is applied to the carriers inside a solid rather than to a particle in a vacuum.
TheoremResult 9.7: the Hall voltage
Conditions
A flat strip carrying a steady current, in a field perpendicular to the face of the strip
$t$ is the thickness of the strip measured along the field, and the voltage is read across the other transverse dimension
Steady state: the sideways drift has already stopped, so the built up electric field exactly cancels the magnetic push
$n$ is the number of mobile carriers per cubic metre and $q$ the magnitude of the charge on each
Send a current through a strip lying in a magnetic field and every carrier gets pushed towards one edge. Charge piles up there, and the electric field of that pile pushes back, until the two forces on a carrier are equal and the sideways drift stops. The voltage left across the strip is the Hall voltage, and because the drift speed hidden inside it can be traded for the current using the definition of current, the answer contains only things a laboratory can measure: the current, the field, the thickness, and the number of carriers per cubic metre.
Proof
Let the carriers drift with speed $v_d$ along the strip. The magnetic force on each one is $qv_dB$, sideways.
Carriers pile up on one edge and leave the opposite edge short, so a transverse electric field $E_H$ builds up between the two edges, pushing the carriers back.
Charge stops moving sideways when the two balance: $qE_H = qv_dB$, so $E_H = v_dB$. The charge cancels, so this step is the same for either sign of carrier.
The transverse field is uniform across the width $w$, so the potential difference between the edges is $V_H = E_Hw = v_dBw$.
Now trade $v_d$ for the current using $I = nqv_dA$ with $A = wt$: this gives $v_d = I/(nqwt)$.
Substituting: $V_H = \dfrac{I}{nqwt}Bw = \dfrac{IB}{nqt}$. The width cancels, which is why the measurement does not need the strip to be cut to any particular shape, only to a known thickness along the field.
A strip of metal carrying a $\textcolor{#d1690a}{\text{current}}$ to the right in a $\textcolor{#1f6feb}{\text{field into the page}}$. The carriers are electrons, so they drift to the left, and the magnetic force pushes them to the top edge; charge piles up there until the electric field it creates across the strip cancels the magnetic push. The $\textcolor{#8250df}{\text{voltage}}$ left between the two edges is the Hall voltage, and its polarity is the experiment that showed the carriers in a metal are negative.
Looks like this, but is not
Positive carriers going one way and negative carriers going the other way give the same current, so they must give the same Hall voltage too. The two are genuinely indistinguishable as far as the current is concerned.
They are not indistinguishable here, and that is the whole point of the measurement. Reversing the sign of the carrier reverses the drift direction as well, and the two reversals cancel in the direction of the force: both kinds of carrier are pushed towards the same edge. But one of them makes that edge negative and the other makes it positive, so the polarity of the Hall voltage is opposite in the two cases. Measuring that polarity is how it was shown that the carriers in an ordinary metal are negative.
The Hall voltage across a copper strip, and why it is so small
A copper strip $20.0\ \mathrm{mm}$ wide and $1.00\ \mathrm{mm}$ thick carries a current of $25.0\ \mathrm{A}$ in a field of $0.800\ \mathrm{T}$ perpendicular to its face. Copper has about one free electron per atom, giving $n = 8.47\times10^{28}\ \mathrm{m^{-3}}$. Find the Hall voltage and the drift speed.
Given
$I = 25.0\ \mathrm{A}$
$B = 0.800\ \mathrm{T}$
$w = 20.0\ \mathrm{mm}$, $t = 1.00\ \mathrm{mm}$ measured along the field
$n = 8.47\times10^{28}\ \mathrm{m^{-3}}$ (given for copper)
$q = 1.60\times10^{-19}\ \mathrm{C}$
Find
the Hall voltage and the drift speed of the electrons
Solution
Voltage and drift speed kept independent, or the check proves nothing.
Independent check that uses the definition rather than the formula: $V_H = v_dBw = (9.22\times10^{-5})(0.800)(2.00\times10^{-2}) = 1.48\times10^{-6}\ \mathrm{V}$. The two calculations meet, and the second one never used the derived expression.
The reason the answer is microvolts is that $n$ sits in the denominator and copper has an enormous number of carriers. Every practical Hall probe is therefore made of a semiconductor instead, where $n$ is smaller by five or six orders of magnitude and the same field gives millivolts.
Counting the carriers in a semiconductor from its Hall voltage
A strip of semiconductor $0.200\ \mathrm{mm}$ thick along the field carries $0.150\ \mathrm{A}$ in a field of $0.650\ \mathrm{T}$, and a Hall voltage of $12.0\ \mathrm{mV}$ is measured across it. Find the number density of the carriers, and compare it with copper.
Given
$I = 0.150\ \mathrm{A}$
$B = 0.650\ \mathrm{T}$
$t = 2.00\times10^{-4}\ \mathrm{m}$ along the field
$V_H = 12.0\times10^{-3}\ \mathrm{V}$
$q = 1.60\times10^{-19}\ \mathrm{C}$
copper for comparison: $n_{\rm Cu}=8.47\times10^{28}\ \mathrm{m^{-3}}$
Find
the carrier density $n$, and its ratio to that of copper
Solution
Rearranged before substituting, so the denominator is built once.
Rearrange for the unknown
$$V_H = \frac{IB}{nqt} \Rightarrow n = \frac{IB}{qtV_H}$$
$n$ is the only quantity not measured directly, so the formula is turned round to make it the subject
copper has a few hundred thousand times as many carriers, which is exactly why its Hall voltage was microvolts while this one is millivolts
Answer $$\boxed{\,n = 2.54\times10^{23}\ \mathrm{m^{-3}}\,,\ \text{about } 3\times10^{5} \text{ times fewer than copper}\,}$$
Check
Feeding $n$ back into $V_H = IB/(nqt)$ would only undo the algebra, so check it against the material instead: $n^{-1/3} = 1.6\times10^{-8}\ \mathrm{m}$, and atoms sit about $0.25\ \mathrm{nm}$ apart, so this is one mobile carrier per $2.5\times10^{5}$ atoms — right for a doped semiconductor and impossible for a metal.
One rearrangement and one substitution, but three negative powers of ten multiplied together in the denominator. Writing that denominator on its own line is worth the space.
Checkpoint
§09.7 — what the polarity tells you●●●○○
Thirty seconds, and no arithmetic at all. Two strips, one of copper and one of a semiconductor whose carriers are positive, carry equal currents in the same direction through the same field, arranged identically.
Given
the current direction is the same in both strips
the field is the same and perpendicular to both strips
in the copper the carriers are electrons; in the semiconductor they are positive
Find
(a) How do the two Hall voltages compare in sign, and which edge goes positive in each case?
Hint 1/4
Work with one real carrier of each kind rather than with the current, because it is the carriers and not the current that the magnetic force acts on.
Hint 2/4
The force on a carrier is $q\vec v\times\vec B$, and reversing the sign of $q$ while also reversing $\vec v$ leaves that product pointing the same way.
Hint 3/4
Here the two carriers have opposite $q$ and opposite drift directions, since both must produce a current in the same direction, and $\vec B$ is the same for both.
Hint 4/4
Both kinds of carrier are driven to the same edge, so that edge goes negative in the metal and positive in the semiconductor: the two Hall voltages have opposite signs.
Show solution
Carriers followed directly; the sign rule's two reversals are the difficulty.
Follow the electron in the metal
$$q<0,\qquad \vec v_d \ \text{opposite to the current}$$
conventional current runs against the electron flow, which is the first of the two reversals
Follow the positive carrier
$$q>0,\qquad \vec v_d \ \text{along the current}$$
the second reversal, relative to the electron case
$$q\vec v\times\vec B \ \text{is the same direction in both}$$
two sign flips multiply to one, so the force lands on the same edge for both kinds of carrier
Read off the polarity
$$\text{same edge, opposite charge} \Rightarrow V_H \ \text{opposite in sign}$$
the edge that collects electrons is at low potential while the edge that collects positive carriers is at high potential
Independent check by a limiting thought: if the polarity did not depend on the sign of the carriers, the Hall measurement could not distinguish them and would never have been used for that purpose. Its historical role is evidence that the sign does flip.
⚠ Putting the width into the Hall formula instead of the thickness
the voltage is measured across the width, so the width feels like the relevant dimension
wrong$$V_H = \frac{IB}{nqw}$$
right$$V_H = \frac{IB}{nqt},\qquad t \ \text{measured along } \vec B$$
⚠ Reading the sign of the carriers from which edge charges up rather than from the polarity
the two sound like the same statement, but both kinds of carrier arrive at the same edge and only the sign of the pile differs
wrong$$\text{electrons go to one edge, positive carriers to the other}$$
right$$\text{both go to the same edge; the polarity of } V_H \text{ is what differs}$$
A charged particle is fired into a field: the five steps
Any question where a charge, an ion or an electron enters a region of magnetic field. The order never changes, and steps 1 and 2 are the ones people skip.
Get the speed before touching the magnetism
If the particle was accelerated electrically, use $\tfrac12mv^{2}=qV$ first. If it was given a kinetic energy in electronvolts, convert with $1\ \mathrm{eV} = 1.60\times10^{-19}\ \mathrm{J}$. This stage is entirely the earlier material and the magnetic field has no say in it.
Split the velocity against the field
Write $v_\perp = v\sin\alpha$ and $v_\parallel = v\cos\alpha$, where $\alpha$ is the angle between the velocity and the field. For perpendicular entry $v_\perp = v$ and $v_\parallel = 0$, which is why the step is invisible in easy problems and fatal in hard ones.
Circle from the perpendicular part
$r = mv_\perp/(|q|B)$ and $T = 2\pi m/(|q|B)$. Use the magnitude of the charge; the sign decides which way round the particle goes, not how wide.
Drift from the parallel part
If $v_\parallel\neq0$ the path is a helix that advances a pitch $p = v_\parallel T$ every turn. If $v_\parallel = 0$ this step is empty and the path closes on itself.
Say what did not change, and check the speed
The speed and the kinetic energy are the same on the way out as on the way in, because the magnetic force does no work. Then check $v/c$: if it is above about a tenth, the formulas here are starting to lie and the problem is out of range.
Where it goes wrong
Putting $v$ where $v_\perp$ belongs, which inflates every radius in an angled entry problem.
Carrying the sign of $q$ into the radius and getting a negative length.
Reporting a change in speed. There is never one.
Forgetting that the period is fixed while the radius is not, and so scaling both when the speed changes.
Getting the direction right, every time
Whenever an answer needs a direction rather than only a size, and especially when the charge is negative or the field goes into the page.
Fix the axes before you start
Write down which way $x$, $y$ and $z$ point and stick to it. On this page $x$ is to the right, $y$ is up and $z$ is out of the page, so a field into the page is $-z$.
Point the cross product as though the charge were positive
Fingers of the right hand along $\vec v$ (or along the current $I\vec L$), curl them into $\vec B$, thumb gives $\vec v\times\vec B$. Do not think about the sign of the charge yet.
Now apply the sign, as a separate step
Positive charge: the force is along the thumb. Negative charge: the force is opposite to the thumb. For a current there is nothing to do, because conventional current already points the way positive carriers would go.
Check with components when it matters
Use $\hat x\times\hat y=\hat z$, $\hat y\times\hat z=\hat x$, $\hat z\times\hat x=\hat y$, reversing sign if the order is swapped. Two independent routes to the same arrow is the only real defence against a mirrored hand.
Sanity test the answer
The force must be perpendicular to both $\vec v$ and $\vec B$. If your arrow has any component along the velocity, it is wrong, because that would change the speed.
Where it goes wrong
Applying the sign of the charge inside the hand rule and then again outside it, which cancels the reversal.
Using the left hand out of habit from a different convention.
Taking the angle to a surface instead of between the two vectors themselves.
Drawing a force with a component along the velocity, which no magnetic force ever has.
A loop or coil in a field: force, torque and energy
Any closed circuit in a uniform field, including motors, galvanometers and single square loops in exam questions.
Write the net force down as zero and move on
In a uniform field the net force on any closed loop is zero. This is a result, not an assumption, and stating it early stops it being recalculated three times.
Build the dipole moment from the loop alone
$\mu = NIA$, with $\vec\mu$ perpendicular to the plane of the loop in the direction your right hand thumb points when your fingers curl along the current. No field appears in this step.
Identify the angle carefully
$\theta$ is between $\vec\mu$ and $\vec B$, not between the plane of the loop and $\vec B$. If the problem describes the loop lying in the field, that means $\theta = 90^{\circ}$ and the torque is at its largest.
Torque and energy from the same two numbers
$\tau = \mu B\sin\theta$ and $U = -\mu B\cos\theta$. Work done by an outside agent turning the coil slowly is $\Delta U$; work done by the field is $-\Delta U$.
Say which way it turns and where it settles
The torque always swings $\vec\mu$ towards $\vec B$. The stable rest position is $\theta = 0$; the position at $\theta = 180^{\circ}$ is an equilibrium that collapses under any disturbance.
Where it goes wrong
Leaving $N$ out of $\mu = NIA$.
Measuring $\theta$ from the plane of the loop, which puts every sine where a cosine belongs.
Looking for a net force on a loop in a uniform field.
Using $W = \tau\theta$ as though the torque were constant during the turn; it is not, which is why the energy formula exists.
An electric field changes the speed
A proton starts at rest and is accelerated through a potential difference of $2.00\ \mathrm{kV}$. Find the work done on it and its final speed.
Given
$q = 1.60\times10^{-19}\ \mathrm{C}$
$m_p = 1.67\times10^{-27}\ \mathrm{kg}$
$\Delta V = 2.00\times10^{3}\ \mathrm{V}$, starting from rest
Find
the work done on the proton and its final speed
Solution
Work energy theorem, not force times distance; no separation given.
The work, which is not zero
$$W = q\Delta V = (1.60\times10^{-19})(2.00\times10^{3}) = 3.20\times10^{-16}\ \mathrm{J}$$
the electric force has a component along the motion at every point, so it can and does deliver energy
The speed that buys
$$\tfrac12 m_pv^{2} = W \Rightarrow v = \sqrt{\frac{2(3.20\times10^{-16})}{1.67\times10^{-27}}}$$
all of the work has gone into kinetic energy because the proton started at rest and nothing else acted
$$v = 6.19\times10^{5}\ \mathrm{m/s}$$
up from zero: the speed has changed, and that is the point of the comparison
Answer $$\boxed{\,W = 3.20\times10^{-16}\ \mathrm{J}\,,\qquad v = 6.19\times10^{5}\ \mathrm{m/s}\,}$$
Check
Check with the electronvolt as a shortcut: a charge of one elementary unit crossing $2.00\ \mathrm{kV}$ gains exactly $2.00\ \mathrm{keV}$, and $2.00\times10^{3}\times1.60\times10^{-19} = 3.20\times10^{-16}\ \mathrm{J}$ as found. Speed check: $v/c = 0.0021$, so the non relativistic expression is safe.
A magnetic field changes only the direction
That same proton, now moving at $6.19\times10^{5}\ \mathrm{m/s}$, enters a uniform field of $0.350\ \mathrm{T}$ at right angles to its velocity. Find the work the field does on it, its speed after a quarter turn, and the radius of its path.
Given
$v = 6.19\times10^{5}\ \mathrm{m/s}$ on entry, perpendicular to $\vec B$
Independent check on the radius through the period: $T = 2\pi m/(qB) = 1.87\times10^{-7}\ \mathrm{s}$, and $2\pi r/v = 2\pi(1.85\times10^{-2})/(6.19\times10^{5}) = 1.88\times10^{-7}\ \mathrm{s}$, agreeing to the rounding.
Same proton, same charge, comparable field strengths, and one number tells the whole story: the electric stage delivered $3.20\times10^{-16}\ \mathrm{J}$ and took the speed from zero to $6.19\times10^{5}\ \mathrm{m/s}$, while the magnetic stage delivered exactly zero joules and left the speed at $6.19\times10^{5}\ \mathrm{m/s}$ while turning the velocity through a right angle.
How to tell them apart
Ask one question of the field before writing anything: can its force have a component along the direction of travel? An electric force can, because $q\vec E$ points along the field wherever the particle happens to be going, so electric fields change speeds. A magnetic force cannot, because $q\vec v\times\vec B$ is built to be perpendicular to $\vec v$, so magnetic fields change directions and nothing else. Any answer in which a magnetic field has changed a kinetic energy should be reread from the first line. An electric field usually changes it, but not always: in the velocity selector the electric force is perpendicular to the velocity too, and does no work either.
Scaffolding comes off
The common skeleton
Say what is given and what the field will and will not do: the speed and the kinetic energy are fixed by whatever happened before the particle entered, and the field cannot change either.
Convert whatever energy is quoted into joules and get the speed from $\tfrac12mv^{2}=K$.
Resolve the velocity against the field into $v_\perp$ and $v_\parallel$, even when one of them is obviously zero.
Equate the magnetic force to the force needed for circular motion, $|q|v_\perp B = mv_\perp^{2}/r$, and solve for the radius.
Get the period from $T = 2\pi m/(|q|B)$, noticing that no speed appears in it.
If $v_\parallel\neq0$, multiply it by the period to get how far the particle advances per turn.
Check: the speed is unchanged, the radius is positive, the answer scales the way the formula says it should, and $v/c$ is small enough for the formulas to be honest.
1 · fully worked
An electron at 3.00 x 10^6 m/s in a field of 0.400 mT
Worked in full, with a reason on every line. An electron enters a uniform field of $4.00\times10^{-4}\ \mathrm{T}$ at right angles to it, moving at $3.00\times10^{6}\ \mathrm{m/s}$. Find the radius of its path, the period and the frequency of its circulation, and its speed after half a turn.
the radius, the period, the frequency, and the speed after half a turn
Solution
We go through the force balance rather than quoting the radius formula, because the balance is what makes the cancellation of one power of $v$ visible, and that cancellation is the reason the period turns out not to depend on the speed two lines later.
Resolve the velocity, even though it is easy here
$$v_\perp = v = 3.00\times10^{6}\ \mathrm{m/s},\qquad v_\parallel = 0$$
the entry is perpendicular, so the split is trivial, but writing it keeps the habit for the problems where it is not
Radius from the force balance
$$evB = \frac{m_ev^{2}}{r} \Rightarrow r = \frac{m_ev}{eB}$$
one power of $v$ cancels, which is why the radius is linear in the speed and not quadratic
the magnetic force is perpendicular to the velocity throughout, so the speed after half a turn, a full turn, or a thousand turns is the same
Answer $$\boxed{\,r = 4.27\ \mathrm{cm}\,,\quad T = 8.94\times10^{-8}\ \mathrm{s}\,,\quad f = 11.2\ \mathrm{MHz}\,,\quad v \text{ unchanged}\,}$$
Check
Independent check of the period: $2\pi r/v = 2\pi(4.27\times10^{-2})/(3.00\times10^{6}) = 8.94\times10^{-8}\ \mathrm{s}$, matching the formula answer although the two share no algebra. Speed check: $v/c = 0.010$.
2 · you write the reasoning
Easier than the last one: no energy conversion, no frequency, three lines. A proton moves at $2.00\times10^{5}\ \mathrm{m/s}$ at right angles to a uniform field of $0.500\ \mathrm{T}$. All three lines below are correct. Before opening the model reasons, say in your own words why each one is allowed, and pay particular attention to what the third line is actually checking.
reasoning
The first line uses the full speed with no sine factor, and it is allowed to because the entry is stated to be at right angles, so $\theta = 90^{\circ}$ and $\sin\theta = 1$. If the problem had said sixty degrees this line would be wrong by a factor of $0.866$, and everything after it would inherit the error. The magnitude of the charge is used, not its signed value: the proton happens to be positive, so nothing shows here, but the habit matters for the next electron.
reasoning
The second line is the radius formula, and the thing to notice is that it did not come from the first line. It comes from setting the magnetic force equal to $mv^{2}/r$ and cancelling one power of $v$. So line two is not a consequence of line one; the two are independent routes out of the same force law, which is exactly what makes the third line worth writing.
reasoning
The third line recomputes the force needed for circular motion from the radius that line two produced, and gets back the force that line one produced. That is a genuine check rather than a repetition: if the radius had been computed wrongly, the two forces would disagree. It is the same test as substituting a root back into the original equation, and it costs one line.
3 · find the buried error
Harder than the last one: a different particle, a charge that is not $e$, and an energy in megaelectronvolts. An alpha particle, of mass $6.64\times10^{-27}\ \mathrm{kg}$ and charge $+2e$, has a kinetic energy of $4.00\ \mathrm{MeV}$ and enters a uniform field of $1.20\ \mathrm{T}$ at right angles. A student produces the four lines below and reports a radius of $0.480\ \mathrm{m}$ and a period of $2.17\times10^{-7}\ \mathrm{s}$. Exactly two of the four lines are faulty. Find them.
the two buried errors (2)
⚠ step 2
The charge of an alpha particle is $+2e = 3.20\times10^{-19}\ \mathrm{C}$, not $e$. With the correct charge the radius is $r = mv/(2eB) = 0.240\ \mathrm{m}$, exactly half of what was reported. The line is written correctly as algebra and wrong as arithmetic, which is what makes it hard to see.
The symbol $q$ is written and then the value of $e$ is typed, because $e$ is the number that has been used in every other problem this week. Nothing in the line looks unusual, the units still work, and the answer is a perfectly plausible length. The only warning sign is that the particle was named as an alpha and its charge was never written down as a number of its own.
right
Write the charge on its own line before using it: $q = 2e = 3.20\times10^{-19}\ \mathrm{C}$. Then $r = (6.64\times10^{-27})(1.39\times10^{7})/[(3.20\times10^{-19})(1.20)] = 0.240\ \mathrm{m}$, and the period, done properly from $T = 2\pi m/(|q|B)$, is $1.09\times10^{-7}\ \mathrm{s}$.
⚠ step 4
The claim confuses two different independences. The period does not depend on the speed, which is true and is the interesting fact. It certainly does depend on the field: $T = 2\pi m/(|q|B)$ is inversely proportional to $B$, so doubling the field halves the period.
The sentence begins with a correct observation, that there is no $v$ in the formula, and then slides one word sideways from speed to field. Both are quantities the problem gives, both feel like inputs, and the formula genuinely does look as though nothing can move it.
right
Read the formula for what it does contain rather than what it does not: $T\propto 1/B$ and $T\propto m/|q|$. Doubling $B$ halves the period; doubling the speed changes nothing. Only one quantity has been eliminated, and it is the speed.
4 · the bare problem
§09.3 — a deuteron with no scaffolding●●●○○
No steps this time, and the particle has been changed again so that the numbers cannot be copied from anywhere above. A deuteron, the nucleus of heavy hydrogen, has mass $3.34\times10^{-27}\ \mathrm{kg}$ and charge $+e$. It is given a kinetic energy of $2.00\ \mathrm{MeV}$ and then enters a uniform magnetic field of $0.900\ \mathrm{T}$ at right angles to its velocity.
Given
$m = 3.34\times10^{-27}\ \mathrm{kg}$
$q = +1.60\times10^{-19}\ \mathrm{C}$
$K = 2.00\ \mathrm{MeV}$
$B = 0.900\ \mathrm{T}$, entry perpendicular to the field
$1\ \mathrm{eV} = 1.60\times10^{-19}\ \mathrm{J}$
Find
(a) Find the speed of the deuteron.
(b) Find the radius of its circular path.
(c) Find the time it takes to go round once, and state what its speed is after ten complete turns.
Hint 1/4
Three quantities are wanted and they come in a fixed order: energy first, then speed, then everything geometric. Nothing about the field is needed until the speed is known.
Hint 2/4
Use $K = \tfrac12mv^{2}$ for the speed, $r = mv/(|q|B)$ for the radius and $T = 2\pi m/(|q|B)$ for the period, and remember that the magnetic force does no work.
Hint 3/4
Here $K = 2.00\times10^{6}\times1.60\times10^{-19}\ \mathrm{J}$, $m = 3.34\times10^{-27}\ \mathrm{kg}$, $q = 1.60\times10^{-19}\ \mathrm{C}$ and $B = 0.900\ \mathrm{T}$.
Hint 4/4
The deuteron moves at $1.38\times10^{7}\ \mathrm{m/s}$ on a circle of radius $0.321\ \mathrm{m}$, taking $1.46\times10^{-7}\ \mathrm{s}$ per turn, and its speed after ten turns is still $1.38\times10^{7}\ \mathrm{m/s}$.
Show solution
The usual spine: energy, speed, radius, then a speedless period.
unchanged, because the only force acting is perpendicular to the velocity and so does no work
Answer $$\boxed{\,v = 1.38\times10^{7}\ \mathrm{m/s},\quad r = 0.321\ \mathrm{m},\quad T = 1.46\times10^{-7}\ \mathrm{s},\quad v \text{ unchanged}\,}$$
Check
Independent check on the period: $2\pi r/v = 2\pi(0.321)/(1.38\times10^{7}) = 1.46\times10^{-7}\ \mathrm{s}$, from a different formula. Speed check: $v/c = 0.046$, small enough for the non relativistic treatment, and a deuteron is twice the mass of a proton with the same charge, so its radius should be larger than a proton of the same energy by $\sqrt2$, which it is.
Full exam-style question
Exam level: a 40 turn coil edge on to a 0.320 T fieldexam format
A rectangular coil of $40$ turns measures $0.150\ \mathrm{m}$ by $0.250\ \mathrm{m}$ and carries a steady current of $2.50\ \mathrm{A}$. It is placed in a uniform magnetic field of $0.320\ \mathrm{T}$ with the plane of the coil containing the field and running parallel to the $0.150\ \mathrm{m}$ sides, so that the coil is edge on to it. Find (a) the magnetic dipole moment of the coil, (b) the torque on it in this position, (c) the net force on it, (d) the force on one of the $0.250\ \mathrm{m}$ sides, and (e) the work an external agent must do to turn the coil slowly into the position it would settle in on its own.
Given
$N = 40$ turns, sides $0.150\ \mathrm{m}$ and $0.250\ \mathrm{m}$
$I = 2.50\ \mathrm{A}$
$B = 0.320\ \mathrm{T}$
the field lies in the plane of the coil, parallel to the $0.150\ \mathrm{m}$ sides, so $\vec\mu$ is perpendicular to $\vec B$
Find
the dipole moment, the torque, the net force, the force on one long side, and the work to turn it to equilibrium
Solution
Parts (a) and (d) are done from different starting points on purpose: (a) from the dipole moment and (d) from the force on a single wire. That is what makes the check at the end worth anything, because a single error in the area or in the number of turns would show up as a disagreement between them.
Independent check tying (b) and (d) together, which were computed by different routes: the two $0.250\ \mathrm{m}$ sides carry $8.00\ \mathrm{N}$ each in opposite directions, and their lines of action are the full width $0.150\ \mathrm{m}$ apart when the coil is edge on. The couple is therefore $F\times$ separation $= (8.00)(0.150) = 1.20\ \mathrm{N\,m}$, matching part (b) exactly. Plausibility: a newton metre is the torque of hanging a $100\ \mathrm{g}$ mass on the end of a metre stick, which is a firm but ordinary twist for a coil this size.
Five parts, and only two of them needed a calculator. Parts (c) and (e) were answered by knowing which result applies, which is what the extra marks in this kind of question are usually for.
Part (e) is the one that catches people, because the word work suggests a positive number. The coil is being allowed to fall into its lowest energy orientation, so the field does the work and the external agent takes energy out. If you are asked instead for the work to turn it from rest at $\theta=0$ to edge on, the same magnitude comes back with a plus sign.
Practice
A · concept 4 questions
1§09.1 — a charge sitting still next to a magnet●●○○○
One mark, and it separates having read the formula from having noticed what is in it. A small charged bead is glued to a bench, and a very strong permanent magnet is brought right up beside it.
Given
the bead carries a charge and is not moving
the magnet is strong and close
no electric field is present
Find
(a) True or false: the magnet exerts a magnetic force on the bead. Give your reason in one sentence.
Hint 1/4
Look at what the force law needs as input, and check whether the situation supplies all of it.
Hint 2/4
The magnetic force is $\vec F = q\vec v\times\vec B$, so it needs a charge, a field and a velocity.
Hint 3/4
Here the charge is present and the field is strong, but the bead is glued down so $\vec v = \vec 0$.
Hint 4/4
False: there is no magnetic force at all, however strong the magnet.
Show solution
Force law inputs checked first; the magnet's strength never enters.
Check the inputs to the force law
$$\vec F = q\,\vec v\times\vec B,\qquad \vec v = \vec 0$$
the situation supplies the charge and the field but not the motion
$$\vec 0\times\vec B = \vec 0 \Rightarrow \vec F = \vec 0$$
a cross product with a zero vector is zero regardless of the other factor
Answer $$\boxed{\,\text{False: } \vec F = \vec 0 \text{ for a charge at rest}\,}$$
Check
Independent check by contrast: an electric field of the same nominal strength would give $F = qE$, which does not vanish at rest. The two force laws differ precisely on this point, which is why the test is a good one.
2§09.2 — the energy after three turns●●○○○
One mark, and it is the single most tested idea in this material. An electron enters a region of uniform magnetic field with kinetic energy $K$ and follows a circular path inside it. Nothing else acts on it.
Given
kinetic energy on entry is $K$
the field is uniform and constant in time
the path is a closed circle, so the electron completes exactly three turns
Find
(a) What is the electron's kinetic energy after three complete turns?
Hint 1/4
Turn the question about energy into a question about the angle between two vectors, because that is all it is.
Hint 2/4
The rate of change of kinetic energy is the power $\vec F\cdot\vec v$, and the magnetic force is perpendicular to $\vec v$ at every instant.
Hint 3/4
Here the force is $-e\,\vec v\times\vec B$, which is perpendicular to $\vec v$ whatever the values of $e$, $v$ and $B$.
Hint 4/4
The kinetic energy is still exactly $K$, and it would be after three million turns as well.
Show solution
Zero power at every instant, so the turn count is irrelevant.
Power delivered by the magnetic force
$$P = \vec F\cdot\vec v = q(\vec v\times\vec B)\cdot\vec v$$
the definition of power, with the magnetic force substituted in
$$= 0$$
the cross product is perpendicular to $\vec v$, so this dot product vanishes identically
Integrate over any interval you like
$$\Delta K = \int P\,dt = 0$$
zero integrand gives zero integral over three turns, or over any other stretch of the path
Answer $$\boxed{\,K_{\rm after} = K\,}$$
Check
Independent check through a different quantity: the radius $r = mv/(|q|B)$ is observed to stay constant while the electron circles. If the speed were dropping, the radius would spiral inwards, and it does not.
3§09.1 — cutting a magnet in two●○○○○
One mark. A bar magnet with a clearly labelled north end and south end is cut cleanly in half across the middle, and the two halves are examined.
Given
the cut is made across the bar, between the two ends
both halves are then tested with a compass
no current and no external field are involved
Find
(a) True or false: one half is now a north pole on its own and the other a south pole on its own. Say in one sentence what the field lines have to do with your answer.
Hint 1/4
Think about what the cut does to the picture of the field rather than about what it does to the metal.
Hint 2/4
Magnetic field lines are closed loops: they run from north to south outside the magnet and from south to north inside it, so none of them begins or ends anywhere.
Hint 3/4
Cutting across the bar cuts through the part of every loop that runs inside the metal, and a cut loop does not become two loose ends; it becomes two smaller loops.
Hint 4/4
False: you get two complete magnets, each with its own north and south end.
Show solution
Closed field lines, not poles; loops have no ends to cut.
What the field looks like before the cut
$$\text{field lines: closed loops, N to S outside, S to N inside}$$
this is the observed pattern of iron filings and the thing the cut has to be reconciled with
What the cut does to those loops
$$\text{each half carries its own complete set of loops}$$
the two new faces become a new south end and a new north end, so each piece is a full magnet
Answer $$\boxed{\,\text{False: two complete magnets, never one isolated pole}\,}$$
Check
Independent check by experiment rather than by argument: hang each half from a thread. Both align north to south, which a single isolated pole would not do, since a lone pole would simply be pulled towards the Earth's field rather than turned by it.
4§09.5 — force and torque on a loop●●○○○
One mark, and it is a distinction that costs marks every year. A closed rectangular loop of wire carries a steady current and sits in a uniform magnetic field, with its plane tilted at some angle to the field.
Given
the loop is closed and rigid
the field is uniform over the whole loop
the plane of the loop is neither along the field nor perpendicular to it
Find
(a) Which statement about the loop is correct?
Hint 1/4
Treat force and torque as two separate questions with two separate answers, because that is exactly what they are here.
Hint 2/4
For a closed loop in a uniform field, $\vec F = I(\oint d\vec L)\times\vec B = \vec 0$, while $\vec\tau = \vec\mu\times\vec B$ with $\mu = NIA$.
Hint 3/4
Here the loop is closed, so the first result applies; and the plane is tilted, so $\theta$ is neither $0$ nor $180^{\circ}$ and the sine in the torque does not vanish.
Hint 4/4
The net force is zero and the net torque is not, which is why a loop in a uniform field turns without drifting.
Show solution
Force and torque on separate lines, since only one vanishes.
The force
$$\vec F = I\left(\oint d\vec L\right)\times\vec B = \vec 0$$
the field comes out of the sum because it is uniform, and the displacement round a closed path is zero
Independent check on a specific case: a square loop with two sides across the field feels equal and opposite forces on those two sides, obviously summing to zero, and just as obviously separated in space, so they form a couple.
B · computation 8 questions
1§09.1 — finding a field from a measured force●●○○○
The definition of the field is a recipe for measuring it, and this is the measurement done backwards. A charge of $2.50\ \mathrm{\mu C}$ is fired at $1.20\times10^{4}\ \mathrm{m/s}$ straight across a uniform magnetic field, and the force on it is measured as $4.50\times10^{-2}\ \mathrm{N}$.
Given
$q = 2.50\times10^{-6}\ \mathrm{C}$
$v = 1.20\times10^{4}\ \mathrm{m/s}$, perpendicular to the field
$F = 4.50\times10^{-2}\ \mathrm{N}$
Find
(a) Find the magnitude of the magnetic field.
(b) State what the measured force would have been if the same charge had been fired along the field instead.
Hint 1/4
The field is the unknown here, so rearrange the definition rather than reaching for anything new.
Hint 2/4
$F = |q|vB\sin\theta$, and for perpendicular firing $\sin\theta = 1$, so $B = F/(|q|v)$.
Hint 3/4
Here $F = 4.50\times10^{-2}\ \mathrm{N}$, $q = 2.50\times10^{-6}\ \mathrm{C}$ and $v = 1.20\times10^{4}\ \mathrm{m/s}$.
Hint 4/4
The field is $1.50\ \mathrm{T}$, and firing along it would have given exactly zero force.
Show solution
The definition rearranged; part two follows with the sine zeroed.
Solve the definition for B
$$F = |q|vB \Rightarrow B = \frac{F}{|q|v}$$
perpendicular firing makes the sine one, which is why this arrangement is the one used to define the field
Plausibility: $1.50\ \mathrm{T}$ is a strong laboratory electromagnet, some thirty thousand times the Earth's field, which is consistent with a measurable force on a charge as small as a few microcoulombs.
2§09.2 — a vector force on an electron●●●○○
Directions are worth as many marks as magnitudes, so this one asks for both. An electron moves in the $+y$ direction at $3.00\times10^{6}\ \mathrm{m/s}$ through a uniform field $\vec B = 0.400\ \hat x\ \mathrm{T}$.
Given
$\vec v = (0,\ 3.00\times10^{6},\ 0)\ \mathrm{m/s}$
$\vec B = (0.400,\ 0,\ 0)\ \mathrm{T}$
$q = -1.60\times10^{-19}\ \mathrm{C}$
axes: $x$ right, $y$ up, $z$ out of the page
Find
(a) Find the force on the electron as a vector, and state its direction in words.
Hint 1/4
Do the geometry and the sign in two separate steps, and write the components out rather than trusting a mental picture.
Hint 2/4
$\vec F = q(\vec v\times\vec B)$, and $\hat y\times\hat x = -\hat z$.
Hint 3/4
Here $\vec v = 3.00\times10^{6}\,\hat y$ and $\vec B = 0.400\,\hat x$, so $\vec v\times\vec B = (3.00\times10^{6})(0.400)(\hat y\times\hat x)$.
Hint 4/4
The force is $1.92\times10^{-13}\ \mathrm{N}$ in the $+z$ direction, out of the page.
Show solution
Components first, charge sign second; two minus signs need room.
The cross product on its own
$$\hat y\times\hat x = -\hat z$$
reversing the order of a cross product reverses its sign, and $\hat x\times\hat y=\hat z$ is the one to remember
$$\vec v\times\vec B = (3.00\times10^{6})(0.400)(-\hat z) = -1.20\times10^{6}\,\hat z$$
units of metres per second times teslas, which will become newtons per coulomb once the charge multiplies in
Multiply by the charge, sign and all
$$\vec F = (-1.60\times10^{-19})(-1.20\times10^{6}\,\hat z) = +1.92\times10^{-13}\,\hat z\ \mathrm{N}$$
two minus signs, one from the order of the cross product and one from the electron, so the force ends up along $+z$
Answer $$\boxed{\,\vec F = 1.92\times10^{-13}\,\hat z\ \mathrm{N},\ \text{out of the page}\,}$$
Check
Independent check of the magnitude alone: $|q|vB\sin 90^{\circ} = (1.60\times10^{-19})(3.00\times10^{6})(0.400) = 1.92\times10^{-13}\ \mathrm{N}$, computed without any reference to direction, and it agrees. Perpendicularity check: the answer is along $\hat z$, at right angles to both $\hat y$ and $\hat x$, as it must be.
3§09.3 — from an accelerating voltage to a radius●●●○○
This is the standard two stage problem, and the two stages must not be mixed. A proton starts from rest, is accelerated through a potential difference of $1.50\ \mathrm{kV}$, and then enters a uniform magnetic field of $0.600\ \mathrm{T}$ at right angles to its velocity.
Given
$\Delta V = 1.50\times10^{3}\ \mathrm{V}$, from rest
$B = 0.600\ \mathrm{T}$, perpendicular entry
$m_p = 1.67\times10^{-27}\ \mathrm{kg}$
$q = 1.60\times10^{-19}\ \mathrm{C}$
Find
(a) Find the speed at which the proton enters the field.
(b) Find the radius of its circular path and the time for one revolution.
Hint 1/4
Two separate physical stages: an electrical one that sets the speed, then a magnetic one that bends the path. Finish the first before starting the second.
Here $\Delta V = 1.50\times10^{3}\ \mathrm{V}$, $B = 0.600\ \mathrm{T}$, $m_p = 1.67\times10^{-27}\ \mathrm{kg}$ and $q = 1.60\times10^{-19}\ \mathrm{C}$.
Hint 4/4
The proton enters at $5.36\times10^{5}\ \mathrm{m/s}$, turns on a radius of $9.33\ \mathrm{mm}$, and takes $1.09\times10^{-7}\ \mathrm{s}$ per revolution.
Show solution
Energy, not force: the plate separation is never given.
The electric stage sets the speed
$$\tfrac12 m_pv^{2} = q\Delta V \Rightarrow v = \sqrt{\frac{2q\Delta V}{m_p}}$$
energy rather than force, because the shape of the accelerating field is not given and does not matter
no speed enters, so this answer would be unchanged if the accelerating voltage were doubled
Answer $$\boxed{\,v = 5.36\times10^{5}\ \mathrm{m/s},\quad r = 9.33\ \mathrm{mm},\quad T = 1.09\times10^{-7}\ \mathrm{s}\,}$$
Check
Independent check on the period: $2\pi r/v = 2\pi(9.33\times10^{-3})/(5.36\times10^{5}) = 1.09\times10^{-7}\ \mathrm{s}$, from a different formula. Speed check: $v/c = 0.0018$, comfortably non relativistic.
4§09.3 — the pitch of a helix●●●○○
When the entry is not perpendicular the path stops being a circle, and the marks are in the resolving rather than in the arithmetic. An electron enters a uniform field of $1.20\times10^{-3}\ \mathrm{T}$ at $5.00\times10^{6}\ \mathrm{m/s}$, with its velocity at $30.0^{\circ}$ to the field.
Given
$v = 5.00\times10^{6}\ \mathrm{m/s}$ at $30.0^{\circ}$ to $\vec B$
the period is the same as it would be for perpendicular entry, because it never contained a speed
$$p = v_\parallel T = (4.33\times10^{6})(2.98\times10^{-8}) = 0.129\ \mathrm{m}$$
uniform motion along the field for one period
Answer $$\boxed{\,r = 1.19\ \mathrm{cm},\qquad p = 12.9\ \mathrm{cm}\,}$$
Check
Independent check by a ratio that avoids both $m$ and $B$: $p/(2\pi r)$ must equal $v_\parallel/v_\perp = \cot 30.0^{\circ} = 1.73$. Here $0.129/(2\pi\times1.19\times10^{-2}) = 1.73$. Speed check: $v/c = 0.017$.
5§09.4 — a wire at an angle to the field●●○○○
Straight from the wire formula, with an angle to make sure the sine goes in the right place. A straight wire carries $6.50\ \mathrm{A}$, and a length of $0.750\ \mathrm{m}$ of it lies in a uniform field of $0.180\ \mathrm{T}$, making an angle of $25.0^{\circ}$ with the field direction.
Given
$I = 6.50\ \mathrm{A}$
$L = 0.750\ \mathrm{m}$ inside the field
$B = 0.180\ \mathrm{T}$
$\theta = 25.0^{\circ}$ between the wire and the field
Find
(a) Find the magnitude of the force on that length of wire.
(b) State how the answer would change if the wire were turned to lie along the field.
Hint 1/4
Compute the largest force this wire could feel first, then reduce it by the angle. Two separate steps means two separate places to check.
Hint 2/4
$F = BIL\sin\theta$, with $\theta$ measured between the wire and the field.
Hint 3/4
Here $B = 0.180\ \mathrm{T}$, $I = 6.50\ \mathrm{A}$, $L = 0.750\ \mathrm{m}$ and $\theta = 25.0^{\circ}$.
Hint 4/4
The force is $0.371\ \mathrm{N}$, and it would fall to zero if the wire lay along the field.
Show solution
Maximum first, angle second, so a misplaced sine breaks the bound.
Bounds check: the answer must sit between zero and $BIL = 0.878\ \mathrm{N}$, and $0.371$ does, at about forty two per cent of the maximum, matching $\sin 25.0^{\circ}$.
6§09.4 — the field needed to support a rod●●●○○
The same balance as the worked rod, with a different unknown, which is the usual way an exam disguises a problem you have already done. A rod of mass $45.0\ \mathrm{g}$ and length $0.350\ \mathrm{m}$ rests across horizontal rails and carries a current of $8.00\ \mathrm{A}$. A horizontal magnetic field runs perpendicular to the rod.
the rails now support the weight and the magnetic push together
Answer $$\boxed{\,B = 0.158\ \mathrm{T}\,}$$
Check
Plausibility: $0.158\ \mathrm{T}$ is a strongish permanent magnet and $8\ \mathrm{A}$ is a heavy but ordinary bench current, which is the right combination for lifting forty five grams. A field of a few thousandths of a tesla would need hundreds of amperes and would melt the rod.
7§09.5 — a circular coil, torque and energy●●●○○
A circular coil this time, so that the area formula has to be supplied rather than read off. A flat circular coil of $60$ turns and radius $4.00\ \mathrm{cm}$ carries a current of $1.50\ \mathrm{A}$ and lies in a uniform field of $0.280\ \mathrm{T}$, with the plane of the coil containing the field.
(a) Find the magnetic dipole moment of the coil and the torque on it.
(b) Find the work done by the field as the coil turns freely into its stable orientation.
Hint 1/4
Build the coil's own property first and only then let the field in, because the first part is reused in the second.
Hint 2/4
$\mu = NIA$ with $A = \pi r^{2}$; $\tau = \mu B\sin\theta$; $U = -\mu B\cos\theta$, and the coil settles where $U$ is least.
Hint 3/4
Here $N = 60$, $r = 4.00\times10^{-2}\ \mathrm{m}$, $I = 1.50\ \mathrm{A}$, $B = 0.280\ \mathrm{T}$, and the field lying in the plane of the coil means $\theta = 90^{\circ}$.
Hint 4/4
The moment is $0.452\ \mathrm{A\,m^{2}}$, the torque is $0.127\ \mathrm{N\,m}$, and the field does $0.127\ \mathrm{J}$ of work as the coil swings into line.
Show solution
$\mu$ from the coil, work from energy, so one product serves both.
$$U(90^{\circ}) = 0,\qquad U(0) = -\mu B = -0.127\ \mathrm{J}$$
the energy is lowest when the moment lines up with the field
$$W_{\rm field} = -\Delta U = +0.127\ \mathrm{J}$$
the field does positive work as the coil falls into its stable orientation, and that energy shows up as rotational kinetic energy unless something holds it back
Torque from the forces instead of the dipole formula: replace the coil by a square of equal area, side $a = \sqrt{A} = 7.09\times10^{-2}\ \mathrm{m}$. Two opposite sides lie across the field with $F = NBIa = 1.79\ \mathrm{N}$ each, $a$ apart, so the couple is $0.127\ \mathrm{N\,m}$. The work repeats those digits because $\mathrm{A\,m^{2}}$ times $\mathrm{T}$ is a joule.
8§09.6 — selector and analyser in sequence●●●○○
Both stages of a mass spectrometer, one after the other, which is how the question is usually set. Ions pass through a velocity selector with $E = 1.50\times10^{5}\ \mathrm{V/m}$ and $B = 0.300\ \mathrm{T}$, and the singly charged ions that emerge are bent by a field of $0.500\ \mathrm{T}$ onto a circle of radius $0.250\ \mathrm{m}$.
Here $E = 1.50\times10^{5}\ \mathrm{V/m}$ and $B = 0.300\ \mathrm{T}$ in the first stage, and $B' = 0.500\ \mathrm{T}$ with $r = 0.250\ \mathrm{m}$ in the second.
Hint 4/4
The ions travel at $5.00\times10^{5}\ \mathrm{m/s}$ and their mass is $4.00\times10^{-26}\ \mathrm{kg}$, about $24.1\ u$.
Show solution
Selector finished first, or the intermediate speed goes unchecked.
Independent check by reversing the second stage: an ion of $24u = 3.98\times10^{-26}\ \mathrm{kg}$ at this speed in $0.500\ \mathrm{T}$ would follow $r = mv/(qB') = 0.249\ \mathrm{m}$, which is the measured radius to three figures. Plausibility: a mass number near twenty four with a single charge is an ordinary light ion, and the radius came out at a quarter of a metre, the scale of a real instrument.
C · exam level 4 questions
1§09.7 — reading a Hall probe●●●●○
Exam level, and it asks for a number and a conclusion, which is the usual shape of a Hall question. A flat strip of a doped semiconductor is $0.400\ \mathrm{mm}$ thick along the field direction. It carries a current of $2.00\ \mathrm{A}$ in a field of $0.750\ \mathrm{T}$ perpendicular to its face, and a Hall voltage of $0.550\ \mathrm{mV}$ is measured across it. The edge towards which the conventional current would be deflected is found to be at the higher potential.
Given
$I = 2.00\ \mathrm{A}$
$B = 0.750\ \mathrm{T}$, perpendicular to the face of the strip
$t = 4.00\times10^{-4}\ \mathrm{m}$, measured along the field
$V_H = 5.50\times10^{-4}\ \mathrm{V}$
$q = 1.60\times10^{-19}\ \mathrm{C}$ per carrier
the edge that $q\vec v\times\vec B$ points towards is at the higher potential
Find
(a) Find the number density of the charge carriers.
(b) State whether the carriers are positive or negative, choosing the option below that gives the reason too.
Hint 1/4
Part (a) is one rearrangement and part (b) needs no arithmetic at all, so do not let them contaminate each other.
Hint 2/4
$V_H = IB/(nqt)$, so $n = IB/(qtV_H)$; and the carriers pile up on whichever edge $q\vec v\times\vec B$ points towards, whatever their sign.
Hint 3/4
Here $I = 2.00\ \mathrm{A}$, $B = 0.750\ \mathrm{T}$, $t = 4.00\times10^{-4}\ \mathrm{m}$ and $V_H = 5.50\times10^{-4}\ \mathrm{V}$, and that same edge is measured to be at the higher potential.
Hint 4/4
The density is $4.26\times10^{25}\ \mathrm{m^{-3}}$, and since the collecting edge is positive the carriers must be positive.
Show solution
Rearrangement and polarity apart; algebra never yields the carriers' sign.
Make n the subject
$$V_H = \frac{IB}{nqt} \Rightarrow n = \frac{IB}{qtV_H}$$
$n$ is the only quantity in the formula that cannot be measured directly, which is why the Hall probe exists
carriers per cubic metre, since every other quantity was in SI units
Read the sign off the polarity, not off the edge
$$\text{both signs of carrier} \to \text{same edge}$$
reversing the charge also reverses the drift direction, and the two reversals cancel in $q\vec v\times\vec B$
$$\text{that edge is at higher potential} \Rightarrow \text{carriers are positive}$$
positive charge collecting somewhere raises its potential; negative charge collecting there would lower it
Answer $$\boxed{\,n = 4.26\times10^{25}\ \mathrm{m^{-3}},\qquad \text{carriers are positive}\,}$$
Check
Substituting $n$ back would only reverse the rearrangement, so check against the material: $n^{-1/3} = 2.9\times10^{-9}\ \mathrm{m}$, and with atoms about $0.25\ \mathrm{nm}$ apart that is one carrier per $1.5\times10^{3}$ atoms. Heavy doping, but real, and impossible for a metal at one carrier per atom.
2§09.5 — a square loop tilted in a field●●●●○
Exam level, three parts, and the angle convention decides the first two. A single square loop of side $0.200\ \mathrm{m}$ carries a current of $4.00\ \mathrm{A}$ in a uniform field of $0.500\ \mathrm{T}$. The normal to the loop makes an angle of $60.0^{\circ}$ with the field.
Given
one turn, square of side $0.200\ \mathrm{m}$
$I = 4.00\ \mathrm{A}$
$B = 0.500\ \mathrm{T}$
$\theta = 60.0^{\circ}$ between $\vec\mu$ and $\vec B$
Find
(a) Find the magnetic dipole moment and the torque on the loop.
(b) Find the potential energy of the loop in this orientation.
(c) Find the work an outside agent must do to turn the loop slowly into its stable orientation, say what that orientation is, and pick that work from the options below.
Hint 1/4
Three parts built on one product, so compute $\mu$ and then $\mu B$ once and reuse them rather than starting each part from scratch.
Hint 2/4
$\mu = NIA$, $\tau = \mu B\sin\theta$, $U = -\mu B\cos\theta$, and the stable orientation is the one where $U$ is least.
Hint 3/4
Here $N = 1$, $A = (0.200)^{2}$, $I = 4.00\ \mathrm{A}$, $B = 0.500\ \mathrm{T}$ and $\theta = 60.0^{\circ}$.
Hint 4/4
The moment is $0.160\ \mathrm{A\,m^{2}}$, the torque $0.0693\ \mathrm{N\,m}$, the energy $-0.0400\ \mathrm{J}$, and turning it to $\theta = 0$ takes $-0.0400\ \mathrm{J}$ of external work, which is to say the field does the work.
Show solution
$\mu B$ computed once and reused, not three separate multiplications.
Independent check by integrating the torque over the turn instead of using the energy: $W_{\rm field} = \int_{0}^{60.0^{\circ}}\mu B\sin\theta\,d\theta = \mu B(1-\cos 60.0^{\circ})$, which is $(0.0800)(0.500) = 0.0400\ \mathrm{J}$, so the external work is $-0.0400\ \mathrm{J}$, as found. Sign check: the loop moves towards lower energy, so the field does positive work and the agent does negative work.
3§09.3 — why the period does not care about the speed●●●●○
Exam level, and the first part is a bookwork derivation that turns up regularly. A particle of mass $m$ and charge $q$ moves in a plane perpendicular to a uniform field $\vec B$.
Given
mass $m$, charge magnitude $|q|$, uniform field $B$
the velocity is entirely perpendicular to the field
for the numerical part: a proton, $m_p = 1.67\times10^{-27}\ \mathrm{kg}$, $q = 1.60\times10^{-19}\ \mathrm{C}$
Find
(a) Starting from the magnetic force and Newton's second law, derive expressions for the radius and the period of the circular path, and state clearly which of the two depends on the speed.
(b) Find the magnetic field strength that would make a proton circulate at exactly $1.00\ \mathrm{MHz}$.
Hint 1/4
Part (a) is not a plug in question: the marks are for saying what each equation is and why it is allowed, not for the final formula.
Hint 2/4
Set the magnetic force $|q|vB$ equal to the force needed for circular motion, $mv^{2}/r$, then get the period from $T = 2\pi r/v$.
Hint 3/4
For part (b), $f = |q|B/(2\pi m)$ with $f = 1.00\times10^{6}\ \mathrm{Hz}$, $m_p = 1.67\times10^{-27}\ \mathrm{kg}$ and $q = 1.60\times10^{-19}\ \mathrm{C}$.
Hint 4/4
The radius depends on the speed and the period does not; and the field needed is $0.0656\ \mathrm{T}$.
Show solution
Derived from the balance, where the marks are: two cancellations.
Set up the force balance
$$F_{\rm mag} = |q|vB$$
perpendicular motion, so the sine is one and the force keeps its full size all the way round
$$F_{\rm needed} = \frac{mv^{2}}{r}$$
the force required to hold any mass on a circle of radius $r$ at speed $v$, from kinematics alone
$$|q|vB = \frac{mv^{2}}{r}$$
the magnetic force is the only force acting, so it must be exactly the one the circle needs
First cancellation: the radius
$$r = \frac{mv}{|q|B}$$
one power of $v$ has gone, leaving a radius proportional to the speed rather than to its square
Independent check of part (b) through the period rather than the frequency: $T = 1/f = 1.00\times10^{-6}\ \mathrm{s}$, and $B = 2\pi m/(|q|T) = 2\pi(1.67\times10^{-27})/[(1.60\times10^{-19})(1.00\times10^{-6})] = 6.56\times10^{-2}\ \mathrm{T}$. Plausibility: this constancy of the period is exactly what allows a particle to be given many small pushes at a fixed frequency while its orbit grows, which is how a cyclotron works.
A period formula with $v$ still in it has missed a cancellation.
4§09.6 — separating the two isotopes of chlorine●●●●○
Exam level, and it wants a separation rather than a radius, which is a step further than most homework asks. Singly charged chlorine ions, of mass numbers $35$ and $37$, leave a velocity selector at $3.00\times10^{5}\ \mathrm{m/s}$ and enter a uniform field of $0.400\ \mathrm{T}$ perpendicular to their velocity. Each follows a semicircle and strikes a detector lying along the entry line.
$m = 35u$ and $m = 37u$ with $u = 1.66\times10^{-27}\ \mathrm{kg}$
each ion strikes the detector a distance $2r$ from its entry point
Find
(a) Find the radius of each ion's path.
(b) Find the distance between the two marks on the detector.
Hint 1/4
Work out one radius properly and get the second by a ratio, then remember that the detector sits a diameter away rather than a radius away.
Hint 2/4
$r = mv/(qB')$, and after half a turn the ion lands $2r$ from where it entered, so the separation of the marks is $2(r_{37}-r_{35})$.
Hint 3/4
Here $v = 3.00\times10^{5}\ \mathrm{m/s}$, $B' = 0.400\ \mathrm{T}$, and the masses are $35(1.66\times10^{-27})$ and $37(1.66\times10^{-27})\ \mathrm{kg}$.
Hint 4/4
The radii are $0.272\ \mathrm{m}$ and $0.288\ \mathrm{m}$, and the marks are $3.11\ \mathrm{cm}$ apart.
Show solution
Second radius by ratio, or rounding swallows the difference.
the half turn brings each ion back to the entry line a diameter away, which doubles the difference and is why the detector is put there
Answer $$\boxed{\,r_{35}=0.272\ \mathrm{m},\quad r_{37}=0.288\ \mathrm{m},\quad \Delta d = 3.11\ \mathrm{cm}\,}$$
Check
Independent check on the separation without either radius: $\Delta d/d_{35} = \Delta m/m_{35} = 2/35 = 5.71\%$, and $d_{35} = 2(0.272) = 0.545\ \mathrm{m}$, giving $\Delta d = 0.0311\ \mathrm{m}$. Plausibility: three centimetres is easily resolved, which is why chlorine's two isotopes were among the first to be separated this way.
When the answer is a small difference, round once at the very end.
D · interleaved 4 questions
1§09.3 — across the gap, then into the field●●●●○
Nothing here says which part of the course this belongs to, and deciding that is most of the work. A proton starts at rest next to the positive plate of a parallel plate capacitor whose plates are held at a potential difference of $800\ \mathrm{V}$. It crosses the gap, leaves through a small hole in the negative plate, and immediately enters a region of uniform magnetic field of $0.250\ \mathrm{T}$ at right angles to its velocity.
Given
$\Delta V = 800\ \mathrm{V}$ across the plates, proton released from rest
$B = 0.250\ \mathrm{T}$, entry perpendicular to the field
$$\vec F\cdot\vec v = 0 \Rightarrow \Delta K = 0$$
the magnetic force is perpendicular to the velocity, so it turns the proton without ever pushing it along
Answer $$\boxed{\,v = 3.92\times10^{5}\ \mathrm{m/s},\quad r = 1.63\ \mathrm{cm},\quad \Delta K = 0\,}$$
Check
Independent check on the radius through the period, an equation with no $r$ in it: $T = 2\pi m_p/(qB) = 2.62\times10^{-7}\ \mathrm{s}$, and one turn at constant speed then needs $r = vT/(2\pi) = 1.63\times10^{-2}\ \mathrm{m}$, as found. Speed check: $v/c = 0.0013$.
2§09.4 — the circuit decides the force●●●○○
Two different weeks of this course are needed here and the join is not signposted. A battery of electromotive force $12.0\ \mathrm{V}$ and internal resistance $0.500\ \mathrm{\Omega}$ is connected through leads of negligible resistance to two horizontal rails, across which lies a rod of resistance $1.50\ \mathrm{\Omega}$ and length $0.400\ \mathrm{m}$. A uniform magnetic field of $0.350\ \mathrm{T}$ runs horizontally, perpendicular to the rod.
Given
battery: $12.0\ \mathrm{V}$ with internal resistance $0.500\ \mathrm{\Omega}$
rod: resistance $1.50\ \mathrm{\Omega}$, length $0.400\ \mathrm{m}$ between the rails
$B = 0.350\ \mathrm{T}$, horizontal and perpendicular to the rod
the rails and leads have negligible resistance
Find
(a) Find the current in the rod.
(b) Find the magnitude of the magnetic force on the rod.
Hint 1/4
Nothing magnetic can be worked out until the current is known, and the current is a question about the circuit only.
Hint 2/4
Circuit: $I = \mathcal{E}/(R+r)$ with the internal resistance in series. Magnetism: $F = BIL\sin\theta$.
Hint 3/4
Here $\mathcal{E} = 12.0\ \mathrm{V}$, $r = 0.500\ \mathrm{\Omega}$, $R = 1.50\ \mathrm{\Omega}$, $B = 0.350\ \mathrm{T}$, $L = 0.400\ \mathrm{m}$ and the field is perpendicular to the rod.
Hint 4/4
The current is $6.00\ \mathrm{A}$ and the force is $0.840\ \mathrm{N}$.
Show solution
Circuit first; the force law needs a current that was not given.
the field is perpendicular to the rod, so the sine is one and only the length between the rails counts
Answer $$\boxed{\,I = 6.00\ \mathrm{A},\qquad F = 0.840\ \mathrm{N}\,}$$
Check
Independent check on the circuit half: the terminal voltage is $\mathcal{E}-Ir = 12.0-(6.00)(0.500) = 9.00\ \mathrm{V}$, and across the rod $IR = (6.00)(1.50) = 9.00\ \mathrm{V}$, which agree. Plausibility: $0.84\ \mathrm{N}$ would lift about eighty five grams, so a light rod on these rails would visibly move.
3§09.2 — two fields, two forces, one electron●●●○○
Both kinds of field act on the same particle here, and the question is which of them matters. An electron moves at $2.00\times10^{6}\ \mathrm{m/s}$ through a region containing an electric field of $5.00\times10^{3}\ \mathrm{V/m}$ and a magnetic field of $1.00\times10^{-2}\ \mathrm{T}$, with the velocity perpendicular to the magnetic field.
Given
$v = 2.00\times10^{6}\ \mathrm{m/s}$, perpendicular to $\vec B$
$E = 5.00\times10^{3}\ \mathrm{V/m}$
$B = 1.00\times10^{-2}\ \mathrm{T}$
$e = 1.60\times10^{-19}\ \mathrm{C}$
Find
(a) Find the magnitude of the electric force and of the magnetic force on the electron.
(b) Find the speed at which the two forces would be equal in size.
Hint 1/4
Two independent forces, so compute each from its own law and only then compare them.
Hint 2/4
$F_E = |q|E$ regardless of the motion, and $F_B = |q|vB$ for perpendicular motion; they are equal when $E = vB$.
Hint 3/4
Here $E = 5.00\times10^{3}\ \mathrm{V/m}$, $B = 1.00\times10^{-2}\ \mathrm{T}$, $v = 2.00\times10^{6}\ \mathrm{m/s}$ and $e = 1.60\times10^{-19}\ \mathrm{C}$.
Hint 4/4
The forces are $8.00\times10^{-16}\ \mathrm{N}$ and $3.20\times10^{-15}\ \mathrm{N}$, and they would be equal at $5.00\times10^{5}\ \mathrm{m/s}$.
Show solution
Each force from its own law; both magnitudes were asked for.
The electric force, which does not care about the motion
$$F_E = eE = (1.60\times10^{-19})(5.00\times10^{3}) = 8.00\times10^{-16}\ \mathrm{N}$$
this force would be the same if the electron were standing still
The magnetic force, which cares about nothing else
Independent check of the ratio without either force: $F_B/F_E = vB/E = (2.00\times10^{6})(1.00\times10^{-2})/(5.00\times10^{3}) = 4.00$, from the given data alone and with no charge in sight. Consistency: the electron here is four times faster than the crossover speed, and the ratio is four, as it must be since $F_B$ is linear in $v$.
Which force wins is a question about speed; the crossover is $v = E/B$.
4§09.7 — building the Hall formula from the current●●●●○
This one is a derivation rather than a substitution, and it needs one result from the section on current alongside everything on this page. A flat strip of width $w$ and thickness $t$ carries a current $I$ made up of carriers of charge magnitude $q$ and number density $n$, and lies in a uniform field $B$ perpendicular to its face.
Given
strip of width $w$ and thickness $t$, the thickness measured along $\vec B$
current $I$, carriers of charge magnitude $q$ and number density $n$
the carriers drift at speed $v_d$ along the strip
from the section on current: $I = nqv_dA$ with $A$ the cross sectional area
Find
(a) Show that in the steady state the transverse electric field is $E_H = v_dB$.
(b) Hence show that the Hall voltage is $V_H = IB/(nqt)$, and explain why the width $w$ does not appear in the answer.
Hint 1/4
The steady state is the key phrase: it means the sideways motion has stopped, so the sideways forces on a carrier must add to zero.
Hint 2/4
Balance $qE_H = qv_dB$ for one carrier, then use $V_H = E_Hw$ and $I = nqv_d(wt)$ to eliminate $v_d$.
Hint 3/4
Here the cross sectional area is $A = wt$, so $v_d = I/(nqwt)$, and the voltage across the width is $V_H = E_Hw$.
Hint 4/4
Substituting gives $V_H = v_dBw = IB/(nqt)$, and the two factors of $w$ cancel.
Show solution
Balance first, drift speed last, so the width enters and leaves.
(a) The steady state balance
$$F_{\rm mag} = qv_dB \ \text{sideways}$$
every carrier moving along the strip is pushed towards one edge
$$\text{charge builds on that edge until } qE_H = qv_dB$$
the pile up creates a transverse field that opposes further pile up, and growth stops when the two forces match
$$E_H = v_dB$$
the charge cancels, so the same relation holds whether the carriers are positive or negative
(b) From field to voltage
$$V_H = E_Hw = v_dBw$$
the transverse field is uniform across the width, so the potential difference is field times distance
(b) Eliminate the drift speed, which cannot be measured directly
Independent check by dimensions. Write $1\ \mathrm{T} = 1\ \mathrm{N\,A^{-1}m^{-1}}$, so the units of $IB/(nqt)$ are $\mathrm{A}\cdot\mathrm{N\,A^{-1}m^{-1}}$ divided by $\mathrm{m^{-3}\cdot C\cdot m}$. That is $\mathrm{N\,m\,C^{-1}} = \mathrm{J/C} = \mathrm{V}$, as required. Check on the cancellation: doubling $w$ halves $v_d$ at fixed current and doubles the distance the field acts over, so the voltage genuinely should not move.
A derivation is marked on its eliminations, not on its final formula.
Mistake ledger (25 entries)
⚠ Expecting a stationary charge to feel a magnetic force
electric fields push charges whether they move or not, and the magnetic case looks like more of the same until you notice the velocity in the formula
wrong$$\vec F_{\rm mag} = q\vec B$$
right$$\vec F_{\rm mag} = q\,\vec v\times\vec B \;\Rightarrow\; \vec F_{\rm mag}=0 \text{ when } \vec v=0$$
⚠ Treating the tesla as a small unit because the numbers in problems are small
fields of a few tenths of a tesla appear in every exercise, so the tesla starts to feel like a modest everyday amount
⚠ Forgetting to reverse the hand rule for a negative charge
the right hand rule is drilled until it feels like the whole answer, and the sign of the charge sits outside the geometry where it is easy to leave behind
⚠ Leaving the number of turns out of the dipole moment
the area and the current are given as single numbers and $N$ arrives separately, so it is easy to treat it as a description of the coil rather than a factor
wrong$$\mu = IA$$
right$$\mu = NIA$$
⚠ Expecting a net force on a loop in a uniform field
a torque is a mechanical effect and it feels as though something must be pushing the loop for it to move at all
⚠ Thinking the selected speed depends on the charge or the mass
every other formula on this page contains one or the other, so a formula containing neither looks incomplete
wrong$$v = \frac{qE}{mB}$$
right$$qE = qvB \Rightarrow v = \frac{E}{B}$$
⚠ Using the selector field in the analyser stage
both stages have a magnetic field and problems often quote the two numbers close together
wrong$$r = \frac{mv}{qB_{\rm selector}}$$
right$$r = \frac{mv}{qB'_{\rm analyser}},\qquad B' \ \text{is the field in the bending region}$$
⚠ Confusing the radius with the distance to the detector
the radius is what the formula gives and the detector distance is what the ruler measures, and the problem asks for whichever one you are not thinking about
wrong$$d_{\rm detector} = r$$
right$$d_{\rm detector} = 2r \ \text{ after a half turn back to the entry line}$$
⚠ Putting the width into the Hall formula instead of the thickness
the voltage is measured across the width, so the width feels like the relevant dimension
wrong$$V_H = \frac{IB}{nqw}$$
right$$V_H = \frac{IB}{nqt},\qquad t \ \text{measured along } \vec B$$
⚠ Reading the sign of the carriers from which edge charges up rather than from the polarity
the two sound like the same statement, but both kinds of carrier arrive at the same edge and only the sign of the pile differs
wrong$$\text{electrons go to one edge, positive carriers to the other}$$
right$$\text{both go to the same edge; the polarity of } V_H \text{ is what differs}$$
⚠ Mixing up the selector field and the analyser field
both stages have a magnetic field and the two numbers are quoted a line apart in the question
wrong$$r = \frac{mv}{qB_{\rm selector}}$$
right$$r = \frac{mv}{qB'_{\rm analyser}}$$
⚠ Reporting the radius when the question asked for the distance to the detector
the formula produces a radius and the instrument measures a diameter, and the question asks for whichever one you were not thinking about
wrong$$d_{\rm detector} = r$$
right$$d_{\rm detector} = 2r$$
⚠ Using the charge of a proton for an alpha particle
the symbol $q$ is written and then the number for $e$ is typed out of habit, and the answer that results is a perfectly plausible length
wrong$$q_\alpha = e = 1.60\times10^{-19}\ \mathrm{C}$$
right$$q_\alpha = 2e = 3.20\times10^{-19}\ \mathrm{C}$$
⚠ Believing the period is independent of the field as well as of the speed
the correct observation that no speed appears in the formula slides one word sideways into a claim about the field
uniform field; here $\theta$ is between $\vec\mu$ and $\vec B$, which is ninety degrees from the angle to the plane of the loop
Energy of a dipole in a field
$$U = -\vec\mu\cdot\vec B = -\mu B\cos\theta$$
zero taken at $\theta = 90^{\circ}$; stable equilibrium at $\theta = 0$, unstable at $\theta = 180^{\circ}$
Velocity selector
$$qE = qvB \ \Rightarrow\ v = \frac{E}{B}$$
$\vec E$ and $\vec B$ perpendicular to each other and to the beam; undeflected passage is the condition imposed
Mass from the analyser radius
$$m = \frac{qB'r}{v}$$
magnetic field alone in the analyser, perpendicular to the beam; $B'$ is the analyser field, not the selector field
Hall voltage
$$E_H = v_dB,\qquad V_H = \frac{IB}{nqt}$$
steady state; $t$ is the thickness along the field, and the width cancels out
Check yourself
Close the page and write out from memory: the force law for a moving charge and how the direction is found for a negative one; the reason a magnetic field can never change a speed; the radius, the period and the frequency of the circular path, and which of them contains the speed; what happens when the entry is at an angle, and what the pitch is; the force on a straight wire, on a bent wire and on a closed loop; the dipole moment of a coil, the torque, the energy, and which orientation is stable; the selected speed in crossed fields and why it contains neither charge nor mass; and the Hall voltage together with what its polarity tells you. Then open the formula card and mark only what you missed.
State what has to be true for a charge to feel no magnetic force at all, and say roughly how many tesla the Earth's field is and how many a laboratory electromagnet is?
c-magnetic-field
Take an electron moving in a stated direction through a stated field and produce the direction of the force in two separate steps, then say without calculating what has happened to its kinetic energy?
c-force-charge
Go from a kinetic energy in electronvolts to a radius and a period, and say what changes and what does not when the speed is doubled and when the field is doubled?
c-circular-motion
Find the current needed to make a rod hover on rails, and say what the force on a semicircular wire is without integrating anything?
c-force-wire
Compute the dipole moment of a coil, get the torque at a stated angle, and say which orientation the coil settles into and how much work it takes to get there?
c-torque-loop
Set the two fields of a velocity selector for a required speed, explain why the setting does not depend on the ion, and then turn a measured radius into a mass?
c-crossed-fields
Derive the Hall voltage from the force balance and the definition of current, and say what the polarity of the measurement tells you that its size does not?
c-hall-effect
Glossary (19 terms)
magnetic fieldmanyetik alan
The field defined at a point by the force it produces on a charge moving through that point: the force is proportional to the charge, to the speed, and to the sine of the angle between the velocity and the field, and it is perpendicular to both. Measured in teslas.
teslatesla
The SI unit of magnetic field, equal to one newton per ampere metre. It is a large unit: the field at the surface of the Earth is a few times ten to the minus five of it, a fridge magnet a few hundredths, and a strong laboratory electromagnet a few whole tesla.
gaussgauss
An older unit of magnetic field still met in quoted values, equal to ten to the minus four teslas. The Earth's field is about half a gauss, which is where the habit of quoting it in gauss comes from.
magnetic polemanyetik kutup
One of the two ends of a magnet, named north or south by which way it turns when free. Unlike electric charge, a pole has never been isolated: cutting a magnet between its poles produces two complete magnets rather than two separate poles.
right hand rulesağ el kuralı
The procedure for pointing a cross product: fingers along the first vector, curled into the second, thumb along the product. Applied to the magnetic force it gives the direction for a positive charge, and the answer is reversed afterwards if the charge is negative.
Lorentz kuvveti
The total electromagnetic force on a charge, the sum of the electric part and the magnetic part. The electric part acts whether the charge moves or not and can change its speed; the magnetic part acts only on moving charge and cannot.
siklotron frekansı
The number of revolutions per second a charged particle makes in a uniform magnetic field, equal to the charge times the field divided by two pi times the mass. It depends on the particle and the field and on nothing about the motion, which is why it can be set in advance.
pitchhelis adımı
The distance a particle following a helical path advances along the field during one complete turn, equal to the component of velocity along the field multiplied by the period.
helical pathhelisel yörünge
The path of a charged particle whose velocity has a component along the field as well as across it: a circle in the plane across the field, combined with steady unaccelerated motion along it.
magnetic dipole momentmanyetik dipol momenti
The single vector that describes everything a uniform field needs to know about a current loop: the number of turns times the current times the area, pointing along the normal to the loop given by the right hand rule. Measured in ampere square metres.
couplekuvvet çifti
A pair of equal and opposite forces whose lines of action do not coincide. Their sum is zero so they push nothing anywhere, but they produce a torque that is the same about every axis, which is what a uniform field does to a current loop.
velocity selectorhız seçici
A device in which an electric field and a magnetic field are crossed so that their forces on a passing charge oppose each other. Only particles whose speed equals the ratio of the two field strengths pass undeflected, whatever their charge or mass.
mass spectrometerkütle spektrometresi
An instrument that first fixes the speed of a beam of ions and then bends it with a magnetic field, so that the radius of the bend is proportional to the mass. The position where an ion lands on the detector is a direct reading of its mass.
charge to mass ratioyük kütle oranı
The combination that governs how a particle moves in given electric and magnetic fields. Experiments of this kind measure it directly, and neither the charge nor the mass separately, which is why it was known for the electron before either of them was.
isotopeizotop
One of two or more forms of the same chemical element differing in mass. Since they are chemically alike and differ only in mass, a magnetic instrument that sorts by mass is the natural way to tell them apart.
atomic mass unitatomik kütle birimi
A unit of mass convenient for atoms and ions, equal to about one point six six times ten to the minus twenty seven kilograms, so that a mass in these units is close to the mass number of the nucleus.
Hall effectHall olayı
The appearance of a voltage across a current carrying strip placed in a magnetic field, caused by carriers being pushed sideways until the charge they pile up pushes back just as hard.
Hall voltageHall gerilimi
The steady potential difference measured across the width of such a strip, equal to the current times the field divided by the carrier density, the carrier charge and the thickness along the field. Its polarity, not its size, reveals the sign of the carriers.
taşıyıcı yoğunluğu
The number of mobile charge carriers per cubic metre in a conductor. It is enormous in a metal and smaller by five or six orders of magnitude in a semiconductor, which is why practical magnetic field probes are made of the latter.
What comes next
§10 · Review and Consolidation II: capacitance, current, DC circuits, and magnetism
Every field on this page arrived as a given: a number attached to a magnet or to a pair of pole pieces, with no account of where it came from. The next stop is a review that puts capacitance, current, circuits and this material side by side and asks you to tell, from the wording of a question alone, which of the four it is.
Sources
D. Giancoli, Physics for Scientists and Engineers with Modern Physics, 5th edition The set textbook. Its treatment of magnetism covers the same ground as this section, and its end of chapter problems run a level harder than the ones here, which is the right next step once this set feels comfortable.
Course syllabus, week 9 line The scope of this section comes from the week line, which reads Magnetism, and from nowhere else. The line carries no chapter number, so no chapter or section number is quoted anywhere on this page, and topics that belong to later week lines have been left out rather than previewed.
Course syllabus, assessment weights and catalogue description The exam note on the card uses only the published weights: two midterms at twenty per cent each, quizzes at ten, homework at five, the final at twenty five and laboratory work at twenty. The catalogue description lists magnetic field among the topics of the course, which is what places this material inside it.
SI values of the physical constants used Elementary charge, electron mass, proton mass, the unified atomic mass unit and the acceleration of gravity, each quoted to three significant figures as listed in the conventions block. The carrier density of copper and the Earth's field strength are quoted as the order of magnitude figures they are, and both are stated inside the problems that use them rather than assumed.