7 concepts18 worked examples30 exercises4 exam-level7 figures
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14Maxwell’s Equations and Electromagnetic Waves
Put a compass beside the wire that feeds a capacitor and the needle turns. Now slide it into the gap between the two plates, where the wire is cut and nothing at all is flowing, and the needle turns by the same amount. Whatever is deflecting it in there is not a current of charge, because there is no charge in the gap to move.
By the end of this section you can take one number about a beam of light or radio — a field , a power, a wavelength — and produce every other number for it: the partner field, the frequency, the , and the force it exerts on a surface.
In 60 seconds
A changing electric field makes a magnetic field just as a current does; put that into the four field laws and they predict a self sustaining disturbance travelling at $1/\sqrt{\varepsilon_0\mu_0}$, which is light.
a beam lands on a surface and a force or a pressure is wanted
Three most common mistakes
Treating the as charge crossing the capacitor gap. Nothing crosses; the term is $\varepsilon_0 d\Phi_E/dt$ and it exists in perfect vacuum.
Using $I/c$ for a mirror. A mirror sends the light back, so the momentum change is doubled and the pressure is $2I/c$.
Mixing peak and rms. $I=\tfrac12\varepsilon_0cE_0^{2}$ carries the one half because $E_0$ is a peak; with $E_{\rm rms}$ the half is already spent and $I=\varepsilon_0cE_{\rm rms}^{2}$.
The syllabus puts 25 per cent of the course on the final and this is the last block of new material before it, so everything here is final material by construction. Nothing on this page tells you how many questions will come from it.
How much time do you have?
10 minutes
You leave able to do the single most common conversion chain in this material: a beam is described by one quantity and you produce the field amplitudes, the intensity and the force it exerts.
In 60 seconds · Formula card · Why a kink in the field has to run, and how fast · How much energy the beam carries · Mistake ledger
45 minutes
You add the two things that make this a physics section rather than a table of formulas: why Ampère's law had to be repaired, and why the repair forces a wave at one fixed speed. You also get the sine wave bookkeeping that most numerical questions are really testing.
In 60 seconds · The current that is not there: completing Ampère's law · The four equations, side by side · Why a kink in the field has to run, and how fast · The sinusoidal plane wave, read off the page · How much energy the beam carries · Scaffolding comes off · Formula card
full reading
Everything above plus the spectrum, radiation pressure, the worked examples that show where the arithmetic goes wrong, and the practice set with the interleaved questions that force you to decide which law a problem is about before you use it.
all seven concepts in order · Scaffolding comes off · Full exam style question · Practice A to D · Check yourself
By the end of this section
Compute the displacement current in a region of changing electric field and use the completed Ampère's law to find the magnetic field inside and outside a charging parallel plate capacitor.
State each of the four field equations, say in one sentence what physical fact it records, and identify which one a given situation is asking about.
Derive the speed of an from the two circulation laws and use $E=cB$ to convert between the electric and magnetic amplitudes of the same wave.
Extract wavelength, frequency, direction of travel and both field amplitudes from a sinusoidal plane wave written as a formula, and write the formula given the physical description.
Place a wave in the electromagnetic spectrum from its wavelength or frequency, convert between the two, and compute travel times and the wavelength change on entering a material.
Relate intensity, field amplitudes and radiated power for a beam and for an , keeping peak and rms values apart.
Calculate the pressure and the force a beam exerts on an absorbing or a reflecting surface, and judge whether that force matters against the other forces present.
Syllabus coverage
Maxwell’s Equations and Electromagnetic Waves
The failure of Ampère's law at a charging capacitor and its repair by the displacement current; Gauss's law for magnetism and the absence of magnetic charge; the four equations gathered as one set; how a changing electric field and a changing magnetic field regenerate each other, giving a whose speed is fixed by $\varepsilon_0$ and $\mu_0$; the sinusoidal plane wave with its wavelength, frequency, and the locked ratio $E_0=cB_0$; light as one member of a spectrum that runs from radio to gamma rays, and the speed of a wave inside a material; the , intensity and the inverse square fall off of an isotropic source; the momentum of a beam and the radiation pressure on absorbing and reflecting surfaces.
covered
wave equation in derivative form
Writing the two circulation laws as partial derivatives, so that any shape $f(x-ct)$, not only a sine, is shown to travel rigidly at speed $c$.
Kept as one short remark inside the speed derivation and never used in a question. It is here because students who have seen the derivative form ask why the rectangle argument gives the same answer, and because it makes clear that the sine is a convenience and not a requirement. You are not responsible for it.
off_syllabus
Recall first
Gauss's law for the electric field
$\oint\vec{E}\cdot d\vec{A}=Q_{\rm encl}/\varepsilon_0$. The outward electric flux through a closed surface counts the charge inside it and nothing else.
It is the first of the four equations and it is the reason the field between capacitor plates is $\sigma/\varepsilon_0$, which is what the displacement current is computed from.
Ampère's law in its original form
$\oint\vec{B}\cdot d\vec{\ell}=\mu_0 I_{\rm encl}$ for steady currents, with the loop and the enclosed current tied by the right hand rule.
The whole first concept is about the two words steady currents, which turn out to be a restriction and not a description.
Faraday's law
$\oint\vec{E}\cdot d\vec{\ell}=-d\Phi_B/dt$. A changing magnetic flux drives an electric field around the loop, in the direction that opposes the change.
It is one half of the feedback loop that makes a wave; the completed Ampère's law is the other half.
Parallel plate capacitor field and charge
$E=\sigma/\varepsilon_0=Q/(\varepsilon_0 A)$ between the plates, $Q=CV$, and $C=\varepsilon_0A/d$ for an empty gap.
Every displacement current example is a capacitor being charged, and the chain runs from the charging current to $dE/dt$ through these.
Dielectric constant
Filling a gap with a material of dielectric constant $K$ multiplies the capacitance by $K$ and divides the field inside by the same factor.
The speed of a wave inside a material is $c/\sqrt{K}$, and the $K$ is the same number met with capacitors.
Magnetic field of a long straight wire
$B=\mu_0 I/(2\pi r)$ at distance $r$ from a long straight wire, circling the wire by the right hand rule.
It is the answer the completed Ampère's law must reproduce outside a charging capacitor, which is how we check the new term is the right size.
Try it yourself first (3 questions)
1§14.1 — where Ampère's law is allowed to be used●●○○○
A steady $4.0\ \mathrm{A}$ runs along a long straight wire. You draw a circle of radius $2.0\ \mathrm{cm}$ around the wire and use Ampère's law on it. Then you keep the same circle but stretch the surface it bounds into a long bag that bulges sideways, so that the wire still pierces the bag once.
Given
a long straight wire carrying a steady $I=4.0\ \mathrm{A}$
an Amperian circle of radius $r=2.0\ \mathrm{cm}$ centred on the wire
two different surfaces bounded by that same circle: a flat disc, and a bag that the wire still pierces once
Find
(a) What does $\oint\vec{B}\cdot d\vec{\ell}$ come to for the bag, compared with the flat disc?
Hint 1/4
The question is about which part of the law changed when you swapped the surface. Look at the two sides separately: one is an integral around a rim, the other counts current through a surface.
Hint 2/4
Ampère's law is $\oint\vec{B}\cdot d\vec{\ell}=\mu_0 I_{\rm encl}$. The left side belongs to the rim alone; the right side belongs to any surface whose rim it is.
Hint 3/4
The rim is the same circle of radius $2.0\ \mathrm{cm}$ in both cases. The wire carries the same steady $4.0\ \mathrm{A}$ and pierces the flat disc once and the bag once.
Hint 4/4
Both surfaces catch the same $4.0\ \mathrm{A}$, so both give $\mu_0 I=5.0\times10^{-6}\ \mathrm{T\,m}$, and the answer is that nothing changed.
Show solution
There is nothing to compute here except the right side, because the left side never mentions a surface at all. That asymmetry is the point of the question.
Separate what belongs to the rim from what belongs to the surface
$$\oint\vec{B}\cdot d\vec{\ell}\ \text{depends only on the circle}$$
the integral runs along the rim; no surface enters its definition, so swapping surfaces cannot change it
$$I_{\rm encl}\ \text{depends on the chosen surface}$$
the enclosed current counts what pierces the surface, so this is the side that could in principle differ
the two sides now agree for both choices, which is what a law is required to do
Answer $$\boxed{\ \text{identical: } 5.03\times10^{-6}\ \mathrm{T\,m}\ \text{for both surfaces}\ }$$
Check
Check it the other way round, from the field: $B=\mu_0I/(2\pi r)$ gives $4.0\times10^{-5}\ \mathrm{T}$ at $r=2.0\ \mathrm{cm}$, and multiplying by the circumference $2\pi r=0.126\ \mathrm{m}$ returns $5.03\times10^{-6}\ \mathrm{T\,m}$, computed without ever mentioning a surface.
Keep the phrase surface independence. When it breaks, in the very next concept, the law has to change rather than the answer.
2§14.3 — what a field in empty space can do●●○○○
Two students argue about a region of empty space, far from any charge or wire, in which an electric field is measured and found to be changing with time. One says nothing can happen there because there is no charge and no current to act as a source.
Given
a region of vacuum with no charge and no conduction current inside it
an electric field in that region whose magnitude is changing with time
Find
(a) True or false: with no charge and no current present, no magnetic field can exist in that region. Give the reason for your answer.
Hint 1/4
You are being asked whether the list of things that can produce a magnetic field is complete. Do not compute anything; ask what is on the list.
Hint 2/4
In the form you learned first, $\oint\vec{B}\cdot d\vec{\ell}=\mu_0I_{\rm encl}$, the only entry on the list is real current.
Hint 3/4
The region has $I_{\rm encl}=0$ by construction but $dE/dt\neq 0$, which is a quantity that does not appear anywhere in that form of the law.
Hint 4/4
The statement is false: a changing electric field is a second entry on the list, and the rest of this section is about what it implies.
Show solution
The quickest honest route is not a calculation but a test of the law against a case it was never checked on, which is how the term was found in the first place.
the deflected compass in the gap is an observation, and one observation is enough to make the claim false
Answer $$\boxed{\ \text{False}\ }$$
Check
The claim also fails a consistency test that needs no experiment: light travels through vacuum, and it carries a magnetic field through regions with no charge in them at all.
Getting this wrong before reading is expected and costs nothing. It is the belief the first concept is built to break.
3§14.5 — what changes when light enters glass●●●○○
A green laser beam of wavelength $532\ \mathrm{nm}$ in air enters a block of glass in which its speed drops to two thirds of the vacuum value. A student is asked what happens to its colour and reports that the frequency drops, since the wave is going slower.
Given
vacuum wavelength $\lambda=532\ \mathrm{nm}$
speed inside the glass $v=\tfrac{2}{3}c$
$\lambda f=v$ holds inside any material, with the local speed
Find
(a) Which quantity is unchanged when the beam crosses into the glass?
Hint 1/4
Three quantities are tied by one equation, so fixing which one is set from outside decides the other two. Ask what the glass can and cannot influence.
Hint 2/4
$\lambda f=v$ holds in any medium with the local speed $v$. The frequency counts crests per second arriving at the boundary.
Hint 3/4
Crests arrive at the boundary at $f$ per second and none pile up or disappear there, so $f$ leaves at the same rate. With $v=\tfrac{2}{3}c$, the wavelength inside is $\tfrac{2}{3}(532\ \mathrm{nm})=355\ \mathrm{nm}$.
Hint 4/4
The frequency is unchanged, the speed and wavelength both drop by a factor of $\tfrac{2}{3}$, and the colour you see is set by the frequency.
Show solution
Start from the quantity the boundary cannot touch. Choosing to start from the wavelength instead would leave you with two unknowns and no way to pick between them.
Get the frequency outside, where the speed is known
A shortcut that avoids the frequency entirely: the wavelength must scale exactly as the speed does, so $\tfrac{2}{3}(532)=355\ \mathrm{nm}$, reached without computing $f$ at all.
Whenever a wave crosses into a new material, write the frequency down first and treat it as fixed. Every other number follows from it.
Notation
symbol
reads as
means
watch out
$\Phi_E$
electric flux
the electric field integrated over a surface, $\int\vec{E}\cdot d\vec{A}$, in $\mathrm{N\,m^{2}/C}$
for a capacitor it is $EA$ with $A$ the plate area, not the area of whatever loop you drew
$I_D$
displacement current
$\varepsilon_0\,d\Phi_E/dt$, measured in amperes
not a flow of charge; the name is historical and describes nothing that moves
$c$
the
$1/\sqrt{\varepsilon_0\mu_0}=3.00\times10^{8}\ \mathrm{m/s}$, the same for every frequency
inside matter the wave is slower and this symbol still means the vacuum value
$k$
wave number
$2\pi/\lambda$, in radians per metre
not the Coulomb constant, which does not appear anywhere in this section
$\omega$
$2\pi f$, in radians per second
$\omega/k$ is the wave speed; dividing $\omega$ by $\lambda$ instead is a common slip
$E_0,\ B_0$
field amplitudes
the peak values of the two fields, locked together by $E_0=cB_0$
$B_0$ is smaller than $E_0$ by a factor of three hundred million in SI units, which is why the magnetic part of a light wave is so hard to detect
$\vec{S}$
the Poynting vector
$(1/\mu_0)\vec{E}\times\vec{B}$, the instantaneous energy flow per unit area, in $\mathrm{W/m^{2}}$
it oscillates twice per cycle; the measurable quantity is its time average
$I$
intensity
the time averaged magnitude of $\vec{S}$, in $\mathrm{W/m^{2}}$
the same letter is used for current in the first concept; the units tell you which one is meant
$P_{\rm rad}$
radiation pressure
force per unit area delivered by a beam, in pascals
written with the subscript throughout so it is never read as power, which is also $P$
$K$
dielectric constant
the pure number by which a material multiplies the capacitance of a gap and divides the square of the wave speed
the speed in the material is $c/\sqrt{K}$, with a square root that is easy to drop
Conventions used here
Which constants this section uses, and to how many digits
$c=3.00\times10^{8}\ \mathrm{m/s}$, $\varepsilon_0=8.85\times10^{-12}\ \mathrm{C^{2}/(N\cdot m^{2})}$, $\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m/A}$, and $g=9.80\ \mathrm{m/s^{2}}$ on the rare occasion a weight is compared with a light force. Answers are quoted to three significant figures. Putting the rounded $\varepsilon_0$ and $\mu_0$ into $1/\sqrt{\varepsilon_0\mu_0}$ returns $2.998\times10^{8}\ \mathrm{m/s}$, and that is why $3.00\times10^{8}$ is written everywhere else.
Two sections of this course using two values of the same constant is the most boring possible source of a wrong answer.
How the axes are set up for every wave picture here
The wave travels along $+x$, the electric field lies along $\pm y$ and the magnetic field along $\pm z$, and the three form a right handed set in that order. The test that survives every rotation of the picture is the cross product: $\vec{E}\times\vec{B}$ points the way the energy goes. If a question rotates the wave, rotate that sentence with it rather than memorising which letter goes up.
Half of the sign errors in this material come from a picture drawn with the axes in a different order than the formula assumed.
What a subscript zero and a subscript rms mean on this page
$E_0$ and $B_0$ are amplitudes, the largest values the field reaches. $E_{\rm rms}=E_0/\sqrt{2}$ and $B_{\rm rms}=B_0/\sqrt{2}$ are the root mean square values. Intensity is always a time average, so it is written without a bar from here on and the averaging has already been done inside whichever formula you use.
The factor of one half in the intensity formula is exactly the average of a squared sine, so writing rms values with the half still in place double counts it.
Which current is which in the completed Ampère's law
$I_{\rm encl}$ is real charge crossing the surface you chose. $I_D=\varepsilon_0\,d\Phi_E/dt$ is not charge and nothing crosses anything; it is a rate of change of electric flux written in amperes so that the two terms can be added. In the arithmetic they behave identically, which is the whole point, but in a sentence they are never interchangeable.
A question that asks what physically passes between capacitor plates has one correct answer, and it is nothing.
How a direction is reported in an answer here
Magnitudes come from the formulas and directions come from a stated rule — the right hand rule for $\vec{E}\times\vec{B}$, Lenz's law for an induced field. A minus sign written in a general law such as $\mathcal{E}=-d\Phi_B/dt$ is a reminder of the rule, not a number to be carried through the algebra. Every numerical answer below is a magnitude plus a direction in words.
Carrying a minus sign through four lines and then reading a direction off it is how a right answer turns into a wrong one at the last step.
What ideal means for the surfaces and sources in this section
A capacitor is two parallel plates whose gap is small enough that the field is uniform inside and negligible outside. A black surface absorbs everything that lands on it, a mirror returns everything straight back, and beams arrive along the normal unless a question says otherwise. A source called isotropic sends equal power in every direction and sits in empty space with nothing to absorb or reflect.
Each of these idealisations removes one factor from a formula, and knowing which one was removed tells you what a real measurement would differ by.
14.1The current that is not there: completing Ampère's law
A changing electric field drives a magnetic field exactly as a real current does, which is what repairs the law.
You have used Ampère's law all term without asking which surface you counted the current through, because for a steady current it never mattered. Charge a capacitor and it starts to matter.
Solvable with what we have
Find $B$ at any distance from a long straight wire.
Find $B$ inside a solenoid or a toroid.
Find $B$ inside a thick wire of uniform current density.
Not solvable yet
Say what $B$ is between the plates of a charging capacitor.
Get one answer for the feeding wire whichever surface you use.
Explain how light carries a magnetic field through empty space.
Draw one circle around the wire feeding the capacitor and apply $\oint\vec{B}\cdot d\vec{\ell}=\mu_0 I_{\rm encl}$ twice. Span it with a flat disc: the wire pierces it, so the circulation is $\mu_0 I$. Now span the same circle with a bag that bulges around the plate and slices the gap. No charge crosses that bag, so the circulation is zero.
Why it fails
The left side never mentions a surface, so it cannot have two values, and the right side just produced two. The fault is in the law. Ampère's law came from steady currents, where charge never piles up and every surface with the same rim catches the same current. A capacitor plate is exactly where charge piles up.
RuleRule 14.1: Ampère's law with the displacement current
Conditions
the loop is any closed curve and the surface is any surface with that curve as its rim
$\Phi_E$ is the electric flux through that same surface, with the normal fixed by the right hand rule on the loop
$I_{\rm encl}$ counts real charge crossing the surface, and is zero for a surface drawn through vacuum
the two terms are added, so a region can have both at once, as inside a leaky capacitor
Walk once around a closed loop adding up the magnetic field along your path. The total is the permeability times two things added: the real current threading the loop, and the rate at which the electric flux through it grows, multiplied by the permittivity so that it too comes out in amperes. Either term alone will make a magnetic field.
Why the extra term is exactly the missing current, and not merely something of the right kind
The repair would be worthless if it patched the contradiction only roughly. It closes it exactly, and the two lines below are worth doing once by hand.
Take the bag surface, the one that slices through the gap. Between the plates the field is $E=\sigma/\varepsilon_0=Q/(\varepsilon_0 A)$, where $Q$ is the charge on one plate and $A$ its area.
The electric flux through the part of the bag inside the gap is $\Phi_E=EA=Q/\varepsilon_0$. The plate area cancels, which is the first sign that this is going somewhere clean.
But $dQ/dt$ is precisely the current arriving along the wire, since that current is what puts charge on the plate. So $I_D=I$.
Now both surfaces give the same answer. The flat disc catches $I$ of real current and no changing flux worth speaking of; the bag catches no real current and a displacement current equal to $I$. The contradiction is gone, and it is gone identically, not approximately.
Off the syllabus but worth one line: written with derivatives instead of integrals the new term is what turns the field equations into a wave equation, which is the subject of the third concept below.
One rim, two surfaces, two different answers from the old law. The $\textcolor{#8250df}{\text{real current}}$ pierces the shaded disc and misses the dashed bag entirely, while inside the gap the $\textcolor{#1f6feb}{\text{electric field}}$ is growing. The new term is what the bag catches instead.
Looks like this, but is not
The displacement current is a faint current of charge crossing the gap, and calling it a current is just being careful about how small it is.
Evacuate the gap completely and the effect is unchanged, because nothing in the formula refers to a charge carrier. The ingredients of $\varepsilon_0\,d\Phi_E/dt$ are a field and a clock. What earns it the word current is the slot it fills in the equation, not anything moving.
$r$
which region
$B$
$1.00\ \mathrm{cm}$
inside the plates
$4.44\ \mathrm{\mu T}$
$2.00\ \mathrm{cm}$
inside the plates
$8.89\ \mathrm{\mu T}$
$3.00\ \mathrm{cm}$
at the plate edge
$13.3\ \mathrm{\mu T}$
$5.00\ \mathrm{cm}$
beyond the plates
$8.00\ \mathrm{\mu T}$
$10.0\ \mathrm{cm}$
beyond the plates
$4.00\ \mathrm{\mu T}$
Rising in proportion to $r$ and then falling as $1/r$, with the peak at the rim: the same shape as the field inside and outside a thick current carrying wire of the same radius. That is not a coincidence, it is the statement that the displacement current is spread uniformly over the plate area exactly as a uniform current density would be.
The field a compass would feel between two charging plates
A parallel plate capacitor has circular plates of radius $3.00\ \mathrm{cm}$ separated by $1.00\ \mathrm{mm}$ of vacuum. It is being charged by a steady current of $2.00\ \mathrm{A}$ delivered along the wires. Find the rate at which the electric field between the plates is growing, and the magnetic field at $r=1.00\ \mathrm{cm}$ from the axis inside the gap and at $r=5.00\ \mathrm{cm}$ outside the plates.
Given
circular plates, radius $R=3.00\times10^{-2}\ \mathrm{m}$, so $A=\pi R^{2}$
vacuum between the plates, gap $1.00\ \mathrm{mm}$ (needed for nothing here, and that is worth noticing)
$dE/dt$ in the gap, and $B$ at $r=1.00\ \mathrm{cm}$ and at $r=5.00\ \mathrm{cm}$ from the axis
Solution
Everything here comes from choosing the Amperian loop to be a circle centred on the axis, because that is the one curve on which $B$ has a single magnitude and runs along the path. Choosing a rectangle would be legal and useless: the left side would then contain an unknown field varying along the path.
Turn the charging current into a rate of change of field
The inside answer can be reached without $dE/dt$ at all. Treat the displacement current as spread uniformly over the plate: the fraction enclosed by a loop of radius $r$ is $r^{2}/R^{2}$, so $B=\mu_0 I r/(2\pi R^{2})=(4\pi\times10^{-7})(2.00)(0.0100)/[2\pi(9.00\times10^{-4})]=4.44\times10^{-6}\ \mathrm{T}$. Different route, different intermediate numbers, same answer.
The gap width was given and never used. That is deliberate: the field between the plates is fixed by the charge per unit area, and the separation only decides what voltage that field corresponds to.
The opening compass is now explained, and the size is explained too. Microtesla fields are what a charging capacitor makes, which is why nobody noticed this term for the first fifty years of electromagnetism.
How fast the voltage climbs when a milliamp charges a microfarad
A $1.00\ \mathrm{\mu F}$ capacitor, initially uncharged, is charged by a constant current of $5.00\ \mathrm{mA}$. Find the displacement current between its plates, the rate at which the voltage across it rises, and the time it takes to reach $12.0\ \mathrm{V}$.
charging current $I=5.00\times10^{-3}\ \mathrm{A}$, constant
target voltage $12.0\ \mathrm{V}$
Find
$I_D$ between the plates, $dV/dt$, and the time to reach $12.0\ \mathrm{V}$
Solution
The displacement current is not computed from the geometry here, because none is given. It is read off the identity from the box, and that shortcut is available whenever the region between the plates is where the whole flux lives.
Name the displacement current without touching the geometry
Check by energy bookkeeping instead: the final charge is $Q=CV=1.20\times10^{-5}\ \mathrm{C}$, and at a constant $5.00\ \mathrm{mA}$ that takes $Q/I=1.20\times10^{-5}/5.00\times10^{-3}=2.40\times10^{-3}\ \mathrm{s}$, arrived at through charge rather than through voltage.
One identity, one derivative of $Q=CV$, one division. No geometry entered at any point.
Whenever a question hands you a capacitor and a charging current and asks for the displacement current, the answer is the charging current. The work in these problems is always somewhere else.
⚠ Putting the loop area into the flux when the loop is outside the plates
inside the plates the enclosed flux really is $E\pi r^{2}$, and the habit carries on past the rim, where there is no field left to enclose
A parallel plate capacitor with square plates is being charged by a steady current of $3.00\ \mathrm{A}$ flowing in the wires. A student places an imaginary flat surface halfway between the plates, parallel to them, large enough to cover the whole gap.
Given
charging current in the wires $I=3.00\ \mathrm{A}$, steady
the surface sits in the vacuum gap, parallel to the plates and covering it entirely
Find
(a) What is the displacement current through that surface?
Hint 1/4
You are being asked for a number in amperes, and one of the results in the box hands it to you without any geometry. Find that result before doing anything else.
Hint 2/4
$I_D=\varepsilon_0\,d\Phi_E/dt$, and for a surface that spans the whole gap of a parallel plate capacitor this equals $dQ/dt$.
Hint 3/4
The plates are being charged at $dQ/dt=3.00\ \mathrm{A}$, and the surface covers the entire gap, so the flux through it is the entire flux.
Hint 4/4
The displacement current is the charging current itself: $3.00\ \mathrm{A}$.
Show solution
Push the flux through symbolically before any number goes in, because the plate area and the permittivity both cancel; substituting first would leave you hunting for plate dimensions the problem never gives.
the permittivity cancels too, leaving the charging current unchanged
Answer $$\boxed{\ I_D=3.00\ \mathrm{A}\ }$$
Check
Consistency check from the other side: if $I_D$ were anything other than $3.00\ \mathrm{A}$, the flat disc and the gap surface would give different values of $\oint\vec{B}\cdot d\vec{\ell}$ for the same rim, which is impossible.
⚠ Calling the displacement current a flow of charge across the gap
the word current is doing the damage; it was chosen in the eighteen sixties for a mechanical picture of the vacuum that nobody holds any more
Four statements, two about sources and two about change, which between them settle every field question in this course.
With the repair in place nothing is left dangling, and the whole term fits on four lines. Writing them together is not ceremony: the gap in the pattern is what produces the rest of this section.
TheoremRule 14.2: Maxwell's equations in free space
Conditions
each closed surface integral runs over a closed surface, each closed line integral around a closed loop with a surface spanning it
$Q_{\rm encl}$ and $I_{\rm encl}$ mean net charge inside the surface and net current through it
written for vacuum; inside a material $\varepsilon_0\to K\varepsilon_0$ and $\mu_0\to \mu$, which is the only change that matters at this level
the field directions follow the right hand rule fixed in the conventions above
Electric field lines start and end on charge. Magnetic field lines never start or end anywhere, so as much magnetic field leaves a closed surface as enters it. A magnetic field that is changing drives an electric field around a loop. A current drives a magnetic field around a loop, and so does an electric field that is changing. Read down the list, the first two say what makes fields, and the last two say that a field which is changing is itself a maker of the other kind.
Reading the missing symmetry, and what the second equation is really claiming
Line up the two circulation laws. Faraday has a changing $\Phi_B$ on the right and no current, because there is no such thing as magnetic current. Ampère has a changing $\Phi_E$ and a current, because electric charge exists and moves.
That is the entire asymmetry of the set, and it traces back to one experimental fact recorded by the second equation: nobody has ever found an isolated magnetic pole.
Cut a bar magnet in half and you do not get a north piece and a south piece; you get two magnets. Cut again and again and you keep getting magnets, down to the individual atom.
Written as a law, that says magnetic field lines close on themselves, so any closed surface has as much flux entering as leaving and the net is exactly zero: $\oint\vec{B}\cdot d\vec{A}=0$.
If a magnetic charge were ever found, the second equation would gain a $Q_{\rm m}$ on the right and the third would gain a magnetic current term, and the set would become symmetric. Searches have been made and nothing has turned up, so the asymmetry stands.
Everything in the rest of this section comes from the last two lines alone, applied to a region with $Q_{\rm encl}=0$ and $I_{\rm encl}=0$. Nothing is left in there except the two flux terms, feeding each other.
The second equation, drawn. The $\textcolor{#d1690a}{\text{magnetic field}}$ loops close on themselves, running from south to north inside the magnet, so a closed surface around one pole has exactly as much flux entering as leaving.
Looks like this, but is not
The north pole of a magnet is a source of magnetic field the way a positive charge is a source of electric field, so a small closed surface around it must have outward flux.
Field lines do stream outward from the north end, and if you only look outside the magnet the analogy holds up. The surface, though, has to close, and to close it must pass through the body of the magnet, where the field runs the other way, from south to north. The returning flux inside cancels the escaping flux outside exactly. A positive charge has no interior return path, which is the whole difference between the first equation and the second.
Flux out of the side of a can placed over the end of a solenoid
A closed cylindrical can, of radius large enough to swallow the solenoid, is slid over the end of a long solenoid so that one flat face sits deep inside the winding and the other face sits well outside, far enough away that the field there is negligible. Inside the solenoid the field is uniform at $0.280\ \mathrm{T}$ along the axis, and the solenoid cross section is $1.20\times10^{-3}\ \mathrm{m^{2}}$. Find the magnetic flux through the curved side wall of the can.
Given
$B_{\rm in}=0.280\ \mathrm{T}$, uniform and along the axis, over the solenoid cross section $A=1.20\times10^{-3}\ \mathrm{m^{2}}$
the inner flat face lies deep inside the winding, the outer flat face lies where $B\approx 0$
outward normals everywhere, as for any closed surface
Find
the magnetic flux through the curved side wall
Solution
The second equation is the whole method. Trying to integrate the fringing field over the curved wall directly would need the field pattern at the mouth of a solenoid, which nobody hands you and which is not needed.
Independent sanity check on the sign, without arithmetic: the field lines that go up the axis inside the solenoid have to get back to the other end somehow, and the only way out of the can is the side wall. Outward is the only possible sign, and the size has to be the whole of what came in, since nothing else escapes.
No integral was performed. The law converted an impossible surface integral into a subtraction.
This is what the second equation is for in problems: it turns the flux through an awkward surface into the flux through the easy surfaces that complete it.
The same sphere around a charge and around a magnet
A sphere of radius $0.200\ \mathrm{m}$ is drawn in empty space. First a point charge of $+3.00\ \mathrm{nC}$ is placed at its centre; find the electric flux out of the sphere and the field at its surface. Then the charge is removed and a small bar magnet, strong enough to be felt across the room, is placed at the centre instead; find the magnetic flux out of the sphere.
Given
sphere radius $r=0.200\ \mathrm{m}$
case one: a point charge $q=+3.00\times10^{-9}\ \mathrm{C}$ at the centre
case two: a bar magnet entirely inside the sphere, of unstated strength
$\Phi_E$ and $E$ at the surface in case one; $\Phi_B$ in case two
Solution
Do the electric case with the flux law rather than by integrating $\vec{E}\cdot d\vec{A}$ over the sphere, then use the integral as the independent check. The reverse order also works but wastes the cheap step.
the field does depend on the radius, which is the difference between a flux and a field and is worth keeping straight
Magnetic flux from the magnet inside
$$\Phi_B=0\ \mathrm{Wb}$$
the second equation gives zero for every closed surface, so the strength of the magnet, its orientation and the radius of the sphere are all irrelevant
Recompute the electric flux the long way, as a surface integral: $E$ is the same everywhere on the sphere and perpendicular to it, so $\Phi_E=E(4\pi r^{2})=(674)(4\pi)(0.0400)=339\ \mathrm{N\,m^{2}/C}$, matching the first line by a route that used the radius twice and cancelled it.
The contrast is the point. Give the first law a source inside and it returns a number that grows with the source; give the second law any source you like and it returns zero, because there is no magnetic source to give it.
⚠ Applying the closed surface law to an open loop
both integrals are written with a ring on the sign, and the difference between $d\vec{A}$ over a closed surface and $d\vec{\ell}$ around a curve is one symbol
wrong$$\oint\vec{B}\cdot d\vec{A}=0\ \Rightarrow\ \Phi_B\ \text{through a loop is zero}$$
right$$\oint\vec{B}\cdot d\vec{A}=0\ \text{only for a closed surface}$$
Checkpoint
§14.2 — picking the right equation●●○○○
A closed loop of wire sits in a region where the magnetic field is being turned up steadily by a nearby electromagnet. No charge is placed anywhere near the loop, and no wire carries current through it. A current appears in the loop.
Given
a closed loop with no source of charge nearby
a magnetic field through the loop that grows steadily with time
no conduction current threading the loop
Find
(a) Which of the four equations accounts for the current that appears in the loop?
Hint 1/4
Sort the four by their left sides first: two are about closed surfaces, two about closed loops. A wire loop with a current in it is a loop, so half the list is gone at once.
Hint 2/4
Of the two loop laws, one has $-d\Phi_B/dt$ on the right and the other has $\mu_0I_{\rm encl}+\mu_0\varepsilon_0\,d\Phi_E/dt$.
Hint 3/4
Here $\Phi_B$ through the loop is growing, there is no conduction current threading it, and no charge has been introduced, so $\Phi_E$ is not changing either.
Hint 4/4
Only the law with $-d\Phi_B/dt$ on the right has a non zero side, so that is the one doing the work.
Show solution
Sort the four equations by their left sides before reading a single right side: the geometry here is a loop, and that alone removes half the list. Testing all four right sides in turn gets there too, at twice the work.
Check the direction independently: Lenz's law says the induced current opposes the growth, so it must circulate to make its own flux point against the electromagnet's. That is a statement about the same equation and it is consistent with a current appearing at all.
⚠ Swapping which flux belongs to which circulation law
both laws now contain a changing flux, so the pair is easy to cross over under time pressure
14.3Why a kink in the field has to run, and how fast
A changing E makes a B whose change makes an E, and the pair survives only by travelling at one particular speed.
Strip the last two equations of every charge and every wire and something is still left on both right hand sides. Two fields, each one made by the other one changing, and no source anywhere.
TheoremTheorem 14.3: an electromagnetic wave, its speed, and the ratio of its fields
Conditions
vacuum, with no charge and no conduction current in the region
a plane wave: the fields are uniform across any plane perpendicular to the direction of travel
the derivation below assumes a flat front, which is what a wave looks like far from whatever made it
for a material of dielectric constant $K$ and negligible magnetism, replace $\varepsilon_0$ by $K\varepsilon_0$ throughout
There is only one speed at which a self supporting pair of fields can move, and it is fixed by two constants that were measured in laboratories with batteries and wires, with no light involved. At every instant and every point of such a wave the electric field in volts per metre is three hundred million times the magnetic field in tesla. The two fields are perpendicular to each other and both are perpendicular to the direction the wave is going, and turning the first into the second by a right hand rule points you along that direction.
The two rectangle derivation, and where the square root comes from
Suppose a flat front moves along $x$ at some unknown speed $v$. Behind it there is a uniform electric field $E$ along $y$ and a uniform magnetic field $B$ along $z$; ahead of it there is nothing. The figure shows this edge on.
First rectangle, in the plane of $E$ and $x$. Put a rectangle of height $\Delta y$ straddling the front, as drawn. Only the far side of it lies in the field region, so $\oint\vec{E}\cdot d\vec{\ell}=E\,\Delta y$.
In a time $dt$ the front advances $v\,dt$, so the area of the rectangle that contains magnetic field grows by $\Delta y\,v\,dt$, and the flux through it grows by $d\Phi_B=B\,\Delta y\,v\,dt$.
Faraday's law, in magnitudes: $E\,\Delta y = B\,\Delta y\,v$, so $\;E=vB$. One relation, and $\Delta y$ has cancelled, which it had to since the answer cannot depend on how big a rectangle you drew.
Second rectangle, in the plane of $B$ and $x$. The same picture rotated ninety degrees about the direction of travel. Now $\oint\vec{B}\cdot d\vec{\ell}=B\,\Delta z$, and the electric flux swept in is $d\Phi_E=E\,\Delta z\,v\,dt$.
The completed Ampère's law with no current in sight: $B\,\Delta z=\mu_0\varepsilon_0 E\,\Delta z\,v$, so $\;B=\mu_0\varepsilon_0 Ev$.
Two relations, two unknowns. Put the first into the second: $B=\mu_0\varepsilon_0 (vB) v=\mu_0\varepsilon_0 v^{2}B$. The field $B$ cancels, provided it is not zero, and what is left is a condition on the speed alone: $\mu_0\varepsilon_0 v^{2}=1$.
Hence $v=1/\sqrt{\varepsilon_0\mu_0}$. Notice what happened: the speed was never assumed, it was forced. A front moving at any other speed would break one of the two equations.
Putting the numbers in: $\varepsilon_0\mu_0=(8.85\times10^{-12})(4\pi\times10^{-7})=1.11\times10^{-17}$, and one over the square root of that is $2.998\times10^{8}\ \mathrm{m/s}$. That number was already known from measurements on light, and it had never before come out of an experiment with batteries.
Off the syllabus, one line: applying the same two laws to infinitesimal rectangles gives $\partial^{2}E/\partial x^{2}=\mu_0\varepsilon_0\,\partial^{2}E/\partial t^{2}$, whose solutions are any shape $f(x-ct)$ at all. The sine used in the next concept is a convenience, not a requirement.
The mechanism in one picture: as the front moves $\textcolor{#2da44e}{\text{to the right}}$ it sweeps a new strip of area into the loop, so the magnetic flux through the loop grows, and by Faraday's law that growth is what supports the $\textcolor{#1f6feb}{\text{electric field}}$ down the far side. Rotate the picture ninety degrees and the same sentence with the fields swapped is the completed Ampère's law.
Looks like this, but is not
Every other wave you have met needs something to wave, so this one needs a medium too, and the constants $\varepsilon_0$ and $\mu_0$ must be properties of that medium.
It is a reasonable expectation and it was held seriously for decades. The derivation above kills it: nothing in either rectangle argument referred to a substance, only to two laws which hold in a region containing no matter. What the constants describe is not a material but the relation between charge and field, and between current and field, in empty space. A wave whose speed is set by those two constants alone has no medium left to be a vibration of.
Getting the speed of light out of two electrostatics constants
Compute $1/\sqrt{\varepsilon_0\mu_0}$ from the tabulated constants and check that its units come out as metres per second. Then find the speed of the same wave inside a slab of glass whose dielectric constant at optical frequencies is $K=2.25$, assuming the glass is not magnetic.
glass with $K=2.25$, magnetically indistinguishable from vacuum
Find
the vacuum speed, its units, and the speed inside the glass
Solution
Do the units separately from the numbers. Mixing them in one line is where people lose track, and the unit check is the only evidence that the formula is not an accident of arithmetic.
Independent check on the glass figure using the ratio rather than the formula: $c/v=\sqrt{K}=1.50$, and $3.00\times10^{8}/1.50=2.00\times10^{8}\ \mathrm{m/s}$. The number $1.50$ is also the value quoted for ordinary glass in optics, where it is called the , so two independent traditions agree.
The unit check took longer than the arithmetic. It is worth it once, and never again.
The rule to carry forward is that $K$ enters under a square root. Doubling the dielectric constant does not halve the speed, it divides it by $\sqrt{2}$.
The magnetic field of a laser beam, and why nobody measures it
A laser beam in air has an electric field amplitude of $1.50\times10^{3}\ \mathrm{V/m}$. Find the amplitude of its magnetic field and compare it with the Earth's field, about $5.0\times10^{-5}\ \mathrm{T}$. Separately, a radio receiver measures a magnetic field amplitude of $1.00\times10^{-9}\ \mathrm{T}$ in the wave reaching it; find the electric field amplitude there.
Given
laser: $E_0=1.50\times10^{3}\ \mathrm{V/m}$, travelling in air, so $c$ applies
radio: $B_0=1.00\times10^{-9}\ \mathrm{T}$
Earth's magnetic field for comparison, about $5.0\times10^{-5}\ \mathrm{T}$
Find
$B_0$ for the laser, $E_0$ for the radio wave, and how the laser's magnetic field compares with the Earth's
Solution
One relation does both parts, in opposite directions. The only decision is which way round to divide, and the units settle it: volts per metre divided by metres per second gives volt seconds per square metre, which is a tesla.
Order of magnitude test on the radio result: a receiving a metre long in a $0.300\ \mathrm{V/m}$ field sees about $0.3\ \mathrm{V}$ across it, which is a large signal for a radio and is the right size for a strong local station. Had the answer come out at microvolts per metre or at kilovolts per metre, the arithmetic would have been wrong.
In SI units the magnetic part of any electromagnetic wave looks tiny next to the electric part, but that is a statement about units, not about importance. The next two concepts show the two fields carrying exactly equal shares of the energy.
⚠ Forgetting the square root when a material is involved
with capacitors the dielectric constant divides the field directly, so the reflex is to divide by $K$ here as well
§14.3 — converting between the two field amplitudes●○○○○
A plane electromagnetic wave travelling through vacuum is measured with a small coil, which reports a magnetic field amplitude of $2.50\times10^{-8}\ \mathrm{T}$. The electric amplitude of the same wave is wanted.
Given
$B_0=2.50\times10^{-8}\ \mathrm{T}$
the wave travels in vacuum, so $c=3.00\times10^{8}\ \mathrm{m/s}$
Find
(a) What is $E_0$?
Hint 1/4
Two field amplitudes of one wave are locked together by a single constant. Decide first which of the two is the larger number in SI units.
Hint 2/4
$E=cB$ at every instant, so $E_0=cB_0$ between the amplitudes.
Hint 3/4
Here $B_0=2.50\times10^{-8}\ \mathrm{T}$ and $c=3.00\times10^{8}\ \mathrm{m/s}$, so multiply rather than divide.
Settle the direction of the operation from the units before touching the arithmetic, since the only way this question goes wrong is dividing where a multiplication belongs. The number itself is one line either way.
the electric field is the larger of the two in SI units, so a multiplication is expected here and a division would signal an error
Answer $$\boxed{\ E_0=7.50\ \mathrm{V/m}\ }$$
Check
Unit check as an independent test: $\mathrm{(m/s)\cdot T}=\mathrm{(m/s)(V\,s/m^{2})}=\mathrm{V/m}$, so the product and not the quotient is the combination that can possibly be an electric field.
⚠ Reading $E=cB$ as though the electric field were more important
$E_0$ is a hundred million times bigger as a number, and size in SI units is mistaken for size in physics
Once the shape is a sine, three numbers carry everything about the wave, and two of them are locked together by the speed.
The last concept fixed the speed but said nothing about the shape. Almost every wave you will be handed is a sine, because that is what an oscillating source makes, so it pays to know exactly which symbol is which.
DefinitionDefinition 14.4: the travelling sinusoidal plane wave
Conditions
travelling along $+x$; for travel along $-x$ the sign inside the bracket becomes a plus
$\vec{E}$ along $y$ and $\vec{B}$ along $z$, ordered so that $\vec{E}\times\vec{B}$ points along $+x$
in vacuum, so that $\omega/k=c$; inside a material the same forms hold with $v$ in place of $c$
a plane wave, meaning the fields do not depend on $y$ or $z$ at all
Both fields swing as the same sine of the same bracket, so wherever one is at a maximum the other is too, and wherever one is zero so is the other. The wave number counts how many radians of the sine you cross per metre and the angular frequency counts how many you cross per second, so their ratio is a speed. That speed is fixed, so choosing a frequency chooses the wavelength and there is no freedom left.
Why the bracket is $kx-\omega t$, and why the fields cannot be out of step
Ask where a particular crest is. A crest sits wherever the bracket equals $\pi/2$, that is where $kx-\omega t=\pi/2$.
Solve for its position: $x=(\pi/2+\omega t)/k$. As $t$ grows, $x$ grows, so the crest moves towards larger $x$. That is what the minus sign inside the bracket buys, and a plus sign there would send the wave backwards.
Its speed is $dx/dt=\omega/k$, which is therefore the speed of the wave, and the previous concept fixed that at $c$. Hence $\omega=ck$, and with $\omega=2\pi f$ and $k=2\pi/\lambda$ this is the same statement as $\lambda f=c$.
Now the phase. Feed $E_y=E_0\sin(kx-\omega t)$ into $\partial E/\partial x=-\partial B/\partial t$, the derivative form of Faraday's law for these directions: the left side is $kE_0\cos(kx-\omega t)$.
For the right side to match at every $x$ and every $t$, $B$ must have exactly the same cosine after differentiating, which forces $B_z=B_0\sin(kx-\omega t)$ with the same bracket and no extra phase.
Matching the coefficients gives $kE_0=\omega B_0$, that is $E_0=(\omega/k)B_0=cB_0$. So the amplitude relation and the in phase relation come out of the same requirement, and neither is a separate assumption.
The contrast worth holding on to: in the oscillating circuit of the previous section the current and the capacitor voltage were a quarter cycle apart, because energy was being handed back and forth between two components. Here the two fields peak together, because neither is a store the other empties into.
The wave of the second worked example, frozen at one instant. The $\textcolor{#1f6feb}{\text{electric field}}$ swings in the vertical plane, the $\textcolor{#d1690a}{\text{magnetic field}}$ swings in the horizontal plane at right angles, and every zero of one is a zero of the other.
Looks like this, but is not
The electric and magnetic fields must be a quarter cycle out of step, since one is always busy creating the other, and in the oscillating circuit of the last section the current and the voltage behaved exactly that way.
In a circuit the energy really does slosh between two places, emptying the capacitor to fill the coil, and a quarter cycle offset is what that looks like. In a wave there is nowhere for the energy to slosh to: it is not stored at a point and released, it moves on. The derivative form of Faraday's law forces the two sines to share one bracket, and the figure shows the consequence, both fields zero at the same places along the beam.
$t$
$E_y$
$B_z$
$E_y/B_z$
$0$
$225\ \mathrm{V/m}$
$7.50\times10^{-7}\ \mathrm{T}$
$3.00\times10^{8}\ \mathrm{m/s}$
$32.7\ \mathrm{ns}$
$159\ \mathrm{V/m}$
$5.30\times10^{-7}\ \mathrm{T}$
$3.00\times10^{8}\ \mathrm{m/s}$
$65.4\ \mathrm{ns}$
$0$
$0$
both zero together
$98.2\ \mathrm{ns}$
$-159\ \mathrm{V/m}$
$-5.30\times10^{-7}\ \mathrm{T}$
$3.00\times10^{8}\ \mathrm{m/s}$
$131\ \mathrm{ns}$
$-225\ \mathrm{V/m}$
$-7.50\times10^{-7}\ \mathrm{T}$
$3.00\times10^{8}\ \mathrm{m/s}$
The last column is the same number in every row where it is defined, and the row where it is not defined is the row where both fields vanish at once. Nothing about this table would be possible if the fields were a quarter cycle apart: there would be instants with a magnetic field and no electric field, and the ratio would swing between zero and infinity.
Wavelength, wave number and angular frequency of an FM station at 98.5 MHz
A radio station broadcasts at $98.5\ \mathrm{MHz}$. Find the wavelength of the wave it sends out, its wave number and its angular frequency, and say how the wavelength compares with the size of a receiving aerial.
the wave travels through air, close enough to vacuum that $c$ applies
Find
$\lambda$, $k$ and $\omega$
Solution
Go through the wavelength first even though $\omega$ could be written down immediately, because $k$ needs $\lambda$ and doing them in the other order means computing the same division twice.
The independent check is built into the pair: $\omega/k=6.19\times10^{8}/2.06=3.00\times10^{8}\ \mathrm{m/s}$, which reproduces $c$ from two numbers computed by different routes. If that ratio ever misses, one of the two factors of $2\pi$ has gone astray.
Three lines, one division each. The only trap is the prefix: mega means $10^{6}$ and a slip there moves the wavelength by a factor of a thousand.
Three metres is roughly a car aerial, and that is not a coincidence: an aerial works best when its length is a simple fraction of the wavelength it is catching. Compare with the oven in the next concept, where the wavelength is small enough to fit inside the box.
Reading everything off $E_y=225\sin(0.0800x-2.40\times10^{7}t)$
A plane wave in vacuum has its electric field given, in SI units, by $E_y=225\sin(0.0800\,x-2.40\times10^{7}\,t)\ \mathrm{V/m}$. Find its wavelength, its frequency, the direction it travels, the magnetic field amplitude, and the direction of $\vec{B}$.
Given
$E_0=225\ \mathrm{V/m}$, along $y$
$k=0.0800\ \mathrm{rad/m}$ and $\omega=2.40\times10^{7}\ \mathrm{rad/s}$, read from the bracket
vacuum
Find
$\lambda$, $f$, the travel direction, $B_0$, and the direction of $\vec{B}$
Solution
Everything comes from matching the given expression against the standard form term by term. Trying to guess the wavelength from the amplitude, or the direction from the sign of $E_0$, are the two ways this goes wrong.
Check the two bracket numbers against each other without using either answer: $\omega/k=2.40\times10^{7}/0.0800=3.00\times10^{8}\ \mathrm{m/s}$. The expression is a consistent vacuum wave, and if this ratio had come out at, say, $2\times10^{8}$, the wave would have been inside a material and $c$ could not have been used for $B_0$.
Five answers, and not one of them needed a derivation. The whole example is pattern matching against the standard form, which is why it is worth memorising that form in exactly one arrangement.
Always run the $\omega/k$ check first when a wave is handed to you as a formula. It costs one division and it tells you whether you are in vacuum, which decides whether $E_0=cB_0$ is even allowed.
⚠ Writing the magnetic field as a cosine when the electric field is a sine
the pattern is imported from oscillating circuits, where the current and the voltage really are a quarter cycle apart
A plane wave in vacuum is written as $E_y=E_0\sin(kx+\omega t)$, with $k$ and $\omega$ both positive numbers. A student says the plus sign is a typing error because waves always travel in the positive direction.
Given
$E_y=E_0\sin(kx+\omega t)$ with $k>0$ and $\omega>0$
the wave is in vacuum, so $\omega/k=c$
Find
(a) In which direction does this wave travel, and at what speed?
Hint 1/4
Pick one feature of the wave, say a crest, and ask where it has to be at a later time. No formula for speed is needed to answer the direction part.
Hint 2/4
A crest sits where the bracket takes a fixed value, so $kx+\omega t=\text{constant}$ traces the crest as time passes.
Hint 3/4
With $k>0$ and $\omega>0$, increasing $t$ in $kx+\omega t=\text{constant}$ forces $x$ to decrease. Solving gives $x=(\text{constant}-\omega t)/k$.
Hint 4/4
The crest moves towards smaller $x$ at rate $\omega/k$, so the wave travels along $-x$ at speed $c$.
Show solution
Track a point of constant phase rather than recalling a sign rule, because the rule is exactly the thing people misremember and the phase argument rebuilds it in two lines every time.
Test it at a single point instead of by differentiating: at $t=0$ the field at $x=0$ is zero and rising with $x$; a moment later the same rising zero is found at a slightly negative $x$. The pattern has shifted left.
⚠ Dividing the angular frequency by the wavelength to get the speed
both are the quantities that appear in the problem statement, and one of the two factors of $2\pi$ is easy to lose
wrong$$c=\frac{\omega}{\lambda}$$
right$$c=\frac{\omega}{k}=\lambda f$$
14.5One family, fifteen decades: the electromagnetic spectrum
Radio waves and gamma rays differ only in frequency, and every one of them moves at the same speed in vacuum.
Nothing in the derivation put any limit on the frequency, so the wave can have any wavelength at all. What that licence produces is a family running from waves the length of a street to waves smaller than an atom.
NoteDefinition 14.5: the two numbers that name a wave, in vacuum and in matter
Conditions
$\lambda f=c$ holds in vacuum for every member of the family, from radio to gamma
in a material of dielectric constant $K$ that is not magnetic, the speed drops to $c/\sqrt{K}$
the frequency is set by the source and does not change when the wave enters a material
the band names are conventions with fuzzy borders, not physical boundaries
Multiply the wavelength of any electromagnetic wave in vacuum by its frequency and you always get the same number, so naming one of them names the other. Push the wave into a material and it slows down by the square root of the dielectric constant, but the number of crests arriving each second cannot change at the boundary, so it is the wavelength that shrinks by the same factor and the frequency that survives.
Why the frequency and not the wavelength is the quantity that crosses a boundary
Stand at the surface of the glass and count crests. Suppose $f$ per second arrive from the air side.
Crests are not created or destroyed at a surface: there is nowhere for one to go and nothing for one to come from. So $f$ per second must leave into the glass.
Therefore $f$ is the same on both sides, and $\lambda f=v$ with a smaller $v$ can only be satisfied by a smaller $\lambda$.
In numbers, for glass with $K=2.25$: $v=c/1.50$, so $\lambda_{\rm glass}=\lambda/1.50$, and green light of $532\ \mathrm{nm}$ in air is $355\ \mathrm{nm}$ inside the glass while remaining the same green.
This is why colour is a statement about frequency. The wavelength of the light entering your eye through the fluid in it is not the wavelength it had in air, and yet the colour does not change as it enters.
Fifteen decades of wavelength on one logarithmic axis, with the frequency that goes with each decade written underneath. The band your eyes respond to is narrower than one decade, which is the honest scale of how much of this family we can see.
Looks like this, but is not
Gamma rays are more energetic than radio waves, so they must move faster; that is what more energetic means.
The derivation of the speed used only $\varepsilon_0$ and $\mu_0$, and neither depends on frequency, so every member of the family travels at the same $c$ in vacuum. What changes across the family is how quickly the fields oscillate, not how quickly the pattern advances. Inside matter the speed does drop, and by a slightly different amount at different frequencies, which is why a prism separates colours, but that is a property of the material and not of the wave.
band
$\lambda$
$f$
comparable size
radio, FM
$3.05\ \mathrm{m}$
$98.5\ \mathrm{MHz}$
a car aerial
microwave
$0.122\ \mathrm{m}$
$2.45\ \mathrm{GHz}$
the width of a hand
infrared
$1.00\times10^{-5}\ \mathrm{m}$
$3.00\times10^{13}\ \mathrm{Hz}$
a human hair is about ten times thicker
visible, green
$5.32\times10^{-7}\ \mathrm{m}$
$5.64\times10^{14}\ \mathrm{Hz}$
a hundredth of the width of a hair
ultraviolet
$1.00\times10^{-7}\ \mathrm{m}$
$3.00\times10^{15}\ \mathrm{Hz}$
a virus
$1.00\times10^{-10}\ \mathrm{m}$
$3.00\times10^{18}\ \mathrm{Hz}$
the spacing between atoms in a crystal
Every row is the same arithmetic, $\lambda f=3.00\times10^{8}$, and the third column was computed from the second in each case. The last column is why each band is used for what it is used for: a wave probes structures about its own size, which is why crystals are studied with X rays and not with radio.
How long sunlight takes to arrive, and a radar pulse to the Moon and back
The Sun is $1.50\times10^{11}\ \mathrm{m}$ away and the Moon is $3.84\times10^{8}\ \mathrm{m}$ away. Find how long light takes to reach us from the Sun, and how long a radar pulse takes to travel to the Moon and return. Both distances are through vacuum.
Given
Sun to Earth $d_{\rm S}=1.50\times10^{11}\ \mathrm{m}$
Earth to Moon $d_{\rm M}=3.84\times10^{8}\ \mathrm{m}$
$c=3.00\times10^{8}\ \mathrm{m/s}$ for both, since radar and light are the same family
Find
the one way time for sunlight and the round trip time for the radar pulse
Solution
The only decision in the problem is one way or two, and it is decided by the word return. Everything else is a division, and the fact that a radio pulse and use the same speed is the physics being tested.
Independent check on the Moon figure from the other direction: light covers $3.00\times10^{8}\ \mathrm{m}$ in one second, and the Moon is a little more than $3.8\times10^{8}\ \mathrm{m}$ away, so one way must be a little more than a second and the round trip a little more than two. It is.
Both numbers are ordinary human durations, which is what makes them memorable: the Sun you see set has already been below the horizon for eight minutes, and a conversation with anyone on the Moon has a two and a half second gap in it that no engineering can remove.
The wavelength inside a microwave oven, and the cold spots it leaves
A microwave oven runs at $2.45\ \mathrm{GHz}$. Find the wavelength of its radiation in air. The waves reflect off the metal walls and form a standing pattern, whose hot spots are half a wavelength apart; find that spacing. Then find the wavelength of the same radiation inside a block of material with dielectric constant $K=2.25$ at that frequency.
Given
$f=2.45\times10^{9}\ \mathrm{Hz}$
in air, close enough to vacuum for $c$
hot spots in a standing wave lie half a wavelength apart
material with $K=2.25$, not magnetic
Find
$\lambda$ in air, the hot spot spacing, and $\lambda$ inside the material
Solution
Work in air first and carry the frequency into the material, rather than recomputing anything from scratch there. The frequency is the quantity that survives the boundary, so it is the one to hold on to.
Check the last answer without going through the speed: the wavelength must shrink by exactly the factor $\sqrt{K}=1.50$, and $12.2/1.50=8.16\ \mathrm{cm}$. Two routes, same number, and the shortcut is the one to use under time pressure.
Three divisions and one square root. The only real decision was to keep the frequency fixed at the boundary.
Six centimetres is a spacing you can measure at home, and it is the reason food heats unevenly in an oven with a broken turntable.
⚠ Changing the frequency instead of the wavelength when light enters glass
the speed drops, and the wavelength feels like the more permanent property of the wave because you can picture it
A detector picks up an electromagnetic wave of frequency $6.00\times10^{14}\ \mathrm{Hz}$ arriving through vacuum. A student is asked which band it belongs to and what its wavelength is.
Given
$f=6.00\times10^{14}\ \mathrm{Hz}$
in vacuum, so $\lambda f=c$
the visible band runs from about $400\ \mathrm{nm}$ to about $700\ \mathrm{nm}$
Find
(a) What is the wavelength, and which band is it in?
Hint 1/4
One equation ties the two numbers that name a wave. Decide which of them you have and which you want before touching a calculator.
Hint 2/4
$\lambda f=c$ in vacuum, so $\lambda=c/f$.
Hint 3/4
With $c=3.00\times10^{8}\ \mathrm{m/s}$ and $f=6.00\times10^{14}\ \mathrm{Hz}$, the leading digits give $0.5$ and the powers give $10^{-6}$.
Hint 4/4
$\lambda=5.00\times10^{-7}\ \mathrm{m}=500\ \mathrm{nm}$, which lands inside the visible band.
Show solution
Convert to a wavelength first and name the band afterwards, rather than trying to recall which frequencies are visible. The band edges are quoted in nanometres, so the wavelength is the form the memory is actually stored in.
Cross check against a known member of the band: green light is quoted at $532\ \mathrm{nm}$ and $5.64\times10^{14}\ \mathrm{Hz}$. A slightly higher frequency should give a slightly shorter wavelength, and $500\ \mathrm{nm}$ against $532\ \mathrm{nm}$ is exactly that.
⚠ Multiplying instead of dividing in $\lambda f=c$
the formula has three symbols and no fraction bar, so the rearrangement is done from memory rather than from the equation
wrong$$\lambda=cf$$
right$$\lambda=\frac{c}{f}$$
14.6How much energy the beam carries
The energy of a beam is split evenly between the two fields and flows in the direction E cross B.
Everything so far describes the shape of the wave and says nothing about what it delivers. A wave that carried no energy would be undetectable, and detecting it is the only reason any of this is useful.
TheoremTheorem 14.6: the Poynting vector, intensity, and an isotropic source
Conditions
$\vec{S}$ is instantaneous; for a sinusoidal wave it oscillates at twice the wave frequency and is never negative
the intensity $I$ is the time average of $|\vec{S}|$ over a whole number of cycles
the amplitude forms carry a factor $\tfrac12$ that is the average of a squared sine; the rms forms do not
the isotropic form assumes equal power in every direction and no absorption on the way out
Cross the electric field into the magnetic field, divide by the permeability, and the result points the way the energy is going and tells you how many watts cross each square metre at that instant. Average it over a cycle and you have the intensity, the quantity a light meter reports. If the source throws its power equally in all directions, spread that power over the sphere the wave has reached, and the intensity falls as one over the distance squared for the plain geometric reason that the sphere is getting bigger.
Why the two fields carry equal energy, and where the factor of one half comes from
You already have the two energy densities separately: $u_E=\tfrac12\varepsilon_0E^{2}$ from the capacitor and $u_B=B^{2}/(2\mu_0)$ from the coil.
Put $B=E/c$ into the magnetic one: $u_B=E^{2}/(2\mu_0c^{2})$.
But $c^{2}=1/(\varepsilon_0\mu_0)$, so $1/(\mu_0c^{2})=\varepsilon_0$, and therefore $u_B=\tfrac12\varepsilon_0E^{2}=u_E$. The two halves are exactly equal at every instant, not on average.
Total: $u=u_E+u_B=\varepsilon_0E^{2}$. The one half has gone, which is worth noticing because it is the source of half the sign of trouble in this material.
Now the transport. Look at the tube in the figure, of face area $A$ and length $c\,dt$. Every joule inside it crosses the face within the time $dt$, because everything in there is moving at $c$ towards the face.
Energy in the tube: $u\,(A\,c\,dt)$. Divide by $A$ and by $dt$: $S=uc=\varepsilon_0cE^{2}$. Using $E=cB$ once more, $S=EB/\mu_0$, which is the magnitude of the cross product in the box.
Finally the average. With $E=E_0\sin(kx-\omega t)$, $S=\varepsilon_0cE_0^{2}\sin^{2}(\ldots)$, and the average of $\sin^{2}$ over a cycle is $\tfrac12$.
So $I=\tfrac12\varepsilon_0cE_0^{2}$. That factor of one half is the time average and nothing else, which is why it disappears when the field is quoted as an rms value: $E_{\rm rms}^{2}=E_0^{2}/2$ has already spent it.
Why the intensity is the energy density times $c$. In a time $dt$ the whole shaded tube passes through the face, so the $\textcolor{#2da44e}{\text{energy flow}}$ per unit area is the energy per unit volume multiplied by the distance covered per unit time.
Looks like this, but is not
In SI units $E_0$ is a hundred million times larger than $B_0$, so almost all the energy of a light wave must be in the electric field and the magnetic part can be neglected.
Compare the energy densities rather than the fields, since the fields are not even measured in the same unit and the comparison is meaningless as it stands. The electric density carries a factor $\varepsilon_0$ and the magnetic one a factor $1/\mu_0$, and those two constants are what close the gap: the derivation above shows $u_E$ and $u_B$ are equal at every instant. Half of the energy in every beam of light is magnetic.
source
$I$
$E_0$
$B_0$
sunlight above the atmosphere
$1360\ \mathrm{W/m^{2}}$
$1010\ \mathrm{V/m}$
$3.37\times10^{-6}\ \mathrm{T}$
a $1.00\ \mathrm{mW}$ pointer in a $2.00\ \mathrm{mm}$ beam
$318\ \mathrm{W/m^{2}}$
$490\ \mathrm{V/m}$
$1.63\times10^{-6}\ \mathrm{T}$
a $50.0\ \mathrm{kW}$ transmitter at $20.0\ \mathrm{km}$
$9.95\ \mathrm{\mu W/m^{2}}$
$0.0866\ \mathrm{V/m}$
$2.89\times10^{-10}\ \mathrm{T}$
Two things are worth reading off. First, the pointer is four times less intense than sunlight, so what makes a laser different is not the watts per square metre at the source but that its beam stays narrow over a distance. Second, the field amplitudes move much less than the intensities do: eight orders of magnitude in intensity become four in the field, because the intensity goes as the square.
The electric and magnetic fields in sunlight
Above the atmosphere the intensity of sunlight is $1360\ \mathrm{W/m^{2}}$. Find the amplitudes of the electric and magnetic fields in it, and their root mean square values.
Given
$I=1360\ \mathrm{W/m^{2}}$, a time averaged intensity
sunlight treated as a plane wave here, which it is at this distance
Find
$E_0$, $B_0$, $E_{\rm rms}$ and $B_{\rm rms}$
Solution
Start from the form with $E_0$ in it rather than the one with $E_0B_0$, because the second contains two unknowns and would need $E=cB$ substituted in anyway. Choosing the one unknown form first saves a line.
Invert the intensity formula for the electric amplitude
a kilovolt per metre, which is about three thousand times weaker than the $3\times10^{6}\ \mathrm{V/m}$ that breaks down dry air, so nothing dramatic happens in the air the sunlight crosses
Independent route back to the intensity through a different formula: $I=E_{\rm rms}B_{\rm rms}/\mu_0=(716)(2.39\times10^{-6})/(1.257\times10^{-6})=1.36\times10^{3}\ \mathrm{W/m^{2}}$, which returns the starting value using the rms numbers and the permeability rather than the amplitudes and the permittivity.
One square root, one division, two divisions by root two. The grouped constant $\varepsilon_0c=2.655\times10^{-3}$ is worth writing on the formula sheet.
A kilovolt per metre and three microtesla is what a bright summer day is, in field terms. Any answer for a light beam that comes out in megavolts per metre or in nanotesla should be recomputed before it is written down.
Field strength 20 km from a 50 kW transmitter
A radio transmitter radiates $50.0\ \mathrm{kW}$ equally in all directions from a mast. Find the intensity and the rms electric field at a receiver $20.0\ \mathrm{km}$ away, ignoring absorption and reflection on the way.
Given
radiated power $P=5.00\times10^{4}\ \mathrm{W}$, isotropic
distance $r=2.00\times10^{4}\ \mathrm{m}$
vacuum assumed, nothing absorbing or reflecting in between
$I$ at the receiver, and $E_{\rm rms}$ and $E_0$ there
Solution
Use the rms form of the intensity here rather than the amplitude form, because the answer wanted is an rms field and going through $E_0$ first means dividing by root two at the end for no reason. When a question asks for rms, start from the rms formula.
Spread the power over the sphere the wave has reached
Go back the other way through the amplitude formula, which uses different constants and the factor of one half: $I=\tfrac12\varepsilon_0cE_0^{2}=\tfrac12(2.655\times10^{-3})(0.0866)^{2}=9.95\times10^{-6}\ \mathrm{W/m^{2}}$, recovering the intensity.
The inverse square is the whole physics; the rest is two square roots. Doubling the distance would quarter the intensity and halve the field.
Fields from broadcast transmitters are tens of millivolts per metre at a few tens of kilometres. That is the scale to expect, and it is why receivers are built around amplifiers.
⚠ Putting the factor of one half into the rms form as well
the half is remembered as part of the intensity formula rather than as the average of a squared sine, so it travels with the formula instead of with the amplitude
A plane wave in vacuum has an electric field amplitude of $200\ \mathrm{V/m}$. A light meter is placed face on to the beam and reports the intensity.
Given
$E_0=200\ \mathrm{V/m}$ (an amplitude, not an rms value)
$\varepsilon_0c=2.655\times10^{-3}$ in SI units
the meter faces the beam squarely
Find
(a) What intensity does the meter report?
Hint 1/4
Decide first whether the number you were given is an amplitude or an rms value, because that decides which of the two intensity formulas is the honest one.
Hint 2/4
For an amplitude, $I=\tfrac12\varepsilon_0cE_0^{2}$. For an rms value the half is already spent and $I=\varepsilon_0cE_{\rm rms}^{2}$.
Hint 3/4
Here $E_0=200\ \mathrm{V/m}$ is an amplitude and $\varepsilon_0c=2.655\times10^{-3}$, so $I=\tfrac12(2.655\times10^{-3})(200)^{2}$.
Hint 4/4
$(200)^{2}=4.00\times10^{4}$, so $I=\tfrac12(2.655\times10^{-3})(4.00\times10^{4})=53.1\ \mathrm{W/m^{2}}$.
Show solution
The field given is a peak value, so start from the amplitude form of the intensity. Converting to rms first and using the other form gives the same answer, at the cost of a division and a multiplication by root two that undo each other.
Independent route through the rms form: $E_{\rm rms}=200/\sqrt2=141\ \mathrm{V/m}$, and $I=\varepsilon_0cE_{\rm rms}^{2}=(2.655\times10^{-3})(2.00\times10^{4})=53.1\ \mathrm{W/m^{2}}$, with the half never written down at all.
⚠ Spreading the power of an isotropic source over a circle instead of a sphere
the picture drawn on paper is a circle, and $\pi r^{2}$ is the more familiar formula of the two
wrong$$I=\frac{P}{\pi r^{2}}$$
right$$I=\frac{P}{4\pi r^{2}}$$
14.7Light pushes, and why the push is so small
A beam carries momentum equal to its energy divided by c, so it presses on whatever stops or turns it.
A wave that carries energy across a boundary also carries momentum across it, and momentum arriving at a surface is a force. That is the last thing this section has to extract from the same beam.
RuleRule 14.7: momentum of a beam and the pressure it exerts
Conditions
the beam arrives along the normal to the surface; at an angle both results pick up a cosine factor
fully absorbed or fully reflected; a real grey surface lies between the two and is not treated here
$I$ is the time averaged intensity, so the pressure obtained is also a time average
the force follows as $F=P_{\rm rad}A$ with $A$ the area the beam actually lands on
Every joule a beam delivers brings with it a momentum of that many joules divided by the speed of light. A black surface takes all of that momentum, so the force on each square metre is the intensity divided by the speed of light. A mirror does more than take the momentum, it sends it back the other way, so the change is twice as large and so is the pressure. Nothing here depends on the frequency or the colour, only on the watts.
Where the factor of two comes from, and why it is the same two as in a bouncing ball
Take a slab of beam carrying energy $\Delta U$ towards the surface, so it brings momentum $\Delta p=\Delta U/c$ pointing forwards.
Black surface. The energy is absorbed and the beam is gone. The momentum it brought has to end up somewhere, and the only place available is the surface, so the surface gains $\Delta U/c$.
Force is momentum per unit time: $F=\Delta p/\Delta t=(\Delta U/\Delta t)/c=\mathcal{P}/c$, where $\mathcal{P}$ is the power landing. Divide by the area and $P_{\rm rad}=I/c$.
Mirror. The same slab arrives with $+\Delta U/c$ and leaves with $-\Delta U/c$, since it now travels backwards with the same energy. Its momentum changed by $2\Delta U/c$, and the surface supplied that change, so by Newton's third law the surface received $2\Delta U/c$.
This is the same factor of two as a ball bouncing elastically off a wall against a ball of putty sticking to it, and it has the same cause: reversing a momentum is twice as big a change as removing it.
One consequence worth stating: a mirror in a beam feels twice the force of a black card in the same beam, even though it absorbs nothing at all and does not warm up.
Left, the factor of two: turning the $\textcolor{#2da44e}{\text{beam}}$ round changes its momentum twice as much as stopping it. Right, the same result put in perspective with a free body diagram drawn to scale, for the sail of the worked example below.
Looks like this, but is not
Sunlight delivers more than a kilowatt to every square metre, and a kilowatt is a lot of power, so the pressure it exerts should be easy to feel.
Power is not force. Turning one into the other divides by the speed of light, and that is a division by three hundred million. Sunlight on a black surface presses with about $4.5\ \mathrm{\mu Pa}$, while the air in the room presses with about $1.0\times10^{5}\ \mathrm{Pa}$, some twenty thousand million times more. Radiation pressure only becomes the dominant force where there is no air and no support, which is why it is a spacecraft topic and not a laboratory bench one.
beam and surface
$I$
$P_{\rm rad}$
sunlight on a black surface
$1360\ \mathrm{W/m^{2}}$
$4.53\ \mathrm{\mu Pa}$
sunlight on a mirror
$1360\ \mathrm{W/m^{2}}$
$9.07\ \mathrm{\mu Pa}$
a $1.00\ \mathrm{mW}$ pointer on a black card
$318\ \mathrm{W/m^{2}}$
$1.06\ \mathrm{\mu Pa}$
a broadcast signal at $20\ \mathrm{km}$
$9.95\ \mathrm{\mu W/m^{2}}$
$3.32\times10^{-14}\ \mathrm{Pa}$
the atmosphere on your hand
not applicable
$1.01\times10^{5}\ \mathrm{Pa}$
The last row is there to be compared with the first: air pressure beats full sunlight by a factor of about twenty thousand million. Nothing in the first four rows is measurable with anything you could build on a bench, and all four become decisive in space, where nothing else is pushing.
A ten thousand square metre solar sail, and whether it goes anywhere
A perfectly reflecting solar sail of area $1.00\times10^{4}\ \mathrm{m^{2}}$ faces the Sun squarely at the Earth's distance, where the intensity is $1360\ \mathrm{W/m^{2}}$. The sail and its payload have a total mass of $100\ \mathrm{kg}$. Find the force the light exerts, the acceleration it produces, and the time to build up a speed of $1.00\ \mathrm{km/s}$ from rest. Then compare the light force with the Sun's gravitational pull on the same craft, which at that distance is $0.593\ \mathrm{N}$.
Given
$A=1.00\times10^{4}\ \mathrm{m^{2}}$, perfectly reflecting, facing the beam squarely
$I=1360\ \mathrm{W/m^{2}}$
$m=100\ \mathrm{kg}$ for sail and payload together
solar gravitational pull on this craft at this distance: $0.593\ \mathrm{N}$, directed towards the Sun
target speed $1.00\times10^{3}\ \mathrm{m/s}$, starting from rest
Find
the light force, the acceleration it alone would give, the time to reach $1.00\ \mathrm{km/s}$, and how it compares with the Sun's pull
Solution
The free body diagram in the figure above is drawn to scale for exactly this craft, and it is worth looking at before the arithmetic: two horizontal forces, opposed, and one clearly longer than the other. Doing the numbers without that picture is how a student ends up reporting an escaping spacecraft.
Check the force by a completely different route, through momentum rather than pressure. The power intercepted is $IA=1.36\times10^{7}\ \mathrm{W}$; the momentum arriving each second is $IA/c=4.53\times10^{-2}\ \mathrm{kg\,m/s}$ per second, and reflection doubles it to $9.07\times10^{-2}\ \mathrm{N}$. Same answer without ever computing a pressure.
Five lines of arithmetic and one comparison. The comparison is the part that changes the answer to the question that was really being asked.
The lesson is not that sails do not work, it is that this one is too heavy. The light force scales with area while the gravitational pull scales with mass, so the design number is square metres per kilogram, and real sails are films a few micrometres thick for exactly that reason.
The push of a laser pointer against the weight of a speck of dust
A $1.00\ \mathrm{mW}$ laser pointer produces a beam $2.00\ \mathrm{mm}$ across. It is aimed at a small black card that absorbs everything. Find the intensity in the beam, the radiation pressure on the card, and the total force. Compare that force with the weight of a speck of dust of mass $1.00\ \mathrm{\mu g}$.
Given
beam power $\mathcal{P}=1.00\times10^{-3}\ \mathrm{W}$
beam diameter $2.00\ \mathrm{mm}$, so radius $1.00\times10^{-3}\ \mathrm{m}$
the card absorbs the whole beam
dust speck mass $1.00\times10^{-9}\ \mathrm{kg}$, $g=9.80\ \mathrm{m/s^{2}}$
Find
$I$, $P_{\rm rad}$, the force on the card, and the ratio to the dust speck's weight
Solution
Compute the force from the pressure and the area even though the shortcut $F=\mathcal{P}/c$ would give it in one line, because the intensity and the pressure are both asked for. The shortcut is then available as an independent check.
The shortcut confirms the force without the area appearing at all: $F=\mathcal{P}/c=1.00\times10^{-3}/3.00\times10^{8}=3.33\times10^{-12}\ \mathrm{N}$. That the area cancels is worth noticing, because it means focusing the beam tighter raises the pressure but not the total force.
Two areas, two divisions, one comparison. The comparison is what makes the number mean anything.
A useful sanity rule falls out of the check: the total force of any fully absorbed beam is just its power divided by $c$, whatever its shape. A one watt beam pushes with about three nanonewtons, always.
⚠ Using the absorbing formula for a mirror
the two formulas differ by one character and the reflecting case is the one that appears less often in worked examples
wrong$$P_{\rm rad}=\frac{I}{c}\quad\text{for a mirror}$$
right$$P_{\rm rad}=\frac{2I}{c}\quad\text{for a mirror}$$
Checkpoint
§14.7 — black card against mirror in the same beam●●○○○
Two small squares of the same size hang side by side in the same beam of light, facing it squarely. One is coated black and absorbs everything that lands on it; the other is a perfect mirror and sends the whole beam back the way it came.
Given
identical areas, same beam, same intensity $I$ on both
one square absorbs everything, the other reflects everything
both face the beam squarely
Find
(a) How does the force on the mirror compare with the force on the black square?
Hint 1/4
Do not think about energy at all. Ask what happens to the momentum the beam brings in each case, and remember that force is the rate of momentum change.
Hint 2/4
$P_{\rm rad}=I/c$ when the beam is absorbed and $P_{\rm rad}=2I/c$ when it is reflected straight back.
Hint 3/4
Both squares receive the same intensity $I$ over the same area, so the only difference between them is the factor in front.
Hint 4/4
The mirror feels exactly twice the force, the same factor of two as an elastic bounce compared with a stick.
Show solution
Compare the two pressures symbolically instead of computing either force, because the intensity and the area are the same on both sides and cancel in the ratio. Working out two forces would only mean doing that same division at the end.
Check it as a mechanics problem with no light in it: a ball of mass $m$ at speed $v$ that sticks to a wall delivers $mv$, while an identical ball that bounces back at $v$ delivers $2mv$. Same factor, same reason.
⚠ Using the beam diameter where the radius belongs
beams are quoted by diameter and areas are computed from radius, and the conversion happens silently
any question that hands you a single quantity about a beam of light or radio, and asks for a different one: amplitude to intensity, power to force, wavelength to frequency.
Get to an intensity
Whatever you were given, turn it into $I$ in $\mathrm{W/m^{2}}$ first. From a power and a beam area, $I=\mathcal{P}/A$. From a total radiated power and a distance, $I=\mathcal{P}/(4\pi r^{2})$. From an amplitude, $I=\tfrac12\varepsilon_0cE_0^{2}$.
Decide peak or rms
If the question says amplitude, peak or maximum, the half belongs in the formula. If it says rms or effective, the half is already spent and $I=\varepsilon_0cE_{\rm rms}^{2}$.
Cross to the other field
$E_0=cB_0$, always, in vacuum. Multiply going from $B$ to $E$ and divide going the other way; the electric number is the big one in SI units.
Wavelength and frequency
$\lambda f=c$, $k=2\pi/\lambda$, $\omega=2\pi f$. If you have $k$ and $\omega$ from a written wave, check $\omega/k$ before anything else: it tells you whether you are in vacuum.
If a surface is involved
$P_{\rm rad}=I/c$ if the surface absorbs, $2I/c$ if it reflects. Then $F=P_{\rm rad}A$ with $A$ the area the beam lands on, and $a=F/m$ if anything is free to move.
Check the size
Sunlight is $1360\ \mathrm{W/m^{2}}$, about $1000\ \mathrm{V/m}$, about $3\ \mathrm{\mu T}$ and about $5\ \mathrm{\mu Pa}$. Anything wildly away from those, for an ordinary beam, is an arithmetic error.
Where it goes wrong
Squaring an rms value in the amplitude formula, which halves the intensity.
Using $\pi d^{2}$ for a beam area when the diameter was quoted.
Using $I/c$ for a mirror.
Spreading an isotropic power over $\pi r^{2}$ instead of $4\pi r^{2}$.
Finding a magnetic field where no wire runs
a magnetic field is wanted in a region with a changing electric field: between capacitor plates, or anywhere a question says no current flows here.
Draw the loop on the symmetry
Take a circle centred on the axis of the field pattern, so that $B$ has one magnitude all along it and runs along it. Then $\oint\vec{B}\cdot d\vec{\ell}=B(2\pi r)$.
Choose the surface and count what it catches
Span the loop with the flat disc. Ask two separate questions: how much real current pierces it, and how much electric flux passes through it.
Differentiate only the flux
$I_D=\varepsilon_0\,d\Phi_E/dt$. If the field is uniform over an area $A_{\rm enc}$ that is the part of the plate area inside your loop, this is $\varepsilon_0A_{\rm enc}\,dE/dt$.
Watch the crossover at the plate rim
Inside the plates $A_{\rm enc}=\pi r^{2}$ and $B$ grows with $r$. Outside, $A_{\rm enc}=\pi R^{2}$ is stuck at the full plate area and $B$ falls as $1/r$, matching the field of the wire.
Use the identity when you can
For a full gap surface, $I_D$ equals the charging current, so if the wire current is given you can skip the field entirely.
Where it goes wrong
Keeping $\pi r^{2}$ in the flux after the loop has passed the plate rim.
Adding the conduction current as well as the displacement current for a loop between the plates, where there is no conduction current at all.
Forgetting that $dE/dt$, not $E$, is what the law contains.
Deciding which of the four equations a question is about
a problem describes a situation and you cannot see which tool it wants; this is most of the interleaved practice at the end.
Look at the geometry named
A closed surface, a box, a sphere or a can points at one of the two flux laws. A loop, a ring, a circuit or a path points at one of the two circulation laws.
Then look at which field is asked for
That splits each pair. Wanting $\vec{E}$ from a closed surface is the first law; wanting $\vec{B}$ from a loop is the fourth.
Ask what is changing
If nothing changes with time, the two circulation laws lose their flux terms and you are back to Ampère and to electrostatics. If something changes, name it: $\Phi_B$ changing points to the third law, $\Phi_E$ changing to the fourth.
Check for a second term
The fourth law is the only one with two terms on the right, and a problem with both a wire and a changing field needs both.
Where it goes wrong
Reading a wire loop as a closed surface because both integrals carry a ring.
Reaching for the flux law for $\vec{B}$ whenever a magnetic field appears, when that law only ever returns zero.
Force on a black disc in a 600 W per square metre beam
A black disc of area $5.00\ \mathrm{cm^{2}}$ is held face on in a beam of intensity $600\ \mathrm{W/m^{2}}$ and absorbs all of it. Find the force on it.
Given
$I=600\ \mathrm{W/m^{2}}$
$A=5.00\times10^{-4}\ \mathrm{m^{2}}$
the disc absorbs everything
Find
the force on the disc
Solution
Go through the pressure and then the area rather than the one line route from the intercepted power, so the pressure is written down where it can be set beside its mirror twin in the next example.
Through the power instead: the disc intercepts $IA=0.300\ \mathrm{W}$, and $F=\mathcal{P}/c=0.300/3.00\times10^{8}=1.00\times10^{-9}\ \mathrm{N}$.
Force on a mirror disc in the same 600 W per square metre beam
A perfectly reflecting disc of the same area, $5.00\ \mathrm{cm^{2}}$, replaces the black one in the same beam of intensity $600\ \mathrm{W/m^{2}}$. Find the force on it.
Given
$I=600\ \mathrm{W/m^{2}}$
$A=5.00\times10^{-4}\ \mathrm{m^{2}}$
the disc reflects everything straight back
Find
the force on the disc
Solution
Deliberately the same route as the black disc, in the same order: only the first line differs, which makes the factor of two the one thing that stands out between the two calculations.
Momentum route: the beam arrives with $IA/c=1.00\times10^{-9}\ \mathrm{kg\,m/s}$ each second and leaves with the same amount reversed, so the change is $2.00\times10^{-9}\ \mathrm{N}$.
Identical beam, identical area, identical arithmetic, and the mirror feels exactly twice the force although it absorbs nothing and does not warm up at all.
How to tell them apart
Ask what leaves the surface. Nothing leaving means the momentum was removed and the pressure is $I/c$; the beam leaving backwards means the momentum was reversed and the pressure is $2I/c$. The energy the surface keeps has nothing to do with it.
Scaffolding comes off
The common skeleton
Reduce whatever you were given to an intensity in watts per square metre.
If a field amplitude is wanted, use $I=\tfrac12\varepsilon_0cE_0^{2}$ and then $E_0=cB_0$.
Decide what the surface does to the beam, and pick $I/c$ for absorbed or $2I/c$ for reflected.
Multiply the pressure by the area the beam actually lands on to get the force.
Check the answer against the beam's power divided by $c$, which is the total force for full absorption whatever the geometry.
1 · fully worked
Mirror of 4.00 square centimetres in a beam of amplitude 800 V per metre
A plane wave in vacuum with electric field amplitude $E_0=800\ \mathrm{V/m}$ falls squarely on a perfect mirror of area $4.00\ \mathrm{cm^{2}}$. Find the magnetic field amplitude, the intensity of the beam, the radiation pressure on the mirror and the force on it.
Given
$E_0=800\ \mathrm{V/m}$, an amplitude
$A=4.00\times10^{-4}\ \mathrm{m^{2}}$
perfect mirror, beam arriving along the normal
$\varepsilon_0c=2.655\times10^{-3}$
Find
$B_0$, $I$, $P_{\rm rad}$ and $F$
Solution
Every step of the skeleton is used once, in order, which is why this is the rung with all the reasoning shown. The only judgement call is at step three, and it is decided by the word mirror.
Independent route to the force, skipping the pressure: the mirror intercepts $IA=0.340\ \mathrm{W}$, and a reflected beam delivers $2\mathcal{P}/c=2(0.340)/3.00\times10^{8}=2.27\times10^{-9}\ \mathrm{N}$.
Two nanonewtons, from a beam bright enough to be uncomfortable to look at. Hold on to that ratio between how bright a beam is and how little it pushes.
2 · you write the reasoning
Now the same skeleton with the first two steps already done for you. A beam of intensity $500\ \mathrm{W/m^{2}}$ falls squarely on a black plate of area $2.00\ \mathrm{cm^{2}}$. Find the radiation pressure and the force. The steps are below; write the reason for each one yourself before opening it.
$P_{\rm rad}=\dfrac{I}{c}$
reasoning
The plate is black, so the beam stops there and its momentum is removed rather than reversed. That is the whole content of choosing this form over the one with a two in it.
Substituting the given intensity and the speed of light. Nothing is squared and nothing is halved here, because the intensity was handed over directly rather than through a field amplitude.
$F=P_{\rm rad}A$
reasoning
Pressure is force per unit area, so multiplying by the area the beam lands on undoes the definition. The plate is smaller than the beam, so the plate's area is the one that counts.
Arithmetic. A third of a nanonewton, which is the right order for a beam a third as bright as sunlight landing on a stamp sized plate.
3 · find the buried error
Harder now, and someone else has done it. A $15.0\ \mathrm{mW}$ laser produces a beam $3.00\ \mathrm{mm}$ in diameter and shines it squarely on a perfect mirror. A student computes the radiation pressure and the force as follows. Exactly two of the four steps are wrong. Find them.
Step 1. $A=\pi d^{2}=\pi(3.00\times10^{-3})^{2}=2.83\times10^{-5}\ \mathrm{m^{2}}$ — the beam area from its diameter.
Step 2. $I=\dfrac{\mathcal{P}}{A}=\dfrac{15.0\times10^{-3}}{2.83\times10^{-5}}=530\ \mathrm{W/m^{2}}$ — the intensity is the power spread over that area.
Step 3. $P_{\rm rad}=\dfrac{I}{c}=\dfrac{530}{3.00\times10^{8}}=1.77\times10^{-6}\ \mathrm{Pa}$ — intensity divided by the speed of light gives the pressure.
Step 4. $F=P_{\rm rad}A=(1.77\times10^{-6})(2.83\times10^{-5})=5.00\times10^{-11}\ \mathrm{N}$ — pressure times the area the beam lands on.
the two buried errors (2)
⚠ step 1
The area was computed as $\pi d^{2}$ instead of $\pi(d/2)^{2}$, so it is four times too large and the intensity that follows is four times too small.
Beams are quoted by diameter and circles are computed from radius, and the halving happens silently between the two. Nothing in the written line looks wrong, because $\pi$ times a length squared is the right shape of formula.
The absorbing formula $I/c$ was used although the surface is a perfect mirror, so the pressure is half what it should be.
The two forms differ by one character, and $I/c$ is the one that gets written in the margin because it is the one the definition of momentum leads to first. The word mirror sits in the question, not in the formula.
right
$P_{\rm rad}=2I/c=2(2.12\times10^{3})/(3.00\times10^{8})=1.41\times10^{-5}\ \mathrm{Pa}$, and then $F=1.00\times10^{-10}\ \mathrm{N}$.
4 · the bare problem
§14.7 — the full chain with no scaffolding●●●●○
A laboratory laser of power $5.00\ \mathrm{mW}$ produces a beam $1.50\ \mathrm{mm}$ in diameter. The beam is aimed squarely at a small black target that absorbs all of it.
Given
beam power $\mathcal{P}=5.00\times10^{-3}\ \mathrm{W}$
beam diameter $1.50\ \mathrm{mm}$
black target, absorbing everything, beam along the normal
Four answers, and each one is the input to the next. Decide the order before starting: nothing here can be reached directly from the power except the intensity.
Hint 2/4
$I=\mathcal{P}/A$ with $A=\pi(d/2)^{2}$, then $I=\tfrac12\varepsilon_0cE_0^{2}$ inverted for $E_0$, then $P_{\rm rad}=I/c$ for an absorber, then $F=P_{\rm rad}A$.
Hint 3/4
The data again: $\mathcal{P}=5.00\times10^{-3}\ \mathrm{W}$, $d=1.50\times10^{-3}\ \mathrm{m}$ so the radius is $0.750\ \mathrm{mm}$, $\varepsilon_0c=2.655\times10^{-3}$. The area comes to $1.77\times10^{-6}\ \mathrm{m^{2}}$.
Hint 4/4
$I=2.83\times10^{3}\ \mathrm{W/m^{2}}$, $E_0=1.46\times10^{3}\ \mathrm{V/m}$, $P_{\rm rad}=9.43\times10^{-6}\ \mathrm{Pa}$ and $F=1.67\times10^{-11}\ \mathrm{N}$.
Show solution
Go through the area rather than reaching for the shortcut $F=\mathcal{P}/c$ straight away, because the intensity and the field are asked for as well. The shortcut is then free as a check on the last answer.
The shortcut checks the last answer without the area appearing: $F=\mathcal{P}/c=5.00\times10^{-3}/3.00\times10^{8}=1.67\times10^{-11}\ \mathrm{N}$. Because the area cancels, focusing this beam to a tenth of its diameter would multiply the pressure by a hundred and leave the force alone.
Whenever a fully absorbed beam is involved and only the total force is wanted, the whole chain collapses to power over $c$.
Full exam-style question
Full question: a green wave given through its magnetic fieldexam format
A plane electromagnetic wave travels through vacuum. Its magnetic field is measured to be $B_z=4.00\times10^{-8}\sin\!\left(1.20\times10^{7}x-\omega t\right)\ \mathrm{T}$, with $x$ in metres and $t$ in seconds. Find (a) the wavelength, the frequency and $\omega$; (b) the electric field amplitude and the direction of $\vec{E}$; (c) the intensity; (d) the power passing through a window of area $25.0\ \mathrm{cm^{2}}$ held face on; (e) the force on a perfect mirror of the same area placed in the beam.
Given
$B_0=4.00\times10^{-8}\ \mathrm{T}$, along $z$
$k=1.20\times10^{7}\ \mathrm{rad/m}$, read from the bracket
vacuum, travel along $+x$ since the bracket is $kx-\omega t$
window and mirror area $A=2.50\times10^{-3}\ \mathrm{m^{2}}$, both face on
wavelength, frequency, $\omega$, $E_0$ and its direction, intensity, transmitted power, and the force on a mirror of that area
Solution
Take the parts in the order given, because each one feeds the next; the only place where a choice exists is part (e), where the word mirror decides between two formulas that differ by a factor of two.
the wave goes along $+x$ and $\vec{B}$ lies along $z$, so the electric field has to lie along $+y$ for $\vec{E}\times\vec{B}$ to point the way the wave is going
Two independent checks. First, $f$ from $\omega$ rather than from $\lambda$: $\omega/2\pi=3.60\times10^{15}/6.283=5.73\times10^{14}\ \mathrm{Hz}$, matching part (a) by a different route. Second, the force straight from the power: a reflected beam delivers $2\mathcal{P}/c=2(4.78\times10^{-4})/(3.00\times10^{8})=3.19\times10^{-12}\ \mathrm{N}$, matching part (e) without the pressure or the area.
Seven answers from two numbers in a bracket and one area. Only two decisions were made in the whole question: which direction $\vec{E}$ points, and whether the surface reflects.
This is the shape most full questions on this material take: a wave handed over as a formula, then a walk down the chain from wave numbers to fields to energy to force. Practise the walk, not the individual formulas.
Practice
A · concept 4 questions
1§14.1 — what the displacement current is made of●●○○○
A capacitor with a perfectly evacuated gap is being charged. A student explains the compass deflection between the plates by saying that a small leakage current of charge must be crossing the gap, too small to measure directly but enough to make the field.
Given
the gap is a perfect vacuum, with nothing in it to carry charge
the capacitor is being charged at a steady rate
a magnetic field is measured between the plates
Find
(a) True or false: the magnetic field between the plates is produced by charge crossing the gap. Say in one sentence what does produce it.
Hint 1/4
Do not compute anything. Read the formula for the new term and list the physical quantities it actually contains.
Hint 2/4
$I_D=\varepsilon_0\,d\Phi_E/dt$. The ingredients are a permittivity, an electric flux and a clock.
Hint 3/4
The gap here is a perfect vacuum, so there is no charge in it at any moment; but the flux through a surface across the gap is growing, because the plates are charging.
Hint 4/4
False. What makes the field is the growing electric flux, and it would do so with the gap emptied of every last atom.
Show solution
Test the claim against the two terms of the law rather than arguing about whether a vacuum can leak. The law names everything that can produce $\vec{B}$, and one of its two terms needs no charge at all.
Independent test: light crosses interstellar space, where the particle density is far lower than in any laboratory vacuum, and it carries a magnetic field the whole way. No carrier is required.
2§14.2 — which law forbids what●●○○○
A student claims to have built a device with a single magnetic pole at its centre, so that magnetic field lines stream outward from it in all directions and never return.
Given
the claimed device has field lines leaving in all directions and none returning
the four field equations as written in this section
Find
(a) Which equation does the claim contradict, and what would the flux out of a sphere around the device be if the claim were true?
Hint 1/4
The claim is about lines leaving a region and never coming back. Which of the four equations is a statement about what leaves a closed region?
Hint 2/4
$\oint\vec{B}\cdot d\vec{A}=0$ says the net magnetic flux out of any closed surface is exactly zero, always.
Hint 3/4
Draw a sphere around the claimed device. Lines leave everywhere and none return, so the outward flux would be positive rather than zero.
Hint 4/4
The claim contradicts the flux law for $\vec{B}$, which demands zero, and would require a non zero flux instead.
Show solution
Turn the claim into a flux through a closed surface before comparing it with anything. The equations are statements about fluxes and circulations, and only in that form can a claim and a law contradict each other.
Translate the claim into a flux
$$\oint\vec{B}\cdot d\vec{A}>0\ \text{for a sphere around the pole}$$
outward everywhere and inward nowhere is exactly what a positive net flux means
$$\text{but the law says}\ \oint\vec{B}\cdot d\vec{A}=0$$
the two statements cannot both hold, and the law is the one supported by every search made for such a pole
Answer $$\boxed{\ \oint\vec{B}\cdot d\vec{A}=0\ \text{is the law contradicted}\ }$$
Check
Cross check against the electric case, where the analogous claim is perfectly legal: a point charge does give a non zero $\oint\vec{E}\cdot d\vec{A}$, and the first equation is written with a source term for exactly that reason. The second has no source term to write.
3§14.6 — how the energy of a beam is shared●●●○○
In a beam of sunlight the electric field amplitude is about $1000\ \mathrm{V/m}$ and the magnetic field amplitude is about $3\ \mathrm{\mu T}$. A student concludes that essentially all the energy in sunlight is carried by the electric field.
Given
$E_0\approx1.0\times10^{3}\ \mathrm{V/m}$ and $B_0\approx3.4\times10^{-6}\ \mathrm{T}$ in sunlight
$u_E=\tfrac12\varepsilon_0E^{2}$ and $u_B=B^{2}/(2\mu_0)$
$c^{2}=1/(\varepsilon_0\mu_0)$
Find
(a) True or false: nearly all the energy in the beam is electric. Justify with the two energy densities.
Hint 1/4
Two quantities can only be compared if they share a unit. Volts per metre and tesla do not, so find two quantities that do.
Hint 2/4
$u_E=\tfrac12\varepsilon_0E^{2}$ and $u_B=B^{2}/(2\mu_0)$, both in joules per cubic metre.
Hint 3/4
Put $B=E/c$ into the magnetic density: $u_B=E^{2}/(2\mu_0c^{2})$, and $1/(\mu_0c^{2})=\varepsilon_0$ because $c^{2}=1/(\varepsilon_0\mu_0)$.
Hint 4/4
That gives $u_B=\tfrac12\varepsilon_0E^{2}=u_E$, so the two shares are equal and the statement is false.
Show solution
Rewrite both energy densities in terms of the same field before comparing them, rather than substituting the given numbers. The two fields look a billion apart only because of their units, and the algebra shows the equality is exact rather than a coincidence of rounding.
the constant that closes the gap is exactly the one the wave speed is built from, which is why the equality is exact and not approximate
Answer $$\boxed{\ u_B=u_E,\ \text{so the statement is false}\ }$$
Check
Check with the numbers rather than the algebra: $u_E=\tfrac12(8.85\times10^{-12})(1.0\times10^{3})^{2}=4.4\times10^{-6}\ \mathrm{J/m^{3}}$ and $u_B=(3.4\times10^{-6})^{2}/(2\times1.257\times10^{-6})=4.6\times10^{-6}\ \mathrm{J/m^{3}}$, equal to the accuracy of the rounded field values.
4§14.4 — orienting the two fields●●●○○
A plane wave in vacuum travels in the $-y$ direction. At the instant considered, its electric field at a certain point lies along $+x$.
Given
direction of travel: $-\hat{y}$
$\vec{E}$ along $+\hat{x}$ at that instant
$\vec{E}\times\vec{B}$ points along the direction of travel
Find
(a) In which direction does $\vec{B}$ point at that instant?
Hint 1/4
One rule fixes the whole geometry, and it survives any rotation of the picture. Write it down before choosing axes.
Hint 2/4
$\vec{E}\times\vec{B}$ points along the direction of travel, and both fields are perpendicular to that direction.
Hint 3/4
Here the travel direction is $-\hat{y}$ and $\vec{E}$ is along $+\hat{x}$, so you need the unit vector $\hat{b}$ with $\hat{x}\times\hat{b}=-\hat{y}$.
Hint 4/4
Since $\hat{x}\times\hat{z}=-\hat{y}$, the magnetic field lies along $+\hat{z}$.
Show solution
Write the cross product condition as an equation in the unknown direction and solve it, rather than turning a right hand in the air. The hand rule reaches the same place, but it is where the sign goes missing when the direction of travel is a negative axis.
Solve the cross product condition
$$\hat{x}\times\hat{b}=-\hat{y}$$
writing the requirement as an equation in the unknown direction stops the guessing that this question is built to catch
checking the standard cyclic products rather than rotating a hand in the air, which is where sign errors come from
Answer $$\boxed{\ \vec{B}\parallel +\hat{z}\ }$$
Check
Test the answer by rebuilding the travel direction from scratch: $\vec{E}\times\vec{B}\propto\hat{x}\times\hat{z}=-\hat{y}$, which is where the wave was said to be going.
B · computation 7 questions
1§14.1 — field between charging plates●●●○○
A parallel plate capacitor has circular plates of radius $5.00\ \mathrm{cm}$ with vacuum between them. It is charged by a steady current of $1.50\ \mathrm{A}$ arriving along the wires.
Given
plate radius $R=5.00\times10^{-2}\ \mathrm{m}$
charging current $I=1.50\ \mathrm{A}$, steady
$\varepsilon_0=8.85\times10^{-12}$, $\mu_0=4\pi\times10^{-7}$ in SI units
Find
(a) the rate at which the electric field between the plates is growing
(b) the magnetic field at $r=2.00\ \mathrm{cm}$ from the axis, between the plates
Hint 1/4
Part (a) is about how fast the field grows, part (b) about the magnetic field a loop inside the plates encloses. Neither needs the plate separation.
Hint 2/4
$dE/dt=I/(\varepsilon_0A)$ with $A=\pi R^{2}$, and inside the plates $B(2\pi r)=\mu_0\varepsilon_0\pi r^{2}\,dE/dt$.
Hint 3/4
The data again: $R=5.00\ \mathrm{cm}$ so $A=7.85\times10^{-3}\ \mathrm{m^{2}}$, $I=1.50\ \mathrm{A}$, and the loop radius is $r=2.00\times10^{-2}\ \mathrm{m}$.
Hint 4/4
$dE/dt=2.16\times10^{13}\ \mathrm{V/(m\cdot s)}$ and $B=2.40\times10^{-6}\ \mathrm{T}$.
Show solution
Use the fraction of the displacement current enclosed rather than working through $dE/dt$ a second time in part (b); it is one line instead of three and it makes the analogy with a thick wire visible.
the separation never enters, because the field between plates is set by charge per unit area
Magnetic field from the enclosed fraction
$$B=\frac{\mu_0 I r}{2\pi R^{2}}=\frac{(4\pi\times10^{-7})(1.50)(2.00\times10^{-2})}{2\pi(2.50\times10^{-3})}$$
inside the plates the loop catches the fraction $r^{2}/R^{2}$ of the displacement current, and that fraction is exactly what turns $\mu_0I/(2\pi r)$ into this
$$B=2.40\times10^{-6}\ \mathrm{T}$$
a few microtesla, the same scale as the worked example with a similar current
Recompute $B$ the long way through the flux, from $B=\tfrac12\mu_0\varepsilon_0 r\,dE/dt$. With $\mu_0\varepsilon_0=1.11\times10^{-17}$, $r=2.00\times10^{-2}\ \mathrm{m}$ and the part (a) answer, this gives $2.40\times10^{-6}\ \mathrm{T}$, reached through different constants.
2§14.1 — displacement current from a rising voltage●●○○○
The voltage across a $4.70\ \mathrm{\mu F}$ capacitor is rising steadily at $250\ \mathrm{V/s}$ while it charges.
Given
$C=4.70\times10^{-6}\ \mathrm{F}$
$dV/dt=250\ \mathrm{V/s}$, constant
the whole gap is spanned by the surface being considered
Find
(a) the displacement current between the plates
(b) the current in the wires feeding the capacitor
Hint 1/4
Both parts want a current, and one identity from the first concept ties them together, so one calculation answers both.
Hint 2/4
$I=dQ/dt=C\,dV/dt$ for a capacitor, and $I_D$ across the whole gap equals that same $dQ/dt$.
Hint 3/4
Here $C=4.70\times10^{-6}\ \mathrm{F}$ and $dV/dt=250\ \mathrm{V/s}$.
Hint 4/4
Both currents are $C\,dV/dt=(4.70\times10^{-6})(250)=1.18\times10^{-3}\ \mathrm{A}$.
Show solution
Get the wire current from $Q=CV$ first and let the identity hand the same number to the displacement current. Building $d\Phi_E/dt$ out of the gap geometry would need a plate area the problem never gives.
the identity from the box, which holds whatever the plate geometry is
Answer $$\boxed{\ I_D=I=1.18\ \mathrm{mA}\ }$$
Check
Sanity check on the size: a microfarad charged at hundreds of volts per second should draw milliamps, since an amp would demand hundreds of kilovolts per second from this capacitance.
3§14.3 — from magnetic amplitude to intensity●●○○○
A plane wave travelling through vacuum is found to have a magnetic field amplitude of $3.20\times10^{-7}\ \mathrm{T}$.
Given
$B_0=3.20\times10^{-7}\ \mathrm{T}$
vacuum, so $c=3.00\times10^{8}\ \mathrm{m/s}$
$\varepsilon_0c=2.655\times10^{-3}$ in SI units
Find
(a) the electric field amplitude
(b) the intensity of the wave
Hint 1/4
Part (b) needs a field amplitude, and part (a) produces one, so the parts are in the right order already.
Hint 2/4
$E_0=cB_0$, then $I=\tfrac12\varepsilon_0cE_0^{2}$.
Hint 3/4
With $B_0=3.20\times10^{-7}\ \mathrm{T}$ and $c=3.00\times10^{8}\ \mathrm{m/s}$, the amplitude is $96.0\ \mathrm{V/m}$, and $\varepsilon_0c=2.655\times10^{-3}$.
Hint 4/4
$E_0=96.0\ \mathrm{V/m}$ and $I=\tfrac12(2.655\times10^{-3})(96.0)^{2}=12.2\ \mathrm{W/m^{2}}$.
Show solution
Cross to the electric field first rather than using $I=cB_0^{2}/(2\mu_0)$ directly, because part (a) asks for $E_0$ anyway and the route then costs nothing extra.
Independent route using the magnetic amplitude only: $I=cB_0^{2}/(2\mu_0)=(3.00\times10^{8})(3.20\times10^{-7})^{2}/(2\times1.257\times10^{-6})=12.2\ \mathrm{W/m^{2}}$, which never uses $E_0$ at all.
4§14.4 — reading a written wave●●●○○
A plane wave in vacuum has its electric field given, in SI units, by $E_y=45.0\sin\!\left(2.10\times10^{7}x-\omega t\right)\ \mathrm{V/m}$.
Given
$E_0=45.0\ \mathrm{V/m}$ along $y$
$k=2.10\times10^{7}\ \mathrm{rad/m}$
vacuum, so $\omega=ck$
$c=3.00\times10^{8}\ \mathrm{m/s}$
Find
(a) the wavelength and the band of the spectrum it belongs to
(b) $\omega$ and the frequency
(c) the magnetic field amplitude
Hint 1/4
Everything here is read off the standard form by matching symbols. Identify which given number is $k$ before computing anything.
The wave again: $E_0=45.0\ \mathrm{V/m}$, $k=2.10\times10^{7}\ \mathrm{rad/m}$, vacuum. Note $2\pi/(2.10\times10^{7})$ is of order $10^{-7}\ \mathrm{m}$, which is already a hint about the band.
Read $k$ straight off the coefficient of $x$ and drive everything from it. Going through the rounded wavelength to reach $\omega$ works but throws away the third digit, and $\lambda$, $\omega$ and $B_0$ each come from $k$ or $E_0$ in a single step.
Independent check with the other relation: $\lambda f=(2.99\times10^{-7})(1.00\times10^{15})=2.99\times10^{8}\ \mathrm{m/s}$, which is $c$ to the rounding of the intermediate values.
5§14.5 — gamma rays and a detector crossing time●●○○○
A gamma ray photon detector is being characterised. The radiation reaching it has a wavelength of $2.50\times10^{-12}\ \mathrm{m}$, and the detector itself is $10.0\ \mathrm{cm}$ deep.
Given
$\lambda=2.50\times10^{-12}\ \mathrm{m}$, in vacuum
detector depth $0.100\ \mathrm{m}$
$c=3.00\times10^{8}\ \mathrm{m/s}$
Find
(a) the frequency of the radiation
(b) how long the radiation takes to cross the detector
Hint 1/4
The two parts have nothing to do with each other except that both use $c$. Do them separately.
Hint 2/4
$f=c/\lambda$ for the first; $t=d/c$ for the second.
Hint 3/4
Here $\lambda=2.50\times10^{-12}\ \mathrm{m}$ and $d=0.100\ \mathrm{m}$, both with $c=3.00\times10^{8}\ \mathrm{m/s}$.
Hint 4/4
$f=1.20\times10^{20}\ \mathrm{Hz}$ and $t=3.33\times10^{-10}\ \mathrm{s}$.
Show solution
Treat the two parts as unrelated, because they share nothing but the constant $c$. Hunting for a link between the frequency and the crossing time is the trap here: a travel time in vacuum has no frequency in it.
Check part (b) against a number worth memorising: light covers about $30\ \mathrm{cm}$ in one nanosecond, so $10\ \mathrm{cm}$ should take about a third of one. It does.
6§14.6 — intensity and field from a lamp●●●○○
A small lamp radiates $60.0\ \mathrm{W}$ of electromagnetic energy equally in all directions. A meter is placed $2.00\ \mathrm{m}$ from it, facing the lamp, with nothing in between and nothing reflecting nearby.
Given
radiated power $60.0\ \mathrm{W}$, isotropic
distance $2.00\ \mathrm{m}$
$\varepsilon_0c=2.655\times10^{-3}$ in SI units
no absorption or reflection
Find
(a) the intensity at the meter
(b) the rms electric field there
(c) the electric field amplitude there
Hint 1/4
First get the watts per square metre, then convert to a field. Decide as you go whether each field asked for is an rms value or a peak.
Hint 2/4
$I=\mathcal{P}/(4\pi r^{2})$, then $I=\varepsilon_0cE_{\rm rms}^{2}$ with no half, and $E_0=\sqrt2E_{\rm rms}$.
Hint 3/4
Here $\mathcal{P}=60.0\ \mathrm{W}$ and $r=2.00\ \mathrm{m}$, so the sphere has area $4\pi(2.00)^{2}=50.3\ \mathrm{m^{2}}$, and $\varepsilon_0c=2.655\times10^{-3}$.
Hint 4/4
$I=1.19\ \mathrm{W/m^{2}}$, $E_{\rm rms}=21.2\ \mathrm{V/m}$ and $E_0=30.0\ \mathrm{V/m}$.
Show solution
Go to the rms field first and multiply up, rather than computing $E_0$ and dividing down; the rms formula has no factor of one half in it and gives fewer chances to lose it.
Close the loop through the amplitude formula, which uses the half: $I=\tfrac12\varepsilon_0cE_0^{2}=\tfrac12(2.655\times10^{-3})(900)=1.19\ \mathrm{W/m^{2}}$, recovering part (a).
7§14.7 — pressure and force on an absorbing plate●●○○○
A beam of intensity $1.20\times10^{3}\ \mathrm{W/m^{2}}$ falls squarely on a black plate of area $0.150\ \mathrm{m^{2}}$, which absorbs all of it. The beam is wide enough to cover the whole plate.
Given
$I=1.20\times10^{3}\ \mathrm{W/m^{2}}$
$A=0.150\ \mathrm{m^{2}}$
the plate absorbs everything and faces the beam squarely
Find
(a) the radiation pressure on the plate
(b) the total force on it
Hint 1/4
Decide what the surface does to the beam before writing any formula, because that single decision changes the answer by a factor of two.
Hint 2/4
For an absorbing surface $P_{\rm rad}=I/c$, and then $F=P_{\rm rad}A$.
Hint 3/4
Here $I=1.20\times10^{3}\ \mathrm{W/m^{2}}$, $c=3.00\times10^{8}\ \mathrm{m/s}$ and $A=0.150\ \mathrm{m^{2}}$, and the plate is black.
Hint 4/4
$P_{\rm rad}=4.00\times10^{-6}\ \mathrm{Pa}$ and $F=6.00\times10^{-7}\ \mathrm{N}$.
Show solution
Settle what the surface does to the beam before writing any formula, because that one word decides between $I/c$ and $2I/c$. Everything after it is one division and one multiplication whichever way it goes.
Through the power instead: the plate intercepts $IA=180\ \mathrm{W}$, and $F=\mathcal{P}/c=180/(3.00\times10^{8})=6.00\times10^{-7}\ \mathrm{N}$, reached without a pressure.
C · exam level 4 questions
1§14.1 — displacement current and the field beyond the rim●●●●○
A parallel plate capacitor has circular plates of radius $4.00\ \mathrm{cm}$ with vacuum between them. While it charges, the electric field between the plates grows uniformly at $6.00\times10^{12}\ \mathrm{V/(m\cdot s)}$. A magnetic field is wanted on a circle of radius $6.00\ \mathrm{cm}$ centred on the axis, still in the plane midway between the plates but outside the plate rim.
Given
plate radius $R=4.00\times10^{-2}\ \mathrm{m}$, so plate area $A=5.03\times10^{-3}\ \mathrm{m^{2}}$
$dE/dt=6.00\times10^{12}\ \mathrm{V/(m\cdot s)}$, uniform between the plates and negligible outside them
loop radius $r=6.00\times10^{-2}\ \mathrm{m}$, outside the plate rim
$\varepsilon_0=8.85\times10^{-12}$, $\mu_0=4\pi\times10^{-7}$ in SI units
Find
(a) the total displacement current between the plates, and the magnetic field on that circle
Hint 1/4
Two answers are wanted. The first is a property of the whole gap; the second depends on where your loop is, and the loop here is beyond the plates.
Hint 2/4
$I_D=\varepsilon_0A\,dE/dt$ over the plate area, and outside the rim $B(2\pi r)=\mu_0 I_D$ with the full $I_D$ enclosed.
Hint 3/4
The data again: $A=5.03\times10^{-3}\ \mathrm{m^{2}}$, $dE/dt=6.00\times10^{12}\ \mathrm{V/(m\cdot s)}$, $r=6.00\times10^{-2}\ \mathrm{m}$, $\varepsilon_0=8.85\times10^{-12}$.
Hint 4/4
$I_D=(8.85\times10^{-12})(5.03\times10^{-3})(6.00\times10^{12})=0.267\ \mathrm{A}$, and $B=\mu_0I_D/(2\pi r)=8.90\times10^{-7}\ \mathrm{T}$.
Show solution
Compute the total displacement current first and only then apply the loop law, rather than carrying $dE/dt$ into the loop equation. Doing it in that order makes the enclosed area question impossible to get wrong: once the loop is outside, the enclosed current is simply all of it.
Continuity check: at the rim itself, $r=R=4.00\ \mathrm{cm}$, the inside formula gives $\mu_0I_DR/(2\pi R^{2})=\mu_0I_D/(2\pi R)=1.33\times10^{-6}\ \mathrm{T}$ and the outside formula gives the same value, so the two expressions meet without a jump. That is the strongest single test of both.
2§14.4 — building a wave from a physical description●●●●○
A transmitter sends a plane wave through vacuum at a frequency of $750\ \mathrm{kHz}$. The wave travels in the $-z$ direction, and at a certain point and instant its electric field points along $+x$ with amplitude $350\ \mathrm{V/m}$.
$\vec{E}$ along $+\hat{x}$, amplitude $350\ \mathrm{V/m}$
vacuum, $c=3.00\times10^{8}\ \mathrm{m/s}$
Find
(a) the wavelength, the magnetic field amplitude and the direction of $\vec{B}$ at that instant
Hint 1/4
Three answers, and they are independent of each other: one from the frequency, one from the field ratio, one from the geometry rule.
Hint 2/4
$\lambda=c/f$; $B_0=E_0/c$; and $\vec{E}\times\vec{B}$ points along the direction of travel.
Hint 3/4
The data again: $f=7.50\times10^{5}\ \mathrm{Hz}$, $E_0=350\ \mathrm{V/m}$ along $+\hat{x}$, travel along $-\hat{z}$. You need $\hat{b}$ with $\hat{x}\times\hat{b}=-\hat{z}$.
Hint 4/4
$\lambda=400\ \mathrm{m}$, $B_0=1.17\times10^{-6}\ \mathrm{T}$, and since $\hat{x}\times(-\hat{y})=-\hat{z}$ the magnetic field lies along $-\hat{y}$.
Show solution
Take the three answers as three independent questions rather than trying to write the wave out in full: the wavelength comes from the frequency, the amplitude from the field ratio, and the direction from a cross product that uses neither number.
Rebuild the direction of travel from the answer: $\vec{E}\times\vec{B}\propto\hat{x}\times(-\hat{y})=-\hat{z}$, which is where the wave was said to go. Separately, $\lambda f=(400)(7.50\times10^{5})=3.00\times10^{8}\ \mathrm{m/s}$ confirms the wavelength.
3§14.6 — how far out an isotropic beacon reaches●●●●○
A beacon radiates $2.00\ \mathrm{kW}$ of electromagnetic power equally in all directions from a point, with nothing absorbing or reflecting anywhere near it. A detector needs an intensity of at least $1.00\times10^{-3}\ \mathrm{W/m^{2}}$ to register a signal.
Given
radiated power $\mathcal{P}=2.00\times10^{3}\ \mathrm{W}$, isotropic
(a) the greatest distance at which the detector still registers, and the electric field amplitude there
Hint 1/4
The first part is the inverse square law solved for the distance rather than for the intensity. The second is the standard intensity to field conversion, and you must decide peak or rms.
Hint 2/4
$I=\mathcal{P}/(4\pi r^{2})$ rearranged gives $r=\sqrt{\mathcal{P}/(4\pi I)}$, and $E_0=\sqrt{2I/(\varepsilon_0c)}$ for the amplitude.
Hint 3/4
The data again: $\mathcal{P}=2.00\times10^{3}\ \mathrm{W}$, $I=1.00\times10^{-3}\ \mathrm{W/m^{2}}$, $\varepsilon_0c=2.655\times10^{-3}$. The bracket under the first root is $2.00\times10^{3}/(1.257\times10^{-2})$.
Hint 4/4
$r=\sqrt{1.59\times10^{5}}=399\ \mathrm{m}$ and $E_0=\sqrt{2(1.00\times10^{-3})/(2.655\times10^{-3})}=0.868\ \mathrm{V/m}$.
Show solution
Solve the inverse square law for $r$ symbolically before substituting, because doing it numerically invites forgetting the square root, which is what one of the wrong options is built from.
Walk the chain backwards: at $399\ \mathrm{m}$ the sphere has area $4\pi(399)^{2}=2.00\times10^{6}\ \mathrm{m^{2}}$, and $2.00\times10^{3}/2.00\times10^{6}=1.00\times10^{-3}\ \mathrm{W/m^{2}}$, the threshold. Then $\tfrac12\varepsilon_0cE_0^{2}=\tfrac12(2.655\times10^{-3})(0.753)=1.00\times10^{-3}\ \mathrm{W/m^{2}}$ as well.
4§14.7 — force on a reflecting sheet in sunlight●●●○○
A perfectly reflecting sheet of area $2.00\ \mathrm{m^{2}}$ is held face on to sunlight in space, where the intensity is $1360\ \mathrm{W/m^{2}}$. To make the answer easier to picture, the force is also to be expressed as the mass whose weight on Earth would be equal to it.
Given
$I=1360\ \mathrm{W/m^{2}}$
$A=2.00\ \mathrm{m^{2}}$, perfectly reflecting, face on
$c=3.00\times10^{8}\ \mathrm{m/s}$
$g=9.80\ \mathrm{m/s^{2}}$ for the comparison
Find
(a) the force on the sheet, and the mass whose weight equals that force
Hint 1/4
Two steps: a force from the beam, and then a comparison that turns that force into a mass. Decide first what the surface does to the beam.
Hint 2/4
For a mirror $P_{\rm rad}=2I/c$, then $F=P_{\rm rad}A$, and $m=F/g$ for the comparison.
Hint 3/4
The data again: $I=1360\ \mathrm{W/m^{2}}$, $A=2.00\ \mathrm{m^{2}}$, perfectly reflecting, and $g=9.80\ \mathrm{m/s^{2}}$.
Hint 4/4
$F=2(1360)(2.00)/(3.00\times10^{8})=1.81\times10^{-5}\ \mathrm{N}$, and $m=F/g=1.85\times10^{-6}\ \mathrm{kg}=1.85\ \mathrm{mg}$.
Show solution
Decide reflecting or absorbing before anything else, since that word alone is worth a factor of two, then go pressure, force, mass. Keeping the division by $c$ and the division by $g$ on separate lines is what stops them being swapped.
Independent route through power: the sheet intercepts $IA=2720\ \mathrm{W}$, and a reflected beam delivers $2\mathcal{P}/c=2(2720)/(3.00\times10^{8})=1.81\times10^{-5}\ \mathrm{N}$, reached without a pressure or an area.
D · interleaved 4 questions
1§14.1 — a charging circuit, watched from inside the gap●●●●○
A $220\ \mathrm{\mu F}$ capacitor, initially uncharged, is connected through a $1.50\ \mathrm{k\Omega}$ resistor to a $12.0\ \mathrm{V}$ battery. The switch closes at $t=0$.
Given
$C=2.20\times10^{-4}\ \mathrm{F}$, uncharged at $t=0$
$R=1.50\times10^{3}\ \Omega$
battery $12.0\ \mathrm{V}$
the surface considered spans the whole gap between the plates
Find
(a) the time constant of the circuit
(b) the displacement current between the plates immediately after the switch closes
(c) the displacement current one time constant later
Hint 1/4
This looks like two topics but it is one: whatever the wire current is at any instant, something in the gap has to match it. Find the wire current first.
Hint 2/4
For a charging circuit $\tau=RC$ and $I(t)=(V/R)e^{-t/\tau}$; and the displacement current across the full gap equals the conduction current at every instant.
Hint 3/4
The data again: $R=1.50\times10^{3}\ \Omega$, $C=2.20\times10^{-4}\ \mathrm{F}$, $V=12.0\ \mathrm{V}$. At $t=\tau$ the exponential factor is $e^{-1}=0.368$.
Hint 4/4
$\tau=0.330\ \mathrm{s}$, $I_D(0)=8.00\ \mathrm{mA}$ and $I_D(\tau)=2.94\ \mathrm{mA}$.
Show solution
Answer the circuit question completely and only then convert, rather than trying to work with fluxes in the gap. The identity from the first concept makes the conversion free, and nothing about the plate geometry is even needed.
Check the last number against the charge instead of the current: at $t=\tau$ the capacitor holds $Q=CV(1-e^{-1})=(2.20\times10^{-4})(12.0)(0.632)=1.67\times10^{-3}\ \mathrm{C}$, so its voltage is $7.58\ \mathrm{V}$ and the resistor carries $(12.0-7.58)/1500=2.95\times10^{-3}\ \mathrm{A}$, matching to the rounding.
2§14.2 — which law applies inside a solenoid whose current is rising●●●●○
A long solenoid with $2000\ \mathrm{turns/m}$ carries a current that is increasing steadily at $150\ \mathrm{A/s}$. A small coil of $50$ turns, each of area $4.00\ \mathrm{cm^{2}}$, sits inside the solenoid with its plane perpendicular to the axis.
Given
solenoid $n=2000\ \mathrm{turns/m}$, field inside $B=\mu_0nI$ and uniform
$dI/dt=150\ \mathrm{A/s}$, steady
small coil: $N=50$ turns, area $4.00\times10^{-4}\ \mathrm{m^{2}}$, perpendicular to the axis
$\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m/A}$
Find
(a) the rate at which the magnetic field inside the solenoid is growing
(b) the magnitude of the induced voltage in the small coil
Hint 1/4
Two of the four equations are candidates. The geometry is a loop and the quantity changing is a magnetic flux, which picks one of them out.
Hint 2/4
$B=\mu_0nI$ inside a long solenoid, so $dB/dt=\mu_0n\,dI/dt$; and the induced voltage in a coil is $N A\,dB/dt$ in magnitude.
Hint 3/4
The data again: $n=2000\ \mathrm{turns/m}$, $dI/dt=150\ \mathrm{A/s}$, $N=50$, $A=4.00\times10^{-4}\ \mathrm{m^{2}}$, $\mu_0=1.257\times10^{-6}$.
Hint 4/4
$dB/dt=(1.257\times10^{-6})(2000)(150)=0.377\ \mathrm{T/s}$, so the voltage is $(50)(4.00\times10^{-4})(0.377)=7.54\times10^{-3}\ \mathrm{V}$.
Show solution
Nothing in this problem needs the new displacement current term, and adding it would be wrong: the electric flux through the coil is not what is changing. Naming which law is in play before computing is the point of putting this question here.
Independent route through flux linkage: after one second the field has grown by $0.377\ \mathrm{T}$, so the flux linkage has grown by $NA\Delta B=(50)(4.00\times10^{-4})(0.377)=7.54\times10^{-3}\ \mathrm{Wb}$, and a weber per second is a volt.
3§14.2 — two closed surface laws applied to one sphere●●●○○
A sphere of radius $0.300\ \mathrm{m}$ is drawn in empty space. Inside it, well away from the surface, sit a point charge of $-5.00\ \mathrm{nC}$ and a small permanent magnet whose field is strong enough to be detected outside the sphere.
Given
sphere radius $0.300\ \mathrm{m}$
point charge $q=-5.00\times10^{-9}\ \mathrm{C}$ inside
(c) what changes in each answer if the sphere's radius is doubled
Hint 1/4
Both parts are closed surface questions, so both of the flux laws are in play and neither circulation law is. Take them one at a time.
Hint 2/4
$\oint\vec{E}\cdot d\vec{A}=Q_{\rm encl}/\varepsilon_0$ and $\oint\vec{B}\cdot d\vec{A}=0$.
Hint 3/4
The data again: $q=-5.00\times10^{-9}\ \mathrm{C}$ enclosed, $\varepsilon_0=8.85\times10^{-12}$, and a magnet of any strength you like enclosed as well.
Hint 4/4
$\Phi_E=-565\ \mathrm{N\,m^{2}/C}$, $\Phi_B=0$, and neither changes when the radius doubles because neither law contains the radius.
Show solution
Answer the two parts with two different laws instead of looking for one calculation that covers both. Part (b) has no arithmetic in it at all, and any effort spent on the magnet's strength is spent on a quantity its law does not contain.
Check the electric answer by integrating instead: at $r=0.300\ \mathrm{m}$ the field magnitude is $kq/r^{2}=(8.99\times10^{9})(5.00\times10^{-9})/0.0900=499\ \mathrm{N/C}$ pointing inward, and $499\times4\pi(0.0900)=565\ \mathrm{N\,m^{2}/C}$ of inward flux, matching in size and sign.
4§14.4 — from an oscillating circuit to the wave it sends out●●●●○
An oscillating circuit with an inductance of $2.00\ \mathrm{\mu H}$ and a capacitance of $3.00\ \mathrm{pF}$ feeds an aerial, which radiates at the circuit's natural frequency into the air above it.
the circuit oscillates freely at its natural frequency, with negligible resistance
the radiated wave travels in air, close enough to vacuum for $c$
Find
(a) the natural angular frequency and the frequency in hertz
(b) the wavelength of the wave radiated, and its band of the spectrum
Hint 1/4
The circuit decides the frequency and the frequency decides the wavelength. Nothing about the wave feeds back into the circuit, so the two halves are done in order.
Hint 2/4
$\omega=1/\sqrt{LC}$ and $f=\omega/2\pi$ for the circuit; $\lambda=c/f$ for the wave.
Hint 3/4
The data again: $L=2.00\times10^{-6}\ \mathrm{H}$ and $C=3.00\times10^{-12}\ \mathrm{F}$, so $LC=6.00\times10^{-18}$ and $c=3.00\times10^{8}\ \mathrm{m/s}$.
Hint 4/4
$\omega=4.08\times10^{8}\ \mathrm{rad/s}$, $f=6.50\times10^{7}\ \mathrm{Hz}$, and $\lambda=4.62\ \mathrm{m}$, which is in the radio band.
Show solution
Go through $\omega$ rather than trying to write a single combined formula for $\lambda$ in terms of $L$ and $C$. The combined form exists but hides the factor of $2\pi$, which is exactly what gets lost.
Check the wavelength without the frequency, using $\lambda=2\pi c\sqrt{LC}=2\pi(3.00\times10^{8})(2.45\times10^{-9})=4.62\ \mathrm{m}$, a single line that skips both intermediate numbers.
Mistake ledger (18 entries)
⚠ Putting the loop area into the flux when the loop is outside the plates
inside the plates the enclosed flux really is $E\pi r^{2}$, and the habit carries on past the rim, where there is no field left to enclose
both integrals are written with a ring on the sign, and the difference between $d\vec{A}$ over a closed surface and $d\vec{\ell}$ around a curve is one symbol
wrong$$\oint\vec{B}\cdot d\vec{A}=0\ \Rightarrow\ \Phi_B\ \text{through a loop is zero}$$
right$$\oint\vec{B}\cdot d\vec{A}=0\ \text{only for a closed surface}$$
⚠ Swapping which flux belongs to which circulation law
both laws now contain a changing flux, so the pair is easy to cross over under time pressure
⚠ Multiplying instead of dividing in $\lambda f=c$
the formula has three symbols and no fraction bar, so the rearrangement is done from memory rather than from the equation
wrong$$\lambda=cf$$
right$$\lambda=\frac{c}{f}$$
⚠ Putting the factor of one half into the rms form as well
the half is remembered as part of the intensity formula rather than as the average of a squared sine, so it travels with the formula instead of with the amplitude
the loop is any closed curve and the surface is any surface with that curve as its rim; $\Phi_E$ is the electric flux through that same surface, with the normal fixed by the right hand rule on the loop; $I_{\rm encl}$ counts real charge crossing the surface, and is zero for a surface drawn through vacuum; the two terms are added, so a region can have both at once, as inside a leaky capacitor
each closed surface integral runs over a closed surface, each closed line integral around a closed loop with a surface spanning it; $Q_{\rm encl}$ and $I_{\rm encl}$ mean net charge inside the surface and net current through it; written for vacuum; inside a material $\varepsilon_0\to K\varepsilon_0$ and $\mu_0\to \mu$, which is the only change that matters at this level; the field directions follow the right hand rule fixed in the conventions above
Theorem 14.3: an electromagnetic wave, its speed, and the ratio of its fields
vacuum, with no charge and no conduction current in the region; a plane wave: the fields are uniform across any plane perpendicular to the direction of travel; the derivation below assumes a flat front, which is what a wave looks like far from whatever made it; for a material of dielectric constant $K$ and negligible magnetism, replace $\varepsilon_0$ by $K\varepsilon_0$ throughout
Definition 14.4: the travelling sinusoidal plane wave
travelling along $+x$; for travel along $-x$ the sign inside the bracket becomes a plus; $\vec{E}$ along $y$ and $\vec{B}$ along $z$, ordered so that $\vec{E}\times\vec{B}$ points along $+x$; in vacuum, so that $\omega/k=c$; inside a material the same forms hold with $v$ in place of $c$; a plane wave, meaning the fields do not depend on $y$ or $z$ at all
Definition 14.5: the two numbers that name a wave, in vacuum and in matter
$\lambda f=c$ holds in vacuum for every member of the family, from radio to gamma; in a material of dielectric constant $K$ that is not magnetic, the speed drops to $c/\sqrt{K}$; the frequency is set by the source and does not change when the wave enters a material; the band names are conventions with fuzzy borders, not physical boundaries
Theorem 14.6: the Poynting vector, intensity, and an isotropic source
$\vec{S}$ is instantaneous; for a sinusoidal wave it oscillates at twice the wave frequency and is never negative; the intensity $I$ is the time average of $|\vec{S}|$ over a whole number of cycles; the amplitude forms carry a factor $\tfrac12$ that is the average of a squared sine; the rms forms do not; the isotropic form assumes equal power in every direction and no absorption on the way out
Rule 14.7: momentum of a beam and the pressure it exerts
the beam arrives along the normal to the surface; at an angle both results pick up a cosine factor; fully absorbed or fully reflected; a real grey surface lies between the two and is not treated here; $I$ is the time averaged intensity, so the pressure obtained is also a time average; the force follows as $F=P_{\rm rad}A$ with $A$ the area the beam actually lands on
non magnetic material of dielectric constant $K$ at the frequency in question
Check yourself
Close the page and write, from memory, the four equations in any order, then next to each one write in a single sentence what physical fact it records. Then write the chain that turns a beam's power into a force on a mirror, naming every quantity you pass through. Whatever you could not produce is where to reread, and the list below says where.
Say what the displacement current is made of, and compute the magnetic field both inside and outside the rim of a charging capacitor?
c-displacement-current
Write all four equations and decide, from a description of a situation, which one it is about?
c-maxwell-equations
Derive $c=1/\sqrt{\varepsilon_0\mu_0}$ from the two circulation laws, and convert between $E_0$ and $B_0$ in both directions without hesitating over which way to divide?
c-wave-speed
Take a wave written as a formula and read off its wavelength, frequency, direction of travel and both field amplitudes, and build the formula back from a physical description?
c-plane-wave
Place a wave in the spectrum from either of its two numbers, and say what changes and what does not when it enters glass?
c-spectrum
Move between intensity, field amplitude, rms field and radiated power for both a beam and an isotropic source, with the factor of one half in the right place every time?
c-poynting
Compute the pressure and force on an absorbing and on a reflecting surface, and judge whether that force matters against the other forces in the problem?
c-radiation-pressure
Glossary (27 terms)
displacement currentyer değiştirme akımı
The quantity $\varepsilon_0\,d\Phi_E/dt$, measured in amperes, which enters the completed Ampère's law in the same slot as a real current and produces a magnetic field in exactly the same way. Nothing moves and no charge is involved.
Maxwell's equations
The four laws that between them describe every electric and magnetic field in this course: two about flux through closed surfaces and two about circulation around closed loops.
manyetik tek kutup
A hypothetical isolated magnetic pole, which would act as a source of magnetic field the way a charge is a source of electric field. None has ever been found, and the flux law for the magnetic field is the statement that none exists.
electromagnetic waveelektromanyetik dalga
A self supporting disturbance of the electric and magnetic fields in which each one is regenerated by the change of the other, travelling through vacuum at $3.00\times10^{8}\ \mathrm{m/s}$ with no medium required.
transverse waveenine dalga
A wave whose oscillation is perpendicular to the direction it travels. Both fields of an electromagnetic wave are transverse and also perpendicular to each other.
plane wavedüzlem dalga
A wave whose fields have the same value everywhere on any plane perpendicular to the direction of travel, so they depend on one spatial coordinate only. It is what any wave looks like far from its source.
dalga cephesi
A surface joining points of the same phase, so a surface of crests or a surface of zeros. For a plane wave these are flat parallel planes advancing at the wave speed.
wave numberdalga sayısı
The quantity $k=2\pi/\lambda$ in radians per metre, counting how much phase the wave advances per metre travelled. It is the spatial partner of the angular frequency.
angular frequencyaçısal frekans
The quantity $\omega=2\pi f$ in radians per second, counting how much phase the wave advances per second. Its ratio to the wave number is the wave speed.
amplitudegenlik
The largest value a field reaches during a cycle, written with a subscript zero. For a sine wave the root mean square value is smaller by a factor of the square root of two.
periyot
The time for one complete cycle of the oscillation at a fixed point, equal to $1/f$ and to $2\pi/\omega$.
speed of light in vacuumışık hızı
The constant $c=1/\sqrt{\varepsilon_0\mu_0}=3.00\times10^{8}\ \mathrm{m/s}$, the same for every frequency and every observer, at which any electromagnetic wave travels through empty space.
electromagnetic spectrumelektromanyetik tayf
The full range of electromagnetic waves ordered by wavelength or frequency, from radio waves metres long to gamma rays smaller than an atomic nucleus, all travelling at the same speed in vacuum.
radio waveradyo dalgası
The longest wavelength part of the spectrum, above roughly a tenth of a metre, produced by currents oscillating in aerials and used for broadcasting and communication.
microwavemikrodalga
The band between about a tenth of a metre and a millimetre, used for radar, mobile telephony and ovens, where the wavelength is comparable with everyday objects.
kızılötesi ışınım
Wavelengths just longer than red light, from about seven hundred nanometres to a millimetre, emitted by any warm object and felt as radiant heat.
visible lightgörünür ışık
The narrow band from roughly four hundred to seven hundred nanometres that the eye responds to, less than one decade wide out of the fifteen shown in the spectrum figure.
morotesi ışınım
Wavelengths just shorter than violet light, from about four hundred nanometres down to ten nanometres, energetic enough to damage biological molecules.
X rayX ışını
Wavelengths from roughly ten nanometres down to a hundredth of a nanometre, comparable with the spacing between atoms in a solid, which is why crystals are studied with them.
gamma raygama ışını
The shortest wavelength part of the spectrum, below about a hundredth of a nanometre, produced by nuclear processes.
index of refractionkırılma indisi
The ratio of the vacuum speed of light to its speed inside a material, equal to $\sqrt{K}$ for a non magnetic medium of dielectric constant $K$.
Poynting vector
The vector $\vec{S}=(1/\mu_0)\vec{E}\times\vec{B}$, whose direction is the direction energy is flowing and whose magnitude is the instantaneous power crossing unit area.
intensityşiddet
The time averaged magnitude of the Poynting vector, in watts per square metre; the quantity a light meter reports and the one every energy formula in this section is written in.
isotropic source
A source radiating equal power in every direction, so that its intensity at distance $r$ is the total power divided by the area $4\pi r^{2}$ of the sphere the wave has reached.
güneş sabiti
The intensity of sunlight above the Earth's atmosphere, about $1360\ \mathrm{W/m^{2}}$, used throughout this section as the reference against which other beams are judged.
radiation pressureışınım basıncı
The force per unit area a beam exerts on a surface, equal to the intensity divided by the speed of light if the beam is absorbed and twice that if it is reflected straight back.
antennaanten
A conductor in which charge is driven back and forth so that the changing fields detach and travel away as a wave, or in which an arriving wave drives a measurable current. It works best when its length is a simple fraction of the wavelength.
What comes next
This is the end of the material. Everything from the first section onward has been building one set of four equations, and this section wrote them down and then found light hiding inside them. What is left is the final, which the syllabus weights at twenty five per cent, and the review sections in the middle of the course are the fastest way back into the earlier half.
Sources
D. Giancoli, Physics for Scientists and Engineers with Modern Physics, 5th edition the set textbook named in the syllabus; the chapter on Maxwell's equations and electromagnetic waves is the one this section corresponds to
SI values of the electric and magnetic constants $\varepsilon_0=8.85\times10^{-12}\ \mathrm{C^{2}/(N\cdot m^{2})}$ and $\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m/A}$, used to three digits throughout
Solar constant and Sun to Earth distance $1360\ \mathrm{W/m^{2}}$ above the atmosphere and $1.50\times10^{11}\ \mathrm{m}$, standard reference values used in the sunlight examples
Band edges of the electromagnetic spectrum the visible band taken as 400 to 700 nm and the other band edges as conventionally quoted; these boundaries are conventions, not physical discontinuities