Ch.10 Electromagnetic Fields and Waves

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Overview

Reference Frames

If something moves at the same velocity as a charge relative to a reference point, that object sees the charge as stationary, while the reference point sees the charge as moving. But that creates two different magnetic field strengths, and two different magnetic forces. Which is correct?

Consider reference point A and reference point B observing a charge C. The velocity of frame B relative to A is v⃗BA\vec{v}_\text{BA}, the velocity of charge C relative to frame A is v⃗CA\vec{v}_\text{CA}, and the velocity of charge C relative to frame B is v⃗CB\vec{v}_\text{CB}. Assume both reference frames are inertial, so v⃗BA\vec{v}_\text{BA} is constant.
We have
v⃗CA=v⃗CB+v⃗BA\vec{v}_\text{CA}=\vec{v}_\text{CB}+\vec{v}_\text{BA}
the Galilean transformation of velocity (using vector addition)
Taking the time derivative, noting that v⃗BA\vec{v}_\text{BA} is constant, we have
dv⃗CAdt=dv⃗CBdt\frac{d\vec{v}_\text{CA}}{dt}=\frac{d\vec{v}_\text{CB}}{dt}
meaning the acceleratiosn are the same. Thus, the force observed by the two reference frames should also be the same.

Now suppose reference frame A establishes a region of uniform magnetic field B⃗A\vec{B}_\text{A} and no electric field. A positive charge qq sent through it at velocity v⃗CA\vec{v}_\text{CA} experiences force F⃗A=qv⃗CA×B⃗A\vec{F}_\text{A}=q\vec{v}_\text{CA}\times\vec{B}_A
But if reference frame B moves at the same v⃗BA=v⃗CA\vec{v}_\text{BA}=\vec{v}_\text{CA}, then the relative velocity of the particle is 0, so there should be no magnetic force, but we know the charge experiences a force. So, there must be an electric field. Since we know F⃗A=F⃗B\vec{F}_\text{A}=\vec{F}_\text{B}, we have
qv⃗CA×B⃗A=qE⃗B⟹E⃗B=v⃗BA×B⃗Aq\vec{v}_\text{CA}\times\vec{B}_A=q\vec{E}_B\implies \vec{E}_B=\vec{v}_\text{BA}\times\vec{B}_A
Thus, whether a field is observed as electric or magnetic depends on the motion of the reference frame relative to the sources of the field.
The same logic can be applied if there is a E⃗A\vec{E}_\text{A} as well, with
E⃗B=E⃗A+v⃗BA×B⃗A\vec{E}_\text{B}=\vec{E}_\text{A}+\vec{v}_\text{BA}\times\vec{B}_\text{A}

If a charge qq is stationary in reference A with B⃗A=0\vec{B}_A=0, then we observe
E⃗A=14πϵ0qr2r^\vec{E}_\text{A}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}
From reference B, because B⃗A=0\vec{B}_\text{A}=0, we have
E⃗B=E⃗A=14πϵ0qr2r^\vec{E}_\text{B}=\vec{E}_\text{A}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}
which shows Coulomb's law is still valid in a frame where the charge moves.
However, from reference B, the charge is moving at a velocity v⃗CB=−v⃗BA\vec{v}_\text{CB}=-\vec{v}_\text{BA}. By Biot-Savart law,
B⃗B=μ04πqr2v⃗CB×r^=−μ04πqr2v⃗BA×r^\vec{B}_B=\frac{\mu_0}{4\pi}\frac{q}{r^2}\vec{v}_\text{CB}\times\hat{r}=-\frac{\mu_0}{4\pi}\frac{q}{r^2}\vec{v}_\text{BA}\times\hat{r}
Factoring out the equation for electric field, we have
B⃗B=−ϵ0μ0v⃗BA×(14πϵ0qr2r^)=−ϵ0μ0v⃗BA×E⃗A\vec{B}_B=-\epsilon_0\mu_0\vec{v}_\text{BA}\times\left(\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}\right)=-\epsilon_0\mu_0\vec{v}_\text{BA}\times\vec{E}_A
Thus, the Biot-Savart law is simply a Coulomb electric field transformed into a moving reference frame. The general transformation equation is
B⃗B=B⃗A+(−ϵ0μ0v⃗BA×E⃗A)\vec{B}_\text{B}=\vec{B}_\text{A}+(-\epsilon_0\mu_0\vec{v}_\text{BA}\times\vec{E}_\text{A})

If we examine the units of ϵ0μ0\epsilon_0\mu_0, we have
T mA⋅C2N m2=(NA m⋅mA)⋅C2N m2=C2A2 m2=s2m2\frac{\text{T m}}{\text{A}}\cdot\frac{\text{C}^2}{\text{N m}^2}=\left(\frac{\text{N}}{\text{A m}}\cdot\frac{\text{m}}{\text{A}}\right)\cdot\frac{\text{C}^2}{\text{N m}^2}=\frac{\text{C}^2}{\text{A}^2\text{ m}^2}=\frac{\text{s}^2}{\text{m}^2}
taking the reciprocal of the square root, the unit is m/s\text{m}/\text{s}, or those of velocity. Later, we will see that
vem=c=1ϵ0μ0=3.00⋅108m/sv_\text{em}=c=\frac{1}{\sqrt{\epsilon_0\mu_0}}=3.00\cdot10^8\text{m}/\text{s}
Using this, we write the Galilean field transformation equations
E⃗B=E⃗A+v⃗BA×B⃗AB⃗B=B⃗A−1c2v⃗BA×E⃗A\vec{E}_\text{B}=\vec{E}_\text{A}+\vec{v}_\text{BA}\times\vec{B}_\text{A}\\\vec{B}_\text{B}=\vec{B}_\text{A}-\frac{1}{c^2}\vec{v}_\text{BA}\times\vec{E}_\text{A}
where v⃗BA\vec{v}_\text{BA} is the velocity of reference frame B relative to reference frame A, and the fields are measured at the same point in space by reference points at rest in each respective frame.
Clearly, E⃗\vec{E} and B⃗\vec{B} are not independent. There must be some electromagnetic field that presents different fields in terms of E⃗\vec{E} and B⃗\vec{B} to different viewers.

Consider two charges, q1q_1 above q2q_2, moving at velocity v⃗CA\vec{v}_\text{CA} right relative to frame A.
Then q2q_2 exerts an electric field at q1q_1 and a magnetic field in frame A by Coulomb's law and Biot-Savart law
E⃗A=14πϵ0q2r2j^\vec{E}_\text{A}=\frac{1}{4\pi\epsilon_0}\frac{q_2}{r^2}\hat{j}
B⃗A=μ04πq2vCAr2k^\vec{B}_\text{A}=\frac{\mu_0}{4\pi}\frac{q_2v_\text{CA}}{r^2}\hat{k}
Then using the transformation equations, we can find the fields in frame B moving at v⃗BA=v⃗CA\vec{v}_\text{BA}=\vec{v}_\text{CA} (putting the charges at rest)
B⃗B=μ04πq2vCAr2k^−1c2(vBAi^×14πϵ0q2r2j^)=μ04πq2vCAr2(1−1μ0ϵ0c2)\vec{B}_B=\frac{\mu_0}{4\pi}\frac{q_2v_\text{CA}}{r^2}\hat{k}-\frac{1}{c^2}\left(v_\text{BA}\hat{i}\times\frac{1}{4\pi\epsilon_0}\frac{q_2}{r^2}\hat{j}\right)\\=\frac{\mu_0}{4\pi}\frac{q_2v_\text{CA}}{r^2}\left(1-\frac{1}{\mu_0\epsilon_0c^2}\right)
Since the charges are stationary in frame B, we must have B⃗B=0\vec{B}_B=0. Since μ0ϵ0=1/c2\mu_0\epsilon_0=1/c^2, the term in the parenthesis is 00, giving B⃗B=0\vec{B}_B=0, which agrees with that.
E⃗B=14πϵ0q2r2j^+vBAi^×μ04πq2vBAr2k^=14πϵ0q2r2(1−vBA2c2)j^\vec{E}_\text{B}=\frac{1}{4\pi\epsilon_0}\frac{q_2}{r^2}\hat{j}+v_\text{BA}\hat{i}\times\frac{\mu_0}{4\pi}\frac{q_2v_\text{BA}}{r^2}\hat{k}\\=\frac{1}{4\pi\epsilon_0}\frac{q_2}{r^2}\left(1-\frac{v^2_\text{BA}}{c^2}\right)\hat{j}
But we established previously that Coulomb's law is still valid for moving charges, so we should have
E⃗B=14πϵ0q2r2j^\vec{E}_B=\frac{1}{4\pi\epsilon_0}\frac{q_2}{r^2}\hat{j}
This is because the Galilean tranformations don't account for Einstein's relativity. Thus, the Galilean transformation equations are only valid for vBA≪c2v_\text{BA}\ll c^2

Faraday's Law Revisited

Recall a conducting loop moving into a magnetic field with velocity v⃗\vec{v} creates a magnetic force that indcues an emf ε=vLB\varepsilon=vLB. How does it appear to a reference frame in which the loop is stationary?
There must be an induced current, but where does it come from? We transform the fields into frame B
E⃗B=E⃗A+v⃗×B⃗A=v⃗×B⃗B⃗B=B⃗A−1c2(v⃗×E⃗A)=B⃗\vec{E}_\text{B}=\vec{E}_\text{A}+\vec{v}\times\vec{B}_\text{A}=\vec{v}\times\vec{B}\\\vec{B}_\text{B}=\vec{B}_\text{A}-\frac{1}{c^2}\left(\vec{v}\times\vec{E}_\text{A}\right)=\vec{B}
There is no magnetic force, becuase the loop is stationary, but there is an electric force.
F⃗E=qE⃗=qv⃗×B⃗=qvB,upward\vec{F}_E=q\vec{E}=q\vec{v}\times\vec{B}=qvB,\text{upward}
which agrees with frame A.
This field is the induced electric field. Faraday's law says that a changing magnetic field creates an electric field, but the magnetic field only changes in frame B, so the electric field is only observable in frame B.

Field Laws Summary

We have Gauss's Law for the flux through a closed loop. This shows that electric field lines start and stop at charges.
(Φe)closed loop=∮E⃗⋅dA⃗=Qinϵ0(\Phi_\text{e})_\text{closed loop}=\oint\vec{E}\cdot d\vec{A}=\frac{Q_\text{in}}{\epsilon_0}
Similarly, magnets are always dipoles that form loop around itself. So, any flux out of a closed lop must come back. Therefore,
(Φm)closed loop=∮B⃗⋅dA⃗=0(\Phi_\text{m})_\text{closed loop}=\oint\vec{B}\cdot d\vec{A}=0
Third, we have Faraday's law, showing that an electric field can be created by a changing magnetic field.
ε=∮E⃗⋅ds⃗=−dΦmdt\varepsilon=\oint\vec{E}\cdot d\vec{s}=-\frac{d\Phi_\text{m}}{dt}
There is an analogous version for magnetic fields, covered next.

Changing Electric Flux


In the diagram above, the surface through which we use Ampere's Law can be S1S_1, which the wire passes through, or S2S_2, which has no current through it. But clearly theres a magnetic field around curve CC, so something is missing.
Using S1S_1, we apply Ampere's Law.
∮B⃗⋅ds⃗=μ0I\oint\vec{B}\cdot d\vec{s}=\mu_0I
Since we can use any closed surface, this should give the correct magnetic field.
In S2S_2, we instead note that there is a changing electric flux that accounts for the magnetic field. From Φe=EA\Phi_\text{e}=EA and Ecap=Q/ϵ0AE_\text{cap}=Q/\epsilon_0 A, we have
Φe=Qϵ0AA=Qϵ0\Phi_\text{e}=\frac{Q}{\epsilon_0 A}A=\frac{Q}{\epsilon_0}
Because we are concerned with the change in electric flux, we have
dΦedt=1ϵ0dQdt=Idispϵ0\frac{d\Phi_\text{e}}{dt}=\frac{1}{\epsilon_0}\frac{dQ}{dt}=\frac{I_\text{disp}}{\epsilon_0}
This quantity Idisp=ϵ0 dΦe/dtI_\text{disp}=\epsilon_0\space d\Phi_\text{e}/dt is the displacement current, from fluid-like model used back then (now abandoned, but the name lives on).
Combining the two, we have Ampere-Maxwell law
∮B⃗⋅ds⃗=μ0(Ithrough+Idisp)=μ0(Ithrough+ϵ0dΦmdt)\oint\vec{B}\cdot d\vec{s}=\mu_0(I_\text{through}+I_\text{disp})=\mu_0\left(I_\text{through}+\epsilon_0\frac{d\Phi_\text{m}}{dt}\right)

Induced Magnetic Field

The induced magnetic field, like the induced electric field, is created from changes in electric flux.
Note that Faraday's Law says the induced magnetic field is in the negative direction but the Ampere-Maxwell equation shows the induced electric field is in the positive direciton.

Maxwell's Equations

The following four equations give a complete description of electric and magnetic fields:
Gauss's Law: charged particles create an electric field
∮E⃗⋅dA⃗=Qϵ0\oint\vec{E}\cdot d\vec{A}=\frac{Q}{\epsilon_0}
Gauss' Law for magnetism: no isolated magnetic poles
∮B⃗⋅dA⃗=0\oint\vec{B}\cdot d\vec{A}=0
Faraday's Law: electric fields can be created by changing magnetic fields
∮E⃗⋅ds⃗=−dΦmdt\oint\vec{E}\cdot d\vec{s}=-\frac{d\Phi_\text{m}}{dt}
Ampere-Maxwell Law: currents and changing electric fields create magnetic fields
∮B⃗⋅ds⃗=μ0Ithrough+ϵ0μ0dΦedt\oint\vec{B}\cdot d\vec{s}=\mu_0I_\text{through}+\epsilon_0\mu_0\frac{d\Phi_\text{e}}{dt}

An additional law, known as the Lorenz Force Law, tells how matter reacts to the electromagnetic fields.
F⃗=q(E⃗+v⃗×B⃗)\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})

Electromagnetic Waves

Maxwell's theories predicted the existance of electromagnetic waves. We first have to assume there are fields with no sources, or in a source-free space.
The equations simplify to having no charges or currents:
∮E⃗⋅dA⃗=∮B⃗⋅dA⃗=0\oint\vec{E}\cdot d\vec{A}=\oint\vec{B}\cdot d\vec{A}=0
∮E⃗⋅ds⃗=−dΦmdt\oint\vec{E}\cdot d\vec{s}=-\frac{d\Phi_\text{m}}{dt}
∮B⃗⋅ds⃗=ϵ0μ0dΦedt\oint\vec{B}\cdot d\vec{s}=\epsilon_0\mu_0\frac{d\Phi_\text{e}}{dt}

From the induced fields being perpendicular, we assume E⃗\vec{E} and B⃗\vec{B} are perpendicular, and are perpendicular to the direction of motion.
We also assume they travel as a plane wave, one that has the same field anywhere in a plane perpendicular to the direction of motion.

Suppose the direction of travel was the xx-axis, E⃗\vec{E} was parallel to the yy-axis, and B⃗\vec{B} to the zz-axis. We check its consistency with the source-free Maxwell's equations
Cosider a box as a Gaussian surface, then the net flux is 0 because E⃗\vec{E} and B⃗\vec{B} enter and exit at the same rate.
If there was an xx-component, then there would be changing field in the direction of motion, so the flux would not exactly cancel out.

Faraday's Law

We use a narrow rectangle with extremely small Δx\Delta x so that BB is essentially constant and height hh.
The flux through the rectangle is
Φm=BzArect=BzhΔx\Phi_\text{m}=B_zA_\text{rect}=B_zh\Delta x
The derivative is
dΦmdt=∂Bz∂thΔx\frac{d\Phi_\text{m}}{dt}=\frac{\partial B_z}{\partial t}h\Delta x
Now, we note that the E⃗\vec{E} field points in the positive yy direction so the line integral in the CCW direction is
∮E⃗⋅ds⃗=∫leftE⃗⋅ds⃗+∫rightE⃗⋅ds⃗=−Ey(x)h+Ey(x+Δx)h=∂Ey∂xhΔx\oint\vec{E}\cdot d\vec{s}=\int_\text{left}\vec{E}\cdot d\vec{s}+\int_\text{right}\vec{E}\cdot d\vec{s}\\=-E_y(x)h+E_y(x+\Delta x)h=\frac{\partial E_y}{\partial x}h\Delta x
Now, by Faraday's Law,
∮E⃗⋅ds⃗=∂Ey∂xhΔx=−dΦmdt=−∂Bz∂thΔx\oint\vec{E}\cdot d\vec{s}=\frac{\partial E_y}{\partial x}h\Delta x=-\frac{d\Phi_\text{m}}{dt}=-\frac{\partial B_z}{\partial t}h\Delta x
⟹∂Ey∂x=−∂Bz∂t\implies \frac{\partial E_y}{\partial x}=-\frac{\partial B_z}{\partial t}
Thus, an electromagnetic wave must have the position rate of change of E⃗\vec{E} must be the opposite as the time rate of B⃗\vec{B} change.

Ampere-Maxwell's Law

Now take a thin rectangle in the xzxz-plane. E⃗\vec{E} is perpendicular, so the rate of change of flux through it is
dΦedt=∂Ey∂tlΔx\frac{d\Phi_\text{e}}{dt}=\frac{\partial E_y}{\partial t}l\Delta x
Using the same logic as before, we see that
∂Bz∂x=−ϵ0μ0∂Ey∂t\frac{\partial B_z}{\partial x}=-\epsilon_0\mu_0\frac{\partial E_y}{\partial t}

Taking the above two and taking partial derivatives in terms of EyE_y, we have
∂2Ey∂x2=−∂2Bz∂t∂x       ∂2Bz∂x∂t=−ϵ0μ0∂2Ey∂t2\frac{\partial^2 E_y}{\partial x^2}=-\frac{\partial^2 B_z}{\partial t\partial x}\space\space\space\space\space\space\space\frac{\partial^2 B_z}{\partial x\partial t}=-\epsilon_0\mu_0\frac{\partial^2 E_y}{\partial t^2}

The wave equation says that any system that obeys
∂2D∂t2=v2∂2D∂x2\frac{\partial^2 D}{\partial t^2}=v^2\frac{\partial^2 D}{\partial x^2}
Combining the second partial derivatives, we see that
∂2Ey∂t2=1ϵ0μ0∂2Ey∂x2\frac{\partial^2 E_y}{\partial t^2}=\frac{1}{\epsilon_0\mu_0}\frac{\partial^2 E_y}{\partial x^2}
The same equation can be fit for BzB_z. This shows that the speed of the wave is c=1/ϵ0μ0=3.00×108 m/sc=1/\sqrt{\epsilon_0\mu_0}=3.00\times10^8\text{ m}/\text{s}

The field strengths are related
Ey=E0sin⁡(kx−ωt)=E0sin⁡[2π(xλ−ft)]Bz=B0sin⁡(kx−ωt)=B0sin⁡[2π(xλ−ft)]E_y=E_0\sin(kx-\omega t)=E_0\sin\left[2\pi\left(\frac{x}{\lambda}-ft\right)\right]\\B_z=B_0\sin(kx-\omega t)=B_0\sin\left[2\pi\left(\frac{x}{\lambda}-ft\right)\right]
where k=2π/λk=2\pi/\lambda, ω=2πf\omega=2\pi f, λf=v=c\lambda f=v=c
By Maxwell's Equations, we must have
∂Ey∂x=2πE0λcos⁡[2π(xλ−ft)]=−∂Bz∂t=2πfB0cos⁡[2π(xλ−ft)]\frac{\partial E_y}{\partial x}=\frac{2\pi E_0}{\lambda}\cos\left[2\pi\left(\frac{x}{\lambda}-ft\right)\right]\\=-\frac{\partial B_z}{\partial t}=2\pi fB_0\cos\left[2\pi\left(\frac{x}{\lambda}-ft\right)\right]
Thus E0=λfB0=cB0E_0=\lambda fB_0=cB_0, and because they oscillate together, this must be true at all points on the wave.