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Part 9 - Electromagnetic Waves

EEM076 Date: April 25, 2023 7 min read

In this part we’ll cover electromagnetic waves - we’ll see why electrical and magnetic fields always appear together.

Displacement current

If we recall Faraday’s law:

∮E⃗⋅ds⃗=−dΦBdt\oint \mathbf{\vec{E}} \cdot d\mathbf{\vec{s}} = -\dfrac{d \Phi_B}{dt}

This says that a change in the magnetic field will produce an electrical field. But how about the converse?

Does a change in an electrical field produce a magnetic field?

If we consider this example with a capacitor being charged:

Using Ampere’s law:

∮B⃗⋅ ds⃗=μ0Ienc\oint \mathbf{\vec{B}} \cdot\ d\mathbf{\vec{s}} = \mu_0 I_{enc}

If we look at S1S_1, we see that Ienc=II_{enc} = I - however, for S2S_2, Ienc=0I_{enc} = 0 - but the current does continue?

So we will need to add an extra term to Ampere’s law:

∮B⃗⋅ ds⃗=μ0(I+Id)\oint \mathbf{\vec{B}} \cdot\ d\mathbf{\vec{s}} = \mu_0(I + I_d)

Where the displacement current, IdI_d:

Id=ε0dΦEdtI_d = \varepsilon_0 \dfrac{d \Phi_E}{dt}

In practice, I=IdI = I_d, thus the choice of the Amperian loop is inconsequential!

Gauss’s Law

For electrostatics, Gauss’s law said that:

ΦE=∯E⃗⋅ dA⃗=qμ0\Phi_E = \oiint \mathbf{\vec{E}} \cdot\ d\mathbf{\vec{A}} = \dfrac{q}{\mu_0}

But in the case for magnetism - magnetic monopoles do not exist (meaning that q=0q = 0). Therefore:

ΦB=∯B⃗⋅ dA⃗=0\Phi_B = \oiint \mathbf{\vec{B}} \cdot\ d\mathbf{\vec{A}} = 0

Maxwell’s Equations

With these new tools in mind - we have now learned the four Maxwell equations that form the basis of electromagnetism!

Let’s write them all down:

Gauss’s Law for electrostatics:

ΦE=∯E⃗⋅ dA⃗=qμ0\Phi_E = \oiint \mathbf{\vec{E}} \cdot\ d\mathbf{\vec{A}} = \dfrac{q}{\mu_0}

Gauss’s Law for magnetism:

ΦB=∯B⃗⋅ dA⃗=0\Phi_B = \oiint \mathbf{\vec{B}} \cdot\ d\mathbf{\vec{A}} = 0

Faraday’s Law:

ε=∮E⃗⋅ds⃗=−dΦBdt\varepsilon = \oint \mathbf{\vec{E}} \cdot d\mathbf{\vec{s}} = -\dfrac{d \Phi_B}{dt}

Ampere-Maxwell Law:

ε=∮B⃗⋅ds⃗=μ0(I+Id)=μ0(I+ε0dΦEdt)\varepsilon = \oint \mathbf{\vec{B}} \cdot d\mathbf{\vec{s}} = \mu_0(I + I_d) = \mu_0\left(I + \varepsilon_0 \dfrac{d \Phi_E}{dt}\right)

If the absence of sources is present (meaning that q = 0 and I = 0) - all of these become quite compact and tidy:

Gauss’s Law for electrostatics:

ΦE=∯E⃗⋅ dA⃗=0\Phi_E = \oiint \mathbf{\vec{E}} \cdot\ d\mathbf{\vec{A}} = 0

Gauss’s Law for magnetism:

ΦB=∯B⃗⋅ dA⃗=0\Phi_B = \oiint \mathbf{\vec{B}} \cdot\ d\mathbf{\vec{A}} = 0

Faraday’s Law:

ε=∮E⃗⋅ds⃗=−dΦBdt\varepsilon = \oint \mathbf{\vec{E}} \cdot d\mathbf{\vec{s}} = -\dfrac{d \Phi_B}{dt}

Ampere-Maxwell Law:

ε=∮B⃗⋅ds⃗=μ0(I+Id)=μ0ε0dΦEdt\varepsilon = \oint \mathbf{\vec{B}} \cdot d\mathbf{\vec{s}} = \mu_0(I + I_d) = \mu_0 \varepsilon_0 \dfrac{d \Phi_E}{dt}

Now we can see, from Faraday’s law and Ampere-Maxwell’s Law that, if we have a change in one field - we’ll see a change in the other!

We’ll see why this picture will make a lot of sense soon:

Inductance

We have seen some aspect of inductances already - the induced current.

Imagine this scenario:

We can see that our B1⃗\mathbf{\vec{B_1}} passes through the second coil, therefore a flux is generated, Φ21\Phi_{21}

As we learned last time, varying I1I_1 will induce an emf!

Which means:

ε21=−N2dΦ21dt\varepsilon_{21} = -N_{2} \dfrac{d \Phi_{21}}{dt}

If we want to write this in terms of I1I_1 we get:

ε21=M21dI1dt\varepsilon_{21} = M_{21} \dfrac{d I_1}{dt}

Where M21M_{21}:

M21=N2Φ21I1M_{21} = \dfrac{N_2 \Phi_{21}}{I_1}

But let’s say that we vary I2I_2, this will also lead to an EMF, but in the first coil:

ε12=−N1dΦ12dt=M12dI2dt\varepsilon_{12} = -N_{1} \dfrac{d \Phi_{12}}{dt} = M_{12} \dfrac{d I_2}{dt}

Then, if we use the theorem of reciprocity - which states that:

Theorem 1 (Reciprocity)

The current at one point in a circuit due to a voltage at a second point is the same as the current at the second point due to the same voltage at the first.

Using this and combining Ampere’s law and the Biot-Savart law - we get:

M12=M21=MM_{12} = M_{21} = M

This means - sending varying current (AC) through coil 1 - will generate an AC current in coil 2 as well!

We can also connect this to voltage, potential drop:

V2V1=N2N1\dfrac{V_2}{V_1} = \dfrac{N_2}{N_1}

Self-Inductance

Say we only have one coil

If we vary, II, an induced EMF will oppose the change in flux, according to Faraday’s law.

We denote this self-inducted EMF with, εL\varepsilon_L.

We can write it as:

εL=−NdΦBdt\varepsilon_L = -N \dfrac{d \Phi_B}{dt}

We can relate this self-inductance with self-inductance, LL:

εL=−LdIdt\varepsilon_L = -L \dfrac{dI}{dt}

We can also write:

L=NΦBI=μ0N2AlL = \dfrac{N \Phi_B}{I} = \dfrac{\mu_0 N^2 A}{l}

Therefore, the inductance, is a measure of an inductor’s resistance to change of the current!

In even simpler terms - inductors oppose change in the current!

So, since inductors oppose changes to the current - work must be done to establish a current in the inductor. Thus, energy must be stored in the magnetic field in an inductor! Similar to electrical fields in a capacitor.

The power, or the rate at an external EMF, εext\varepsilon_{ext}, works to overcome the self-induced EMF, εL\varepsilon_L, to pass the current II:

PL=dWextdt=Iεext=−IεL=ILdIdtP_L = \dfrac{d W_{ext}}{dt} = I \varepsilon_{ext} = -I \varepsilon_L = IL \dfrac{dI}{dt}

This assumes that only εext\varepsilon_{ext} and the inductor is present.

The total work done by an external source to increase the current from 0 to II is:

Wext=∫dWext=∫0IIL dI=12LI2W_{ext} = \int dW_{ext} = \int_0^I IL\ dI = \boxed{\dfrac{1}{2}LI^2}

Electromagnetic waves

We’ll now cover a very cool phenomena - we’ll cover why electromagnetic waves travel at light speed!

To prove this we’ll need to rewrite our definition for E⃗\mathbf{\vec{E}} and B⃗\mathbf{\vec{B}}:

E⃗=Ey(x,t)J⃗=E cos(kx−ωt)J⃗\mathbf{\vec{E}} = E_y(x, t)\mathbf{\vec{J}} = E\ cos(kx - \omega t)\mathbf{\vec{J}} B⃗=Bz(x,t)K⃗=B cos(kx−ωt)K⃗\mathbf{\vec{B}} = B_z(x, t)\mathbf{\vec{K}} = B\ cos(kx - \omega t)\mathbf{\vec{K}}

Using Faraday’s law:

∮E⃗⋅ ds⃗=−ddt∬B⃗⋅ dA⃗\oint \mathbf{\vec{E}} \cdot\ d\mathbf{\vec{s}} = - \dfrac{d}{dt} \iint \mathbf{\vec{B}} \cdot\ d\mathbf{\vec{A}}

If we take our ring integral inside a arbitrary rectangle - on the xy-plane:

Ey(x+Δx)−Ey(x)=−ddt∬B⃗⋅dxdyE_y(x + \Delta x) - E_y(x) = - \dfrac{d}{dt} \iint \mathbf{\vec{B}} \cdot dxdy

We get that:

∂Ey∂x=−∂Bz∂t\dfrac{\partial E_y}{\partial x} = - \dfrac{\partial B_z}{\partial t}

Let’s calculate both:

∂Ey∂x=−kE sin(kx−ωt)\dfrac{\partial E_y}{\partial x} = -kE\ sin(kx - \omega t) ∂Bz∂t=ωB sin(kx−ωt)\dfrac{\partial B_z}{\partial t} = \omega B\ sin(kx - \omega t)

Which, finally, means:

EB=ωk=c\boxed{\dfrac{E}{B} = \dfrac{\omega}{k} = c}

EMFs in circuits

Now that we have learned this beauty of electromagnetic waves - let’s tie it back to our electrical circuits.

As we learned with electrical fields and capacitor:

Recall

Electrical field opposes change in voltage

C=ε0AdU=12C∣V2∣I(t)=CdVdtC = \varepsilon_0 \dfrac{A}{d} \newline U = \dfrac{1}{2} C|V^2| \newline I(t) = C \dfrac{dV}{dt}

We have now seen that, magnetic fields and inductors:

Recall

Magnetic field opposes change in the current

L=μ0N2AlU=12LI2V(t)=LdIdtL = \mu_0 N^2 \dfrac{A}{l} \newline U = \dfrac{1}{2} LI^2 \newline V(t) = L \dfrac{dI}{dt}

We can now use this knowledge to understand LC and RLC circuits - and see how and why the energy oscillates between the electrical and magnetic field!