\(\)
We will take a critical look at Ampere’s law.
Wire and capacitor
Assume current through a wire and a circular plate capacitor. We will get a changing electric field inside the capacitor.
Recall: the electric field inside the capacitor
$$ \begin{align*} E &= \frac{\sigma_f}{\kappa\varepsilon_0}, & \left(\sigma=\frac{Q}{A}\right) \\ &= \frac{Q_f}{\pi R^2\kappa\varepsilon_0} \end{align*} \nonumber $$
Since we run a current, \(Q_f\) is building up all the time, so the current by definition is \(I=dQ/dt\). The changing electric field $$ \begin{align} \frac{dE}{dt} &= \frac{d}{dt} \left(\frac{Q_f}{\pi R^2\kappa\varepsilon_0}\right) \nonumber \\ &= \frac{1}{\pi R^2\kappa\varepsilon_0} \frac{dQ}{dt}, & \left(I=\frac{dQ}{dt} \right) \nonumber \\ &= \frac{I}{\pi R^2\kappa\varepsilon_0} \label{eq:dEdt} \end{align} $$
Note: only when the current is zero, is there no changing field inside.
Magnetic fields
How does this affect the magnetic field? Let’s take
- a point \(P1\) at distance \(r\) from the wire. If you’re from the capacitor, it is hard to believe that Ampere’s law would not give the right answer.
- a point \(P2\) above the capacitor at the same distance \(r\). The is no current going through the capacitor, so you expect a magnetic field at \(P2\) to be a little lower than at \(P1\).
We can calculate the magnetic fields. Biot-Savart could handle it, but we don’t know how to do that, because if there’s a current in the wire, there is also a current going up on the plates.
Let’s try Ampere’s law. For this cylindrical symmetric problem, we choose a closed loop which is a circle with radius \(r\)
For point \(P1\)
We attached a flat open surface, and apply Ampere’s law. Everywhere on the closed loop, the magnetic field has the same strength based on symmetry, so we get $$ \begin{align*} \oint \vec B \cdot d\vec l &= \mu_0 I_\rm{encl} \\ B \oint d\vec l &= \mu_0 I_\rm{encl}, & (\text{I penetrates}) \\ B\,2\pi r &= \mu_0 I \\ \Rightarrow B &= \frac{\mu_0 I}{2\pi r} \end{align*} $$
We find, as have seen before $$ \shaded{ B = \frac{\mu_0 I}{2\pi r} } \nonumber $$
For point \(P2\)
Let’s try the same for point \(P2\)
We’re in for a shock, because \(2\pi r\) is not changing, but there is no current that penetrates that surface. So, I have to conclude that the magnetic field at point \(P2\) is zero, which is absurd. Can’t be.
Even worse
We can make the situation even worse, by taking a different surface for \(P1\). We don’t have to choose a flat surface. We may choose a surface that goes through the capacitor plates
We apply Ampere’s law, but there is no current penetrating that surface. So we find that the magnetic field at \(P1\) is now also zero. There is something wrong. Ampère’s law is inadequate.
Maxwell
Faraday and Ampère were both perfectly aware of this, but it was James Clerk Maxwell, a Scottish mathematician, who zeroed in on this. He argued that any open surface that you assign to a closed loop should give you exactly the same result. He suggested that amend Ampere’s law.
He asked himself the question: “What is so special about in-between the capacitor plates?”. What is special is the changing electric field. Maxwell reasoned
- Faraday’s law tell me that a changing magnetic flux gives rise to an electric field.
- So, maybe a changing electric flux gives rise to a magnetic field.
Remember electric flux through an open surface \(S\)
$$ \phi_E = \iint_S \vec E\cdot d\vec A \nonumber $$
Maxwell suggested that we add a term which contains the derivative of the electric flux. This is called the displacement current. $$ \mu_0\kappa \frac{\phi_E}{dt} = \mu_0\kappa \frac{d}{dt} \iint_S \vec E\cdot d\vec A \nonumber $$
He wrote $$ \shaded{ \oint \vec B \cdot d\vec l = \mu_0\left(I_\rm{encl} + \varepsilon_0\kappa \frac{d}{dt} \iint_S \vec E\cdot d\vec A \right) } \tag{Maxwell add.} \label{eq:maxwelladd} $$
This is almost the last of Maxwell’s equations, but it still needs a little adjustment (that we will do later when we deal with radio waves and the propagation of electromagnetic radiation).
Using this new law
Let’s using this new law to clean up the mess. We’ll revisit point \(P1\) by first having a flat surface, and compare it to the surface that goes in-between the plates.
Point \(P1\) with flat surface
With the flat surface, there is a current penetrating but no electric flux going through the surface. So the displacement term is zero for this flat surface. So the answer stays $$ \shaded{ B = \frac{\mu_0 I}{2\pi r} } \nonumber $$
With the surface that goes in-between the plates,
Applying equation \(\eqref{eq:maxwelladd}\) $$ B 2\pi r = \mu_0\left(0 + \varepsilon_0\kappa \frac{d}{dt} \iint_S \vec E\cdot d\vec A \right) \label{eq:p1} $$
Assuming no fringe fields, the change in electric flux \(d\phi_E/dt\) is easy to calculate because \(\) $$ \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} \phi_E &= \iint_S\vec E\cdot d\vec A, & \left(\vec E\parallelsum d\vec A\right) \\ &= E \iint_S dA, & \left(A = \pi r^2\right) \\ &= E\, \pi r^2 \end{align*} $$
Taking the derivative on both sides and substituting \(dE/dt\) from equation \(\eqref{eq:dEdt}\) $$ \begin{align*} \frac{d\phi_E}{dt} &= \pi R^2 \frac{dE}{dt} \\ &= \pi r^2 \frac{I}{\pi R^2\kappa\varepsilon_0} \\ &= \frac{I\pi r^2}{\pi R^2\kappa\varepsilon_0} \end{align*} $$
Substituting this in equation \(\eqref{eq:p1}\) $$ \begin{align*} B 2\pi \bcancel{r} = \mu_0\left(0 + \bcancel{\varepsilon_0\kappa} \frac{I\cancel{\pi} r^{\bcancel{2}}}{\cancel{\pi} R^2\bcancel{\kappa\varepsilon_0}} \right) \end{align*} $$
So, the magnetic field is $$ \shaded{ B = \frac{\mu_0 I r}{2\pi R^2} } \nonumber $$
This is proportional with \(r\), where we had it proportional with \(1/r\) before. A plot of the magnetic field inside the capacitor plates
So, at the edge of the capacitor \(r=R\), the magnetic field is the same as we found using the flat surface $$ \shaded{ B = \frac{\mu_0 I}{2\pi r} } \nonumber $$
Now we have a tool to calculate the magnetic field even inside the capacitor while we are charging
The top part of the graph is not correct, because we have assumed that there is no fringe field. So, when you get close to the edge of the capacitor, this is not correct.
Radio waves
Maxwell had introduced his displacement current term. He predicted that as a consequence of that term that radio waves should exist. That was at a time we didn’t know that radio waves existed. Not only did he predict them, he was even able to calculate what their speed was going to be. We call that the speed of light. We will do that later.
In 1879, the year that Maxwell died (at 48), the German physicist Hermann von Helmholtz asked one of his students, Heinrich Hertz, who was 22 years at the time, to try to demonstrate that radio waves indeed exist.
Hertz declined, because he argued that the equipment that was available at the time was not good enough. But seven years later, when new equipment came out, he accepted the challenge. It took him two years, but then he indeed as able to demonstrate that radio waves do exist. Imagine the victory! Hertz died 5 years after his great experiments (at 37).
“Displacement current”
Why did Maxwell called that term “displacement current”? In the presence of a dielectric, a changing electric field will indeed cause a current in-between the plates. Because the polarization will change all the time. You get a rearrangement of those induced charges, so there is indeed a current.
In vacuum there shouldn’t be any current. Any electric field changing or not changing, will not cause a current in vacuum. But Maxwell believed that vacuum in a way behaves like any other dielectric. Just a special dielectric, happened to be a dielectric with \(\kappa=1\). And so he really believed that there was an actual current going between the plates. even though we now know that that is not the case.
The name “displacement current” as perhaps not a very lucky one but the term is a must. It completes the theory of electricity and magnetism.