Electromagnetic induction; Faraday’s law



My notes of the excellent lecture 16 by “Walter Lewin. 8.02 Electricity and Magnetism. Spring 2002. The material is neither affiliated with nor endorsed by MIT, https://youtube.com. License: Creative Commons BY-NC-SA.”

Conceptually, this is one of the harder things to wrap your head around. You may have to watch the lecture and/or do the problems several times.

Electromagnetic Induction

Remember that Ørsted in 1819 discovered that

A steady current produces a steady magnetic field. This connects electricity with magnetism.

Michael Faraday, an English scientist, therefore suggested that maybe a steady magnetic field produces a steady current. He did many experiments, but it turned out not to be so.

In one of the experiments, he tried a battery, a switch and a solenoid. When the switch is closed, a current flows that creates a magnetic field in the solenoid.

He added circuit ② by putting a loop around the solenoid, but he never saw a current in circuit ②. He concluded that a steady magnetic field, as produced by the solenoid circuit ①, doesn’t produce a steady current in circuit ②.

Changing magnetic field

One day, Faraday noticed that as he closed the switch, he saw a current in ②. And when he opened the switch, he again saw a current in ②. Therefore, he concluded that

a changing magnetic field causes a current.

A current, therefore an electric field, can be produced by a changing magnetic field. That is called electromagnetic induction.

This profound discovery changed our world and contributed largely to the technological revolution of the late nineteenth and early twenty century.

Lenz’s law

In 1834, the Russian physicist Heimrich Lenz, formulated that

The direction of the electric current induced in a conductor by a changing magnetic field is such that the magnetic field created by the induced current opposes changes in the initial magnetic field.

The short version of Lenz’s law is

The current wants to oppose the change in the magnetic field.

Lenz’s law is very powerful in always determining in which direction the induced currents will run.

Increasing magnetic field

For example, we approach a round wire loop with a bar magnet (as shown on the left)

As the bar magnet moves towards the loop, we see a current through the loop. That current is in such a direction that it opposes the change in magnetic field $$\Longrightarrow$$ the current loop produces a magnetic field that is up, as visualized by the blue magnetic field lines in the illustration on the right.

Decreasing magnetic field

If you move the bar magnet out, then the magnetic field in the loop goes down. That current is in such a direction that it opposes the change in magnetic field $$\Longrightarrow$$ the current in the loop will reverse. The current loop will produce a magnetic field that is down, as visualized by the blue magnetic field lines in the illustration on the right.

Magnetic flux and EMF

In another of Faraday’s experiments, he generated a changing magnetic field with a switch and solenoid and measured the current in a second loop

He found that the Electromagnetic Force (EMF) in ② is proportional to the change in magnetic field in ① and the area of loop in circuit ② \begin{align*} \mathcal E_2 &\propto \frac{dB_1}{dt} \\ \mathcal E_2 &\propto \rm{area}_2 \end{align*}

From that he concluded that the EMF is the result of the change of the magnetic flux through the surface of ②.

Magnetic flux

Remember electric flux (also see multivariable calculus)

$$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \phi = \iint_R \vec E \cdot d\vec A \nonumber$$

The magnetic flux is very similar

The magnetic flux through open surface $$R$$ is defined as $$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \shaded{ \phi_B = \iint_R \vec B \cdot d\vec A } \tag{Magnetic flux} \label{eq:flux}$$

Electromagnetic force

To measure magnetic flux, we need a surface that the magnetic field passes through. Assume there is a closed wire, that we assign a simple simple flat surface to.

If there is a magnetic field coming out of the screen, and it is growing, then Lenz’s law determines the direction of the current $$I$$ $$\Longrightarrow$$ the wire loop produces a magnetic field that opposes the change in magnetic field (it goes into the screen).

It is the flux change of that magnetic field through the flat surface $$R$$, that determines the EMF. $$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \shaded{ \mathcal E = -\frac{d\phi_B}{dt} } \tag{EMF} \label{eq:emf}$$

The minus sign expresses that it is always opposing the change of the magnetic flux from Lenz’s law.

Faraday’s law (Maxwell’s eq #3)

When we combine equations $$\eqref{eq:emf}$$ with $$\eqref{eq:flux}$$ we get $$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \mathcal E = -\frac{d\phi_B}{dt} = -\frac{d}{dt}\iint_R \vec B \cdot d\vec A \label{eq:emf1}$$

If you put yourself inside the conductor, and you marched around with the current, you will see an electric field everywhere. (Otherwise, there would be no current flowing.) If you stay in the wire, $$\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \vec E \parallelsum d\vec l$$

If you go once around the wire, then that EMF must also be $$\vec E\cdot d\vec l$$ over the closed loop $$C$$. $$\mathcal E = \oint_C \vec E\cdot d\vec l \label{eq:emf2}$$

Combining equations $$\eqref{eq:emf1}$$ and $$\eqref{eq:emf2}$$ $$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \shaded{ \mathcal E = -\frac{d\phi_B}{dt} = -\frac{d}{dt}\iint_R \vec B \cdot d\vec A = \oint_C \vec E\cdot d\vec l } \nonumber$$

This brings us to the third of four Maxwell’s equations: Faraday’s law from 1831 $$\shaded{ \oint_C \vec E\cdot d\vec l = -\frac{d}{dt}\iint_R \vec B \cdot d\vec A } \tag{Faraday’s law} \label{eq:faraday2}$$

Induced current

There is no battery in the circuit. There is only a change in magnetic flux through surface $$R$$, and we get an induced EMF. The EMF in turn produces an induced current given by Ohm’s law $$\shaded{ I_{ind} = \frac{\mathcal E}{R} } \tag{Induced current} \label{eq:current}$$ where $$R$$ is the resistance of the entire closed conductor.

(Recipe)

Method 1

1. You define the loop first.
2. Define the direction in which you want to march around that circuit.
3. Assign an open surface $$R$$ to that closed loop.
4. On that entire surface determine the $$\int_R \vec B\cdot d\vec A$$
5. This gives the magnetic flux.
6. If you know the time change of that magnetic flux, then you know minus the EMF.

Method 2

1. Go around the circuit and measure the electric fields $$\vec E$$ everywhere.
2. then the integral $$\int \vec E\cdot d\vec l$$ around the loop, will give you the EMF.

This is less complicated as it seems, because the loop is always a conducting wire in your circuit. The minus sign is never an issue, because with a href=”#lenz”>Lenz’s law, you always know in which direction the EMF is. (You may not even need to look at the minus sign.)

Convention for surface$$\text{ }R$$

Surface $$R$$ is used to determine the magnetic flux. Its border will be the conducting wire. As we will see, the direction of $$d\vec A$$ is set by convention, but the surface can have any open shape.

Direction of$$\text{ }d\vec A$$

To be mathematically correct, you should follow the convention for the direction of surface element $$d\vec A$$.

Similar as in Ampere’s law, we apply the right-hand corkscrew:

• if you march around clockwise, then $$d\vec A$$ will go into the screen.
• if you march around counter-clockwise, then $$d\vec A$$ will come out of the screen.

Any open surface

Similar to Ampere’s law, you are free to choose a surface. It can be flat, or e.g. like a bakers hat. Think of the magnetic field lines as a flow of water (or spaghetti). If there is some kind of flow through this opening, then it’s got to come out somewhere. So it always comes out of this surface. Therefore, you’re really free to choose that surface.

Multiple windings

If we redo Faraday’s experiment with a modern sensitive amp meter, it will show a current in circuit ② when the switch is being closed or opened.

If circuit ② has one winding, we can simply assign a flat $$R$$. The figure on the right illustrates this, along with the magnetic field $$\vec B_1$$ produced by circuit ①.

If circuit ② has three windings, we can assign a surface $$R$$ that looks like a spiral staircase as illustrated on the right. That implies that the same magnetic field $$\vec B_1$$ will pass through the surface three times $$\Longrightarrow$$ the EMF $$\mathcal E$$ will be about three times that of the one winding circuit.

Non-conservative electric fields

If an electric field is conservative, and you go from point $$A$$ to point $$B$$, the potential difference is independent of the path $$V_A – V_B = \int_A^B \vec E\cdot d\vec l \nonumber$$

Kirchhoff’s rule was very intuitive. He said if you go around a circuit

$$\oint \vec E\cdot d\vec l = 0 \nonumber$$

But, if you have a changing electric flux, the electric fields inside the conducting wires become non-conservative as shown by $$\eqref{eq:faraday2}$$

$$\oint_C \vec E\cdot d\vec l = -\frac{d}{dt}\iint_R \vec B \cdot d\vec A \nonumber$$

As we saw with multiple windings, if you go around once with this experiment, you get a certain EMF. But when you go three times around, you get a different value. You path is now different, and that’s very non-intuitive.

You’re dealing with non-conservative fields for which we have very little feeling.

Experiment

We will compare conservative fields, where we can use Kirchhoff’s voltage rule, with non-conservative fields. The findings are quite astounding.

Conservative fields

Given a circuit with a battery and two resistors

The current follows from Ohm’s law $$I = \frac{\mathcal E}{R_1+R_2} = \frac{1}{100+900} = 1\,\rm{mA} \nonumber$$

The potential difference between point $$D$$ and $$A$$ (over $$R_2$$) $$V_D – V_A = I\,R_2 = 10^{-3} 900 = +0.9\,\rm{V} \nonumber$$

Going the other way, is of course the same $$V_D – V_A = \mathcal E – I\,R_1 = +0.9\,\rm{V} \nonumber$$

Kirchhoff’s rule works, the closed loop integral $$\oint \vec E\cdot d\vec l$$, going from $$D$$ to $$A$$, and back to $$D$$ is zero.

Non-conservative fields

We take the battery out, and place a solenoid in the middle of the circuit. The solenoid produces an increasing magnetic field that is going out of the screen. The magnetic field is local to the loop.

Lenz’s law determines that the direction of current $$I$$ is clockwise. The magnetic flux change $$d\phi/dt$$ at a particular moment in time, happens to be $$\mathcal E_{ind}=1\,\rm{V}$$. What is the current?

Using equation $$\eqref{eq:current}$$, we find the induced current $$I = \frac{\mathcal E_{ind}}{R_1+R_2} = \frac{1}{100+900} = 1\,\rm{mA} \nonumber$$

The potential difference between point $$D$$ and $$A$$ (over $$R_2$$) $$V_D – V_A = I\,R_2 = 10^{-3} 900 = +0.9\,\rm{V} \nonumber$$

But .. going the other way (over $$R_1$$) .. $$V_D – V_A = – I\,R_1 = -0.1\,\rm{V} \nonumber$$

Both volt meters are connected to $$D$$ and $$A$$, but

• the volt meter on the right reads $$+0.9\,\rm{V}$$, while
• the volt meter on the left reads $$-0.1\,\rm{V}$$

It is hard to digest, because we don’t intuitively know how to handle non-conservative fields.

More details can be found in the handout for lecture 16.

Path dependence

If you go through a non-conservative field from e.g. $$A$$ to $$D$$, the potential difference is no longer independent of the path $$V_D – V_A = \int_A^D \vec E\cdot d\vec l$$

You will find different answers depending on the path

Faraday has no problems with that. Faraday’s law always holds. Kirchhoff voltage’s rule is simply a special case of Faraday’s law when $$d\phi/dt$$ is zero.

Going around

Suppose you go from $$D$$ to $$A$$ and back to $$D$$.

• If we go through $$R_2$$, we know that $$V_D – V_A = +0.9\,\rm{V}$$.
• Now we go back through $$R_1$$, and $$V_A – V_D = +0.1\,\rm{V}$$

If we add them up, we find $$V_D – V_D = 1\,\rm{V} \nonumber$$

This is exactly the EMF of $$1\,\rm{V}$$. It is the closed loop integral of $$\vec E\cdot d\vec l$$ around that loop. It is no longer zero.

Therefore, if you define potential difference, if you do that in the way of $$\int \vec E\cdot d\vec l$$, keep in mind that with non-conservative fields, it depends on the path. That is very non-intuitive.