magnetism

Materials; 4th Maxwell equation

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My notes of the excellent lectures 21 and 22 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.”

When discussing dielectrics we learned

Electric fields can induce electric dipoles in materials. When the molecules or atoms themselves are permanent electric dipoles, an external electric field will try to align them. The degree of success depends entirely on how strong the external electric field is, and on the temperature. If the temperature is low, there is very little thermal agitation, making it easier to align those dipoles.

We have a similar situation with magnetic fields. An external magnetic field can induce magnetic dipoles in materials. It induces magnetic dipoles at the atomic scale. When the atoms/molecules themselves have a permanent magnetic dipole moment, the external field will try to align these dipoles. Again, the degree of success depends on the strength of the external field, and again on the temperature. The lower the temperature, the easier it is to align them.

So the material modifies the external field. We often call this external field, the vacuum field. When you bring material into a vacuum field, the field changes. The field inside is different from the external field.

Magnetic dipole moment

If we have a current in a loop, and the current is running clockwise as seen from below, and the area is \(A\), then the magnetic dipole moment \(\vec\mu\) $$ \vec \mu = I\,\vec A \tag{dipole moment} \label{eq:dipolemoment} $$

We define \(\vec A\) according to the righthand corkscrew rule. If we come from below clockwise, then the \(\vec A\) is perpendicular to the surface. So the magnetic dipole moment \(\vec\mu\), is also pointing upwards.

If we have \(N\) of these loops, then the magnetic dipole moment will be \(N\) times larger.

Diamagnetism

When you expose any material to a permanent external magnetic field, it will to some degree, oppose that external field. On an atomic scale, the material will generate an EMF that opposes the external field. This has nothing to do with Lenz law. It has nothing to do with the free electrons in conductors which produce an eddy current in a changing magnetic field.

In other words: when we apply a permanent magnetic field, to any material, a magnetic dipole moment is induced to oppose that field. This can only be understood with quantum mechanics. Here, we’ll make no attempt to explain it, but we will accept it.

The magnetic field inside the material, is always a little bit smaller than then external field, because the dipoles oppose the external field.

Paramagnetism

There are many substances where the atoms/molecules themselves have magnetic dipole moments. You can think of them as being little magnets. If you have no external field, then these dipoles are completely chaotically oriented. So the net magnetic field is zero. They are not permanent magnets.

If you bring a paramagnetic material in a magnetic field, its atomic magnetic dipoles will move their north poles a little bit in the direction of the external magnetic field. They align a little with the external field. The degree of success depends on the strength of that field and the temperature. The lower the temperature, the easier it is. If you remove the external field, immediately there is complete, total chaos. There is no permanent magnetism left.

In non-uniform field

If you bring a paramagnetic material in a non-uniform magnetic field, it will be pulled towards the strong side of that field. Suppose we have a magnet, and we bring paramagnetic material in its field. Let’s consider just one atom of the material.

Very much not to scale

The atoms magnetic dipole moment would like to align in the direction to support the field. So if we look from above, the current in this atom/molecule runs clockwise direction. That would be the ideal alignment. This current loop will be attracted to the magnet. Let’s look at a point on the left. At that point, the current goes into the screen, and the magnetic field goes diagonal.

  

The Lorenz force will be in the direction \(\vec I\times\vec B\), and put it up towards the left. At a point on the right, the force will be to the right. So everywhere around this loop, there is a force that is pointing diagonally outwards up. Clearly, there is a net force up. The matter wants to go towards the magnet.

Essential is that the external magnetic field is non-uniform. (In diamagnetic material, the current would be running in the opposite direction, because it opposes the external field.)

The exception

Diamagnetic materials are always pushed towards the weak part of the field. Paramagnetic and ferromagnetic materials that experience a force towards the strong part of the field, if the field itself is non-uniform.

There is one interesting exception. Oxygen at one atmosphere and 300 °K, has a \(\chi_m\) of \(2\times10^{-6}\). But liquid oxygen at 90 °K, the \(\chi_m\) is 1,800 times larger. Why?

Liquid, in general, is about thousand times denser than gas at one atmosphere. So you have a thousand times more dipoles per cubic meter that, in principle, can align. So you expect a 1:1 correspondence between the density, and the value of \(\chi_m\). Indeed, you see that this value is substantially larger. It is more than a factor 1000 larger is that the temperature is also lower. That gives us another factor of two.

Even though the value of \(\chi_m\) is extraordinarily high for a paramagnetic material, notice that the field inside would only be 0.35% higher than the vacuum field. But that is enough for liquid oxygen to be attracted by a very strong magnet, provided that it has a very non-uniform field outside the magnet. So the force with which liquid oxygen is pulled towards a magnet can be made larger than the weight of the liquid oxygen.

Ferromagnetism

Again the atoms have permanent magnetic dipole moments. But this time, for reasons which can only be understood with quantum mechanics, there are domains the size of about \(0.1\) to \(0.3\,\rm{mm}\).

Domains

In the domains, the dipoles are all aligned. The number of atoms involved in such a domain is typically \(10^{17}\ldots10^{21}\) atoms. The domains are uniformly distributed throughout the ferromagnetic material. So there may not be any net magnetic field.

When we apply an external field, these domains will be forced to align themselves with the magnetic field. The domains a a whole can flip. The degree of success depends on the strength of the field and the temperature. The lower the temperature, the better, because there is less thermal agitation with adds a certain randomness to the process. Inside the ferromagnetic material, the magnetic field can thousands of times stronger than the external field..

If we remove the external field, some of those domains may stay aligned in the direction that the external field was forcing them. Undoubtedly some domains will flip back, because of the temperature, but some may remain oriented and therefore the material, once it has been exposed to an external magnetic field, may have become permanently magnetic.

The only way you can remove that permanent magnetism could be to bang on it with a hammer. Then these domains get very nervous and randomize themselves. Or you can heat them up to undo the orientation of the domains. The domains themselves will remain, but then they average out not to produce any permanent magnetic field.

In non-uniform field

For the same reason, that paramagnetism is pulled towards the strong field, ferromagnetism will also be pulled towards the strong field. Except in the case of ferromagnetism the forces with which the material is pulled towards the magnet, is much larger than in with paramagnetic material.

If we take a paperclip, you can hang it on the south pole of your magnet, or the north pole of your magnet. Ferromagnetic material is always pulled towards the strong field. If we hang a few of those paperclips on there, and you carefully and slowly remove them, you may notice that after you move them, that the paperclips themselves have become magnetic. You can not do this paramagnetic material, because the forces involved are only a few percent of the weight of the material itself. So if you try it with aluminum, it will not stick to the magnet.

Paramagnetic material will be pulled, with huge forces, towards a strong magnetic field, provided that the magnetic field is non-uniform. When it as a strong gradient. it is pulled towards the strong side.

Dependencies

Relative permeability

In all cases, whether we have diamagnetic, paramagnetic or ferromagnetic material, the magnetic field inside is different from what the field would be without the material. In many cases, but not all, is the field inside the material proportional to the vacuum field. We call this proportionality constant is called the relative permeability, \(\kappa_m\). $$ \vec B_\text{inside} = \kappa_m \, \vec B_\text{vac} \tag{\(\kappa_m\)} \label{eq:kappam} $$

Now we can look at these values for the relative permeability, and we can understand the difference between diamagnetic material, paramagnetic material and ferromagnetic material.

Magnetic susceptibility

In the case of diamagnetic and paramagnetic material, the \(\vec B\)-field inside is only slightly different from the vacuum field. It is common to express \(\kappa_m\) in terms as \(1\) plus something that we call magnetic susceptibility, \(\chi_m\). Because if it is very close to \(1\), it is easier to simply list chi of M. $$ \kappa_m = 1 + \chi_m \nonumber $$

Diamagnetic materials

For diamagnetic materials, the values for \(\chi_m\) are all negative. It is slightly smaller than \(1\), because these induced dipoles oppose the external field.

Diamagnetic materials
Material \(\chi_m\)
\(\rm{Bi}\) \(-1.7\times10^{-4}\)
\(\rm{Cu}\) \(-10^{-5}\)
\(\rm{H_2O}\) \(-10^{-5}\)
\(\rm{N_2}\,(1\,\rm{atm})\) \(-7\times 10^{-9}\)

Paramagnetic materials

For paramagnetic, the \(\chi_m\) is positive, and again the values are small. Inside the material, the magnetic fields is a little larger than the vacuum field.

Diamagnetic materials
Material \(\chi_m\) Temperature
\(\rm{Al}\) \(+2\times10^{-5}\) \(\approx 300\,^oK\)
\(\rm{O_2\,(1\,\rm{atm})}\) \(+2\times10^{-6}\) \(\approx 300\,^oK\)
\(\rm{O_2\,(\rm{liquid})}\) \(+3.5\times10^{-3}\) \(90\,^oK\)

Ferromagnetic materials

For ferromagnetic materials, it would be absurd to list \(\chi_m\) because it is so large that you can forget about the \(1\). So \(\chi_m\) is about the same as \(\kappa_m\) $$ \chi_m \approx \kappa_m \approx 10^2 \ldots 10^5 \nonumber $$ If \(\kappa_m\) is ten thousand, you would have a field inside the ferromagnetic material that is \(10,000\times\) larger than the vacuum field. There is a limit, that we discuss next time.

The three most common ferromagnetic materials are cobalt, nickel and iron. Gadolinium is ferromagnetic in the winter, when the temperature is below 16 °C, but is paramagnetic in the summer.

Curie point

So paramagnetic and ferromagnetic properties depend on the temperature. (Diamagnetic properties do not depend on the temperature.)

At very low temperatures, there is very little thermal agitation. So we can then easier align those dipoles. So the values for \(\kappa_m\) will be different. If you cool ferromagnetic material, you expect the \(\kappa_m\) to go up. You get a stronger field inside.

If you make the material very hot, then it can lose its ferromagnetic properties completely. At a very precise temperature the domains fall apart. The domains themselves no longer exist. That is also something that you need quantum mechanics for to understand. We call this the Curie temperature. For iron this is 1043 °K, or 770 °C, where all of a sudden all the domains disappear and the material becomes paramagnetic.

In other words, if ferromagnetic material would be hanging on a magnet and you heat it up above the Curie point, it will fall off.

Maximum dipole moment

As we have seen: in paramagnetic and ferromagnetic, the \(\kappa_m\) is the result of the intrinsic dipoles of atoms/molecules are aligning with the external field.

The question is: how large can the magnetic dipole moment of a single atom be? How strong can we have a field inside ferromagnetic material? In other words, if we were able to align all the dipole moments of all the atoms, what is the maximum that we can achieve.

Bohr magnetron

To calculate the magnetic dipole moment of an atom, you have to do some quantum mechanics, and that is beyond our scope. We will derive it in a classical way.

Assuming a hydrogen atom, with a proton in the center, and an electron with orbit radius \(R\). The electron \(e^-\) moves with velocity \(v\), so the current goes in the opposite direction.

The mass, charge and Bohr radius of an electron $$ \begin{align*} m_e &= 9.1\times10^{-31}\,\rm{kg} \\ e &= 1.6\times 10^{-10}\,\rm{C} \\ R &= 6\times 10^{-11}\,\rm{m} \end{align*} $$

The current running around the proton, creates a magnetic field, so the dipole moment \(\vec u\) is upwards

Recall the magnetic dipole moment from equation \(\eqref{eq:dipolemoment}\)

$$ \vec\mu = I \vec A \nonumber $$

The area \(A\) is simply $$ A = \pi R^2 \approx 8\times10^{-21}\,\rm{m} \nonumber $$

For the current we have to combine knowledge Newtonian mechanics and electro magnetics. The electron goes around, because the proton and electron attract each other. So there is a Coulomb force \(\vec F\).

The Coulomb force is $$ \begin{align*} F &= \frac{q_1\,q_2\,K}{4\pi\varepsilon_0\,r^2} \\ &= \frac{e^2}{4\pi\varepsilon_0\,r^2} \end{align*} $$

From Newtonian mechanics we know that, the centripetal force (the force that holds it in orbit)

$$ F = m\,v^2 \nonumber $$

Combining these, allows us to calculate the velocity of the electron $$ \begin{align*} \frac{e^2}{4\pi\varepsilon_0\,R^{\cancel{2}}} &= \frac{m\,v^2}{\cancel{R}} \\ \implies v &= \sqrt{\frac{e^2}{m\,4\pi\,\varepsilon_0\,R}} \\ &\approx 2.3 \times 10^{6}\,\rm{m/s} \end{align*} \nonumber $$

To find the current, we first find the time \(T\) that it takes the electron to go around $$ \begin{align*} T &= \frac{2\pi\,R}{v} \\ &\approx 1.4\times10^{-16}\,\rm{sec} \end{align*} $$

At every one point on the radius, every \(1.4\times10^{-16}\) seconds, the electron goes by. The definition of current is: charge per unit time $$ \begin{align*} I &= \frac{e}{T} \\ &= 1.1\times 10^{-3}\,\rm{A} \end{align*} $$

This is a lot. One electron going around a proton represents a current of a milliampere! Now we have the magnetic moment \(\mu\) $$ \begin{align*} \mu &= I\,A \\ &\approx (1.1\times 10^{-3})(8\times10^{-21}) \end{align*} $$

This \(\mu_b\) is called the Bohr magneton $$ \shaded{ \mu_b\approx 9.3\times 10^{-24}\,\rm{Am^2} } \tag{Bohr magnetron} $$

What we can’t understand with our current knowledge, but will with quantum mechanics, is that the magnetic moment of all electrons in orbit can only be a multiple of this number, nothing in between. It is quantization. It includes even zero, which is even harder to understand.

Spin

In addition to a dipole moment due to the electron going around the proton, the electron itself is a charge which spins around its own axis. That means that a charge is going around on the spinning scale of the electron. That magnetic dipole moment is always the value \(\mu\).

So the net magnetic dipole moment of an atom or molecule is now the vectorial sum of all these dipole moments, of all these electrons going around, the orbital dipole moments, and now you have to add these spin dipoles.

Some of these cancel each other out. The net result is that most atoms/molecules have dipole moments which are either one Bohr magneton or two Bohr magnetons. That is what we will use discussing how strong a field we can create if we align all those magnetic dipoles.

The magnetic field that is produced inside a material when we expose it to an external field, that magnetic field \(\vec B\) is the vacuum field that we can create with a solenoid, plus what we will call \(\vec B^\prime\) $$ \vec B = \vec B_{vac} + \vec B^\prime \tag{net field} \label{eq:netfield} $$

Here \(\vec B^\prime\) is the result of aligning these dipoles. The degree of success depends on the strength of the external field and of course the temperature.

A big if

If, a big if, they \(\vec B^\prime\) is linearly proportional to \(\vec B_{vac}\), then we can write what we saw earlier $$ \begin{align*} B^\prime &= \chi_m\,\vec B_{vac}, & (\vec B = \vec B_{vac} + \vec B^\prime) \\ \implies \vec B &= (1 + \chi_m) \vec B_{vac} \\ &= \kappa_m\,\vec B_{vac} \end{align*} $$

This is only a meaningful equation if the sum of the alignment of all these dipoles can be written as linearly proportional with the external field. Let’s explore that in more detail.

Saturation

With paramagnetic material, the linearity always holds up, but with ferromagnetic material that is not the case. With ferromagnetic material is relative easy to align those dipoles, because they already group in domains and the domains flip in unison.

As we will see, with ferromagnetic material we can go into what we call saturation, where all the dipoles are aligned in the same direction. That leaves the question: how strong would that field be?

We will make a rough calculation that gives a pretty good feeling for the numbers. We choose a material whereby the magnetic dipole moment is two Bohr magnetrons $$ \mu = 2\,\mu_b \nonumber $$

We take the situation where they’re all aligned. The illustration shows the electron path around the nuclei in a solid material, so the atoms are nicely packed

All the dipole moments are nicely aligned, so all the magnetic fields support each other. We want to find the magnetic field in one atom. Note that this looks like a solenoid, where you have windings and currents going around.

Remember for a solenoid

$$ B = \mu_0\,I\,\underline{\frac{N}{l}} \nonumber $$

We need to figure out what would be the factor \(\frac{N}{l}\). Let’s take a material where the atom density \(\mathcal N\) is $$ \mathcal N = 10^{29}\,\rm{atoms/m^3} \nonumber $$

Now we have to introduce the magnetic moment, the two Bohr magnetons.

We take a length of one meter. Each loop has area \(A\), so the volume of this “solenoid” is $$ \rm{vol} = A\,\rm{m^2}\times 1\,\rm{m} = A\,\rm{m^3} \nonumber $$

But the number of atoms per cubic meter is \(mathcal N\), so the number of atoms (windings) in this solenoid per meter is $$ \rm{windings} = A\,\mathcal N \nonumber $$

Now the factor \(A\mathcal N\) is our factor \(\frac{N}{l}\), so for this assumed material we can write $$ \begin{align*} B &= \mu_0 \, \underline{I \, A}\, \mathcal N, & (IA = \mu = 2\mu_b) \\ &= u_0\,2\mu_b\,\mathcal N \\ &\approx (1.25\times10^{-6})\,2\,(9.3\times10^{-24})(10^{29}) \\ &\approx 2.3\,\rm{T} \end{align*} $$

Let’s use this number and understand what’s going to happen in that ferromagnetic material.

Inside the ferromagnetic material

If we take ferromagnetic material and expose it to an external field, a vacuum field. So we stick it in a solenoid, and we choose the current through the solenoid. The vacuum field is linearly proportional to the current through the solenoid

The strength of the solenoid’s vacuum field is

$$ B = \mu_0\,I\,\underline{\frac{N}{l}} \nonumber $$ Don’t confuse this the atomic scale magnetic field.

When we stick the ferromagnetic material inside the solenoid, the magnetic field there is \(B\)

We can’t plot it on a 1:1-scale because \(\kappa_m\) for ferromagnetic material is so large, say \(1000\), because the field inside will be \(1000\times\) higher. If it were to scale then \(\tan\alpha = \kappa_m \approx 10^3\).

Not on a 1:1-scale

At the beginning we get a nice linear curve, but now slowly we’re beginning to reach saturation, where all these dipoles are going to be aligned. What you see then is that this curve bends over and over and will finally reach \(2.3\,\rm{T}\) that we just calculated for our imaginary material plus \(B_{vac}\).

We will now call the \(2.3\,\rm{T}\) field \(B^\prime\). This is the field that is the result of the alignment of all those dipoles. So when I increase the vacuum field, (B^\prime\) goes into saturation and settle for \(2.3\,\rm{T}\) and can no longer increase because all the magnetic dipoles are aligned.

So at point \(A\) the field is no longer \(1000\times\) stronger than the vacuum field. We’re no longer in the linear part. You could also think of it as \(\kappa_m\) being smaller than a thousand. It is no longer proportional to the value of one thousand.

If the temperature of the material is lower, it’s easier to align them, so you will achieve saturation earlier and the cure will go steeper towards saturation.

Not on a 1:1-scale

When it reaches point \(B\), when \(B^\prime\) goes into saturation, we can only increase the \(B\)-field in the material, by increasing the vacuum field. Because \(B^\prime\) is not going to go up again. It will go up very slowly, because the huge multiplication factor of \(1000\times\) is gone now. So the growth is very slow.

Hysteresis curve

What happens once we have driven the material into saturation? What happens when we change the current and make the vacuum field zero again? Now you get a very unusual behavior.

Let’s assume that positive current, creates a vacuum field to the right, and a negative (reversed) current creates a vacuum field to the left.

Varying the external field

  1. O to B – This part of the curve is called “virginal”. Again, we start with zero current and increase it until saturation. At that point \(B\), all the domains have flipped in the direction of the vacuum field. The material itself is now magnetic. We have created permanent magnetism.
  2. B to P – We reduce the current to zero. The vacuum field is still to the left. At point \(P\)
    • \(B_{vac}=0\), but
    • \(B^\prime\) is still to the right, so
    • the net field \(B\) is to the right.
    Here we have something bizarre: the vacuum field is to the left, but there is no magnetic field inside the material.
  3. P to Q – We reverse the current, and slowly increase it from zero. So the vacuum field is now to the left. This brings us to location \(Q\), where we have something bizarre: the vacuum field is to the left, but there is no magnetic field inside the material.
    • \(B_{vac}\) is to the left, but
    • \(B^\prime\) is still to the right (because the domains are still aligned to the right), so
    • the net field \(B\) is zero
  4. Q to R – We increasing the (reverse) current. So the vacuum field remains to the left. At point \(R\), the material goes into saturation again.
  5. R to B – We bring the current back to zero and we arrive at point \(S\). There we have permanent magnetism again because some domains stay aligned towards the left.
    • the vacuum field is zero, and
    • \(B^\prime\) points to the right.
    • Again we have permanent magnetism.
  6. S to P – We reverse the current back to the clockwise direction, and increase the current until the material is back in saturation.

Memory

We call this the hysteresis curve. For a given value of the current / vacuum field, we have two possibilities for the magnetic field \(B\). When I expose this material to an external field, I can’t calculate what the magnetic field inside the material will be. It depends on the history of the material.

\(\kappa_m\)

Looking at point \(Q\) and \(Q\), and you ask for \(\kappa_m\), it is almost a ridiculous question. Because \(\kappa_m\) is defined in equation \(\eqref{eq:kappam}\)

$$ \vec B_\text{inside} = \kappa_m \, \vec B_\text{vac} \nonumber $$

At those points we have a vacuum field, but no net field inside, so \(\kappa_m\) has to be zero. In the second and fourth quadrant it is even more bizarre, because \(\kappa_m\) has to be negative.

Removing the permanent magnetism

To make the material virginal again

  • By taking the material out and heating it up above the Curie point, so the domains completely fall apart. Then you cool it again below the Curie point.
  • The other way is banging it with a hammer and hope for the best.
  • The other way is called demagnetization. That is what happens when you steal a book in the library and the alarm goes off. Someone hasn’t demagnetized the magnetic strip in the book. To demagnetize, one would slowly reduce an AC current through the solenoid.

Effect on external field

If we bring ferromagnetic material in the vicinity of a magnet, we change the magnetic field configuration. Suppose we have a magnet and we bring ferromagnetic material close by

The material will see the vacuum field, so its domains try to align a bit. It will get a south and north pole and create a field in the same direction. It will support the external field. The net result is that the field inside the material becomes very strong.

The external field lines will get sucked into the ferromagnetic material. The external field elsewhere will weaken.

The last Maxwell’s equation

Let’s look at the Maxwell equations as we have them so far.

  1. Gauss’s law $$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \color{blue} \phi = \oiint_S \vec E \cdot d\vec A = \frac{\sum_i Q_i}{\kappa\,\varepsilon_0} \nonumber $$ The electric flux through a closed surface is equal to all the charge inside divided by \(\kappa \varepsilon_0\). With electric fields, the \(\kappa\) always lowers the field inside the material. But nothing going to change here.
  2. Gauss’s law for magnetism $$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \color{brown} \oiint_S \vec B \cdot d\vec A = 0 \nonumber $$ This tells us that magnetic monopoles don’t exist. (Or at least we think they don’t exist.)
  3. Faraday’s law $$ \color{green} \oint_C \vec E\cdot d\vec l = -\frac{d}{dt}\iint_R \vec B \cdot d\vec A \nonumber $$ When you move conducting loops in magnetic fields, you create electricity. This doesn’t require any adjustment in terms of \(\kappa_m\) either.
  4. Ampère’s law amended by Maxwell $$ \color{purple} \oint \vec B \cdot d\vec l = \mu_0\left(I_\rm{encl} + \varepsilon_0\kappa \frac{d}{dt} \iint_S \vec E\cdot d\vec A \right) \nonumber $$ Tells us the magnetic field for vacuum. Now we know that’s not true anymore.

Ampère’s law with Maxwell’s addition needs to be adjusted by a factor of \(\kappa_m\), the relative permeability. $$ \shaded{ \oint \vec B \cdot d\vec l = \textcolor{purple}{\kappa_m}\, \mu_0\left(I_\rm{encl} + \varepsilon_0\kappa \frac{d}{dt} \iint_S \vec E\cdot d\vec A \right) } \tag{Ampère/Maxwell add} \label{eq:maxwell4} $$

This \(\kappa_m\) is perfectly kosher for paramagnetic and diamagnetic materials. But with ferromagnetic material, you have to be very careful as we have seen with the hysteresis phenomenon. There are even situations where \(\kappa_m\) is negative; where \(\kappa_m\) is zero; and where \(\kappa_m\) can as high as \(10^3\). So we have to be very careful when applying this equation without thinking.

This moment is very special, because we have all four Maxwell’s equations in place. Hope you can appreciate them.

celebrateva.org

Magnetic field energy

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My notes of the excellent lecture 20 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.”

Magnetic field energy

Let’s start by calculating how much heat is produced as the current goes from a maximum value to zero.

At any moment in time, the current is producing heat in the resistor. If my voltage difference becomes zero at \(t=0\), then the amount of heat is produced as the current dies out $$ Q = \int_0^\infty i^2 R\,dt $$

Substituting \(I\) from \(\eqref{eq:lroff}\) $$ \begin{align*} Q &= R \int_0^\infty \left(i_{max}\,e^{-\frac{R}{L}t}\right)^2 \,dt \\ &= i_{max}^2 R \int_0^\infty e^{-\frac{2R}{L}t}, & \left( \int \rm{e}^{cx}dx=\frac{1}{c}\rm{e}^{cx} \right) \\ &= i_{max}^2 R \Big[ -\frac{L}{2R} e^{-\frac{2R}{L}t} \Big]_0^\infty \\ &= i_{max}^2 \bcancel{R} \frac{L}{2\bcancel{R}} \end{align*} $$

The heat produced is $$ \shaded{ Q =\frac{1}{2}L\, i_{max}^2 } \label{eq:lrheat} $$

Density

We can now calculate how much energy there was in the magnetic field per cubic meter. This field was exclusively inside that solenoid.

We know

  • The energy that is ultimately coming out from equation \(\eqref{eq:lrheat}\) $$ Q = \frac{1}{2}L\, i_{max}^2 \nonumber $$ you can drop the max. this simply tells you then that any moment in time that i am running a current through a solenoid, that the energy that is available in the solenoid in the forms of magnetic energy is one \(\frac{1}{2}LR^2\) $$ Q = \frac{1}{2}L\, I \label{eq:roundup1} $$
  • Using Ampère’s law we derived $$ \begin{align} B &= \frac{\mu_0 I N}{l} \nonumber \\ \implies I &= \frac{B\,l}{\mu_0\,N} \label{eq:roundup2} \end{align} $$
  • We know \(L\) from equation \(\eqref{eq:l2}\) $$ L = \mu_0\, \pi r^2 \frac{N^2}{l} \label{eq:roundup3} $$

The energy stored in the solenoid follows as $$ \begin{align*} Q &= \frac{1}{2}L\, I \\ &= \frac{1}{2} \left(\mu_0\, \pi r^2 \frac{N^2}{l}\right) \left(\frac{B\,l}{\mu_0\,N}\right)^2 \\ &= \frac{ B^2}{2\mu_0}\underbrace{\pi r^2 l}_\text{volume} \end{align*} $$

The term \(\pi r^2 l\) is the volume where the magnetic field exists, since we assumed that the magnetic field is zero elsewhere.

The magnetic field energy density, how much energy there is per cubic meter, is $$ \shaded{ \frac{Q}{\text{volume}} = \frac{ B^2}{2\mu_0} }\quad \left[\rm{\frac{J}{m^3}}\right] \tag{magnetic field density} $$

In principle, if we know the magnetic field everywhere in space, then you can integrate over all space and calculate how much energy is present in the magnetic field.

Parallel with electric field density

Earlier, we did something similar for electric fields. We calculated the electric field energy density.

$$ \frac{W}{\rm{volume}} = \frac{1}{2} \varepsilon_0\kappa E^2 \nonumber $$

Note:

  • In the case of the electric field, it represents the work that we have to do to arrange the charges in a certain configuration.
  • In the case of a magnetic field, it represents the work that I have to do to get a current going inside a pure self-inductor. That means the resistance of the self-inductor is zero, and it takes work because the solenoid opposed the building up of current.

Self-inductance; RL circuit

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My notes of the excellent lecture 20 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.”

We will quantify the ability of a circuit to fight the electric flux that is produced by the circuit itself.

If you run a current through a circuit, then you create some magnetic fields. If the current is changing, then the magnetic fields are changing. There will be an induced EMF in that circuit that fights the change. We express that in terms of self-inductance \(L\).

The magnetic flux that is produced by the circuit is always proportional to the current $$ \shaded{ \phi_B = L\, I } \tag{self-inductance} \label{eq:L} $$

Therefore the induced EMF $$ \shaded{ \mathcal E = – \frac{d\phi}{dt} = -L \frac{dI}{dt} } \tag{EMF self-inductance} \label{eq:Lemf} $$

This \(L\) is only a matter of geometry. It is not a function of the current itself.

Solenoid

We will calculate the self-inductance of a solenoid with radius \(r\), length \(l\) and \(N\) windings.

Using Ampère’s law we derived

$$ B = \frac{\mu_0 I N}{l} \label{eq:B} $$

We assign an open surface to this closed loop. The surface is complicated to imagine, but has a spiral staircase-like surface inside the solenoid. That magnetic field penetrates that surface \(N\) times, because we have \(N\) loops. Assuming the magnetic field is constant inside the solenoid, and zero outside, the magnetic flux is $$ \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} \phi_B &= \iint_R \vec B\cdot d\vec A, & \left(\vec B\parallelsum\vec A\right) \\ &= B \iint_R dA \\ &= B\, N\, \pi r^2 \end{align*} $$

Substituting \(B\) from equation \(\eqref{eq:B}\) $$ \begin{align*} \phi_B &= B\, N\, \pi r^2 \\ &= \frac{\mu_0 I N}{l} N\, \pi r^2 \\ &= \frac{\mu_0 I N^2 \pi r^2}{l} \end{align*} $$

This must equal the definition of \(\eqref{eq:L}\) $$ \frac{\mu_0 \bcancel{I} N^2 \pi r^2}{l} = L\,\bcancel{I} \nonumber $$

The self-inductance of the solenoid follows as $$ \shaded{ L = \mu_0\, \pi r^2 \frac{N^2}{l}\quad \Big[\rm{H}\Big] } \tag{\(L\) solenoid} \label{eq:l2} $$

The SI unit for inductance is Henry, named for the American physicist Joseph Henry.

Every circuit has a finite value for the self-inductance, however small that may be. Sometimes it’s so small that we ignore it.

RL Circuit and DC (step response)

Let’s look at a circuit with a battery, a switch, a self-inductor in series with a resistor.

Note that we use the lowercase \(i\) for current to indicate that it is not a constant. It is a function of time.

Battery provides energy

The switch is in position \(a\) and there is no current. Then, we flip the switch to position \(b\)

  1. Before we flip the switch there is no current running. At time \(t_0\) the current \(i=0\).
  2. As we close the switch the current wants to increase, but the self-inductance fights the change in current.
  3. If we wait long enough, the self-inductance loses the fight, and the current reaches a maximum value dictated by Ohm’s law, because the (ideal) self-inductance has no resistance.

We can’t use Kirchhoff’s rule, because there is a changing electric flux. The electric fields inside the conducting wires become non-conservative, and the closed loop integral around the circle is not zero. (see the lecture supplement)

Recall

$$ \oint_C \vec E\cdot d\vec l = -\frac{d\phi_B}{dt} \nonumber $$

Substituting \(\phi_B\) from equation \(\eqref{eq:L}\) \(\phi_B = L\, i\) $$ \oint_C \vec E\cdot d\vec l = -L\frac{di}{dt} \nonumber $$

Going around in the direction of the current, the electric fields are

\(U\)Instead of \(\Delta V\), we use the European symbol for voltage difference \(U\).

Applying Faraday’s law $$ \begin{align*} -U + 0 + i R &= -L\frac{di}{dt} \\ \implies U – L\frac{di}{dt} &= i\, R \end{align*} $$

Since \(di/dt\) is positive and growing with time, the induced EMF, \(-L\frac{di}{dt}\), is always opposing the voltage difference of the battery. That is where Lenz’s law is all about. It is not until you wait long enough, and \(di/dt\) has become zero, that \(U=iR\).

Solve using the antiderivative

Details

$$ \require{physics} \begin{align*} L\dv{i}{t} + Ri &= U \\ \implies \dv{i}{t} &= \frac{U – Ri}{L} \\ \implies \frac{L}{U – Ri}\dd i &= \dd t \\ \implies \int \frac{L}{U – Ri}\,\dd i &= \int \dd t + C\\ \end{align*} $$

Substitute \(v\equiv U-Ri\) \(\implies\) \(\dv{v}{i}=-R\) \(\implies\) \(\dd i = -\frac{\dd v}{R}\) $$ \newcommand{dv}[2]{\frac{\dd #1}{\dd #2}} \begin{align*} \int \frac{L}{\underline{v}} \left(\underline{-\frac{\dd v}{R}}\right) &= \int \dd t + C \\ \implies -\frac{L}{R} \int \frac{1}{v}\,\dd v &= \int \dd t + C \\ \implies -\frac{L}{R} \ln v &= t + C \\ \implies \ln v &= -\frac{R}{L} t + C_2 & \text{(subst. back)} \\ \implies \ln U-Ri &= -\frac{R}{L} t + C_2 \\ \implies U-Ri &= C_3\,\rm{e}^{-\frac{R}{L} t} \\ \implies Ri &= U – C_3\,\rm{e}^{-\frac{R}{L} t} \\ \implies i &= \frac{U}{R} – C_4\,\rm{e}^{-\frac{R}{L} t}, & \left(i(0)=0\right) \\ &= \frac{U}{R} \left( 1 – \rm{e}^{-\frac{R}{L} t} \right) \end{align*} $$

Solve using Laplace transform

Details

Use a Laplace transform to solve this differential equation $$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \begin{align*} L\frac{di}{dt} + Ri &= \gamma(t)\,U & \left(\laplace\right) \\ \implies \mathfrak{L} \left\{ L\frac{d\,i(t)}{dt} + R\,i(t) \right\} &= \mathfrak{L} \left\{ \gamma(t)\,U \right\} \\ \implies L\left(sI(s) – i(0^+)\right) + R\,I(s) &= \frac{U}{s} & \left(i(0^+)=0 \right) \\ \implies I(s) &= \frac{U}{s(Ls + R)} = \frac{U}{L}\frac{1}{s\left(s + \frac{R}{L}\right)} \\ &= K\frac{1}{s\left(s – p\right)}, \quad K=\frac{U}{L},\quad p=-\frac{R}{L} \\ \end{align*} $$

Split up this complicated fraction into forms that are in the Laplace Transform table. According to Heaviside, this can be expressed as partial fractions. [swarthmore, MIT-cu] $$ I(s) = K\frac{1}{s\left(s – p\right)} \equiv\frac{c_0}{s}+\frac{c_1}{s-p} \label{eq:heavi} $$

Multiply with respectively \(s\) and \((s-p)\) $$ \left\{ \begin{align*} K\frac{\cancel{s}}{\cancel{s}\left(s – p\right)} &\equiv \frac{\cancel{s}\,c_0}{\cancel{s}}+\frac{s\,c_1}{s-p} \\ K\frac{\cancel{s-p}}{s\cancel{(s-p)}} &\equiv \frac{c_0(s-p)}{s}+\frac{c_1\cancel{(s-p)}}{\cancel{s-p}}\\ \end{align*} \right. $$

Given that these equations are true for any value of \(s\), choose a convenient value that help us find \(c_0\), and substitute \(K\) and \(p\) back $$ \left\{ \begin{align*} c_0 &= \left.\frac{K-sc_1}{s-p}\right|_{s=0} = -\frac{K}{p} = \frac{U}{R} \\ c_1 &= \left.\frac{K – (s-p)c_0}{s}\right|_{s=p} = -\frac{U}{R} \\ \end{align*} \right. $$

Transfer equation \(\eqref{eq:heavi}\) back to the time domain, and fill in the values for \(c_0\), \(c_1\) and \(p\) $$ \begin{align*} i(t) &= \mathcal{L}^{-1} \left\{ \frac{c_0}{s} \right\} + \mathcal{L}^{-1} \left\{ \frac{c_1}{s-p} \right\}, & t\geq 0 \\ &= c_0 + c_1\,\rm{e}^{pt} \\ &= \frac{U}{R} – \frac{U}{R}\rm{e}^{-\frac{R}{L}t} \end{align*} $$

Solution

The current in the RL circuit as a result of the step-function $$ \shaded{ i(t) = i_{max}\left( 1 – e^{-\frac{R}{L}t} \right) }, \quad i_{max} = \frac{U}{R} \tag{\(LR\) battery} \label{eq:lron} $$

Plotting \(i(t)\)

Magnetic field supplies energy

All of a sudden, we’re making the voltage difference zero by flipping the switch to position \(a\). The current is still running, and at a new time \(t_0\), the voltage difference, \(U\), is zero.

The self-inductance is going to fight the change of the current going down. You expect that the current will not die off right away.

The time behavior can be calculated by going back to our differential equation and making \(U\) zero. $$ \begin{align*} L\frac{di}{dt} + Ri &= 0 \end{align*} $$

Solve using the antiderivative

Details

$$ \newcommand{dv}[2]{\frac{\dd #1}{\dd #2}} \begin{align*} L\dv{i}{t} + Ri &= 0 \\ \implies \dv{i}{t} &= -\frac{Ri}{L} \\ \implies \frac{1}{i}\dd i &= -\frac{R}{L}\dd t \\ \implies \int \frac{1}{i}\dd i &= -\frac{R}{L} \int \dd t + C\\ \implies \ln i &= -\frac{R}{L}t + C \\ \implies i &= C\,\rm{e}^{-\frac{R}{L}t}, & \left( i(\infty)=\tfrac{U}{R}\right) \\ &= \frac{U}{R}\,\rm{e}^{-\frac{R}{L}t} \end{align*} $$

Solve using Laplace transform

Details

Use a Laplace transform to solve this differential equation $$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \begin{align*} L\frac{di}{dt} + Ri &= 0 & \left(\laplace\right) \\ \implies \mathfrak{L} \left\{ L\frac{d\,i(t)}{dt} + R\,i(t) \right\} &= \mathfrak{L} \left\{ 0 \right\} \\ \implies L\left(sI(s) – i(0^+)\right) + R\,I(s) &= 0 & \left( i(0^+)=\frac{U}{R} \right) \\ \implies L\left(sI(s) – \frac{U}{R}\right) + R\,I(s) &= 0 \\ \implies sLI(s) + R\,I(s) &= \frac{UL}{R} \\ \implies I(s) &= \frac{UL}{R} \frac{1}{sL + R} = \frac{U}{R}\frac{1}{s + \frac{R}{L}} \\ &= K\frac{1}{s – p}, \quad K=\frac{U}{R},\quad p=-\frac{R}{L} \\ \end{align*} $$

This is already in a form that we can find in the Laplace Transform table. Transfer this equation back to the time domain, and fill in the value for \(p\) $$ \begin{align*} i(t) &= \mathcal{L}^{-1} \left\{ K\frac{1}{s-p} \right\}, & t\geq 0 \\ &= K\,\rm{e}^{pt} \\ &= \frac{U}{R}\rm{e}^{-\frac{R}{L}t} \end{align*} $$

Solution

The current in the RL circuit as a result of the step-function $$ \shaded{ i(t) = i_{max}\,e^{-\frac{R}{L}t} }, \quad i_{max} = \frac{U}{R} \tag{\(LR\) magnetic field} \label{eq:lroff} $$

Plotting \(i(t)\)

Note

  • For \(t=0\), we find \(i(t)=U/R\)
  • For \({t\to\infty}\), we find \(i(t)\to 0\)
  • For \(t=\frac{L}{R}\) seconds, then \(i(t)\approx 0.37\, i_{max}\)

This is all the consequence of the circuit is capable of fighting the change its own magnetic flux that it is creating.

With the battery still connected, the battery was providing the heat that was dissipated in the resistor. Now, with no battery, there is still current. While the current is dying, there is still heat produced in that resistor. That energy must come from the magnetic field in the solenoid.

The idea that energy comes out as heat, that was there earlier in the form of a magnetic field, allows us to evaluate what we call the magnetic energy field density.

RL Circuit and AC (frequency response)

This time, let’s look at a circuit with an AC power source, and a self-inductor in series with a resistor.

The AC power source has angular frequency \(\omega\) $$ U = U_0 \sin\omega t $$

Applying Faraday’s law is the same as before, except that \(U = U_0 \cos\omega t\) $$ L\frac{di}{dt} + i\, R = U_0 \sin\omega t $$

Solve using Fourier Series

Details

Solve using Fourier Series $$ L \frac{di}{dt} = Ri = U_0\sin\omega t \label{eq:dv} $$

Let \(i(t)\) $$ \begin{align} i(t) &= \sum_{n=\infty}^\infty c_n\, e^{\rm{j}nt} \label{eq:i} \\ \implies \frac{di}{dt} &= \sum_{n=\infty}^\infty \rm{j}n\,c_n\, e^{\rm{j}nt} \nonumber \end{align} $$

Substitute \(i(t)\) and \(\frac{di}{dt}\) in the differential equation \(\eqref{eq:dv}\), and express the sine term using Euler’s formula $$ \begin{align*} L \sum_{n=\infty}^\infty jn\,c_n\, e^{jnt} + R\sum_{n=\infty}^\infty c_n\, e^{jnt} &= U_0\left( \frac{\mathrm{e}^{j\omega t} – \mathrm{e}^{-j\omega t}}{2j} \right) \\ \implies \sum_{n=\infty}^\infty c_n\, e^{\rm{j}nt} \left( R + \rm{j}nL \right) &= -j\,\tfrac{U_0}{2}\,\mathrm{e}^{j\omega t} + j\,\tfrac{U_0}{2}\,\mathrm{e}^{-\rm{j}\omega t} & \text{(only \(n=\omega\) matches)} \\ \end{align*} $$

The only values for \(n\) that fit are \(\omega\) and \(-\omega\) $$ \left\{ \begin{align*} c_\omega\, e^{j\omega t} ( R + j\omega L) &= -j\tfrac{U_0}{2}\,\mathrm{e}^{j\omega t} \\ c_{-\omega}\, e^{j\omega t} ( R + j\omega L) &= j\tfrac{U_0}{2}\,\mathrm{e}^{-j\omega t} \\ \end{align*} \right. $$

Now, we need to find the coefficients \(c_\omega\) and \(c_{-\omega}\). Let \(c_\omega=a_\omega + j\,b_\omega\) $$ \begin{align*} (a_\omega + j\,b_\omega)( R + j\omega L) &= j\tfrac{U_0}{2} \\ \implies (a_\omega R – b_\omega\omega L) + j(a_\omega\omega L + b_\omega R) &= j\tfrac{U_0}{2} \end{align*} $$

The real and imaginary parts should match $$ \begin{align} a_\omega R – b_\omega\omega L &= 0 \implies a_\omega = b_\omega\omega \frac{L}{R} \label{eq:re}\\ a_\omega\omega L + b_\omega R &= \frac{U_0}{2} \label{eq:im} \end{align} $$

To find \(b_\omega\), substitute \(a_\omega\) from equation \(\eqref{eq:re}\) in \(\eqref{eq:im}\) $$ \begin{align} b_\omega\omega \tfrac{L}{R} \omega L + b_\omega R &= \tfrac{U_0}{2} \nonumber \\ \implies b_\omega(\omega^2 \tfrac{L}{R} L + R) &= \tfrac{U_0}{2} \nonumber \\ \implies b_\omega &= \tfrac{U_0R}{2(\omega^2 L^2 + R^2)} \label{eq:bw} \\ \end{align} $$

To find \(a_\omega\), substitute \(b_\omega\) from equation \(\eqref{eq:bw}\) in \(\eqref{eq:re}\) $$ \begin{align*} a_\omega &= \frac{U_0R}{2(\omega^2 L^2 + R^2)} \omega\frac{L}{R} \\ &= \frac{U_0 L \omega}{2(\omega^2 L^2 + R^2)} \\ \end{align*} $$

Putting the real an imaginary parts back together, \(c_\omega=a_\omega + j\,b_\omega\) $$ \begin{align*} c_\omega &= \frac{U_0\omega L}{2(\omega^2 L^2 + R^2)} + j\left(\frac{U_0R}{2(\omega^2 L^2 + R^2)}\right) \\ &= \frac{U_0}{2(\omega^2 L^2 + R^2)}(L \omega + jR) \\ \end{align*} $$

Since \(i(t)\) is real, \(c_{-\omega} = c_\omega^* = a_\omega – j b_\omega\) $$ \begin{align*} c_{-\omega} &= \frac{U_0}{2(\omega^2 L^2 + R^2)}(L \omega – jR) \end{align*} $$

Subst. \(c_\omega\) and \(c_{-\omega}\) back in equation \(\eqref{eq:i}\) $$ \begin{align} i(t) &= c_\omega\, \rm{e}^{\rm{j}\omega t} + c_{-\omega}, \rm{e}^{-\rm{j}\omega t} \nonumber \\ &= \frac{U_0}{2(\omega^2 L^2 + R^2)}(L \omega + jR)\, \rm{e}^{\rm{j}\omega t} + \frac{U_0}{2(\omega^2 L^2 + R^2)}(L \omega – jR), \rm{e}^{-\rm{j}\omega t} \nonumber \\ &= \frac{U_0}{2(\omega^2 L^2 + R^2)}\, \left( (L\omega + jR)\,\rm{e}^{\rm{j}\omega t} + (L\omega – jR)\,\rm{e}^{\rm{j}\omega t} \right) \nonumber \\ &= \frac{U_0}{\omega^2 L^2 + R^2}\, \left( \omega L \frac{\rm{e}^{\rm{j}\omega t} + \rm{e}^{-\rm{j}\omega t}}{2} – R \frac{\rm{e}^{\rm{j}\omega t} – \rm{e}^{-\rm{j}\omega t}}{2} \right) \nonumber \\ &= \frac{U_0}{\omega^2 L^2 + R^2}\, \left( \underline{\omega L \cos\omega t – R\sin\omega t} \right) \label{eq:i2} \end{align} $$

Examine the term: \(\omega L \cos\omega t – R\sin\omega t\), using the Sinusoidal identity $$ \begin{align} \omega L \cos\omega t – R\sin\omega t &= A\sin(\omega t – \phi) \label{eq:i3} \\ &= A\sin(\omega t) \cos(-\phi) + A\cos(\omega t) \sin(-\phi) \nonumber \end{align} $$

Sine and cosine terms should match $$ \left\{ \begin{align*} \omega L \cancel{\cos(\omega t)} &= A\cancel{cos(\omega t)} \sin(-\phi) \\ – R\cancel{\sin(\omega t)} &= A\cancel{\sin(\omega t)} \cos(-\phi) \end{align*} \right. $$

Angle of \(\phi\), and modules \(A\) $$ \left\{ \begin{align*} \tan\phi &= -\tfrac{\sin(-\phi)}{\cos(-\phi)} = \tfrac{\omega L}{R} \implies \underline{\phi = \atan\tfrac{\omega L}{R}} \\ A^2\left( \sin(-\phi)^2+\cos(-\phi)\right) &= (\omega L)^2 + (-R)^2 \implies \underline{A = \sqrt{\omega^2 L^2 + R^2}} \end{align*} \right. $$

Substituting \(A\) back in equation \(\eqref{eq:i3}\) $$ \begin{align*} \omega L \cos\omega t – R\sin\omega t &= A\sin(\omega t – \phi) \\ &= \sqrt{\omega^2 L^2 + R^2}\,\sin(\omega t – \phi) \end{align*} $$

Substituting this term back in \(\eqref{eq:i2}\) $$ \begin{align*} i(t) &= \frac{U_0}{\omega^2 L^2 + R^2}\,\sqrt{\omega^2 L^2 + R^2}\,\sin(\omega t – \phi) \\ &= \frac{U_0}{\sqrt{\omega^2 L^2 + R^2}}\,\sin(\omega t – \phi),\quad \phi = \atan\tfrac{\omega L}{R} \end{align*} $$

Solve using Laplace transform

Details

Let’s approach this as a LR-filter, where output \(y(t)\) depends on input \(u(t)\). The transfer function $$ H(s) = \frac{Y(s)}{U(s)} = \frac{R}{sL + R} = \frac{R}{L} \frac{1}{s+\frac{R}{L}} \nonumber $$

The denominator is a first-order polynomial. The root is called the system’s pole $$ \shaded{ H(s) = K \frac{1}{1-p} },\quad K=\frac{R}{L},\quad p=-\frac{R}{L} \nonumber $$

The frequency response \(y_{ss}(t)\) is defined as the steady state response to a sinusoidal input signal \(u(t)=\sin(\omega t)γ(t)\). In Evaluating Transfer Functions, we have proven that

$$ y_{ss}(t)=|H(j\omega)|\,\sin(\omega t+\angle H(j\omega))\,\gamma(t) \nonumber $$

Based on Euler’s formula, we can express \(H(s)\) in polar coordinates $$ \left\{ \begin{align} H(s) &= |H(s)|\ e^{j\angle{H(s)}}\nonumber \\ |H(s)| &= K \frac{\prod_{i=1}^m\left|(s-z_i)\right|}{\prod_{i=1}^n\left|(s-p_i)\right|} =K \frac{1}{\left|s-p\right|}\nonumber \\ \angle{H(s)}&=\sum_{i=1}^m\angle(s-z_i)-\sum_{i=1}^n\angle(s-p_i) =-\angle(s-p)\nonumber \end{align} \right. $$

The transfer function with pole \(p\), evaluated for \(s=j\omega\) can be visualized with a vector from the pole to \(j\omega\).

own work

The length of the vector corresponds to \(|(H(j\omega)|\), and minus the angle with the real axis corresponds to phase shift \(\angle H(j\omega)\). $$ \left\{ \begin{align} |H(j\omega)| &=K \frac{1}{\left|j\omega-p\right|}= K \frac{1}{\sqrt{\omega^2+p^2}}\nonumber \\ \angle{H(j\omega)}&=-\angle(j\omega-p)=\mathrm{atan2}(\omega,-p)\nonumber\\ &=-\arctan\frac{\omega}{-p},\ p\lt0\land p\in\mathbb{R}\nonumber \end{align} \right. \label{eq:polar1} $$

Substitute \(K=\frac{R}{L}\) and \(p=-\frac{R}{L}\) $$ \left\{ \begin{align*} |H(j\omega)| &=\frac{R}{L} \frac{1}{\left|j\omega-p\right|} = \frac{R}{\sqrt{(\omega L)^2+R^2}} \\ \angle{H(j\omega)} &= -\angle\left(j\omega+\frac{R}{L}\right) = -\arctan\left(\frac{\omega L}{R}\right) \end{align*} \right. $$

The output signal \(y_{ss}\) $$ \begin{align*} y_{ss}(t) &= \frac{U_0 R}{\sqrt{(\omega L)^2+R^2}}\,\sin(\omega t – \phi)\,\gamma(t)\nonumber \\ \phi &= \arctan\left(\frac{\omega L}{R}\right) \end{align*} $$

Solution

The steady state current \(i_{ss}=y_{ss}/R\) for a sinusoidal input signal \(U_0\sin(\omega t)γ(t)\) $$ \shaded{ i_{ss}(t) = \frac{U_0}{\sqrt{(\omega L)^2+R^2}}\,\sin(\omega t – \phi) },\quad \tan\phi = \frac{\omega L}{R} \tag{Current AC} \label{eq:currentac} $$

The term in front of the cosine, is the maximum possible current. The cosine term is just oscillating between \(+1\) and \(-1\).

  • The \(\omega L\) has the role of resistance. Its dimension is \(\Omega\). If \(\omega\) is very high, then the resistance becomes very high, so the current becomes very low. If \(\omega\) is high, then the changes \(di/dt\) are very high. The induced EMF is going to be high, so the current will be low.
  • Also, if \(L\) is very high, then the system is capable of fighting back very hard, so it puts up a large resistance.
  • If \(\omega\) is very low, say zero, you simply get \(i(t)=U_0/R\)

The current comes later than the driving voltage, because the self-inductance is fighting the change in current. It is not so surprising that there’s a delay. Looking at the phase shift \(\tan\phi=\omega L/R\)

  • If the self-inductance is very large, the system has a strong ability to fight back. It can delay that current by a large amount.
  • Same when \(\omega\) is high.

Levitation

Posted on

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My notes of the excellent lectures 19 and 20 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.”

Using superconductors

Superconductivity was discovered by Kamerlingh Onnes, a Dutch physicist in 1911. He discovered how to make liquid helium, that he used to cool various substances. He discovered that when you cool mercury to \(4^o\,\rm{K}\), that the electrical resistance goes down to zero. He received the Nobel Prize in 1913.

Heike Kamerlingh Onnes (1913)
Les Prix Nobel 1913, p. 52 / Nobel foundation / PD

At higher temperature

We can only understand superconductivity using quantum mechanics, and even quantum mechanics has a major problems explaining all the phenomena about superconductivity. The problem started in 1986, when two scientists in Zurich, Muller and Bednorz, discovered that certain alloys can be made superconducting at a temperature as high as \(35^o\,\rm{K}\). Theorist earlier had proven that it was impossible to ever get superconductivity at that temperature. This was such as splash in the community, that they were awarded the Nobel Price within one year (1987).

As of today, theorists still cannot explain fully why there is what’s called high-temperature superconductivity. Today’s record is \(135^o\,\rm{K}\). Since liquid nitrogen has a temperature of \(77^o\,\rm{K}\), anyone can now play with superconducting materials, because liquid nitrogen is easy to come by.

If you have zero resistance in a material, you can run an extremely high current through it and therefore get magnetic fields as high as \(6\,\rm{T}\). Such superconducting coils are used in the colliders that we talked about earlier.

No electric field can exist in a superconductor. If there were an electric field, there would be a potential difference over the superconductor. Ohm’s law tells you that if the potential difference is not zero, and the resistance is zero, then the current would go to infinity. $$ \mathcal E = I\textcolor{red}{\uparrow^\infty}\,R_\textcolor{red}{=0} \nonumber $$

Levitating magnet

If we approach a superconducting disk with a magnet, there will be a change in magnetic flux. Based on Faraday’s law there will be an EMF generated in the disk.

However, that EMF must remain zero because the resistance \(R\) of the superconductor is zero. No matter what the current is, there cannot be an EMF $$ \mathcal E = I\textcolor{red}{\uparrow^\infty}\,R_\textcolor{red}{=0} \nonumber $$

As the magnet approaches, eddy currents are going to flow inside the superconductor. These eddy currents create a magnetic field themselves. These eddy current make sure that the net change in magnetic flux \(d\phi_B/dt\) is always zero. $$ \frac{d\phi_B}{dt} = 0 \nonumber $$

Since there was no magnetic flux when the magnet was high up, and the eddy current cancel out any change in magnetic flux. there will never be a net magnetic flux inside the superconductor.

Magnetic pressure

The sketch on the left shows, the magnetic field from the bar magnet and of the eddy currents in the superconductor. The top of the superconductor acts as a north pole, repelling the north pole of the magnet.

Magnetic field of magnet and from induced current
  
Net magnetic field

The sketch on the right shows, the superposition of the two fields. You get a squeezed field between the magnet and the superconductor. When you have such a squeezed magnetic field, there is magnetic pressure because the north poles repel each other. The magnetic pressure \(P\) can be expressed as $$ \shaded{ P = \frac{B^2}{2\mu_0} } \quad \left[\frac{\rm{N}}{\rm{m}^2}\right] \tag{magnetic pressure} $$

The end result is that the magnet is repelled. Pushed up by the superconductor. Even when you start rotating the magnet, the eddy currents will instantaneously adjust to always repel the magnet.

A superconductor levitating a permanent magnet
Julien Bobroff &Frederic Bouquet / CC BY-SA 3.0 / supraconductivite.fr

The eddy current never dissipates any heat. There is no \(I^2R\), because \(R\) is zero. So you never lose the eddy current. The currents never die out! This is different with the next form of levitation.

Using velocity

Another for of levitation is where you move a magnet fast over a conducting surface.

As the magnet overs over the conducting plate, the magnetic flux through that plate will change. Lenz’s law says it will run an eddy currents so that its magnetic field opposes that of the magnet.

2BD Fieldlines thicker
Magnetic field of magnet and induced field
  
2BD Fieldlines thicker 2
Net magnetic field

The north pole of the eddy current opposes the north pole of the magnet. If the magnet has an high enough speed, so that the change in magnetic flux \(d\phi/dt\) is high, the train can float. Many of tons of weight can be made to float.

For the train to float, it has to keep going. If the trains stops the eddy current will die out. There’s no longer a \(d\phi/dt\), but there is resistance in the conductor. So you get Ohmic dissipation \(I^2R\). The heat will dissipated in the conductor, and the train will just plunge down.

That was not the case with the superconductor, because superconductors don’t dissipate any heat. But the idea is the same, as you get a squeezed magnetic field that causes magnetic pressure.

Transrapid 09 in Norddeutschland.
Állatka / PD / wikipedia.org

Japan and Germany are leaders in the world of magnetic levitation trains. There is an enormous reduction in friction if you can have a train that is not in contact with the rails. Speeds have been recorded up to \(550\,\rm{km/h}\). It is no more expensive as building a 4-lane highway at \(30\,\rm{M$/mile}\).

Using alternating current

There is a third form of magnetic levitation whereby we don’t need any speed or superconductors. We feed alternating current (AC) to a solenoid placed over a conducting plate.

Magnetic field from solenoid

Say, at one moment in time the magnetic field is shown below and is increasing. Then of course the magnetic field turns around, up, down, up, down, because it’s AC. We have this continuous magnetic field change, so we have a continuous change in magnetic flux in that plate.

At the moment in time, that the \(\vec B\)-field is down and increasing, we’re going to get an eddy current which will create an opposing magnetic field. Again, you have two opposing north poles. So again, the eddy current in the conducting plate is responsible for a magnetic field and the two repel each other.

Induced magnetic field

A little later in time, the magnetic field strength will decrease. When that happens the eddy current will reverse direction, and the two will attract each other. It now seems quite reasonable that half the time they will attract each other and the other half of the time they will repel each other. That however is not the case. There will be a net repelling force, that we will explain that in the next lecture.

Levitating Barbecue! Electromagnetic Induction
Veritasium / Palais de la Découverte, Paris

AC Levitation – part 2

(Continuing “Levitation using AC current“”)

Induced magnetic field

The behavior depends on the AC current’s sine wave

  • Between the 0° and 90° of the current, the magnetic field points down and is increasing in strength. Because it is increasing in strength, the eddy currents will be in the opposite direction. The magnetic fields will repel each other. (As shown in the illustration above.)
  • Between the 90° and 180° of the the current, the magnetic field of the inductor still points down, but is decreasing in strength. Because it is decreasing, \(d\phi/dt\) is negative and the EMF flips over. Now, the eddy currents are in the same direction as the current through the solenoid. The magnetic fields will attract.

One would expect that half the time the solenoid and the conductor attract, and half the time they repel. So the net effect would be no levitation. However, it is not that simple ..

The secret

As we will see in RL Circuit with AC current, the secret lies in the self-inductance. The eddy current runs over a path which has a resistance \(R\) and self-inductance \(L\). So in the conductor we get an induced current that is delayed over the induced EMF, driven by equation \(\eqref{eq:currentac}\). $$ \phi = \arctan\frac{\omega L}{R} \nonumber $$

The red curve is the current for the coil. When it is positive, it is clockwise, otherwise anti-clockwise. The green curve is the EMV which is induced in the conductor.

Notice when the magnetic field increases, the current goes up in the coil, then the EMF in the conductor is in such a direction that it opposes the change in that magnetic field.

Now when the magnetic field decreases, the current in the coil decreases, immediately the EMF flips over. Therefore if the induced current and the induced EMF were in phase with each other, half the time you would have attraction, and half the time you would have that the two repel each other.

When we add a blue curve representing the induced current,

If there is no phase shift between the induced EMF and the induced current, half the time the blue and red curve are in opposite direction.

Now, if I have a phase delay, so that the induced current comes later than the EMF, say 90°

Now, the red curve and the blue curve are always in opposite directions. Now all the time there is a repelling force.

Even if the phase delay is less than 90°, the net result is that on average you get a repelling force. So the secret of the repelling force, lies in the fact that there is a finite self-inductance in the conductor.

If \(R\) is zero, then of course we have a superconductor, then \(\phi\) is always 90°. The floating magnet above a super conductor, was such an ideal case.

Aurora Borealis

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My notes of the excellent lecture 19 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.”

If we have a magnetic field \(\vec B\), and a positive charged particle with velocity \(\vec v\)

Then the Lorenz force on that particle is $$ \vec F = q\,\left(\vec v\times\vec B\right) \nonumber $$

Decompose the velocity vector, into parallel and perpendicular to the magnetic field

We can rewrite the Lorenz force as below, taking into account that the force is zero for motion parallel to the magnetic field. $$ \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} \vec F &= q\left(\vec v_\parallelsum + \vec v_\perp\right)\times\vec B, & \left( \vec v_\parallelsum\times\vec B = 0\right) \\ &= q \left(\vec v_\perp\times\vec B\right) \end{align*} $$

Path

The Lorenz force makes the charged particle circle around with radius \(R\). The parallel component of the velocity is unaffected, and so the charge spirals along a field lines.

Remember: this radius is

$$ R = \frac{mv_\perp}{qB} \nonumber $$

The magnetic field of the Earth is not a straight line, but is curved. Charges will be trapped in a spiral orbit along magnetic field lines.

The field lines come down on earth near the magnetic poles. This increases the field strength in the direction of motion, causing a force to slow the charges. Some even reverse their direction and get trapped forming two radiation belts above the atmosphere.

Solar wind

Our sun emits a plasma. Plasma is highly ionized; electrons and protons. We call that the solar wind. Sometimes it’s strong, and sometimes it’s weak.

When the solar wind reaches the Earth, it ionizes the upper atmosphere of the Earth. The Auroras are a result of these electrons and protons from the solar wind colliding with the Nitrogen and Oxygen atoms in the upper atmosphere. Let’s assume it is electrons \(\text e\). The collision bumps these atoms \(X\) in an “excited” electronic state \(X^*\) at the expense of the kinetic energy of the electron $$ X + \rm{e} \to X^* + \text{e} \nonumber $$ The \(X^*\) relaxes to the ground state by emitting a photon, and produces light $$ X^* \to X^* + \text{h} f \nonumber $$

Color

The color frequency \(f\) follows from quantum mechanics, and depends on whether the nitrogen molecules in the atmosphere or oxygen molecules are being excited. It also depends on the height in the atmosphere where the ionization occurs.

molecule altitude color
\(\rm O_2\) \(\lt 250\,\rm{km}\) Green
\(\rm O_2\) \(\lt 250\,\rm{km}\) Red
\(\rm N_2\) \(\lt 100\,\rm{km}\) Blue
\(\rm N_2\) \(\lt 100\,\rm{km}\) Purple/violet

The light is very faint, and can only be seen at night. That light is called Aurora, and here we call it Northern Lights. When the sun is very active, it can be breathtaking. The Aurora can change very fast on time scales of seconds to minutes. It is strongest near the magnetic poles.

Aurora Borealis – above Bear Lake
Senior Airman Joshua Strang / U.S. Air Force

Ring

For reasons that are not easy to understand, the maximum light comes from a ring that has a radius of about \(500\,\rm{km}\) from the magnetic pole.

Aurora Borealis – Ring in UV overlaid on Earth
unknown source
Aurora Borealis – beads seen from ISS
nasa.gov

Maxwell’s addition to Ampère’s law

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My notes of the excellent lecture 18 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.”

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

  1. Faraday’s law tell me that a changing magnetic flux gives rise to an electric field.
  2. 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

Lec.17 12:42 LIKE electric flux ex6d with B

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.

Motional EMF; AC motors

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My notes of the excellent lectures 17 and 18 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.”

Assume a conducting loop in a uniform magnetic field \(\vec B\).

Remember: when we assign a flat surface and choose \(d\vec A\) upwards, the magnetic flux is defined as

$$ \phi_B = \iint_R \vec B \cdot d\vec A \nonumber $$

The surface element \(dA = dx\,dy\), and \(d\vec A=\hat n\,dx\,dy\). The dot-product \(\vec B\cdot\hat n\) equals \(|\vec B|.1.\cos\theta\) $$ \begin{align} \phi_B &= \iint_R B\,\cos\theta\,dx\,dy \nonumber \\ &= B\,\cos\theta\int_0^y\int_0^x dx\,dy \nonumber \\ &= xy\,B\,\cos\theta \label{eq:phib} \end{align} \nonumber $$

According to Faraday, the time derivative of \(\phi_B\) determines the EMF (\(\mathcal E\)).

If the flux is positive and is getting larger positive, then based on Lenz’s law the current will be clockwise. This current will create an induced magnetic field \(\vec B_{ind}\) that is down.

Changing \(\text{ }d\phi_B/dt\)

We can change \(d\phi_B/dt\) by changing

  • the magnetic field strength, (\(B\))
  • the angle (\(\theta\))
  • the area (\(xy\))

While looking at Lenz’s law, we already examined a change in magnetic field strength. Here we will examine a change in angle, and a change in area.

Changing the angle\(\text{ }\theta\)

To change the angle \(\theta\), we will rotate the loop around the \(y\)-axis with angular frequency \(\omega = 2\pi/\text{period}\)

The angle \(\theta\) is a function of \(\omega t\). We choose \(\theta_0\) so that at \(t=0\) the angle \(\theta=0\) $$ \theta = \bcancel{\theta_0} + \omega t \nonumber $$

Substituting this \(\theta\) in equation \(\eqref{eq:phib}\), and assigning the area \(A=xy\) $$ \begin{align*} \phi_B &= xy\,B\,\cos \omega t \\ &= A\,B\,\cos \omega t \end{align*} $$

The EMF is minus the derivative of \(\phi_B\) $$ \begin{align*} \mathcal E &= -\frac{d\phi_B}{dt} \\ \Rightarrow \mathcal E(t) &= -A\,B\,\frac{d}{dt}\cos \omega t \end{align*} $$

The EMF as a result of rotating the loop follows as $$ \shaded{ \mathcal E(t) = A\,B\,\omega\,\sin\omega t } \tag{EMF rotating} \label{eq:emfrotating} $$

Note that both the amplitute and period depend on the angular frequency. As the loop rotates, the current is going to alternate in a sinusoidal fashion. We call this alternating current (AC).

The induced current in the wire with resistance \(R\) is a result of the EMF \begin{align*} i(t) &= \frac{\mathcal E(t)}{R} \\ &= \frac{AB}{R}\omega\,\sin\omega t \end{align*}

This is the idea behind an alternating current generator. It has a permanent magnet, and you rotate conductor loops (windings), through the magnetic field. The higher the magnetic field, the faster you rotate, the more windings, bigger area of the loop .. the higher the EMF.

Multiple windings

If it was a closed loop with two windings, we would assign a surface with two layers. One is lower, and the other comes out at top. So, the magnetic flux will double, because the magnetic field penetrates the surface twice. You get twice the EMF.

On other words: if you have \(N\) windings in one closed loop, then the EMF would be \(N\) times larger.

Changing the area \(\text{ }xy\)

We can also change the area using a crossbar. Assuming a uniform magnetic field \(\vec B\)

We find the magnetic flux using on equation \(\eqref{eq:phib}\). We are varying the width (\(l\)) while keeping the angle \(\theta\) is zero. $$ \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} \phi_B &= xl\,B\cos\theta, & \left(\vec B\parallelsum d\vec A\right) \\ &= xl\,B \end{align*} $$

With \(\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \vec B\parallelsum d\vec A\) and pointing in the same direction, the flux will be positive. Assuming the crossbar moves to the right with velocity \(v\), the flux will increase. According to Lenz’s law, the current will be clockwise

The EMF is minus the derivative of \(\phi_B\) $$ \begin{align*} \mathcal E &= -\frac{d\phi_B}{dt} \\ \Rightarrow \mathcal E(t) &= -lB\frac{dx}{dt}, & \left(\tfrac{dx}{dt} = v\right) \\ &= lBv \end{align*} $$

The EMF as a result of moving the crossbar follows as $$ \shaded{ \mathcal E(t) = l\,B\,v } \tag{EMF crossbar} $$

EMF without using Faraday’s law

We can alse determine the EMF without using Faraday’s law.

Looking at the left rod, the current is coming out of the screen, with the magnetic field pointing up. So, the Lorenz force is in the direction of \(\vec I\times \vec B\) (to the left)

With \(\vec I \perp \vec B\), the magnitude of the Lorenz force \(F_L\) is $$ F_L = I\,l\,B \nonumber $$

This is force that we have to apply to pull the crossbar to the right $$ F_{wl} = I\,l\,B \label{eq:fwl} $$

With \(\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} F_{wl} \parallelsum \vec v\), we have to do work to move the crossbar to the right. This work comes out in the form of heat, in the resistance of this conductor. Because we’re creating an EMF, a current will flow and the dissipate the power $$ P = \mathcal E\,I = I^2 R \nonumber $$

If we change the direction by pushing the crossbar to the left, the current is going to change direction. So the Lorenz force will also flip over, and the force for us will flip over. Again we have to do positive work $$ \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} W &= \vec F\cdot d\vec x, & \left(P = \tfrac{dW}{dt}\right) \\ \Rightarrow P &= \vec F\cdot \frac{d\vec x}{dt}, & \left(\tfrac{d\vec x}{dt} = \vec v\right) \\ &= \vec F\cdot \vec v, & \left(\vec F\parallelsum\vec v\right) \\ &= F\,v \end{align*} $$

Substituting the force from equation \(\eqref{eq:fwl}\), the power that we generate is $$ \begin{align*} P &= I\,l\,B\,v, & \left(P = \mathcal E I\right) \\ \mathcal E\,\bcancel{I} &= \bcancel{I}\,l\,B\,v \end{align*} $$

This EMF is exactly what we found before $$ \shaded{ \mathcal E(t) = l\,B\,v } \tag{EMF crossbar} $$

Eddy currents

If we move a conducting disk into a magnetic field, then in the part of the disk moving under the magnet, the magnetic field will increase

This creates a clockwise electric field in the disk. This field induces a clockwise flow of electric current. So it produces an magnetic field in the opposite direction to oppose the change in magnetic flux

These currents are called eddy currents, after the localized water eddies in fluid dynamics. These currents produces heat $$ P = \mathcal E\,I = I^2 R \nonumber $$

The Lorenz force is in the direction of \(\vec I\times \vec B\), to the left. It opposes the motion. As a consequence, the disk will slow down, at the expense of kinetic energy. Heat is produced and it won’t go as fast through this field as then there was no magnetic field. We call that magnetic braking.

3-phase Motors

Generator

Remember: that if you rotate conducting loops in magnetic fields, that you create induced EMFs.
As we rotate about the axis, we’re going to get an induced EMF. We derived the induced EMF as equation \(\eqref{eq:emfrotating}\) $$ \mathcal E(t) = xy\,B\,\omega\,\sin\omega t \nonumber $$

Now we are going to add two more loops, which are not electrically connected to each other.

If you look from the front, you will see the first loop (1). The second loop (2) that is rotated 120°, and the third loop (3) again rotated 120°

Each one of those will give a sinusoidal EMF, but they are offset in phase by 120 degrees. We call this three phase current

If the period of (1) is \(60\,\rm{Hz}\), then the period of (2) and (3) is also \(60\,\rm{Hz}\).

Synchronous motor

Suppose, you’re looking down onto a horizontal table, and we have one solenoid (1) that we feed the current (1) going clockwise. Then we have another one (2), which is rotated 120°. We are going to feed that current (2). The same for solenoid (3), that we feed current (3).

The three phase current will produce a rotating magnetic field.

  1. At the moment the current through number (1) reaches a maximum, the current in (2) and (3) is two times smaller. But the vectorial sum of the magnetic field produced by (2) and (3) is in the same direction, so the net magnetic field is in the same direction.
  2. One third of a period later in time: now the current in (2) reaches a maximum, and the vectorial sum of (1) and (3) is in the same direction as well.
  3. Again one third of a period later: now the current in (3) reaches a maximum, and the vectorial sum of (1) and (2) is in the same direction as well.

We have created is now a rotating magnetic field. In one period, it rotates 360°. It rotates around in the period of the alternating current. If we put a magnet in the middle, then this magnet will to follow the magnetic field. We call this a synchronous motor.

So, the rotor of such a synchronous motor is magnet, and it rotates with the frequency of the alternating current.

Induction motor

For the rotor, we can also use a conducting sphere. The rotating magnetic field causes a continuous magnetic flux change with the surface of the sphere, and it’s going to run eddy currents. The eddy currents going around will give rise to a magnetic field, that causes a torque on the current. Similar to when we discussed earlier the idea of a synchronous motor.

So, it starts to torque up the sphere. You’re going to get a torque which is always be in the same direction, and the sphere object will start to rotate. We call this an induction motor

The speed of the induction motor depends on the conducting object. If it is a sphere, it will probably come very close to frequency the current, because a sphere has many possibilities for eddy currents to run around. Whereas if you take a ring, and you try to spin that in a rotating magnetic field, then the paths that are available for eddy currents are very limited. They can only go around in a ring.

Many of the stationary tools that you find in people’s workshops are induction motors. A table saw, drill press and electric grass mowers are induction motors.

2-phase motor

We can also make an induction motor that runs on 2-phase AC current. This is done using two solenoids, 90° apart.

The magnetic field will rotate 360° during one period of the AC current

  1. When the magnetic field is maximum in (1), it is zero in (2).
  2. When it is maximum in (2), it is zero in (1).
  3. When it is minimum in (1), it is zero in (2)
  4. When it is minimum in (2), it is zero in (1).

If we put a conductor in the middle, we’re going to have eddy currents, because we have change of magnetic flux all the time. These eddy currents will cause a torque on the conductor. The torque will always be in the same direction and the conductor is going to rotate.

Electromagnetic induction; Faraday’s law

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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.

Faraday’s experiments

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.

Steady magnetic field

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)

\(\vec B\) increasing
  
\(\vec B\) increasing, induced \(\vec B_{ind}\)

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.

\(\vec B\) decreasing
  
\(\vec B\) decreasing, induced \(\vec B_{ind}\)

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.

\(\vec B_1\) and circuit ② with \(1\) winding
  
\(\vec B_1\) and circuit ② with \(3\) windings

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.

Ampère’s law

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My notes of the excellent lecture 15 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.”

Ampère’s law relates the magnetic field of a closed loop to the electric current passing through that loop.

Ampère’s law

A current going into the paper, causes a magnetic field at distance \(r\) tangentially to the circle

Using using Biot-Savart’s law, we saw that integrating over \(d\vec l\) along the wire, gives the magnetic field strength at distance \(r\)
$$ B = \frac{\mu_0 I}{2 \pi r} \nonumber $$

Now, carve the circle up in little elements \(d\vec l\),

The closed circle integral, with \(\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \vec B \parallelsum d\vec l\) $$ \begin{align*} \oint \vec B \cdot d\vec l &= B\oint dl, & \left(\oint dl=2\pi r\right) \\ &= B\, 2\pi r, & \left(B = \frac{\mu_0 I}{2 \pi r}\right) \\ &= \frac{\mu_0 I}{\bcancel{2 \pi r}}\, \bcancel{2\pi r} \\ &= \mu_0I \end{align*} $$

Note:

  • Note the different \(d\vec l\). One is along the wire, other other around the circle.
  • It doesn’t matter what distance \(r\), the integral is always \(\mu_0 I\)

Ampère first realized that you could walk around in any closed crooked path,

The closed path around a current wire, is $$ \shaded{ \oint \vec B \cdot d\vec l = \mu_0 I_\rm{encl} } \tag{Ampère’s law} \label{eq:amperelaw} $$ As we will see shortly, this is part of the fourth of Maxwell’s equations.

Enclosed current definition

In the formula above, \(I_\rm{encl}\) stands for enclosed current. We define this as:

If we have a very strange looking closed path. To that closed loop, we have to attach an open surface. We’re free to choose it. It can be flat, or e.g. like a bakers hat.

If the current goes in the curve and penetrates the surface, the current is said to be enclosed.

By convention, we applying right-hand corkscrew notation for the connection between the magnetic field and current.

  • If we follow current \(I_1\), the direction is clock-wise and \(I_1\gt 0\) and \(I_2\lt 0\)
  • If we follow current \(I_2\), the direction is counter clock-wise and \(I_1\lt 0\) and \(I_2\gt 0\)
How you go around the path is your choice, but it defines the sign of the current.

Since you are free to choose a surface attached to the loop, if you can get away with it, you can use a flat surface.

Now we can calculate the magnetic field inside a wire that draws a current using Ampère’s Law.

Straight wire

Let’s assume an uniform current throughout the wire. What is the magnetic field everywhere?

  • For \(r \gt R\): Because of the cylindrical symmetry, we choose a circle as the path. So the magnetic field strength is the same everywhere.
    Since the current is coming towards me, we know the magnetic field direction, and we choose to march in the same direction. We choose a flat open surface (the paper). Apply Ampère’s Law $$ \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} \oint \vec B \cdot d\vec l &= \mu_0 I & \left(B \parallelsum d\vec l\right) \\ B \oint dl &= \mu_0 I & \left(\oint dl=2\pi r\right) \\ B\, 2\pi r &= \mu_0 I \end{align*} $$ We find the same result, as when we applied Biot and Savart $$ \shaded{ B = \frac{\mu_0I}{2\pi r} } \tag{straight, \(r \gt R\)} $$
  • For \(r \lt R\): again, because of the cylindrical symmetry, we choose a circle as the path. So the magnetic field strength is the same everywhere.
    Once more, we choose to march in the direction of the magnetic field. We choose a flat open surface (the paper). Only a fraction of the current \(I\) penetrates our surface. $$ \frac{A_{\circ r}}{A_{\circ R}} = \frac{\bcancel{\pi} r^2}{\bcancel{\pi} R^2} \nonumber $$ Applying Ampère’s Law, $$ \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} \oint \vec B \cdot d\vec l &= \mu_0 \frac{r^2}{R^2}I & \left(B \parallelsum d\vec l\right) \\ B \oint dl &= \mu_0 \frac{r^2}{R^2}l & \left(\oint dl=2\pi r\right) \\ B\, 2\pi \bcancel{r} &= \mu_0 \frac{r^{\bcancel{2}}}{R^2}l \end{align*} $$ So, this is linear in \(r\) $$ \shaded{ B = \frac{\mu_0Ir}{2\pi R^2} } \tag{straight, \(r\lt R\)} $$

If you substitute \(r=R\), for the magnetic field right at the surface, you find exactly the same result as for the case where \(r\gt R\).

Plotting \(B(r)\)

Solenoids

A solenoid is a cylindrical coil of wire that acts as a magnet.

Remember, the circular current loop.

Looking at its cross section and its magnetic field configuration

If you now add other loops next to it, these loops also pull the field lines “to come through” its circle. So you’re beginning to get a near-constant magnetic field. The more tightly these loops are wound, the more constant the magnetic field will be.

We see that inside the magnetic field is fairly constant, and outside it is very low.

Magnetic field inside

A solenoid with length \(L\), radius \(R\), current \(I\), there are \(N\) windings and looking from the left the windings are clock-wise. We assume the field outside the solenoid is approximate zero. What is the magnetic field inside the solenoid?

We as a (surprising) path: a rectangle with length \(l\).

In applying Ampère’s law, we split the path in four pieces.

  • Path 1: we assumed the magnetic field is zero, so \(\int\vec B\cdot d\vec l=0\)
  • Path 2,3: \(\vec B \perp d\vec l\), so the dot-product is zero
  • Path 4: is the only path that contributes to the closed-loop integral. Here we assumed \(B\) is constant, and \(\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \vec B \parallelsum d\vec l\). $$ \int \vec B\cdot d\vec l = B \int dl = B\,l \nonumber $$ Now, we attach a flat surface to the closed-loop, and calculates the current that penetrates that surface. The number of windings that poke through the surface is $$ B\,\bcancel{l} = \mu_0\,\frac{\bcancel{l}}{L}N\, I \nonumber $$

For as long as \(L \gg R\) $$ \shaded{ B = \frac{\mu_0 I N}{L} } \tag{inside solenoid} $$

Note: that the magnetic field is not proportional to the number of loops, but to the number of loops per unit length. The shape of the magnetic field of one loop is like a dipole field, and spreads. it falls off so rapidly that the other end of the solenoid doesn’t notice it. It matters how close together the windings are, so the windings per unit length.

Biot-Savart law; Gauss’s law for magnetism

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My notes of the excellent lecture 14 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.”

Biot-Savart

Assume a current going through a straight wire. We know that pieces of magnetite line up in a circle

From magnetic field experiments, we learned

For a circle has radius \(R\), the magnetic field strength \(B\) $$ B \propto \frac{I}{R} \nonumber $$

For electric fields, we know

For a straight wire with uniformly distributed with positive charge, the electric field \(E\) at distance \(R\) is proportional as $$ E \propto \frac{1}{R} \nonumber $$ This is because, according to Coulomb’s law the electric field for electric monopoles (individual charges) falls off as \(1/R^2\). As you integrate that over the length of the wire, you get the \(1/R\) electric field.

Even tough the direction is different, magnetic fields like electric fields fall off as \(1/R\). By analogy, it would be plausible that if magnetic monopoles existed, that the magnetic field would also fall off as \(1/R^2\).

The simple fact that the magnetic field around a current wire falls off as \(1/R\), suggest that if you cut this wire up in little elements \(dl\), that each one of those elements contributes to the magnetic fields in inverse R-squared law. By integrating out over the whole wire, would then get the \(1/R\) fields.

Biot and Savart formalized this

If you have a little current element \(d\vec l\), and the current \(I\) is in the same direction,

Then at distance \(r\), the contribution to the magnetic field \(d\vec B\) is $$ \shaded{ d\vec B = \frac{\mu_0 I}{4 \pi r^2}\,(d\vec l\times \hat r) } \tag{Biot-Savart} \label{eq:BiotSavart} $$

Where

  • \(\hat r\) is to get the direction right
  • \(\mu_0\) is the permeability of free space \(\approx 1.26\times 10^{-6}\,\rm{H/m}\)

Examples

Straight wire

We can apply Biot-Savart’s law on a straight wire to find the magnetic field at distance \(R\)

Divide the wire in tiny segments \(d\vec l\), at distance \(r\). Then calculate \(dB\), and integrate it over the whole wire.

Using the right-hand rule, \(d\vec l\cdot\hat r\) points out of the page for any element along the wire. Therefore, \(d\vec B\) for all \(d\vec l\) have the same direction. This means we can calculate the net field at \(P\) by evaluating the scalar sum of \(dB\) $$ \left|d\vec l\times\hat r\right| = (dl)(1)\sin\theta = \sin\theta\,dl \nonumber $$

Distance \(r\) and the angle \(\theta\) using trig $$ \left\{ \begin{align*} r &= \sqrt{l^2 + R^2} \\ \sin\theta &= \frac{R}{\sqrt{l^2 + R^2}} \end{align*} \right. $$

Substituting these in \(\eqref{eq:BiotSavart}\) $$ \begin{align*} d\vec B &= \frac{\mu_0\,I}{4\pi r^2}\,\sin\theta\,dl \\ &= \frac{\mu_0\,I}{4\pi(l^2+R)^2}\,\frac{R}{\sqrt{l^2 + R^2}}\,dl \\ &= \frac{\mu_0 I R}{4\pi}\, \frac{1}{(l^2+R^2)^{3/2}}\,dl \\ \Rightarrow B &= \frac{\mu_0 I R}{4\pi} \int_{-\infty}^{\infty}\frac{1}{(l^2+R^2)^{3/2}}\,dl \\ \end{align*} $$

There is symmetry in \(O\) $$ \begin{align*} B &= \frac{\mu_0 I R}{4\pi} 2 \int_0^{\infty}\frac{1}{(l^2+R^2)^{3/2}}\,dl \\ &= \frac{\mu_0 I R}{2\pi} \int_0^{\infty}\frac{1}{(\underline{l}^2+R^2)^{3/2}}\,\underline{dl} \\ \end{align*} $$

Let \(u = \rm{atan}\frac{l}{R}\) $$ \begin{align*} u &= \rm{atan}\frac{l}{R} \\ \Rightarrow \underline{l} &= R\tan u \\ \Rightarrow \frac{dl}{du} &= \frac{d}{dl} R\tan u = R\,\rm{sec}^2 u \\ \Rightarrow \underline{dl} &= R\,\rm{sec}^2 u\,du \end{align*} $$

Substituting \(l = R\tan u\) and \(dl = R\,\rm{sec}^2 u\,du\) in the integral $$ \begin{align*} B &= \frac{\mu_0 I R}{2\pi} \int_0^{\infty}\frac{1}{\left[ (\underline{R\tan u})^2+R^2 \right]^{3/2}}\,\underline{R\,\rm{sec}^2 u\,du} \\ &= \frac{\mu_0 I R}{2\pi} \int_0^{\infty}\frac{R\,\rm{sec}^2\,u}{ R^3\,\left( \tan^2 u+1 \right)^{3/2} }\,du \\ &= \frac{\mu_0 I \bcancel{R^2}}{2\pi R^{\bcancel{3}}} \int_0^{\infty}\frac{\rm{sec}^2\,u}{\rm{sec}^3\,u}\,du = \frac{\mu_0 I}{2\pi R} \int_0^{\infty}\frac{1}{\rm{sec}\,u}\,du \\ &= \frac{\mu_0 I}{2\pi R} \int_0^{\infty}\cos u\,du = \frac{\mu_0 I}{2\pi R} \Big[ \sin u \Big]_0^{u=\infty} \\ \end{align*} $$

Substitute back \(u = \rm{atan}\frac{l}{R}\) $$ \begin{align*} B &= \frac{\mu_0 I}{2\pi R} \Big[ \sin\left(\rm{atan}\frac{l}{R}\right) \Big]_0^{l=\infty} \\ &= \frac{\mu_0 I}{2\pi R} \Big[ \sin\frac{\pi}{2} – \sin 0 \Big]_0^{l=\infty} \\ &= \frac{\mu_0 I}{2\pi R} \end{align*} $$

So, the magnetic field at distance \(R\) is from the wire $$ \shaded{ B = \frac{\mu_0 I}{2\pi R} } \tag{straight wire} \label{eq:straightwire} $$

If you take a radius of \(R=0.1\,\rm{m}\) and you have a current \(I=100\,\rm A\), then $$ \begin{align*} B &= \frac{\mu_0 I}{2\pi R} = \frac{(1.26\times10^{-6}) 100}{2\pi\,0.1} \\ &\approx 2\times10^{-4}\,\rm{T} \end{align*} $$

Compared to the Earth’s magnetic field of \(0.5\times10^{-4}\,\rm T\), it is not that significant. If you would go at a radius of \(R=1\,\rm m\), the field would be 10 times lower, \(2\times10^{-5}\,\rm T\). Then the magnetic field of the Earth already dominates substantially.

Circular wire loop

At center

We can apply Biot-Savart’s law on a current loop to find the magnetic field at the center

Divide the wire in tiny segments \(d\vec l\). Using the right-hand rule, \(d\vec l\cdot\hat r\) at the center points up for any element along the wire. Since \(d\vec l\perp \hat r\) $$ d\vec l\times\hat r = dl.1.\sin\frac{\pi}{2} = dl \nonumber $$

Substituting these in \(\eqref{eq:BiotSavart}\) $$ \begin{align*} d\vec B &= \frac{\mu_0\,I}{4\pi R^2}\,dl \\ \Rightarrow B &= \frac{\mu_0\,I}{4\pi R^2}\int_\rm{circle}\,dl \\ &= \frac{\mu_0\,I}{4\bcancel\pi R^{\bcancel 2}}\,2\bcancel\pi \bcancel{R} \\ &= \frac{\mu_oI}{2R} \end{align*} $$

So, the magnetic field at the center of a wire loop is $$ \shaded{ \vec B = \frac{\mu_oI}{2R}\,\hat z } \tag{wire loop center} \label{eq:wireloopcenter} $$

When you very far away, the electric fied configuration is very similar to that of a electric dipole.

Along axis

We may extend this approach to calculate the field anywhere along the center (\z\)-axis of the loop.

Based on the cylindrical symmetry, any outward component of \(d\vec B\) will be exactly balanced by an opposite-directed outward component from an opposing wire fragment. The magnetic field of the wire loop as a whole only has a \(z\)-component. In calculating the total field, we only need to integrate the z-components. $$ \begin{align*} d\vec l\times\hat r &= dl.1.\sin\theta, & \left( \sin\theta=\frac{R}{r} \right) \\ &= \frac{R}{r}\,dl & \end{align*} $$

Substituting this in \(\eqref{eq:BiotSavart}\) $$ \begin{align*} d\vec B &= \frac{\mu_0\,I}{4\pi (z^2 + R^2)}\, \frac{R}{r}\,dl, & \left(r=\sqrt{z^2+R^2}\right) \\ &= \frac{\mu_0\,I}{4\pi (z^2 + R^2)}\, \frac{R}{\sqrt{z^2 + R^2}}\,dl \\ &= \frac{\mu_0\,I\,R}{4\pi (z^2 + R^2)^{3/2}}\, dl \\ \Rightarrow B &= \frac{\mu_0\,I\,R}{4\pi (z^2 + R^2)^{3/2}}\, \int_\rm{circle} dl, & \left( \int_\rm{circle} dl = 2\pi R\right) \\ &= \frac{\mu_0\,I\,R}{\bcancel{4}_2\bcancel{\pi} (z^2 + R^2)^{3/2}}\, \bcancel{2}\bcancel{\pi} R \end{align*} $$

So, the magnetic field along the center axis of a wire loop is $$ \shaded{ \vec B = \frac{\mu_0\,I\,R^2}{2 (z^2 + R^2)^{3/2}} \hat z } \tag{wire loop axis} \label{eq:wireloopaxis} $$

Field lines

We would expect the magnetic field lines as

Gauss’s law for magnetism

Looking from far away, ignoring what happens between the charges: the loops of electric field lines from a dipole, as similar to the magnetic field lines from the current loop

Magnetic field lines of current loop
  
Electric field lines of dipole

Gauss’ law tells us that the closed surface \(S\) integral of the electric flux is the charge inside the box divided by \(\varepsilon_0\) $$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \iint_S \vec E \cdot d\vec A = \frac{\sum_i Q_i}{\varepsilon_0} \nonumber $$

For the electric field, there is clearly a charge inside the box.

No matter where in the magnetic field you make a closed surface \(S\), there is never any magnetic flux going through that surface. Because there (as far as we know) are no magnetic monopoles.

This brings us to the second of four Maxwell’s equations: Gauss’s law for magnetism $$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \shaded{ \oiint_S \vec B \cdot d\vec A = 0 } \tag{Gauss’s law} $$

Moving charge; Cyclotron

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My notes of the excellent lecture 13 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.”

Path of a moving charge in \(\vec B\)

Remember: Lorenz force \(\vec F\) depends on the velocity of the charge and the magnetic field that it experiences

$$ \vec F = q\,\left(\vec v\times\vec B\right) \nonumber $$

If we have the following charge \(q\) and velocity \(\vec v\), it will experience \(\vec F\)

This causes the charged particle to go around in a perfect circle. The Lorenz force can’t change the speed (can’t change the kinetic energy), because \(\vec F\perp \vec v\), but it can change the direction of the velocity.

The radius \(R\) of the circle follows from the force. With \(\vec F\perp \vec v\), the \(\sin\) of the angle between them is \(1\). That is now the centripetal force \(\frac{mv^2}{R}\), where \(m\) is the mass of the particle $$ F = q\,v\,B = \frac{mv^2}{R} \nonumber $$

So the radius $$ \shaded{ R = \frac{m\,v}{q\,B} } \label{eq:R} $$

Here \(mv\) is the momentum.

Numerical example

\(U\)Instead of \(\Delta V\), we often use the European symbol for voltage difference: \(U\). The letter ‘u’ stands for “Potentialunterschied”. Do not confuse it with potential energy, that also uses \(U\).

Assume a proton with kinetic energy \(1\,\rm{MeV}\) \(=1.6\times10^{-13}\,\rm J\) in a constant magnetic field \(\vec B\) $$ \begin{align*} q\,\Delta V &= 1.6\times10^{-13}\,\rm J \\ m_p &= 1.7\times 10^{-27}\,\rm{kg} \\ q_p &= 1.6\times 10^{-19}\,\rm{C} \end{align*} $$

The speed of the proton $$ \begin{align} q\,\Delta V &= \frac{1}{2}m\,v^2 \nonumber \\ \Rightarrow v &= \sqrt{\frac{2\,q\Delta V}{m_p}} \nonumber \\ &\approx \sqrt{\frac{2\,(1.6\times10^{-13})}{1.7\times 10^{-27}}} \nonumber \\ &\approx 1.4\times 10^7\,\rm{m/s} \approx 0.05\,\rm c \label{eq:onemkvproton} \end{align} $$

That is 5% of the speed of light, comfortably low, so we don’t have to make any relativistic corrections.

If this proton now enters the magnetic field \(B=1\,\rm T\). Based on equation \(\eqref{eq:R}\), it will circle with radius \(R\) $$ \begin{align*} R &= \frac{mv}{qB} \\ &= \frac{(1.7\times10^{-27})(1.4\times 10^7)}{(1.6\times 10^{-19})(1)} \\ &\approx 1.5\,\rm m \end{align*} $$

Eliminate \(\text{ }\vec v\)

In equation \(\eqref{eq:R}\), the velocity \(v\) is commonly replaced by the potential difference \(\Delta V\) over which we accelerate these particles.

Find \(v=f(\Delta V)\) $$ \begin{align*} q\,\Delta V &= \frac{1}{2}m\,v^2 \\ \Rightarrow v &= \sqrt{\frac{2\,q\,\Delta V}{m}} \end{align*} $$

Substituting this in equation \(\eqref{eq:R}\) $$ \begin{align*} R &= \frac{m\,\sqrt{\frac{2\,q\,\Delta V}{m}}}{qB} \end{align*} $$

That simplifies to $$ \shaded{ R = \sqrt{\frac{2\,m\,\Delta V}{q\,B^2}} } \tag{Radius} $$

Approaching \(\text{ c}\)

When the speed approaches the speed of light, we have to apply special relativity. That is not part of this course, but we will briefly touch upon that

Things go sour, when we have a \(500\,\rm{keV}\) electron, then $$ \begin{align*} \frac{1}{2}m\,v^2 &= q\,\Delta V \\ \Rightarrow v &= \sqrt{\frac{2\,q_e\,\Delta V}{m_e}} \\ &= \sqrt{\frac{2\,(1.6\times10^{-19})\,(500\times10^3)}{9.1\times 10^{-31}}} \\ &\approx 4.2\times10^8\,\rm{m/s} \gt \rm c \end{align*} $$

This is larger than the speed of light, so clearly not possible. If you make relativistic corrections, you will find the actual speed. Just to show you, we’ll make these corrections below.

Relativistic correction

Start with same kinetic energy \(q\,\Delta V\), but it is no longer \(\frac{1}{2}m\,v^2\) $$ q\,\Delta V = (\gamma – 1)\,m\,c^2 \nonumber $$

Here the Lorenz factor \(\gamma\) is defined as $$ \gamma = \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}} \tag{Lorenz factor} $$

The Lorenz factor follows as $$ \begin{align*} q\,\Delta V &= (\gamma – 1)\,m\,c^2 \\ \Rightarrow \gamma &= \frac{q\,\Delta V}{m\,c^2} + 1 \\ &\approx \frac{(500\times 10^3)(1.6\times 10^{-19})}{(9.1\times10^{-31})\,(300\times10^6)^2} + 1 \\ &\approx 1.977 \end{align*} $$

The speed \(v\) $$ \begin{align*} \sqrt{1 – \frac{v^2}{c^2}} &= \frac{1}{\gamma} \Rightarrow \frac{v^2}{c^2} \\ \Rightarrow v &= c\,\sqrt{1 – \frac{1}{\gamma^2}} \\ \Rightarrow v &= 300\times10^6 \sqrt{1 – \frac{1}{1.977^2}} \\ &\approx 2.6\times 10^8\,\rm{m/s} \end{align*} $$

The radius \(R\) in a \(1\,\rm T\) magnetic field $$ \begin{align} R &= \gamma\,\frac{m\,v}{q\,B} = \sqrt{\frac{(\gamma+1)\,m\,\Delta V}{q\,B^2}} \label{eq:corrR} \\ &= 1.977\,\frac{(9.1\times10^{-31})\,(2.6\times 10^8)}{(1.6\times10^{-19})\,1} \nonumber \\ &\approx 2.9\,\rm{mm} \nonumber \end{align} $$

Some examples

Proton accelerator
KE v [m/s] B [T] R
\(1\,\rm{MeV}\) \(1.4\times 10^7\) \(1\) \(0.15\,\rm{m}\)
\(50\,\rm{MeV}\) \(9.4\times 10^7\) \(1\) \(1.04\,\rm{m}\)
Fermilab \(500\,\rm{GeV}\) \(2.99999\times10^8\) \(1.5\) \(1.1\,\rm{km}\)
LHC \(7,000\,\rm{GeV}\) \(\approx\rm c\) \(5.5\) \(4.2\,\rm{km}\)

Electron accelerator
KE v [m/s] B [T] R
\(15\,\rm{eV}\) \(2.3\times 10^6\) \(0.5\times10^{-4}\) \(26\,\rm{cm}\)
\(100\,\rm{keV}\) \(1.6\times 10^8\) \(1\) \(1.1\,\rm{mm}\)
\(500\,\rm{keV}\) \(2.6\times10^8\) \(1\) \(2.9\,\rm{mm}\)

Practical use

In 1945, the Americans needed Uranium \({}^{235}\rm U\) to build an atomic bomb. This has 143 neutrons, instead of the more common 146. To separate them, Ernest Lawrence of Berkeley built a mass spectrometers that could separate \({}^{235}\rm U\) and \({}^{238}\rm U\).

In a mass spectrometer, the uranium is heated so that it ionizes. Let’s assume it looses one electron, so it’s positively charged with one unit charge. It then accelerates it over a potential difference, so they get a certain speed.

In a constant magnetic field, the charged particles go around a circle and hit a collector. The radius of the circle is proportional with \(\sqrt{m}\). The mass of 238 is 1.2% larger than the mass of 235. So they land is a collector slightly separated from the other one.

Accelerating protons

In 19939, the UC Berkeley physicist, Ernest Laurence received a Nobel Prize for Physics. He invented the cyclotron that accelerates protons to \(730\,\rm{MeV}\), almost the speed of light.

Cyclotron by Ernest Lawrence
Credit: berkeley.edu

Cyclotron

In the early days accelerating was done in a cyclotron. It consists of two conducting chambers \(D\) in vacuum. Seen from the top and side:

Suppose we release a \(1\,\rm{MeV}\) proton. That comes out a speed of \(1.4\times10^7\,\rm{m/s}\) \(\eqref{eq:onemkvproton}\). In a \(1\,\rm{T}\) field, based on the earlier table, the radius is going to be \(15\,\rm{cm}\).

The proton start to make a circle. But when it gets halfway, a potential difference is introduced between these two \(D\)’s. The high potential on the right \(D\), and the low potential on the left \(D\). In the gap between the \(D\)’s, you get an electric field \(\vec E\)

The field accelerates the proton. If the potential difference is \(20\,\rm{kV}\), the proton will gain \(20\,\rm{keV}\) kinetic energy. When it crossed the gap, it has \(1.02\,\rm{MeV}\). If \(V\) is 2% higher, then based on equation \(\eqref{eq:corrR}\), the radius is 1% higher than \(15\,\rm{cm}\). It comes out a little higher.

Once, it gets to the top, the potential difference is reversed, so the \(\vec E\) points to the right, and it is again accelerated by \(20\,\rm{keV}\).

Very gradually, it spirals out the largest radius. So for every rotation it gains \(40\,\rm{keV}\). The electric fields are doing the work, they accelerate the particles. The magnetic field changes the direction, but they can’t do work on the particles. It confines the particles.

If we go 1225 full rotations, where each time the kinetic energy increases by \(40\,\rm{keV}\). Multiplying these you find that it gained \(49\,\rm{MeV}\). Plus the initial \(1\,\rm{MeV}\), it now has \(50\,\rm{MeV}\). Referring to the the earlier table, the radius is one meter.

If we don’t have to make relativistic corrections, the time to go around is independent on the proton’s speed. With the radius proportional to \(v\), so the time is independent of \(v\) $$ T = \frac{2\pi R}{\rm{speed}} = \frac{2\pi\,m}{q\,B} \nonumber $$

The time to go around once, is only \(66\,\rm{ns}\). To go around 1,225 times will take only \(80\,\rm{msec}\). That means the switching frequency becomes about \(30\,\rm{MHz}\).

Because of relativistic correction, the time to go around once, becomes $$ T = \frac{2\pi\,m}{q\,B}\,\gamma = 6.6\times 10^{-8}\,\rm{sec} \nonumber $$

If you go to very high energies, that time is not constant for a full rotation. So you have to adjust the switching frequency. We call those instruments synchrotrons or synchrocyclotrons.

Modern accelerators

Modern accelerators have constant radii. They are rings. The only way to keep the particles in the ring when they have a low energy, and when they have a high energy, is by gradually increasing the magnetic field. By making the magnetic field go up, you can keep the protons in that ring.

Fermilab

Fermilab near Chicago. One of the modern accelerators, occasionally called colliders. A diameter of \(2.2\,\rm{km}\), accelerates up to \(1000\,\rm{GeV} = 10^{12}\,\rm{eV}\). The beams of high energy protons are made collide with other nuclei to undercover the inner workings of nuclear physics.

Fermilab near Chicago
Credit: fnal.gov

Large Hadron collider

The tunnel of the largest ring in the work, at CERN in Geneva, has a radius of \(4.3\,\rm{km}\). Using superconducting magnets, they can go to about \(5\,\rm T\). The Large Hadron collider accelerates protons to the unprecedented energy of \(6500\,\rm{GeV} = 6.5\times 10^{12}\,\rm{eV}\).

Large Hadron collider at CERN
Credit: home.cern

Higgs boson

In July 2012, the LHC made one of the most important discoveries in particle physics. They proved the existence of the Higgs boson, which was hypothesized about 45 years earlier by Peter Higgs.

The Higgs boson is part of what’s called the Standard Model of particle physics. A set of rules that lays out our understanding of the fundamental building blocks of the universe. The Higgs boson has a mass of about \(125\,\rm{GeV/c^2}\), about \(133\) proton masses.

In 2013, Peter Higgs and François Englert were awarded the Nobel Prize in physics for their work in identifying and discovering the Higgs boson.

Magnets; charges in motion and Lorenz fore

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My notes of the excellent lecture 11 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.”

Magnetic fields

In the 5th century B.C., the Greeks discovered rocks that attract bits of iron. They are named after the area Magnesia where they are plentiful. The rocks contain iron oxide, which we call magnetite.

Magnetites have two places of maximum attraction, the poles. Let’s call them A and B.

  • A and A repel each other,
  • B and B repel each other, but
  • A and B attract each other

There is a huge difference between electricity and magnetism

  1. Electric monopoles exist. If you have a plus charge or minus, that is an electric monopole. If you have a plus and minus of equal strength, you have an electric dipole.
  2. Magnetic poles come in pairs. Magnetic monopoles do not exist, as far as we know. You always have a magnetic dipole.

Poles

The earth is a giant magnet. By convention the part of a magnet that points to the North Canada is called the north pole.

Credit: sos.noaa.gov

Since A and B attract, that means that North Canada is the magnetic South pole of the earth.

Discovery

In 1819, the Danish physicist Hans Christian Ørsted, discovered that a magnetic needle responds to an electric current in a wire. This linked magnetism with electricity. In this important discovery, Ørsted concluded that the current in the wire produces a magnetic field, and that the magnetic needle moves in response to that.

This discovery caused an explosion of activity in the 19th century, by André-Marie Ampère, Michael Faraday and Joseph Henry. It culminated in the brilliant work of Scottish theoretician James Clerk Maxwell. Maxwell composed a unified field theory which connects electricity with magnetism. The heart of this course. You will see all “vier” (4) equation at the end of this course.

Coulomb’s law already described the force between electric charges, at rest or in motion. But, these physicists found that the magnetic force is simply an extra electric force acting between moving electric charges.

Moving charges

Remember:

A potential difference generates an electric field along and inside the conductor, that makes the free electrons move. We call this move of charge: current.

Notation convention

Let the wire with a electric current be perpendicular to your screen, so the current goes into the screen.

We put ⊗ to indicate the rear of an arrow. If it would come out of the screen we would the arrow point ⊙.

Current exerts a force on magnets

If we put magnetized needles in the vicinity of the wire with an electric current, they align to form a circle.

This is how we define the magnetic field and its direction, \(\vec B\)

When the current goes into the screen ⊗, by convention, the magnetic field is in the clockwise direction. To remember that, you can use the righthand corkscrew rule: if you turn it clockwise it goes into the screen.

Magnet exerts a force on wire

Action equals reaction. If the current exerts a force on magnet, then the magnet must exert a force on the wire.

It is an experimental fact that the force on the wire \(\vec F\) is always the vectorial cross-product $$ \shaded{ \hat F = \hat I \times \hat B } \nonumber $$

Force between two wires

If I run a current through two wires, it will product a magnetic field. Current \(\vec I_1\) will produce \(\vec B_1\).

A current \(\vec I_2\), based on the righthand rule for the cross-product, \(\hat F = \hat I \times \hat B\), the force will be up.

Since action equals minus reaction, the top wire will come down.

Lorenz force

With electricity we defined the strength of an electric field \(\vec E\) as the force \(\vec F\) on an electric charge \(q\)

$$ \vec F_{el} = q\,\vec E \nonumber $$

It would be nice if we could define the strength of the magnetic force \(\vec F_B\) as a magnetic charge \(q_B\) times the magnetic field \(\vec B\) $$ \bcancel{ \vec F_B = q_B\,\vec B } \nonumber $$

But there are no magnetic monopoles \(q_B\)

Force on a charge

Instead, the magnetic field is defined based on an electric charge \(q\) moving with velocity \(\vec v\) through a magnetic field \(\vec B\).

It is an experimental fact, that the force \(\vec F_B\) is always perpendicular to \(\vec v\). The magnitude of \(\vec F_B\) is proportional to the magnitude of \(\vec v\), and proportional to the charge \(q\) $$ \begin{align*} \vec F_B &\perp \vec v \\ F_B &\propto v \\ F_B &\propto q \end{align*} $$

We define the magnetic field strength as $$ \shaded{ \vec F_B = q\,\left(\vec v \times \vec B\right) } \tag{Lorenz force} $$

This is called the Lorenz force, after the Dutch physicist Hendrik Antoon Lorentz.

The unit is \(\frac{N\,s}{C\,m}\) that we call Tesla, \(\rm T\). We occasionally use the unit Gauss \(\rm G=10^{-4}\,\rm T\).

If you have both electric and magnetic fields, the total force on a particle \(\vec F\) $$ \shaded{ \vec F = q\,\left(\vec E + \vec v\times\vec B\right) } \tag{Lorenz force} $$ This is also called the Lorenz force.

An electric field can do work on a charge $$ W = q\,\Delta V \nonumber $$ It can change the kinetic energy of the charge.

Magnetic fields can never do work on a moving charge, because the force \(F_B\) is always perpendicular to the velocity \(\vec v\). You can change the direction, but you can’t change the kinetic energy.

Force on a whole wire

Calculate the force on a wire with current \(I\) through magnetic field \(\vec B\). In the wire is charge \(dq\) with drift velocity \(\vec v_d\)

Thought experiments

  • If the current is zero, at room temperature, the electrons have a huge speed 3 million meters per second, but they are in all chaotic directions (random thermal motion). So on each individual charge there will be a force, but they average out to zero.
  • It is not until I run a current, that they charges are going to walk through with a very slow drift velocity. Now, the net force will not be zero.

The angle \(\theta\) will come in later, because the force is the cross-product between \(\vec v\) and \(\vec B\).

Side note: in reality negative charges (electrons) move in the opposite direction as the current. However, a negative charge going opposite to the current, is mathematically the same as a positive charge going with the current.

On charge \(dq\) there is a force \(d\vec F_B\) $$ \begin{align*} d\vec F_B &= dq \, \left(\vec v_d \times \vec B\right), & \left(I = \frac{dq}{dt}\right) \\ &= I\,dt \, \left(\vec v_d \times \vec B\right) \end{align*} $$

The drift velocity \(\vec v_d\) is the derivative of the distance \(d\vec l\) in time,

With \(v_d = \dfrac{d\vec l}{dt}\) \(\Longrightarrow\) \(d\vec l = \vec v\, dt\) $$ \shaded{ d\vec F_B = I\, \left( d\vec l \times \vec B \right) } \tag{Force} \label{eq:forcedq} $$

This is the force of a small segment of the wire, with length \(d\vec l\). \(I\) is the current through the wire, and \(\vec B\) is the local magnetic field at location \(d\vec l\).

If you want to know the entire force on the wire, you have to take the integral along the whole wire. At every potion \(d\vec l\), you have to determine what \(\vec B\) is. That will give you a force vector that you have to add those vectors vectorially. Not easy, but that’s the idea. $$ \begin{align*} \int_\rm{wire} d\vec F_B &= \int_\rm{wire} I\, \left( d\vec l \times \vec B \right) \end{align*} $$

Example

Simplifying the geometry so we can solve the integral. Assume the magnetic field was constant over the portion of the wire.

With \(\vec B\perp d\vec l\), so \(\sin\theta=1\). And \(\vec B\) is constant \((B)\). $$ \begin{align*} F_B &= \int_\rm{wire} I\, \left( d\vec l \times \vec B \right) \\ &= I\, \left( \int_\rm{wire} d\vec l \right) \times B \end{align*} $$

The current \(\vec I\) and magnetic field \(\vec B\) are constant, so only \(d\vec l\) stays in the integral $$ \shaded{ F_B = I\, l \, B } \tag{Force on wire} \label{eq:forcewire} $$

Filling in the values $$ \begin{align*} F_B &= I\, \left( \int_\rm{wire} d\vec l \right) \times B \\ &= I\, l \, B \\ &= (300)(0.1)(0.2) = 6\,\rm N \end{align*} $$

DC Motors

Current meter

A constant current loop in a constant magnetic field. The force \(\vec F\) is in the direction \(\vec I\times\vec B\)

36:45

If the wire has length \(a\), then \(B\) with \(\vec B\perp \vec I\), the force is what we just derived in equation \(\eqref{eq:forcewire}\) $$ \begin{align*} F &= I\,a\,B \end{align*} $$

On the front and rear sections, the force is zero, because the magnetic field and current are parallel, so their cross-product is zero.

As a result of the forces, there is a torque that wants to rotate it in the counter clockwise direction. The the distance between the forces \(b\), the magnitude of the torque at this moment in time $$ \tau = 2F\,\frac{1}{2}b = I\,a\,b\,B \nonumber $$

As it rotates, the forces come closer, so the torque will be come less. There will come a time, 90° later, that the torque is zero.

At that time, the whole loop is perpendicular to the magnetic field. The forces follow from the cross-product rule

At that point in time, there is again no net force on the system, but now there is no torque either. At it reaches 90°, it has enough inertia, so that it rotates a little further. Now the torque will reverse, causing it to rotate back.

Commutator

How do you get over this torque reversal and the wires to \(A\) and \(D\) intertwining?

The solution is slippery contacts, called brushes.

This not only keeps the wires from tangling up, it also reverses the potential. It it rotates 180° then A will be on the minus side of the potential. Now every half a rotation, the current will reverse direction. We call that a commutator.

Now, the torque reversal will not occur. The torque will always want to rotate the loop in exactly the same direction.

Copyright © 1996-2022 Coert Vonk, All Rights Reserved