Contents

1 Basic concepts 3

2 Lorentz force 4 2.1 Experimental basis and formulation . . . . . . . . . . . . . . . 4 2.2 The direction of the Lorentz force . . . . . . . . . . . . . . . . 5 2.3 A force without work . . . . . . . . . . . . . . . . . . . . . . . 6 2.4 The right-hand-rule . . . . . . . . . . . . . . . . . . . . . . . . 7 2.5 Example: force on a wire . . . . . . . . . . . . . . . . . . . . . 7 2.6 Current loops & Magnetic dipoles . . . . . . . . . . . . . . . 8

3 Biot-Savart’s law 9 3.1 A wire with a current . . . . . . . . . . . . . . . . . . . . . . . 10 3.2 Two parallel wires . . . . . . . . . . . . . . . . . . . . . . . . 12 3.3 Knowledge and Skills . . . . . . . . . . . . . . . . . . . . . . . 13

4 Intermezzo: Current densities 14

5 Ampere’s law 16 5.1 Thick wire with a current . . . . . . . . . . . . . . . . . . . . 17 5.2 Plane with a homogeneous current density . . . . . . . . . . . 18 5.3 The Solenoid . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 5.4 Knowledge and Skills . . . . . . . . . . . . . . . . . . . . . . . 21

6 Field equations 21 6.1 Stokes’ theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 21 6.2 Stokes theorem: example . . . . . . . . . . . . . . . . . . . . . 23 6.3 The curl of B⃗ . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 6.4 The divergence of B⃗ . . . . . . . . . . . . . . . . . . . . . . . 25 6.5 Summary: Field equations for E⃗ and B⃗ fields . . . . . . . . . 26 6.6 Knowledge and Skills . . . . . . . . . . . . . . . . . . . . . . . 27

7 Magnetic fields in matter 27 7.1 Paramagnetism: electron spin . . . . . . . . . . . . . . . . . . 28 7.2 Paramagnetism: Electron orbit . . . . . . . . . . . . . . . . . 29 7.3 Diamagnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 7.4 Magnetization and Bound Currents . . . . . . . . . . . . . . . 30 7.5 Linear materials . . . . . . . . . . . . . . . . . . . . . . . . . . 32 7.5.1 Example: Infinite bar . . . . . . . . . . . . . . . . . . . 33 7.5.2 Example: Infinite bar with free current . . . . . . . . . 33

7.6 Ferromagnetism . . . . . . . . . . . . . . . . . . . . . . . . . . 33 7.7 Knowledge and Skills . . . . . . . . . . . . . . . . . . . . . . . 34

1 Basic concepts

Magnetostatics aims at the description of all phenomena that involve nonchanging magnetic fields. Qualitative knowledge about magnetostatics has been around for many centuries: it is possible to use a magnet needle to make a compass without knowing how to quantify magnetic processes. In this section some basic concepts of magnetism are discussed.

In electrostatics you are familiar with the fact you have positive and negative charges. The charges for magnetostatics always come in pairs. In a magnet one side is the magnetic-positive side, while the other side is the magnetic-negative side. The magnetic-positive side we like to call the ’northpole’ and the magnetic-negative side the ’south-pole’. Off course this naming is derived from our own earth, which itself is a permanent magnet. Forces between magnets are such that a north- and a south pole attract each other, while there is a repelling force between two poles of the same kind.

Now you could decide to cut a magnet in half, just between the north- and the south pole, in order to obtain a north-pole and a south-pole separately. The result of your experiment, however, will be that you will have indeed two magnets, both with a north AND a south pole (see Fig. 1). You can repeat

Figure 1: Cutting a magnet in two pieces will only result in creating two magnets, each with a magnetic north- and south pole. You can repeat cutting, but you will not create a magnetic monopole.

this as many times as you like, but the only thing you will succeed in, is to get many more small magnets. As far as we know no magnetic monopoles exist!

Also it is well known that compared to gravity the magnitude of the magnetic force is huge. It is very common for even small magnets to lift little (or sometimes big) pieces of iron, against the gravitational force from the whole earth!

In addition there exists an intricate relation between electrical currents and magnetic field, which will become clear further in these lecture notes. The unit of a magnetic field is called the Tesla, and it can be described in terms of other units as:

[T esla] = [N]/([A][m]) = [N][s]/([C][m]) = [kg]/([C][s])  (1)

with [A] = [C]/[s] an electrical current, and [N] the unit of force.

2 Lorentz force

2.1 Experimental basis and formulation

The Dutch physicist Hendrik Antoon Lorentz was the first to quantify the movement of an electrically charged particle inside a constant magnetic field. Just as Coulomb’s law that describes forces in electrostatics, the law for the force caused by a magnetic field is based on experimental data:

1. The Lorentz force is proportional to the strength of the magnetic field;

2. The Lorentz force is proportional to the electric charge;

3. The Lorentz force is proportional to the velocity of the object it acts on;

4. The Lorentz force is proportional to the sine of the angle between the velocity vector and the speed vector;

5. The Lorentz force is perpendicular to both the velocity direction and the direction of the magnetic field;

This list of experimental data can be elegantly translated into the following mathematical statement, that fully describes the force of a magnetic field B⃗ on a moving object with charge q and a velocity v⃗:

F⃗ = qv⃗ × B⃗  (2)

The × in equation 2 represents the ’cross-product’ between two vectors. With our knowledge of the cross product we can prove that the Lorentz force actually obeys the experimental data as listed above. The magnitude of the Lorentz force is given by:

|F⃗ | = q|v⃗ × B⃗)|
= q|v⃗||B⃗| sin θ  (3)

with θ the angle between the magnetic field lines and the velocity vector. From the last equation it can be seen that the first four experimental requirements on the Lorentz force have been fulfilled.

2.2 The direction of the Lorentz force

When it comes to proving that the Lorentz force fulfills the last requirement things are slightly more complicated, since we now have to calculate the direction of the force. Remember that for electrostatics the electrical force was just proportional to the size and direction of the electrical field. For the Lorentz force this is not the case, since we are dealing with a cross-product of two vectors. Now we have too look into the detail of the cross-product: what is the direction of the result vector? First we will work out mathematically what is the direction of the field. The cross product between any two vectors v⃗ = (v_{x}, v_{y}, v_{z}) and B⃗ = (B_{x}, B_{y}, B_{z}) can be calculated by evaluating the following determinant (try this yourself and you will be a commander of cross-products!):

v⃗ × B⃗ = | \begin{matrix}x̂ & ŷ & ẑ \\ v_{x} & v_{y} & v_{z} \\ B_{x} & B_{y} & B_{z} \end{matrix} |
= (v_{y}B_{z} − v_{z}B_{y}) x̂ + (v_{z}B_{x} − v_{x}B_{z}) ŷ + (v_{x}B_{y} − v_{y}B_{x}) ẑ
= ( \begin{matrix}v_{y}B_{z} − v_{z}B_{y} \\ v_{z}B_{x} − v_{x}B_{z} \\ v_{x}B_{y} − v_{y}B_{x} \end{matrix} )  (4)

This general equation might not tell the full story of the direction of the Lorentz force directly, but we can prove that the direction of the force is both perpendicular to the velocity vector and the magnetic field vector. Two vectors are perpendicular if their inner product equals zero (remember: A⃗ · B⃗ = |A||B| cos θ), so we have to validate that:

F⃗ · v⃗ = 0  (5)

We can explicitly calculate the inner product using equation 4 as follows:

F⃗ · v⃗ = q(v⃗ × B⃗) · v⃗

= q(v_{y}B_{z} − v_{z}B_{y}) v_{x} +
q(v_{z}B_{x} − v_{x}B_{z}) v_{y} + q(v_{x}B_{y} − v_{y}B_{x}) v_{z}
= qB_{x}(v_{z}v_{y} − v_{y}v_{z}) + qB_{y}(−v_{z}v_{x} + v_{z}v_{y}) + qB_{z}(v_{y}v_{x} − v_{x}v_{y})
= 0  (6)

Following similar tactics you can now prove yourself that the Lorentz force is also perpendicular to the magnetic field In fact you then have proved a general property of cross-products between vectors A⃗ and B⃗:

A⃗ · (A⃗ × B⃗) = 0  (7)
B⃗ · (A⃗ × B⃗) = 0  (8)

Note that the second statement follows directly from the first, because of another property of the cross-product:

A⃗ × B⃗ = −B⃗ × A⃗  (9)

Prove this!

2.3 A force without work

A very important consequence of the fact that the Lorentz force is perpendicular to the velocity of an object, is that the Lorentz force does not do any work. To understand this a bit more in detail we have to recall our definition of work W being done by a force. It has been defined as the displacement in the direction of the force, or as an equation:

W = ∫_{line} F⃗ · d⃗l  (10)

This can be rewritten with a change of variables as:

W = ∫_{line} F⃗ · v⃗dt  (11)

The last equation is clearly zero due to equation 5. The consequence of this is that if a charged particle moves through a magnetic field its energy does not change, since that requires work done. So the only thing that does change is the direction of the velocity. This is really special to the Lorentz force: all other forces you have thus far encountered, like the Coulomb force or gravitational force, do actually work.

2.4 The right-hand-rule

Now it is time to get a clearer understanding of the direction of the Lorentz force by studying a simple example. Let’s look at a configuration with a magnetic field B⃗ = (0, 0, B) pointing in the ẑ direction and a particle with charge q moving with velocity v⃗ = (v, 0, 0) in the x̂ direction. The equation now becomes much more explicit and clear:

F⃗ = −qvBŷ  (12)

You can verify that all requirements on the Lorentz force as mentioned before are still fulfilled (sin \frac{π}{2} = 1!) and also you can see that the direction of the Lorentz force can be obtained without calculation, by using the (in)-famous ’right hand rule’. Stretch out your hand and make your thumb point in the direction of v⃗. Point your other fingers in the direction of the magnetic field B⃗. The cross-product v⃗ × B⃗ is now pointing out of the palm of your hand. Exercise this a couple of times and you will never make any mistakes in pointing out the direction of cross-product. Be however careful that a negative electric charge does flip the direction of the Lorentz force by 180^{∘}.

2.5 Example: force on a wire

What happens if you place a wire of length L carrying a current I, inside a homogeneous magnetic field, B (see Fig. 2)? The direction of the force

Figure 2: A wire with length L is placed inside a homogeneous magnetic field B with field lines pointing out of the paper.

can be found using the right hand rule. The electrons move in the opposite direction as the current, but they have a negative charge. So the direction of the force is the direction of the cross product of I⃗ and B⃗. The size of the force is calculated by integrating:

|F⃗ | = ∫_{line} dq(v⃗ × B⃗)
= ∫_{line} λdl(v⃗ × B⃗)
= ∫_{line} dl(I⃗ × B⃗)
= LIB  (13)

We have used the fact that the line charge density λ multiplied by the velocity of the electrons gives the electrical current I. So we now have an equation that tells us that the Lorentz force of a magnetic field on a wire carrying a current I is equal to:

dF⃗ = dl(I⃗ × B⃗)  (14)

2.6 Current loops & Magnetic dipoles

Now let us consider a square wire loop with sides of length L carrying a current I, placed in a homogeneous magnetic field B⃗ (see Fig. 3). The forces

Figure 3: A square wire loop with sides L and carrying a current I is placed inside a homogeneous magnetic field B.

on each of the sides of the loop can be calculated in the same way as shown in the previous paragraph:

F⃗ = L(I⃗ × B⃗)  (15)

It can be seen that the net force on the wire loop is zero, since the forces on top and bottom of the loop cancel each other, just as the forces on both of the sides. However, there is a torque on the current loop! The torque, τ, exerted on for example the left side of the loop is just the size of the Lorentz force times the ’arm’ the force. So:

τ_{L} = (LIB)\frac{L}{2} sin θ  (16)

In he right side of the loop the Lorentz force points in the opposite direction, but the arm of the force also points in the opposite direction. So the total torque becomes:

τ = L^{2}IB sin θ  (17)

Notice that the bottom and top sides of the current loop do not contribute to the torque. The Lorentz force is certainly pulling of the wire there as well, but it is trying to stretch the wire loop instead of trying to rotate it (try to understand this yourself).

At this point it is instructive to rewrite equation17 as the following crossproduct:

τ⃗ = m⃗ × B⃗  (18)

where m⃗ is the magnetic dipole moment of the current loop. The direction of the magnetic dipole moment is perpendicular to the plane spanned by the wire loop. Its size is just the current multiplied by the surface area:

|m⃗| = L^{2}I  (19)

The torque exerted on the wire loop tries to rotate it such that it is perpendicular to the magnetic field lines. In terms of a magnetic dipole moment we can make a more general statement: a magnetic field tries to set magnetic dipoles parallel to the magnetic field lines.

3 Biot-Savart’s law

Now we have learned in the previous section how to calculate the force exerted by a magnetic field on a moving charge, we have to know how to actually calculate a magnetic field. Magnetic fields are caused by electric currents. An expression that relates the magnetic field, B⃗_{P}, in a point P to an electric current, I, is called the law of Biot-Savart:

B⃗_{P} = \frac{μ_{0}I}{4π} ∫_{line} \frac{dl⃗× r̂}{r^{2}}  (20)

The integration is done along the current path in the direction of the flow; dl⃗ is an element of length along the wire and r̂ is the direction vector from Figure 4: Variables needed to calculate the magnetic field in a point P due to an infinitesimal wire element with length dl⃗ at a distance r⃗.

the location of dl⃗ and the point P as is sketched in Fig. 4. Try to verify by using the right-hand-rule that the magnetic field due to the infinitesimal wire element in Fig. 4 is pointing into the paper. When using Eq. 20 you should realize that along your integration path both dl⃗, r̂, and r may change, both in magnitude and direction! To master this equation practice is mandatory.

The constant μ_{0} is called the magnetic permeability which is defined as:

μ_{0} ≡ 4π 10^{−7} \frac{N}{A^{2}}  (21)

A remarkable thing happens when the magnetic permeability is multiplied with the electrical permittivity ϵ_{0}. Remember that ϵ_{0} was defined as:

ϵ_{0} ≡ \frac{1}{4π 10^{−7}c^{2}} \frac{C^{2}}{Nm^{2}}  (22)

The multiplication leads to:

\frac{1}{ϵ_{0}μ_{0}} = c^{2}  (23)

with c the speed of light. This amazing results indicates that there might be a special relationship between electric and magnetic field. This relationship will be unveiled toward the end of this course.

3.1 A wire with a current

With Biot-Savart’s law we are now able to calculate the magnetic field for any current configurations. So let us start with the calculation of the magnetic field of a relatively simple example: an infinitely long wire carrying a current I (see Fig. 5). We could start calculating like blind chickens at this point, but

Figure 5: Wire carrying a current I. The magnetic field is calculated at a point P at a distance r from the wire.

we can also use symmetry arguments to argue what the direction of the field must be. The cylindrical symmetry of this example excludes the existence of a radial component to the magnetic field. We can try to argue that this is impossible by making a ’gedanken’ experiment: we can rotate the infinitely long wire by 180^{∘} so the current is now pointing in the downward direction. As can be seen in the law of Biot-Savart changing the sign of the current would result in flipping the sign of the B⃗ field. However, the rotation of the wire leaves the radial component of the magnetic field pointing in the same direction as before. So there can exist no radial magnetic field in our example. The component of the magnetic field pointing along the direction of the wire drops out because the contribution coming from the current at z < 0 is canceled exactly by the contribution from the current at z > 0 (try to prove and visualize this yourself).

So the only component of the magnetic field that is not equal to zero is the component in the azimuthal, ϕ, direction. Its size is calculated with Biot-Savart’s law 20, but we have to be extremely careful how to fill in the variables in the equation. Let us consider the magnetic field caused by an infinitesimal line element d⃗z first (see Fig. 5): the magnitude of the magnetic field in point P due to this line element is:

|dB⃗_{P} | = \frac{μ_{0}I}{4π} \frac{|dz⃗| sin α}{r^{2} + z^{2}}  (24)

There are a couple of variables in this equation that at first might seem a bit strange. First there is the r^{2} + z^{2} term, while you would expect a r^{2} term if you look at the Biot-Savart law. But in fact the r^{2} + z^{2} is correct since it is the distance between the line element d⃗z and the point P. So the r in Biot-Savart’s law is a very different r than is used to indicate the radius in cylindrical coordinates. The same is the case for r̂ which indicates the unit vector in the direction of the line element d⃗z and the point P, and thus also has nothing to do with the radius in cylindrical coordinates. You must think very well before just using a formula! The sin α comes from the size of the cross product:

|d⃗z × r̂| = dz sin α  (25)

where α is the angle between d⃗z and r̂. It can be geometrically worked out to be:

sin α = \frac{r}{\sqrt{r^{2} + z^{2}}}  (26)

Now we can calculate the magnetic field adding the contributions from all line elements to the magnetic field.

|B⃗_{P} | = |∫_{line} dB⃗_{ϕ}|
= \frac{μ_{0}I}{4π} ∫_{−∞}^{+∞} \frac{dz}{r^{2} + z^{2}} \frac{r}{\sqrt{r^{2} + z^{2}}}
= \frac{μ_{0}I}{4π} \frac{z}{r\sqrt{r^{2} + z^{2}}} |_{−∞}^{+∞}
= \frac{μ_{0}I}{2πr}  (27)

The integral over z you do not have to be able to solve yourself. If you encounter such an integral at any point we will provide you with the answer. However you can verify the answer by differentiating the one but last equation.

3.2 Two parallel wires

We can now consider a configuration with two parallel wires with length L at a distance R⃗_{12} each carrying an electrical current, I_{1} and I_{2},respectively (see Fig. 6). We can calculate the magnetic field at the position of the second wire, due to Figure 6: Two wires with electric currents both generate a magnetic field. The magnetic fields result in a Lorentz force on the wires.

the current in the other wire using equation 27. The equation for the Lorentz force on a wire (Eq. 13) tells us that:

|F⃗_{2}| = L|(I⃗_{2} × B⃗_{1})|  (28)
= \frac{μ_{0}LI_{2}I_{1}}{2πR_{12}}  (29)

The calculation of the size of the cross product is fairly easy since there are only 90^{∘} angles involved in this example. I leave it up to you to show that the Lorentz force between the wires is attractive if the currents point in the same direction.

3.3 Knowledge and Skills

The knowledge and skills you should have acquired during reading of the previous can be summarized as follows:

• Magnets always have a north- and a south pole. Magnetic monopoles do not exist;

• The Lorentz force of a magnetic field on a moving charge is:

F⃗_{L} = qv⃗ × B⃗  (30)

Make sure you can figure out both the size and the direction of the Lorentz force. For the direction make sure that you are comfortable with the right-hand-rule;

• The Lorentz force on a straight wire segment is:

F⃗ = L(I⃗ × B⃗)  (31)

;

• The torque on a current loop is:

τ = m⃗ × B⃗  (32)

where m⃗ is the magnetic dipole moment;

• The law of Biot-Savart can be used to calculate magnetic fields that are caused by line currents:

B⃗_{P} = \frac{μ_{0}I}{4π} ∫_{line} \frac{dl⃗× r̂}{r^{2}}  (33)

4 Intermezzo: Current densities

In the previous section we used currents all the time. Just as for electric charges it is often instructive to talk about current densities instead of currents. We can speak of line, surface, and volume current densities (see Fig. 7). A line current is defined as the charge density per meter, λ, times the velocity

Figure 7: Different types of current densities: a) shows a line current density,

b) a surface current density, and c) shows volume current density.

of the charges:

I⃗ = λv⃗ → [I] = C/s = A (Ampere)  (34)

The size of the current is just the magnitude of the vector I⃗. The corresponding Lorentz forces are calculated as:

F⃗_{L} = ∫_{line} I⃗ × B⃗dl = I ∫_{line} I⃗ × B⃗  (35)

For surface currents the situation is similar (see Fig. 7b):

K⃗ ≡ \frac{dI⃗}{dl_{⊥}} → [K] = A/m  (36)
= σv⃗  (37)

To obtain a current one now has to multiply the surface current density, K⃗, with the length of the surface. Or to be more precise one has to integrate the current density along a surface if the current density is not a constant. The Lorentz force is now:

F⃗ = ∫_{surface} K⃗ × B⃗do  (38)

Biot-Savart’s law can be used again to calculate the magnetic field resulting from a surface current density:

B⃗ = \frac{μ_{0}}{4π} ∫_{surface} \frac{K⃗(r′) × d̂}{d^{2}}do′  (39)

where the integration is over the surface carrying the current, and d⃗ = r⃗ − r′.

Most frequent you will encounter volume currents, J⃗, which are defined as:

J⃗ ≡ \frac{dI⃗}{do_{⊥}} → [J] = A/m^{2}  (40)
= ρv⃗  (41)

The differentiation with respect to do_{⊥} gives a vector pointing through a surface: the size of J⃗ is the current per unit area perpendicular to the flow. For example take a wire with radius R and a current I. If we now assume that the current density is homogeneous, we can obtain the current density J⃗, by simply dividing I by the surface πR^{2} of the wire. The direction of J⃗ is along the direction of current flow. The Lorentz force can be calculated through the volume integral:

F⃗ = ∫_{volume} J⃗ × B⃗dV  (42)

Try to prove this last equation yourself. Once more Biot-Savart’s law can be used to calculate the magnetic field resulting from a volume current density:

B⃗ = \frac{μ_{0}}{4π} ∫_{volume} \frac{J⃗(r′) × d̂}{d^{2}}dV′  (43)

where the integration is over the volume carrying the current, and d⃗ = r⃗ −r′.

5 Ampere’s law

We now know how to calculate magnetic fields in any configuration using the Biot-Savart law in principle. We can have a closer look at the properties of magnetic fields to see if there are laws, like Gauss’s law for electrical fields, that can simplify our calculation of magnetic fields in certain cases. Let us go back to the magnetic field of a wire carrying a current:

B_{ϕ} = \frac{μ_{0}I}{2πr}, B_{r} = B_{z} = 0  (44)

We can take a circular path around the wire, so that the line integral:

∫_{line} B⃗ · dl⃗ = μ_{0}I  (45)

where the ∫ represents an integral in a path around the wire. In fact the path does not have to be a circle; any path around the wire will give exactly the same answer (can you prove this one? see Griffiths section 5.3.1). We can generalize this law a bit further to obtain:

∫_{line} B⃗ · dl⃗ = μ_{0} ∫_{surface} J⃗ · do⃗  (46)

This equation is known as the law of Ampere: it states that the integral of the magnetic field along a closed path is equal to the total current that is enclosed by the path. We have seen that it is true for a thin wire carrying a current. It can be explicitly proven, using the Biot-Savart law and the superposition principle, that it is true in general for any current density causing a magnetic field. (see Griffiths section 5.3.2).

The law of Ampere is always true, but for calculation of magnetic fields it is useful only if there is a high degree of symmetry in the configuration you want to solve. The reason for this is that in that case the line integral ∫ B⃗d⃗l can become easy to solve. This is especially true if you know that the magnetic field over your integration path is constant, since the magnetic field then drops out of the integral, or in an equation:

∫_{line} B⃗d⃗l → B⃗ ∫_{line} d⃗l  (47)

In the next sections we show a selection of examples where the magnetic fields can be calculated in a fairly easy way, using the law of Ampere.

5.1 Thick wire with a current

In the previous section you have already seen how Ampere’s law can be used to calculate the magnetic field outside a wire. Now let us consider a wire with a radius R carrying a current I⃗ distributed homogeneously over the wire (see Fig. 8). The current density J⃗ in this case is just the total current divided

Figure 8: A thick wire of radius R carrying a current I that is homogeneously distributed.

by the surface of a cross-section through the wire:

J⃗ = \frac{I⃗}{πR^{2}}  (48)

From the symmetry of the problem we can conclude that the magnetic field only has a component pointing in the azimuthal (ϕ) direction, and that it only depends on the radius r. Ampere’s law inside the wire now reduces to:

∫_{line} B⃗d⃗l = μ_{0} ∫_{surface} J⃗d⃗o  (49)
B · 2πr = \frac{μ_{0}Ir^{2}}{R^{2}}  (50)

so that the magnetic field in the ϕ direction as a function of r now becomes:

B(r < R) = \frac{μ_{0}Ir}{πR^{2}}  (51)

Outside the wire you can show (see previous section) that the magnetic field is the same as for an infinitely thin wire. Off course if you do not see the cylindrical symmetry in a problem like this you can always use the law of Biot-Savart to calculate the magnetic field. The advantage is that you are always right to use this law, the disadvantage is that you will have to calculate much more complicated integrals. Remember for example how much work it was to find the magnetic field of a wire with the Biot-Savart law!

5.2 Plane with a homogeneous current density

As a next example of the Ampere’s law we calculate the magnetic field of an infinite plane carrying a current density K⃗ (A/m) as shown in Fig. 9. The

Figure 9: An infinitely big plane with homogeneous current density K⃗.

plane is lying in the y − z plane and the current is flowing in the y direction.

Since we know that magnetic fields to be perpendicular to the currents causing them, the magnetic field cannot have a component in the y direction. From a clever symmetry argument we can also show that the component of the B-field perpendicular to the plane must also be zero. Suppose there were a component of the magnetic field in the x direction. Then rotate the whole (infinite!) plane 180^{∘} around the x axis. The current is now pointing in the opposite direction, while the x component of the B-field is unaffected. This is a clear contradiction and thus the x component of the magnetic field must be zero as well. We are only left with a magnetic field parallel to the plane in the z direction.

Now we can use Ampere’s law to calculate the magnitude of the field, by constructing an imaginary rectangular path as shown in Fig. 9. Along this path we can again calculate the integral ∫_{line} B⃗dl⃗. The sides of the rectangle that are perpendicular to the plane do not contribute to the line integral since we just showed that the magnetic field does not have a component perpendicular to the plane, so we are left with the two sides that are parallel to the plane. So:

∫_{line} B⃗dl⃗ = 2aB_{z}  (52)

Ampere’s law tells us that this integral should be equal to μ_{0} times the enclosed current. So we can now solve the B⃗ field:

2aB_{z} = μ_{0}Ka  (53)
⇓  (54)
B_{z} = \frac{μ_{0}K}{2}  (55)

So the field is homogeneous on either side of the plane. Only the sign of the magnetic field flips (try to confirm the direction yourself by using the right-hand-rule).

5.3 The Solenoid

An even more complicated configuration that is made simple by Ampere’s law is the solenoid. A solenoid is an infinitely long cylinder of radius R over which a wire carrying a current I is wrapped around with N windings per meter (see Fig. 10). We can start-off with a couple of symmetry arguments to reduce some of the components of the magnetic field to zero, just as we did before for the infinite plane. First there can be no component in the ϕ direction, since it is parallel to the current. Secondly, suppose there is a radial component to the field. Then by rotating the cylinder around the x Figure 10: A solenoid with radius R and current I running through N windings per meter.

axis over 180^{∘} the current is now going in the opposite direction and the radial component of the B field is unaffected: contradiction! The radial component of the field can only be zero, so we are left with only a z component.

Let’s first calculate the field outside the solenoid. We can take the line integral around a path indicated by 1 in Fig. 10. There is no enclosed current so Ampere’s law reduces to:

∫_{line} B⃗dl⃗ = 0  (56)

We can safely assume that the magnetic field should go to zero when the far end of the integration path goes to infinity. In that case we only get a contribution to the path integral from the line piece at r = r. Since the total integral needs to be zero, this can only be true if the field itself is zero. So outside a solenoid the magnetic field is zero!

The field inside we can obtain by integrating the B-field along the path indicated by 2 in Fig. 10. The only contribution to the integral is from the path inside the solenoid parallel to the z axis. The enclosed current is simply:

I_{enclosed} = aIN  (57)

Ampere’s law is now written as:

∫_{line} B⃗dl⃗ = μ_{0} ∫_{surface} J⃗do⃗  (58)
B_{z}a = μ_{0}aIN  (59)
⇓  (60)
B_{z} = μ_{0}IN  (61)

A surprisingly simple answer for such a complicated configuration.

5.4 Knowledge and Skills

The knowledge and skills you should have acquired during reading of the previous can be summarized as follows:

• You should be familiar with line, surface and volume currents;

• You should be able to calculate the Lorentz force on these different kinds of current densities;

• You should be able to write down and use Ampere’s law to calculate magnetic fields in situations with a high degree of symmetry.

6 Field equations

In this section we will give an overview of the equations that give a full description of electro- and magnetostatics. In order to do so there is one more bit of mathematics you need to know.

6.1 Stokes’ theorem

The last bit of math you need to know before we can complete the theory of electro- and magneticstatics is the theorem of Stokes. In electrostatics Gauss’s law (see section ??) gave a relation between surface integrals of the electrical field and a volume integral of a charge density. It turns out that for the line integrations you encounter in magnetostatics a similar type of relation can be found. For any vector field the integral around a surface is equal to the curl of the vector field integrated over the surface:

∫_{line} A⃗ · dl⃗ = ∫_{surface}(∇ × A⃗)do⃗  (62)

We will not rigorously prove Stokes theorem, but we will try to give a more hand-waving argument to make it acceptable to you. Assume you have a vector field A⃗(x, y, z) and you want to calculate the path integral of the vector field around an infinitesimal small loop in the plane z = 0 as shown in Fig. 11. The line integral ∫_{line} A⃗dl⃗ is then split up in four parts, with a contribution from each of the sides of the little square. For each of the sides we must think what dl⃗ should be, especially paying attention to the plus and minus signs. Along path 1 dl⃗ = (dx, 0, 0), along path 2 we move in the positive y direction so dl⃗ = (0, dy, 0). Paths 3 and 4 are in the −x and −y Figure 11: Vector field integrated along a path enclosing an infinitesimal loop in the xy plane.

direction, respectively, so they pick up a minus sign with respect to paths 1 and 2. So the line integration becomes:

∫_{line} A⃗dl⃗ = ∫_{x}^{x+dx} A_{x}(x′, y)dx′ + ∫_{y}^{y+dy} A_{y}(x, y′)dy′ −  (63)
∫_{x+dx}^{x} A_{x}(x′, y + dy)dx′ − ∫_{y+dy}^{y} A_{y}(x′, y + dy)dx′ (64)

Notice that I have left out the z index everywhere since z = 0 anyhow. The first component can be calculated explicitly:

∫_{x}^{x+dx} A_{x}(x′, y)dx′ = \frac{1}{2}(A_{x}(x, y) + A_{x}(x + dx, y))dx  (65)

If you do this to all four terms and then collect the A_{x} and A_{y} terms together (go ahead and try) you will find that:

∫_{line} A⃗dl⃗ = dxdy(\frac{∂A_{x}}{∂y} − \frac{∂A_{y}}{∂x})  (66)
= (∇ × A⃗)_{z}do  (67)
= (∇ × A⃗) · do⃗  (68)

For a non-infinitesimal path in the x−y plane we can write the surface integral as the sum of many infinitesimal surface integrals as shown in Fig. 12. Using the theorem we just derived for infinitesimal path integrals, you can see from the figure each side of the infinitesimal loop borders to another infinitesimal loop that gives exactly an opposite contribution to the path integral, except when the loop is at the border of your integration area. That is how the Figure 12: Many infinitesimal path integrations make up a big one. Notice how only the boundary integration remains.

relation between a total surface integration and the path integral around the surface comes about. In this context it is instructive to think of the the curl of a vector as a little whirlpool around it: then it becomes easy to see why Stokes theorem makes sense. You have to notice that we really did not give any general proof of the theorem. We restricted ourselves to integration in the xy plane, and certainly not any surface. For a more rigorous proof please refer to any vector analysis textbook.

6.2 Stokes theorem: example

At this point it is worth to go through an example in which we can apply Stokes’ theorem. Suppose we want to calculate the path integral over a circular path with radius R and center x = y = 0 for a vector field A⃗ that is parametrized as:

A⃗ = \frac{(−y, x, 0)}{\sqrt{x^{2} + y^{2}}}  (69)

In Fig. 13a you see what the vector field looks like. So what about the line integral? The vector field is just the unit vector in the ϕ direction so:

∫_{line} A⃗ · dl⃗ = ∫_{circle} dl = 2πR  (70)

According to Stokes theorem we could also calculate the curl of this vector field and then do the integration over the surface spanned by the path. So first the curl of the vector field:

∇ × A⃗ = | \begin{matrix}x̂ & ŷ & ẑ \\ ∂_{x} & ∂_{y} & ∂_{z} \\ A_{x} & A_{y} & A_{z} \end{matrix} | = ẑ(\frac{∂}{∂x} \frac{x}{\sqrt{x^{2} + y^{2}}} + \frac{∂}{∂y} \frac{y}{\sqrt{x^{2} + y^{2}}} ) (71)

Figure 13: In Fig. a) you see a drawing of the vector field A⃗, in Fig. b) a drawing of ∇ × A⃗.

= \frac{ẑ}{\sqrt{x^{2} + y^{2}}} ≡ \frac{ẑ}{r}  (72)

In Fig. 13b the curl of the vector field A⃗ is drawn. Now let’s calculate the surface integral of ∇ × A⃗.

∫_{disk} ∇ × A⃗ = ∫_{r=0}^{r=R} ∫_{ϕ=0}^{ϕ=2π} \frac{1}{r}rdrdϕ = 2πR  (73)

Notice the factor r coming from the integration in cylindrical coordinates. The answer you get either way is exactly the same, though the work has been considerably more extensive when calculating the surface integral.

6.3 The curl of B⃗

We can now use Stokes theorem to find an elegant differential formulation of the law of Ampere, just as we could use Gauss law in electrostatics to go from Gauss’s law in integral form to Gauss’s law in differential form. Recall Ampere’s law:

∫_{line} B⃗dl⃗ = μ_{0} ∫_{surface} J⃗ · do⃗  (74)

Stokes theorem (equation 62) now tells us that the path integral on the left hand side of equation 74 can be written as the surface integral of the curl of B⃗. So equation 74 becomes:

∫_{surface}(∇ × B⃗) · do⃗ = μ_{0} ∫_{surface} J⃗ · do⃗  (75)

This can only be true for any B⃗ if:

∇ × B⃗ = μ_{0}J⃗  (76)

Please realize that equation 76 is equivalent to Ampere’s law in integral form (equation 74); there is really nothing new in this equation, except for the notation.

Now a little example. We have calculated the magnetic field for a thick wire with a current in section 5.1 to be:

\{ \begin{matrix}B⃗ = \frac{μ_{0}Ir}{2πR^{2}} ϕ̂ = \frac{μ_{0}I}{2πR^{2}} (−y, x, 0) & r < R \\ B⃗ = \frac{μ_{0}I}{2πr} ϕ̂ & r > R \end{matrix}  (77)

We can explicitly calculate the curl of B⃗ inside the wire:

∇ × B⃗ = | \begin{matrix}x̂ & ŷ & ẑ \\ ∂_{x} & ∂_{y} & ∂_{z} \\ −y & x & 0 \end{matrix} | = \frac{2μ_{0}I}{2πR^{2}} ẑ = μ_{0}J⃗  (78)

Can you prove that outside the wire ∇ × B⃗ = 0?

6.4 The divergence of B⃗

We have an expression for the curl of the magnetic field. What about its divergence? For electrical fields we had Gauss’s law in differential form which stated that ∇ · E⃗ = ρ/ϵ_{0}. The divergence of the E⃗ field at any point is equal to the charge density at that point. For the the divergence of any magnetic field we could get a similar expression, by taking the divergence of the law of Biot-Savart (equation 20), since Biot-Savart’s law is true for magnetic fields caused by any current distribution. For such a proof see for example Griffiths chapter 5.3.2. However, we can actually predict what ∇ · B⃗ should be by comparing it to the same expression for the electrical field. In the equation for ∇ · E⃗ = ρ/ϵ_{0} you see that on the right hand side of the equation you find the electric charge density. Since magnetic charges do not seem to exist, we can guess that:

∇ · B⃗ = 0  (79)

This should be true for any magnetic field and it is indeed the answer you find by following the explicit proof in Griffiths. Let’s see if it is true for the magnetic field of our ’good old’ thick wire (see above). Inside the wire we have:

∇ · B⃗ = ∇ · \frac{μ_{0}Ir}{2πR^{2}} ϕ̂  (80)
= \frac{μ_{0}I}{2πR^{2}} (−∂_{x}y + ∂_{y}x) = 0  (81)

Outside the wire:

∇ · B⃗ = ∇ · \frac{μ_{0}I}{2πr}ϕ̂  (82)
= \frac{μ_{0}I}{2π} (−∂_{x} \frac{y}{r^{2}} + ∂_{y} \frac{x}{r^{2}} ) = 0  (83)

Notice here that we would run into serious problems if we would have chosen to calculate ∇ · B⃗ for a wire with zero thickness, due to the singularity in the B⃗ field at r = 0.

Now remember that Gauss’s law for electrostatics could be either written in the differential form as shown above or in integral form:

∫_{surface} E⃗do⃗ = \frac{1}{ϵ_{0}} ∫_{volume} ρdV  (84)

Or in words: the integration of the E⃗ field over a closed surface (i.e. flux) is equal to the total charge enclosed. Since we know now that ∇ · B⃗ = 0, we can write for the magnetic flux through any surface with help of the law of Gauss:

Φ_{B} = ∫_{surface} B⃗do⃗  (85)
= ∫_{volume} ∇ · B⃗dV = 0  (86)

So the magnetic flux through any closed surface is equal to zero. This is another way of stating that there exist no magnetic charges.

6.5 Summary: Field equations for E⃗ and B⃗ fields

At this point we know all there is to know about electrostatics and magnetostatics. For any stationary charge distribution we can in principle calculate the corresponding electric field and for any steady current we can calculate the resulting magnetic field (though the calculations are not always easy). In addition we know how a charged particle behaves in these fields, since we have the equation for the Lorentz force on a charged particle:

F⃗_{L} = q(E⃗ + v⃗ × B⃗)  (87)

There are just four equations needed to calculate all electrical and magnetic fields. In integral form:

∫_{surface} E⃗ · do⃗ = \frac{1}{ϵ_{0}} ∫_{volume} ρdV (Gauss law)  (88)

∫_{line} E⃗ · dl⃗ = 0  (89)
∫_{surface} B⃗ · do⃗ = 0  (90)
∫_{line} B⃗ · dl⃗ = μ_{0} ∫_{surface} J⃗ · do⃗ (Ampere’s law)  (91)

Using Stokes theorem and the divergence theorem - just mathematical trickery - these integral equations can be re-written as the following set of differential equations:

∇ · E⃗ = \frac{ρ}{ϵ_{0}}  (92)
∇ × E⃗ = 0  (93)
∇ · B⃗ = 0  (94)
∇ × B⃗ = μ_{0}J⃗  (95)

That is all there is to know about electro- and magneto- statics; nothing more nothing less.

6.6 Knowledge and Skills

The knowledge and skills you should have acquired during reading of the previous can be summarized as follows:

• You should be familiar with Stokes’ theorem relating line integrals to surface integrals

• Ampere’s law in differential form:

∇ × B⃗ = μ_{0}J⃗  (96)

• The divergence of the magnetic field is zero:

∇ · B⃗ = 0  (97)

• You should know the four equations describing electrostatics and magnetostatics

7 Magnetic fields in matter

In this section we will try to explain what happens if matter is placed inside magnetic fields.

7.1 Paramagnetism: electron spin

Since every electron in matter has intrinsic spin it carries a little magnetic dipole moment with a dipole moment:

m = \frac{q_{e}h}{2m_{e}}  (98)

You can imagine the electron as a spinning electric charge, or a current loop if you want (see section 2.6). Without a magnetic field (see Fig. 14a) these dipoles are randomly oriented, but if a magnetic field is applied the dipole moments - and thus the spin vectors - try to align themselves with the magnetic field. As a result of the aligned dipole moments the magnetic field is enhanced: enhancement of a magnetic field in matter due to an applied magnetic field we call paramagnetism. It seems that since every material con-

Figure 14: Alignment of electron spins in a magnetic field as a cause of para-magnetism.

tains electrons paramagnetism should be a universal effect for all substances. However the Pauli exclusion principle (quantum mechanics) does not allow electrons in the same orbit to have their spins aligned. So as a rule of thumb only substances with an odd number of electrons exhibit paramagnetism due to spin alignment, while for substances with even number of electrons the effect cancels out. Even when a substance is paramagnetic the alignment of spins is usually (room temperatures) far from complete due to thermal fluctuations. A last point to remember is that as soon as the external magnetic field disappears, the paramagnetic effect disappears, and the dipole moments are randomly distributed once more.

7.2 Paramagnetism: Electron orbit

Just as the electron has a magnetic moment, so does the orbiting electron around the nucleus. Again a current loop! This effect only gives a minute contribution to the paramagnetism, since it turns out to be much harder to ’turn’ an entire orbit than it is to turn a spin.

Figure 15: Alignment of electron orbits resulting in para-magnetism.

7.3 Diamagnetism

There is a second, more subtle, effect of a magnetic field on the orbit of an electron, for which we will have to look a bit deeper into the centripetal acceleration that holds an atom together. Without a magnetic field the centripetal acceleration is caused by the Coulomb attractive force alone. So:

\frac{1}{4πϵ_{0}} \frac{e^{2}}{R^{2}} = m_{e} \frac{v^{2}}{R}  (99)

Now suppose there is a magnetic field, that is perpendicular to the plane of orbit of the electron and opposite to the direction of the magnetic moment of the atomic orbit (situation shown to the right side of Fig. 16). In that case the centripetal force is sustained by both the Coulomb and the Lorentz forces. Equation 99 now becomes:

\frac{1}{4πϵ_{0}} \frac{e^{2}}{R^{2}} + ev′ = m_{e} \frac{v^{′2}}{R}  (100)

By subtraction of the above two equation, and assuming that the change in velocity is small, we get for the change in velocity (see Griffiths section Figure 16: Changing velocity of the electrons in a magnetic field resulting in diamagnetism.

6.1.3):

v − v′ =≈ \frac{eRB}{2m_{e}}  (101)

When a magnetic field is turned on the electron speeds up and thus the magnetic moment increases. If the magnetic moment would have been in the opposite direction as shown to the left side in Fig. 16, the velocity of the electron would decrease, and thus the magnetic moment would decrease.

In both cases there is a change in magnetic dipole moment that is opposite to the direction of the applied magnetic field. This effect is called diamagnetism. All substances show diamagnetism, but in general it is much weaker than paramagnetism, so in general it is only visible when paramagnetism is absent. In general this was shown to be the case for materials with even numbers of electrons. Furthermore, the calculation as shown here does not always give a reliable result, it is just made to make the argument clear. To get quantitative answers on diamagnetism a full quantum mechanically correct calculation is required.

7.4 Magnetization and Bound Currents

In the previous section we saw the effects of magnetic field on matter on a microscopic scale. Now we will give a macroscopic description of magnetic fields in matter. Let us first consider a cylinder of para-magnetic material that is placed inside a magnetic field (see Fig. 17a). The magnetic field - on average - aligns the magnetic dipoles in the material. So there is an average dipole moment per volume element, which is defined as

M⃗ ≡ \frac{dipole moment}{volume}  (102)

Figure 17: Effect of a magnetic field on a macroscopic object.

We call M⃗ the magnetization of the material. If you remember that a magnetic dipole is in fact nothing else than a little current loop you can have a look on top of the cylinder (see Fig.17b). You see all the current loops lying side-by-side and effectively only a current is running over the edge of our cylinder. So in case we have a uniformly magnetized object, it can be described as if there was a current running over the edge of the object. The magnetization of an object can thus be interpreted as a surface current.

We can make the above argument more quantitative by considering an example in which a slab of para-magnetic material is placed in a magnetic field (see Fig. 18a). We can have a look at the dipole moment for each little block in the slab by multiplying the magnetization by the volume of the little block . This should be equal to the effective current going around the block multiplied by its area (remember that m⃗ = I⃗Area). So:

d|M⃗|Area = I_{block}Area  (103)

Now it can be clearly seen from the figure that neighboring blocks have sides carrying opposite currents, at least in case the magnetization is homogeneous. As a result there is an effective boundary current which happens to be exactly I_{block}. So:

I_{boundary} = I_{block} = |M⃗|d  (104)

Figure 18: Magnetization can effectively be described by bound current densities.

The surface current density K is just the current per unit length:

K = \frac{|M⃗|d}{d} = |M⃗|  (105)

Figure 18b shows the equivalent situation for our magnetized slab of material. In a more general notation our current density can now be written as the cross-product of the magnetization vector and the normal vector to the surface:

K⃗_{mag} = M⃗ × n̂  (106)

Finally if the magnetization is non-homogeneous there will also be an effective volume current density (see Griffiths section 6.2), which is written as:

J⃗_{mag} = ∇ × M⃗  (107)

The vectors K⃗_{mag} and J⃗_{mag} are what we call the bound surface and volume currents, respectively. Please notice that these are some kind of effective currents, since no real charge is moved from one place to another.

7.5 Linear materials

A material is called a linear material its magnetization is proportional to the magnetic field. In such cases M⃗ is written as:

M⃗ = \frac{χ_{m}}{μ_{0}(1 + χ_{m})}B⃗  (108)

where χ_{m} is called the magnetic susceptibility, which is a measured quantity for each material. Please note that B⃗ is not just the external magnetic field, but the field including the modifications due to the magnetization (tricky):

B⃗ = B⃗_{mag} + B⃗_{0}  (109)

where B⃗_{mag} is the field due to the magnetization and B⃗_{0} is the externally supplied field.

7.5.1 Example: Infinite bar

Consider a long para-magnetic (linear material) cylinder inside an externally supplied magnetic field B⃗_{0} as in Fig. 17. The surface bound current is written as:

K_{mag} = M⃗ × n̂  (110)

The magnetic field due o the magnetization can be calculated with Ampere’s law assuming that the cylinder is infinitely long (at this point you should be able to do the calculation yourself, so I don’t do it):

B⃗_{mag} = μ_{0}|K⃗_{mag}|ẑ  (111)

Now we can use Eq. 108 to rewrite this equation in terms of its susceptibility as:

B⃗_{mag} = χ_{m}B⃗_{0}  (112)

So we can see that the total magnetic field inside the cylinder is modified with a factor (1 + χ_{m}).

7.5.2 Example: Infinite bar with free current

We are now considering the infinite cylinder once more, but now we are generating the external magnetic field with a wire running around the cylinder (also see section 5.3). Suppose we have n windings per meter and a free current I then the magnetic field due to the free current is:

B⃗_{0} = nIẑ = K_{free}ẑ  (113)

Now suppose we fill the interior of the cylinder with a linear magnetic material with a susceptibility χ_{m}, then we know from the previous example that:

B⃗ = (1 + χ_{m})B⃗_{0} = (1 + χ_{m})nIẑ  (114)

So by inserting some material inside a magnet we have a way of amplifying (para-magnetic) or damping (dia-magnetic) the resulting field.

7.6 Ferromagnetism

Finally we come to the special case of ferromagnetism. In ferro-magnetic materials electron spins really love to align with each other. Whereas for the normal paramagnetic materials the alignment was far from perfect, for a ferromagnetic material the alignment can be almost 100%. Usually there are little areas - so called Weiss domains - in which the spins are aligned (see Fig. 19), and usually the effect of these areas cancel out giving no noticeable external magnetic field. However, once an external magnetic field is applied

Figure 19: In a ferro-magnetic material there are domains where spins spontaneously align to each other. These are the so called Weiss areas.

all these domains can spontaneously align themselves to the magnetic field. A small externally applied field can then result in a huge magnetic field from the material. Another interesting property of ferromagnetism is what happens when the external magnetic field is switched off. The alignment of spins might become less, but it does not disappear: you have created a permanent magnet!

If for example our solenoid from the previous example is filled with a ferromagnetic material, the resulting B⃗ field may be 100 times larger than the externally applied field. The effect is highly non-linear as illustrated in Fig. 20, which shows the magnetic field as a function of the externally applied field. First the B⃗ field grows rapidly as a function of the external field. Then at a certain point the dipole moments of all domains are aligned and there is no further increase in B⃗ (saturation). If the external field is removed, there is still quite some magnetic field remaining due to the ferromagnetism. Only once the external field is fully reversed, the B⃗ field finally changes sign.

7.7 Knowledge and Skills

The knowledge and skills you should have acquired during reading of the previous can be summarized as follows:

• In para-magnetic material the magnetization is in the same direction as the externally applied magnetic field; Figure 20: If the external magnetic field is removed from a ferro-magnetic material the magnetization does not directly disappear. This non-linear behavior is called hysteresis.

• In dia-magnetic materials the magnetization is in the opposite direction as the externally applied magnetic field;

• A magnetized object can be effectively described with bound surface current density K⃗_{mag} and a volume current density J⃗_{mag}, with:

K⃗_{mag} = M⃗ × n̂  (115)
J⃗_{mag} = ∇ × M⃗  (116)

;

• Linear materials are materials for which the magnetization is proportional to the magnetic field:

M⃗ = \frac{χ_{m}}{μ_{0}(1 + χ_{m})}B⃗  (117)

where χ_{m} is the magnetic susceptibility of the material.