The goal of this course is to provide an understanding of the behaviour of em walves in real materials and in artificial structures (such as waveguides and antennae).

Practically all applications of em thoery in the real world involve materials and structures.

The course:

- Revision of vacuum theory.
- The effect of materials on electric fields (here we shall introduce the
D -field). - The effect of materials on magnetic fields (introducing the
H -field). - Rederive Maxwell's equations in presence of materials.
- Propagation of em waves in idealised and in real materials.
- Behaviour of em waves at interfaces (e.g. reflection).
- Propagation of em waves in man made structures (e.g. cables, waveguides).
- Transmission and reception of em waves (aerials).

Elements of Electromagnetism by M N O Sadiku Oxford University Press

- Electric Field
E - Electrostatic Potential V
- Electric Dipole
p - Electric flux and the Gauss theorem in electrostatics

- Electromotive force (EMF)
- Current I
- Current density
J

- Magnetic dipoles (
m ) - The magnetic
B -field - Magnetic flux and the Gauss theorem in magnestatics
- Magnetic fields set up by electrical currents
- The Biôt-Savart law
- The Ampère Law

- The Faraday-Lenz law

- The displacement current
- Maxwell's equations
- The em wave equation (radio, light, ...)

grad | scalar field | ∇V |

div | vector field | ∇. |

curl | vector field | ∇× |

∇^{2} | scalar | ∇^{2}V |

Throughout this section, we will be assuming that everything is in a vacuum.

We define the

Force on a charge q at X is (Qq)/(4πε_{0}r^{2}).

ε_{0} = permittivity of free space.

Hence, the

The lines of an electric field

The work done in moving test charge q in the presence of the charge Q from infinity to an arbitrary point is the potential V of this arbitary point.

_{ext} is the force an external agent requires to hold charge q in position against force

When q moves by d_{ext} is:

But F_{ext} = F = qE, so:

Hence, in qoing from A to B, the work done by the external agent is (by adding together the work done for each little bit d

But E = Q/(4πε_{0}r^{2}), so:

...which is independent of the route taken (and is thus a conservative force).

If q is a unit charge,

...which we write as:

...where

...which is the potential at the point concerned, and is the work required to bring a unit test charge (q=1) from ∞ (r_{A}=∞) to the point concerned (r_{B}).

for unit charge

Suppose A and B coincide (close loop). Then,

$V$Hence,

This is only correct in electrostatics.

From earlier:

$dW$If q=1, dW_{ext} = dV

Hence,

$dV\; =\; -E$i.e.

Since

Two point charges +q and -q separated by a distance a constitute a dipole.

Usually we assume a is small.

The dipole moment

The potential at X due to a dipole is given by:

From this diagram one sees that:

But, if you drop a perpendiculous from +q to r_{-}, and assume that r>>a, you can approximate that r_{-} - r_{+} ≅ a cos θ and r_{-}r_{+} ≅ r^{2}, so:

...where ds = rdθ is a deplacement in the tangential direction.

Hence:

To find the electric field

Hence:

Couple (Torque) acting on a dipole is denoted by the greek letter Γ and is:

$\Gamma \; =\; (Eq)\times (a\; sin\; \alpha )\; \Gamma \; =\; pE\; sin\; \alpha $In vector notation:

$\Gamma =p\times E$Electric Flux passing through dS is defined as:
(normal Component of E) × (area)
= E_{n} dS
= E cos α dS
=

For a large surface, the flux passing through a surface is:

...were E may be a function of the position on dS.

For a closed surface we can show that:

i.e. Electrical Flux leaving a closed surface is equal to the charge enclosed divided by ε_{0}

This is the Gauss Theorem.

ρ(v) = charge per unit volume at the point concerned.

Clearly, the charge enclosed is given by:

The Gauss Theorem becomes:

The divergence div(

Let's consider the divergence of the electrical field

Flux of _{x} dydz (negative because leaving in -x direction)

At RHFACE, the field is (E_{x}+dE_{x}), so the flux leaving RHFACE = (E_{x}+dE_{x})dydz

Adding these two expressions, we get the net flux leaving in x-direction:

$dE$...where dE_{x} is:

Substituting:

Similarly for other directions, giving:

The flux leaving per unit volume is that divided by dv, thus:

...which can be written ∇.

Now the net flux leaving is equivalent to the charge enclosed divided by ε_{0}:

...giving:

$divE=\; \rho /\epsilon $Consider a charged capacitor:

Inside the capacitor there is an electric field E_{Q} due to the charges.

(This is all electrostatics -- no moving charges.)

Consider now an electrochemical cell.

Owing to chemical interactions, - charges are driven to A, and + charges are driven to B.

This continues until the electric field E_{Q} due to charge separation balances the field set up by the chemical interaction.

Therei s then no net electric field inside the cell.

The A terminal is at a higher potential than terminal B (as evidenced by the fact that current flows).

The difference V_{A} - V_{B} is called the EMF ε of the cell (on an open circuit).

But:

...because the two fields cancel each other inside the cell. Hence:

The general definition of the EMF in a circuit is therefore:

Note that the circuit could be fictional, this is just any mathematical closed path.

There are only magnetic dipoles — e.g. small permanent magnets, small coils of current carrying wire. — no magnetic monopoles. (At least, not in the macroscopic world.)

For small coils carrying a circuit I the magnetic dipole moment is given by:

$m\; =\; I\; \times \; area$Direction is that in which a freely suspended dipole points (a compass). Magnitude is obtained by measuring the torque needed to hold the compass needle at π/2 rads to the field.

If α = π/2, Γ = mB, ∴ B = Γ/m (where m is the magnetic dipole magnitude).

In E-fields,

...and by analogy, the magnetic dipole case is similar:

μ_{0} = permeability of free space.

Magnetic flux passing through d_{n}dS = B cos α dS =

Magnetic flux leaving a closed surface is zero.

What is the field d

We find that:

...and it is into the diagram at the point B, so:

We can integrate this law to give:

B = μ_{0}nI inside (uniform and independent of radius).

The line integral of B around a closed path is equal to μ_{0} times the current enclosed.

This is still just for a vacuum.

Current passing through area d

The lines of B are circles, by symmetry. Around a circle of radius h, B is a constant. Therefore:

But ∫_{0} × (current enclosed) = μ_{0}I
∴ B 2πh = μ_{0}I
and B = μ_{0}I/(2πh) as before.

We perform the line integral in steps.

Using the Ampère theorem,

We perform the line integral in steps.

$\begin{array}{ccc}1\to 2& {\int}_{1}^{2}B.dl=Bl& \\ 2\to 3& {\int}_{2}^{3}B.dl=0& Bat\frac{\pi}{2}todl\\ 3\to 4& {\int}_{3}^{4}B.dl=0& B=0\text{outside}\end{array}$Thus, if we make ${Z}_{s}=\frac{{Z}_{0}^{2}}{{Z}_{1}}$, all power from the source enters the line.

We may approximate: $B={\mu}_{r}{\mu}_{0}H$ $D={\epsilon}_{r}{\epsilon}_{0}E$ (linear approximation).

Mnemonic: DeeDee.

Mnemonic: p-e-chi-e. Pixie.

Mnemonic: mmm-chi-'h. Mickey.

Mnemonic: Neh.

The EMF is given by:

$\epsilon =\frac{-\partial \Phi}{\partial t}$where

$\Phi =\iint B.ds$ $\epsilon =\frac{-\partial \Phi}{\partial t}$