# 物理代写|电动力学代写electromagnetism代考|ELEC3104

## 物理代写|电动力学代写electromagnetism代考|SOME APPLICATIONS

In general, electroconduction is the most familiar of all the field systems we have considered. Some typical applications include resistors, conducting wire, fuse wire and heating elements in electric fires, cookers and immersion heaters. As these examples are so familiar, we will not consider them. Instead, we will examine the formation of a resistor in semiconductor material.

When designing analogue circuits, resistors are often used. If the circuits are fabricated on printed circuit board, the designer can choose between wire-ended or surface-mount resistors. When producing an analogue integrated circuit, designers have no choice but to fabricate the resistors out of semi-conductor material.

Figure $4.8$ shows the basic form of an integrated circuit (i.c.) resistor. It is formed by depositing electrodes on to the silicon and doping the material to the required conductivity. (Doping involves diffusing impurities into the silicon to alter the material conductivity.)

When working with i.c. resistors, the surface resistance of the material is used. If we consider a square of material, with electrodes on opposite sides, the resistance will be given by
$$R=\frac{l}{\sigma \times \text { area }}$$
where $l$ is the length of one side of the square, $\sigma$ is the material conductivity and area is the cross-sectional area of the resistor.

## 物理代写|电动力学代写electromagnetism代考|FORCE FIELDS

In all the field systems we have considered, we started from the basic premise that like charges or poles repel and unlike charges or poles attract. This led to the observation that the force decreases as the square of the distance between the charges or poles, i.e.,
Electrostatics
$$F=\frac{q_1 q_2}{4 \pi \varepsilon r^2}$$
Magnetostatics
$$F=\frac{p_1 p_2}{4 \pi \mu r^2}$$
Electroconduction
$$F=\frac{q_1 q_2}{4 \pi \varepsilon r^2}$$
As we can see, there is a great deal of similarity in the general forms of these equations – they all describe a force field of some sort. We should note that these equations assume point sources and, as we have seen with magnetism, this can be open to question. In spite of this fact, this does help us to visualize the field systems.
Once we accept that we are studying a force field, it is a simple step forward to define field strength as the force on a unit charge or pole. Thus,
Electrostatics
$$E=\frac{q_1}{4 \pi \varepsilon r^2}$$

Magnetostatics
$$H=\frac{p_1}{4 \pi \mu r^2}$$
Electroconduction
$$E=\frac{q_1}{4 \pi \varepsilon r^2}$$
Let us take a moment to study these equations in detail. We should be fairly happy with the force field in an electrostatic system – we have all rubbed a balloon and felt the hairs on the back of our hands stand up. Anyone who has played with permanent magnets has seen the effect of the force field in a magnetic system. However, the idea of a force field in electroconductive systems is rather difficult to accept. This is because we are generally taught, at an early stage, a model of current flow that has the potential of a source as the force that drives current around a circuit. This idea is often reinforced by the term electromotive force for the potential of a source. (A water analogy is also often used. Although this analogy can be useful, its pull is very seductive and we must be very careful.) As this model is very simple, it is generally very hard to accept that it is wrong. Potential does not force current around a circuit: it is the electric field set up in a conductor that causes current to flow.

So, each field system has a force field that repels or attracts charges or poles. This raises the question of what radiates from the charge or pole generating the force field. This is where we introduce the ideas of flux and flux density.

## 物理代写|电动力学代写electromagnetism代考|SOME APPLICATIONS

$$R=\frac{l}{\sigma \times \text { area }}$$

## 物理代写|电动力学代写electromagnetism代考|FORCE FIELDS

$$F=\frac{q_1 q_2}{4 \pi \varepsilon r^2}$$

$$F=\frac{p_1 p_2}{4 \pi \mu r^2}$$

$$F=\frac{q_1 q_2}{4 \pi \varepsilon r^2}$$

$$E=\frac{q_1}{4 \pi \varepsilon r^2}$$

$$H=\frac{p_1}{4 \pi \mu r^2}$$

$$E=\frac{q_1}{4 \pi \varepsilon r^2}$$

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