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\begin{bf}
\noindent
Institutt for fysikk, NTNU

\noindent
TFY4155/FY1003: Elektrisitet og magnetisme

\noindent
Spring 2005

\end{bf}

\begin{center}
\begin{large}
Summary, week 11 (March 15 and 16)
\end{large}
\end{center}

\vspace{0.5cm}
\noindent
{\bf Electric current}

\noindent
[FGT 26.1;  YF 25.1; TM 25.1; AF 24.1, 24.2; LHL 21.1; DJG 5.1.3]

\vspace{0.3cm}
\noindent
Electric current = amount of (positive) charge that passes through a cross section of
a conductor pr unit time.

\noindent
In a {\em metal}, the mobile charges are {\em electrons}, with charge $-e$. Then the particle
current and the electric current go in opposite direction.

\vspace{0.2cm}
\noindent
With charge $\Delta Q$ passing a cross section $A$ in time $\Delta t$:
$$I = \frac{\Delta Q}{\Delta t} \stackrel{\Delta t \rightarrow 0}{\rightarrow}
\frac{dQ}{dt}$$

\vspace{0.2cm}
\noindent
Unit for current: $[I] = [Q/t] =$ C/s = A (ampere)

%%FIGURE
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\centerline{\epsfxsize=3.0in \epsfbox{uke42_fig1.eps}}
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%%END FIGURE

\vspace{0.2cm}
\noindent
With $n=\Delta N/\Delta V$ mobile charges pr unit volume, each with mean
{\em drift velocity} $\bold{v}$ and charge $q$:
$$\Delta Q = q\Delta N = nq\Delta V = nq \Delta x A$$
$$\Rightarrow I = \frac{\Delta Q}{\Delta t} = nqA \frac{\Delta x}{\Delta t} = nqAv$$

\vspace{0.2cm}
\noindent
Current density = current pr unit area:
$$j = \frac{I}{A}$$
Hence:
$$j = nqv$$
Both current density $j$ and drift velocity $v$ are vectors:
$$\bold{j} = nq\bold{v}$$

\vspace{0.2cm}
\noindent
If we also consider the cross section $A$ as a vector, then $I$ becomes a {\em scalar} quantity:
st{\o}rrelse:
$$I = \bold{j} \cdot \bold{A}$$
The current $I$ then has direction only relative to the conducting wire (positive or negative).

\vspace{0.2cm}
\noindent
Generalization, if $\bold{j}$ is not constant over the cross section of the conductor:
$$I = \int_S \bold{j} \cdot d\bold{A}$$

%%FIGURE
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\vspace{1.0cm}
\noindent
{\bf Ohm's "law"} 

\noindent
[FGT 26.3;  YF 25.2,25.3; TM 25.2; AF 24.3, LHL 21.2, DJG 7.1.1]

\vspace{0.2cm}
\noindent
We need a {\em driving force} $\bold{F}$ to obtain a current through the conductor. If
$$I \sim \bold{F} \sim \bold{E} \sim V$$
(i.e.: if $I$ is proportional to the driving force, and hence proportional to
the electric field, and hence also proportional to the potential difference $V$)
then we have so-called {\em linear response}:
$$I = \frac{1}{R} V$$
$$\Rightarrow V = R I$$
which is Ohm's law.

\vspace{0.2cm}
\noindent
Unit for {\em resistance}: $[R] = [V/I] =$ V/A = $\Omega$ (ohm)

\vspace{0.2cm}
\noindent
Ohmic materials: Obey Ohm's law for large variations in $I$.

\noindent
Non-ohmic materials: Considerable deviations from linear relation between $I$ and $V$.

\vspace{0.2cm}
\noindent
Circuit symbol for resistor:

%%FIGURE
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%%END FIGURE


\vspace{1.5cm}
\noindent
{\bf Electric conductivity}

\noindent
[FGT 26.2,26.3;  YF 25.2,25.3; TM 25.2; AF 24.4, LHL 21.2, DJG 7.1.1]

\vspace{0.2cm}
\noindent
Random movement (diffusion) of mobile charges through the conductor, {\em plus}
a net drift due to the field $\bold{E}$

%%FIGURE
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%%END FIGURE

\vspace{0.2cm}
\noindent
Mean drift velocity along $-\bold{E}$: \hspace{0.2cm} $\bold{v}$

\noindent
Particle velocity associated with the temperature in the conductor: $v_T \sim \sqrt{\frac{3 k_B T}{m}} \gg v$

\vspace{0.2cm}
\noindent
{\small Comment: A correct quantum mechanical description of the electrons in a metal will in fact
result in an even larger particle velocity. This is because electrons are a type of elementary
particles called {\em fermions} that obey the so-called {\em Pauli principle}, saying that it is not
possible to have more then a single fermion in each allowed "state". This forces electrons into states
with higher energy than what they would have had if they were classical particles. You can learn more
about this in courses on quantum mechanics and solid state physics and so on!}

\vspace{0.2cm}
\noindent
For ohmic material: $\bold{v} \sim \bold{E}$ 

\vspace{0.2cm}
\noindent
This gives a linear relation between current density and electric field:
$$\bold{j} = \sigma \bold{E}$$
which defines the {\em conductivity} of the material, $\sigma$. This is also Ohm's law.

\vspace{0.2cm}
\noindent
A conducting wire of length $l$, (constant) cross section $A$ and conductivity $\sigma$ has resistance
$$R = \frac{1}{\sigma} \frac{l}{A}$$
Proof:
$$I = jA = \sigma E A = \sigma \frac{V}{l} A = \frac{1}{R} V \;\; \Rightarrow \;\; R = \frac{l}{\sigma A}$$
Here, we have assumed that $E$ is constant in the whole conductor (which is OK, see for example
Griffiths, Example 7.3), and therefore equal to the voltage drop across the length of the conductor, $V$,
divided by the length $l$. The final equality in the equation above is simply Ohm's law, i.e., the definition of $R$.

\vspace{0.2cm}
\noindent
We may now introduce {\em conductance}:
$$G = \frac{1}{R}$$
and {\em resistivity}:
$$\rho = \frac{1}{\sigma}$$

\vspace{0.2cm}
\noindent
$\sigma$ and $\rho$ are material constants

\noindent
$R$ and $G$ also depend on the size and shape of the conductor

\vspace{0.2cm}
\noindent
Units:

\noindent
$[G] = \Omega^{-1}$

\noindent
$[\sigma] = [l/RA] = \Omega^{-1}$ m$^{-1}$

\noindent
$[\rho] = \Omega$ m


\vspace{1.0cm}
\noindent
{\bf The temperature dependence of $\rho$}

\noindent
[FGT 26.3; YF 25.2; TM 25.2; LHL 21.2]

\vspace{0.2cm}
\noindent
Increase in temperature $T$ results in stronger lattice vibrations and thereby
more frequent collisions between the electrons and the lattice. This results in
a reduced drift velocity $v$ and a reduced conductivity $\sigma$, i.e., an increased resistivity $\rho$.

\noindent
Empirically, we have, over some temperature interval, for metals:
$$\rho(T) = \rho_0 [1 + \alpha (T-T_0)]$$
Here, $T_0$ is a chosen reference temperature, e.g., 300K, $\rho_0 = \rho(T_0)$ is the
measured resistivity at temperature $T_0$, and $\alpha$ is the measured temperature
coefficient, i.e., the slope of $\rho(T)/\rho_0$ plotted as a function of $T$.


\vspace{1.0cm}
\noindent
{\bf Electric effect}

\noindent
[FGT 26.7;  YF 25.5; TM 25.3; AF 24.5, LHL 22.2, DJG 7.1]

\vspace{0.2cm}
\noindent
Change in potential energy, $\Delta U$, for a charge $\Delta Q$ that goes through a voltage
drop $V$:
$$\Delta U = \Delta Q \cdot V$$

\noindent
Energy conservation:

\noindent
Without collisions: We would get acceleration of the mobile charges, and thereby increased
kinetic energy.

\noindent
With collision (which we actually have!), i.e., resistance $R$: $\Delta U$ is "lost" as
{\em heat} in the resistor.

\vspace{0.2cm}
\noindent
Lost effect = lost energy pr unit time:
$$P = \frac{\Delta U}{\Delta t} = V \frac{\Delta Q}{\Delta t} = V \cdot I$$
If we have an ohmic material (i.e. $V = RI$):
$$P = RI^2 = \frac{V^2}{R}$$
Unit for effect:
$$[P] = [\frac{U}{t}] = \frac{{\rm J}}{{\rm s}} = {\rm W} \;\; {\rm (watt)}$$


\newpage
\noindent
{\bf Several resistors coupled together}

\noindent
[FGT 26.4; YF 26.1; TM 25.4; AF 24.6, LHL 21.3]

\vspace{0.2cm}
\noindent
$N$ resistors $R_i$, $i = 1,..,N$ connected in series:
$$R = \sum_{i=1}^{N} R_i$$
$N$ resistors $R_i$, $i = 1,..,N$ connected in parallel:
$$\frac{1}{R} = \sum_{i=1}^{N} \frac{1}{R_i}$$

\vspace{0.2cm}
\noindent
From {\o}ving 9, {\bf several capacitors coupled together}

\noindent
[FGT 25.4; YF 24.2; TM 24.4; AF Ex. 25.8, LHL 20.2]:

\vspace{0.2cm}
\noindent
$N$ capacitors $C_i$, $i = 1,..,N$ connected in series:
$$\frac{1}{C} = \sum_{i=1}^{N} \frac{1}{C_i}$$
$N$ capacitors $C_i$, $i = 1,..,N$ connected in parallel:
$$C = \sum_{i=1}^{N} C_i$$

\vspace{0.3cm}
\noindent
In these expressions, $R$ and $C$ represent, respectively, the equivalent resistance
and equivalent capacitance if we replace all the elements connected in series or parallel
with a single resistor or capacitor.


\vspace{2cm}
\noindent
A couple of comments!
\begin{itemize}
\item
Didn't we agree that inside an electric conductor, the electric field is zero? Yes,
but then we had {\em electrostatic equilibrium}! If a current runs through the conductor,
we have no longer electrostatic equilibrium. And if we don't have electrostatic equilibrium,
the electric field doesn't have to be zero any more.
\item
In the lectures, I stated, without proof, that if we have a (straight) wire with constant cross section,
carrying a {\em stationary} (i.e., time independent) electric current $I$, then the electric field
$\bold{E}$ is {\em uniform} everywhere inside the conductor. I won't prove this here.
(But see example 7.3 in Griffiths, if you are interested.) However, consider the
{\em consequences} of having a uniform electric field inside such a conductor. This means that
there can be no net charge anywhere inside the conductor, just as we found earlier (in
electrostatic equilibrium). A uniform electric field means that no matter what kind of volume
element we choose, big or small, the same amount of electric flux passes in and out of the volume
element, and hence, by Gauss' law, the net charge inside is zero. Conclusion: The potential difference
across the length of the conductor, and the uniform electric field inside the conductor, are created
by charges that must be on the surface of the conductor. Exactly how these charges distribute
themselves, well that's not so easy to tell.
\item
Different materials have very different values of the resistivity. Examples:
Silver has $\rho = 1.59 \cdot 10^{-8}$ while diamond has $\rho = 2.7$ (both in the unit
$\Omega$m and at room temperature). Various types of glass have typically resisitivity
values in the range $10^{10} - 10^{14}$. Earlier, we treated insulators like glass as
materials with no mobile charges, and therefore zero conductivity, i.e., infinite
resistivity. So, this is only almost true! At "finite" temperature (i.e., not zero temperature),
there will be a few electrons that are free to move around and may contribute to electric
current if we apply a voltage. But the conductivity of glass is really small! The ratio of
the resistivity of glass and silver can be as high as $10^{22}$. A resistor in an electric
circuit is typically made of a material with considerably higher resistivity than the
metal in the connecting wires. We may therefore, with good approximation, consider the
metallic connecting wires as equipotentials, i.e., with zero voltage drop across them,
and thereby also zero electric field inside them. Since we have the relation
$\bold{j} = \sigma \bold{E}$, we see that zero electric field together with a non-zero
current density must imply $\sigma \rightarrow \infty$. Then we say that we have a
{\em perfect conductor}. And correspondingly, a "perfect insulator" means a material
with infinite resistivity, or $\sigma = 0$. Then we see that $j=0$ {\em always}, even 
if $E \ne 0$.
\end{itemize}


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