\documentclass[12pt]{article}
\usepackage{epsfig}
\topmargin=-0.5cm
\oddsidemargin=-0.5cm
\evensidemargin=-0.5cm
\textheight=23cm
\textwidth=17cm
\begin{document}
\newcommand{\bold}[1]{\mbox{\boldmath $#1$}}
\newcommand{\ems}{{\cal E}}
\begin{bf}
\noindent
Institutt for fysikk, NTNU

\noindent
TFY4155/FY1003: Elektrisitet og magnetisme

\noindent
Spring 2005

\end{bf}

\begin{center}
\begin{large}
Summary, week 14 (April 5 and 6)
\end{large}
\end{center}


\vspace{0.5cm}
\noindent
{\bf Magnetic interaction} 

\noindent
[FGT 28, 29; YF 27, 28; TM 26, 27; AF 22, 24B; LHL 23; DJG 5]

\vspace{1.0cm}
\noindent
{\bf Magnetism as relativistic phenomenon (orientational only)} 

\noindent
[DJG 12.3.1]

\vspace{0.2cm}
\noindent
See "Notater."

\vspace{0.5cm}
\noindent
{\bf Charged particle in uniform magnetic field}

\noindent
[FGT 28.3; YF 27.4; TM 26.2; AF 22.3; LHL 23.1, 23.4; DJG 5.1.2]

\vspace{0.2cm}
\noindent
Force on charge $q$ with velocity $\bold{v}$ in magnetic field $\bold{B}$:
$$\bold{F} = q\bold{v}\times\bold{B}$$

\vspace{0.2cm}
\noindent
With angle $\theta$ between $\bold{v}$ and $\bold{B}$:
$$F = qvB\sin\theta$$
If $\bold{v}\perp\bold{B}$:
$$F = qvB$$

\vspace{0.2cm}
\noindent
We always have $\bold{F}\perp\bold{v}$ and $\bold{F}\perp\bold{B}$.
When $\bold{F}\perp\bold{v}$, the particle trajectory becomes a {\em circle} with
constant $v = |\bold{v}|$:
\begin{eqnarray}
& &\bold{F}\perp\bold{v} \;\;\Rightarrow\;\; \bold{F}\perp\frac{d\bold{l}}{dt} \nonumber \\
&\Rightarrow& dW = \bold{F} \cdot d\bold{l} = 0 \nonumber
\end{eqnarray}
i.e., $\bold{F}$ performs zero work
$$\Rightarrow v = \; {\rm konstant} \; \; {\rm og} \;\; T = \frac{1}{2}mv^2 = \; {\rm konstant}$$

\vspace{0.2cm}
\noindent
Centripetal acceleration:
\begin{eqnarray}
a &=& \frac{v^2}{r} \nonumber \\
\Rightarrow F &=& ma = m\frac{v^2}{r} = qvB \nonumber \\
\Rightarrow r &=& \frac{mv}{qB} \nonumber
\end{eqnarray}
where $r$ is the radius of the circular path.

\vspace{0.2cm}
\noindent
The {\em angular frequency} of the circular movement:
$$\omega = \frac{v}{r} = \frac{qB}{m} \equiv \omega_c \;\;\; {\rm (syklotronfrekvensen)}$$
Angular frequency = angle pr unit time

\vspace{0.2cm}
\noindent
Frequency = number of full circles performed pr unit time:
$$f = \frac{\omega}{2\pi}$$

\vspace{0.2cm}
\noindent
Period = time for one circle:
$$T = \frac{1}{f} = \frac{2\pi}{\omega}$$

\vspace{0.5cm}
\noindent
With both electric field $\bold{E}$ and magnetic field $\bold{B}$ present,
the charge is influenced by the {\em Lorentz force}:
$$\bold{F} = q\bold{E} + q\bold{v}\times\bold{B}$$
We see that the unit for magnetic field must be
$$[B]=\frac{[F]}{[qv]}=\frac{{\rm N}}{{\rm C}{\rm m/s}}$$
In the SI system, this has its own unit:
$$[B] = {\rm T}$$
or {\em tesla}. An alternative unit for magnetic field is {\em gauss} (G).
1 tesla equals 10000 gauss. The earth magnetic field is ca 0.5 G, so a magnetic field
of 1 T is pretty high.


\vspace{1.0cm}
\noindent
{\bf Magnetic field from electric current}

\noindent
[FGT 29.4; YF 28.2; TM 27.2; AF 24.11; LHL 23.5; DJG 5.2]

\vspace{0.3cm}
\noindent
The Biot--Savart law (empirical, i.e., experimentally found):

%%FIGURE
\vspace{0.5cm}
\centerline{\epsfxsize=3.0in \epsfbox{uke44_fig1.eps}}
\vspace{0.5cm}
%%END FIGURE

\vspace{0.2cm}
\noindent
The contribution $d\bold{B}$ to the magnetic field in the point in distance
$\bold{r}$ from the wire element $d\bold{l}$, when the wire carries a stationary
(i.e. time independent) current $I$, is
$$d\bold{B} = \frac{\mu_0}{4\pi} \frac{I\;d\bold{l}\times\hat{r}}{r^2} = 
\frac{\mu_0}{4\pi} \frac{I\;d\bold{l}\times\bold{r}}{r^3}$$
The superposition principle is valid for the magnetic field, so the magnetic field
from a complete closed current loop becomes
$$\bold{B} = \oint d\bold{B} = \frac{\mu_0 I}{4\pi}\oint\frac{d\bold{l}\times\hat{r}}{r^2}$$
Since charge cannot be created and does not disappear, one (typically) needs a closed loop
to maintain a constant electric current $I$.

\vspace{0.5cm}
\noindent
In principle, we may use the Biot--Savart law to find the magnetic field in an arbitrary position,
due to an arbitrary current loop.

\noindent
In practice, we are only able to solve the integral in the Biot--Savart law analytically for
certain special cases, e.g., straight conductor, or on the symmetry axis of a circular or
quadratic current loop etc.

\noindent
(Problems that cannot be solved analytically may be solved with numerical methods.)

\vspace{1.0cm}
\noindent
{\bf Magnetic field lines}

\noindent
[FGT 29.2; YF 27.3; TM 26.1; LHL 23.1]

\vspace{0.3cm}
\noindent
For visualizing the magnetic field in a region. These are defined just like we did
for electric field lines:
\begin{itemize}
\item Direction: $\bold{B}$ parallell to the field lines everywhere.
\item Strength: $|\bold{B}|$ proportional to the density of field lines (i.e. number of
field lines pr unit area)
\end{itemize}
NOTE that we always have {\em closed field lines} for $\bold{B}$. That is because there
are no magnetic {\em monopoles} in nature. (Of course, {\em electric} monopoles, i.e.,
negative and positive charges do exist!)

\vspace{0.2cm}
\noindent
However, nature consists of...

\vspace{1.0cm}
\noindent
{\bf Magnetic dipoles} 

\noindent
[FGT 28.5, 29.4; YF 27.7; TM 26.3; AF 22.7; LHL 23.3, 26.2; DJG 5.4.3]

\vspace{0.3cm}
\noindent
Current loop = Magnetic dipole:

%%FIGURE
\vspace{0.5cm}
\centerline{\epsfxsize=3.5in \epsfbox{uke44_fig2.eps}}
\vspace{0.5cm}
%%END FIGURE

\noindent
Magnetic dipole moment (for {\em plane} current loop):
$$\bold{m} = I \bold{A} = I A \hat{n}$$
Unit: $[m] = [IA]$ = Am$^2$

\vspace{1.0cm}
\noindent
{\bf Magnetic dipole moment of elementary particles and atoms}

\noindent
[FGT 31.2; YF 28.8; TM 27.5; AF 22.7, 23.7; LHL 26.2; DJG 5.4.3, Problem 5.56]

\vspace{0.3cm}
\noindent
Classical picture of atom: Electrons in (circular) orbits around the nucleus.

\vspace{0.3cm}
\noindent
Current $I$ in the circular path for charge $q$ with velocity $v$ in circular path
with radius $r$:
$$I = \frac{q}{\tau} = \frac{q}{2\pi r/v} = \frac{qv}{2\pi r}$$
where $\tau$ = time needed for one cycle (i.e. the period).

\noindent
Enclosed area is:
$$A = \pi r^2$$
Magnetic dipole moment becomes:
$$|\bold{m}| = IA = \frac{qv}{2\pi r} \cdot \pi r^2 = \frac{1}{2} qvr$$
May be expressed in terms of the particle's {\em orbital angular momentum} $\bold{L}$:
$$\bold{L} = m \bold{r} \times \bold{v}$$
where $m$ = mass of the particle.

%%FIGURE
\vspace{0.5cm}
\centerline{\epsfxsize=2.6in \epsfbox{uke44_fig3.eps}}
\vspace{0.5cm}
%%END FIGURE


\noindent
For circular path, $\bold{r} \perp \bold{v}$, so that
$$L = |\bold{L}| = mrv$$
Hence:
$$|\bold{m}| = \frac{1}{2}qvr = \frac{q}{2m}L$$
On vector form:
$$\bold{m} = \frac{q}{2m} \bold{L}$$
For electron (with $q=-e$ and $m = m_e$):
$$\bold{m} = -\frac{e}{2m_e}\bold{L}$$

%%FIGURE
\vspace{0.5cm}
\centerline{\epsfxsize=2.7in \epsfbox{uke44_fig4.eps}}
\vspace{0.5cm}
%%END FIGURE

\vspace{0.3cm}
\noindent
Electrons (and protons and neutrons etc) also have "inner angular momentum",
i.e. {\em spin} $\bold{S}$.

\noindent
Classical picture of spin for electron: Rotating charged sphere. Classically, we expect
contribution
$$-\frac{e}{2m_e}\bold{S}$$
from the spin movement to the total magnetic dipole moment of the electron.

\vspace{0.2cm}
\noindent
{\em Qualitatively}, these classical pictures are OK: Elementary particles like electrons,
protons and neutrons do have magnetic dipole moment $\bold{m}$ that can be expressed in
terms of the total angular momentum $\bold{J} = \bold{L} + \bold{S}$ = the vector sum of
the orbital angular momentum $\bold{L}$ and the spin $\bold{S}$.

\vspace{0.2cm}
\noindent
{\em Quantitatively}, the classical model breaks down: It is totally necessary with a
{\em quantum mechanical} description in order to calculate the magnetic dipole moment
of the various particles.

\vspace{0.3cm}
\noindent
We cannot go into the quantum mechanical description here. However, we may have convinced
ourselves that atoms are in face small current loops, and therefore small magnetic dipoles.
In other words, matter around us consists of many small atomic magnetic dipoles:

%%FIGURE
\vspace{0.5cm}
\centerline{\epsfxsize=3.0in \epsfbox{uke44_fig5.eps}}
\vspace{0.5cm}
%%END FIGURE


\vspace{1.5cm}
\noindent
Earlier in this course, we have studied electric dipoles with dipole moment $\bold{p}$,
and we have seen how these are subject to a torque $\bold{\tau} = \bold{p}\times\bold{E}$
in an electric field $\bold{E}$, so that they are aligned along the external field. We have seen
that this is relevant for understanding how dielectric materials behave in an electric field,
because a dielectric consists of atomic or molecular electric dipoles, i.e., atoms or molecules
with electric dipole moment $\bold{p}$.

\vspace{0.5cm}
\noindent
Later, we will see how magnetic dipoles with magnetic dipole moment $\bold{m}$ in the same way
are subject to a torque $\bold{\tau} = \bold{m}\times\bold{B}$ in a magnetic field $\bold{B}$,
and are aligned along the external field. This is relevant for understanding how materials
behave in a magnetic field, because matter consists of atomic magnetic dipoles, i.e., atoms (or
molecules) with magnetic dipole moment $\bold{m}$.

\vspace{0.3cm}
\noindent
{\em Magnetization} and magnetism is in magnetostatics with {\em polarization} is in electrostatics.

\vspace{0.3cm}
\noindent
Week 15: Force on electric current in magnetic field. (Necessary to understand this if we want to
understand how a magnetic dipole is influenced by a magnetic field.) Also, Ampere's law, which may
be used in cases with "suitable symmetry" to find the magnetic field very easily.

\end{document}
