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

\noindent
TFY4155/FY1003: Elektrisitet og magnetisme

\noindent
V{\aa}r 2005

\end{bf}

\begin{center}
\begin{large}
{\O}ving 13
\end{large}
\end{center}

\noindent
Guidance: Thursday April 14 and Friday April 15

\noindent
To be delivered by: Monday April 18

\vspace{0.7cm}
\noindent
{\em Exercise 1}

\vspace{0.2cm}
\noindent
In the lectures, we showed that atoms may be viewed as small current loops, i.e.,
as small magnetic dipoles with magnetic dipole moment $\bold{m} = I \bold{A}$,
where the current $I$ goes in a loop which encloses a (planar) area $A$.
("The vector area" is then $\bold{A} = A\;\hat{n}$, where $\hat{n}$ is a unit
vector perpendicular to the enclosed surface, with the positive direction
determined by the right hand rule.)

\noindent
Here, we will use a {\em quadratic} current loop as a model for such an atomic
magnetic dipole and look closer at how it will behave in a magnetic field
$\bold{B}$. (We could have used a circular loop, but the quadratic one is a little
simpler when it comes to the calculations...)

\noindent
The current loop has edges with length $a$ and transports a current $I$. It is placed
in a {\em homogeneous} magnetic field $\bold{B} = B\;\hat{z}$ and is allowed to
rotate freely around the $y$-axis, which in our case passes through the
centre of the current loop, as shown in the figure:

%%FIGURE
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%%END FIGURE

\noindent
The orientation of the current loop is defined through the angle $\theta$
between the $z$-axis and the surface normal $\hat{n}$. (Positive$\theta$ {\em
counterclockwise}, as shown in the figure.)

\vspace{0.3cm}
\noindent
$a)$ What is the magnetic dipole moment $\bold{m}$ of this current loop?
What is the total force due to $\bold{B}$ on the current loop?

\vspace{0.3cm}
\noindent
$b)$ Find the torque $\bold{\tau}$ on the loop with respect to the
$y$-axis and show that it can be written in the form
$\bold{\tau} = \bold{m} \times \bold{B}$.

\noindent
[Hint: Find the force on each of the four straight elements of the loop and
use the face that torque equals "arm times force".]

\vspace{0.3cm}
\noindent
$c)$ Determine the potential energy $U(\theta)$ of such a magnetic dipole in
the field $\bold{B}$. Draw a sketch of $U(\theta)$. Which orientation of
the dipole with respect to $\bold{B}$ represents a stable and an unstable
equilibrium, respectively?

\vspace{0.3cm}
\noindent
$d)$ In iron, each atom has a magnetic dipole moment $\bold{m}_{\rm Fe}$
which is made up of two parallel electron spins, so that $m_{\rm Fe} = 2 \mu_B$. 
Here, $\mu_B = e\hbar/2m_e$ is the magnetic dipole moment of a single electron 
spin, the so-called Bohr magneton, which has the value $9.27\cdot10^{-24}$ Am$^2$.

\noindent
What is then the maximum density of magnetic dipole moment, i.e., the maximum
magnetic dipole moment pr unit volume, in iron?

\noindent
[Comment: Magnetic dipole moment pr unit volume is, by definition, the quantity
{\em magnetization}. In electrostatics, we introduced {\em polarization}, which
by definition is electric dipole moment pr unit volume. More about magnetism and
magnetization in the lectures!]

\vspace{0.2cm}
\noindent
Given information: Molar mass of iron: 55.9 g/mol. Mass density
of iron: 7.9 g/cm$^3$. 1 mol = $6.02\cdot10^{23}$.

\vspace{1.0cm}
\noindent
{\em Exercise 2}

\vspace{0.2cm}
\noindent
Show, by using Ampere's law, that the magnetic field $\bold{B}$ from
a uniform "surface current" $\bold{i} = i\;\hat{y}$ flowing in the
(complete) $xy$-plane in the positive $y$ direction is
$$\bold{B} = \left\{
\begin{array}{ll}
-(\mu_0 i/2) \; \hat{x} & {\rm for} \; z<0 \\
+(\mu_0 i/2) \; \hat{x} & {\rm for} \; z>0
\end{array} \right. $$
(I.e., independent of the distance from the $xy$ plane, just like we
found for the electric field from an infinitely large uniformly charged plane.)
Here, $i$ is the current {\em pr unit length} of the $x$ direction. In other
words, in a "stripe" of width $\Delta x$ runs a current 
$\Delta I = i \cdot \Delta x$.


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\noindent
Hint:
\begin{itemize}
\item You have already been informed that the $y$ and the $z$ component
of $\bold{B}$ are both zero. However, spend some time to convince yourself
that it has to be like that! Such an "investigation" of the symmetry of
the problem is completely {\em essential} if you want to take advantage of
Ampere's law in order to determine the magnetic field. Often, you then need
to go back to the Biot-Savart law and look at the consequences of "current
elements" $I\;d\bold{l}$ that give contributions $d\bold{B} \sim I\;d\bold{l}\times\hat{r}$ 
to the total magnetic field.
\item In this particular problem, you will perhaps convince yourself that
a sensible choice of "amperian loop" is a rectangular curve with surface
normal in the current direction. If so, you are on the right track!
\end{itemize}


\vspace{1.0cm}
\noindent
{\em Exercise 3}

\vspace{0.2cm}
\noindent
In this exercise, we shall use a classical model of an atom and take a closer look
at how an external magnetic field $\bold{B}$ will influence the orbital movement of
the electron around the nucleus. Such a {\em diamagnetic response} is present in all
atoms. (More about different types of magnetism in the lectures.) Let's for simplicity
have a hydrogen atom in our thoughts, with a single electron with charge $-e$
in a circular orbit (in the $xy$ plane) with radius $R$ around a nucleus
with charge $+e$.

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\vspace{0.3cm}
\noindent
$a)$ Without an external magnetic field, the velocity of the electron
is $v_0$. Show that uniform circular motion in the Coulomb field from
the nucleus then results in a circular orbit with radius
$$R = \frac{e^2}{4\pi\varepsilon_0 m_e v_0^2}$$
What is the angular momentum $\bold{L}_0$ and the magnetic dipole moment
$\bold{m}_0$ of this electron?
(Here, we ignore the spin of the electron.)

\vspace{0.3cm}
\noindent
$b)$ Next, we turn on a magnetic field $\bold{B}$, for simplicity with direction
perpendicular to the circular orbit of the electron.

%%FIGURE
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%%END FIGURE

\noindent
Now, the electron will be affected by an additional force (i.e., in addition to
the Coulomb force from the proton in the nucleus), namely the magnetic force
$\bold{F}_m = -e \bold{v} \times \bold{B}$ which of course must result in
a modified equation of motion. Consequently, we must find a different relation
between the velocity $v$ of the electron and the radius of the circular orbit,
$R$. Let us further assume that the magnetic field only changes the velocity
(and not the radius $R$), and determine the new velocity $v$.
Also find the new value $\bold{m}$ of the magnetic dipole moment of the electron
and show that the {\em change}
$$\Delta \bold{m} = \bold{m} - \bold{m}_0$$
will always be directed {\em opposite to} the magnetic field $\bold{B}$. Show
that this will be the case both for $\bold{B}$ "up" and "down" with respect to
the original magnetic dipole moment $\bold{m}_0$ of the electron.

\vspace{0.2cm}
\noindent
Comments:
\begin{enumerate}
\item
We have earlier stated that a static magnetic field never does any work on a moving
charge since $\bold{F}_m \perp \bold{v}$. Thus, a static magnetic field cannot
change the magnitude of the velocity of the charge, apparently in conflict with
what we have just found above. However, the point is that here we start with
$B=0$ and {\em turn on} a magnetic field. This means that we do not have a static
magnetic field at all times, but rather a field that during a certain period of time
must change from zero to its final value. As we will see later in the lectures, a time
dependent magnetic field will create ("induce") an electric field. (This is
Faraday's law of electromagnetic induction.) And, as you know, an electric field
may very well change the velocity of an electron. This resolves the apparent paradox.
\item
The sign of the diamagnetic response is actually an expression of
{\em Lenz' law}, which you may have heard of earlier, and which we will
come back to in the lectures: The "response" of the system to a change in
the magnetic field is in a direction so that it attempts to reduce the
effect of the applied change.
\item
Strictly speaking, it is necessary with a {\em quantum mechanical} description
in order to explain diamagnetism "properly". In fact, there is a {\em theorem}
in statistical physics which states that for a system of classical charged particles
in thermal equilibrium in an external magnetic field, the induced magnetic dipole
moment is exactly zero (Bohr - van Leeuwen's theorem). In other words:
Diamagnetism is actually a purely quantum mechanical effect!
Still, the simple classical model described above, with one atom only, gives a
useful qualitative picture of the effect.
\end{enumerate}



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