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

\noindent
TFY4155/FY1003: Elektrisitet og magnetisme

\noindent
V{\aa}r 2007

\end{bf}

\begin{center}
\begin{large}
{\O}ving 14
\end{large}
\end{center}

\noindent
Guidance: Week 16

\noindent
To be delivered by: Monday April 23

\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:

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\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
An infinitely long, thin solenoid is located with its central axis on the $x$ axis.

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\noindent
In the lectures, we argued that if there is a magnetic field on the outside of the
solenoid, it must be directed {\em along} the direction of the solenoid, i.e.,
$\bold{B} = B \, \hat{x}$. Using Ampere's law, we then showed that $B=0$. This can be
explained by the fact that some of the turns on the solenoid contribute with negative
$x$ component to $\bold{B}$ while others contribute with positive $x$ component.

\noindent
In a distance $y = 50$ cm from the axis of a solenoid with 1000 turns pr meter, how many
turns will contribute with a negative $x$ component to $\bold{B}$? (The rest of the turns,
an infinite number, will contribute with positive $x$ component to $\bold{B}$.) Witout telling
you the exact answer, I can say that if you obtain somewhat more than 700, you have
probably done this one correctly. (Or: Done the same mistakes as I have...)

\vspace{0.3cm}
\noindent
Hint: Each turn on the solenoid may be regarded as an {\em ideal} magnetic dipole $\bold{m} = m \, \hat{x}$,
i.e., we assume that the radius of the solenoid is small compared to the distance $y$. In that case,
the magnetic field in a distance $\bold{r}$ from a specific turn is given by
$$\bold{B}_{\rm dipol} = \frac{\mu_0}{4 \pi r^3} \left[ 3 \left(\bold{m} \cdot \hat{r} \right) \hat{r}
- \bold{m} \right]$$
With a current $I$ in the solenoid wire and a cross section of area $A$, we have the relation $m=IA$.
However, you don't need $I$ and $A$ to solve this problem.



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