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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}
Solution to {\o}ving 13
\end{large}
\end{center}

\noindent
Guidance Thursday April 14 and Friday April 15


\vspace{0.5cm}
\noindent
{\em Exercise 1}

\vspace{0.2cm}
\noindent
$a)$ The magnetic dipole moment of the loop:
$$\bold{m} = IA\;\hat{n} = I a^2\;\hat{n}$$
The loop consists of 4 straight wires of length $a$, pairwise with the current
in the same direction. Hence, the magnetic force,
$$\bold{F} = I \int \;d\bold{l} \times \bold{B} = I \bold{L}\times\bold{B}$$
has opposite direction, but is equal in magnitude, for such pairs of straight wires
("current carrying conductor elements"). Thus, the total force on the current loop is zero.
Some details are included in this figure:

%%FIGURE
\vspace{0.5cm}
\centerline{\epsfxsize=2.8in \epsfbox{los13_fig1.eps}}
\vspace{0.5cm}
%%END FIGURE

\noindent
We see from the figure that the forces from the magnetic field would
have deformed the current loop if that had been a possibility. For a
macroscopic current loop, this is usually a negligible effect, but if
the current loop is a classical model of an electron in an orbit around
a nucleus, we see that in addition to an alignment of the current loop
(which is the topic of this exercise), the magnetic field will influence
the orbital motion of the electron around the nucleus. In other words:
The magnetic moment changes both in direction and in absolute value.
The first effect is {\em paramagnetism}, the second effect is {\em diamagnetism}.
A classical model of diamagnetism is the topic in exercise 3, here we concentrate
on the orientation of $\bold{m}$.

\vspace{0.2cm}
\noindent
$b)$ From the figure above, we see that the two currents running parallel to
the $xz$-plane are influenced by forces in positive and negative $y$ direction,
respectively. These forces will then not contribute to the torque around the
$y$ axis.

\noindent
The forces acting on the currents running parallel to the $y$ axis results all together
in a torque (see figure below and above)

%%FIGURE
\vspace{0.5cm}
\centerline{\epsfxsize=2.5in \epsfbox{los13_fig2.eps}}
\vspace{0.5cm}
%%END FIGURE

\begin{eqnarray}
\bold{\tau} &=& \sum \bold{r}\times\bold{F} \nonumber \\
&=& - r_1 F_1 \sin\theta\;\hat{y} - r_2 F_2 \sin\theta\;\hat{y} \nonumber \\
&=& -2 \cdot \frac{a}{2} \cdot IaB \sin\theta \;\hat{y} \nonumber \\
&=& -Ia^2 \cdot B \sin\theta \;\hat{y} \nonumber \\
&=& -m \cdot B \sin\theta \;\hat{y} \nonumber \\
&=& \bold{m} \times \bold{B} \nonumber
\end{eqnarray}
The change of sign in the last line is because we had chosen a positive
angle $\theta$ between the $z$ axis and $\hat{n}$, i.e., between $\bold{B}$
and $\bold{m}$. The cross product $\bold{m} \times \bold{B}$ is, by definition,
$m$ times $B$ times the sine of the angle between $\bold{m}$ and $\bold{B}$,
i.e., $mB\sin(-\theta) = -mB\sin\theta$.

\vspace{0.2cm}
\noindent
$c)$ In exercise 3c in {\o}ving 6 we derived a {\em general} relation between
the torque $\tau$ and the corresponding potential energy $U$, namely that a rotation
through an angle $d\alpha$ under the influence of a torque $\tau$ results in a change
$dU$ in potential energy given by
$$dU = - \tau\;d\alpha$$
The derivation of this relation did not depend upon what {\em kind} of forces and
torques we are talking about, and must therefore be valid also for our magnetic dipole
in a magnetic field. Thus:
\begin{eqnarray}
U(\theta) &=& \int_{\theta_0}^{\theta} dU \nonumber \\
&=& -\int_{\theta_0}^{\theta} \tau(\alpha)\;d\alpha \nonumber \\
&=& m B \int_{\theta_0}^{\theta} \sin\alpha\;d\alpha \nonumber \\
&=& m B \left( \cos\theta_0 - \cos\theta \right) \nonumber \\
&=& - m B \cos\theta \nonumber \\
&=& - \bold{m} \cdot \bold{B} \nonumber
\end{eqnarray}
Here, I chose $U(0) = - m B$, i.e., $\theta_0 = \pi/2$.

\noindent
Sketch:

%%FIGURE
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\centerline{\epsfxsize=2.8in \epsfbox{los13_fig3.eps}}
\vspace{0.5cm}
%%END FIGURE

\noindent
We have minimal $U$, and thus a stable equilibrium for $\theta = 0$, i.e.,
when the dipole is oriented so that $\bold{m}$ is parallel with $\bold{B}$.
We have maximum value of $U$, and therefore an unstable equilibrium
for $\theta = \pm \pi$, i.e., when the dipole is oriented so that
$\bold{m}$ is parallel with $-\bold{B}$.

\noindent
This is the analogy in magnetism to polarization of dielectric media in an external
electric field: Magnetic dipoles align in the external magnetic field. As mentioned
above, this is what we call {\em paramagnetism}. We have talked about different
kinds of magnetism in the lectures -- here is a brief summary:

\noindent
Materials that consist of atoms which have an atomic magnetic dipole moment which
is {\em not} zero, and where the dipole moments on atoms close to each other to
{\em not} interact with each other, are {\em paramagnets}. {\em Without} an external
magnetic field, the atomic magnetic dipole moments will be oriented in random
directions, so that the average {\em magnetization}, i.e., the average
magnetic dipole moment pr unit volume (see item $d)$), becomes zero everywhere
in the material. We had exactly the same situation concerning average polarization
in a dielectric when we had zero external electric field. {\em With}
an external magnetic field, we obtain a tendency of alignment of magnetic dipole moments
along the external field, and thereby an average magnetization different from zero.
Examples of paramagnetic materials are aluminum (Al) and magnesium (Mg).

\noindent
Materials consisting of atoms with {\em zero} atomic magnetic dipole moments have
nothing to align in an external magnetic field. However, as mentioned above,
the {\em orbital motion} of the electrons around the nucleus will be affected
by an external magnetic field, so that we have an {\em induced} magnetic dipole
moment in each atom. Such materials are {\em diamagnets}.
We will look qualitatively on this effect in exercise 3 below and find that the
induced magnetic dipole moment will always be directed {\em opposite} to the
external field. Diamagnetism is a much weaker effect than paramagnetism. Also in
paramagnets, where we do have permanent atomic magnetic dipole moments, we have
this diamagnetic "response" in an external magnetic field.
However, the diamagnetic response will be almost negligible in a paramagnet. In order
to be able to measure diamagnetism, we need a material with zero atomic magnetic
dipole moments at the outset (i.e., before we switch on the external field).
Examples of diamagnets are gold (Au), silver (Ag) and copper (Cu).

\noindent
In some materials, we have atoms with magnetic dipole moments that {\em interact}
with the dipole moments on the neighbouring atoms. For example, the interaction
may be such that it is energetically favored that neighbouring atoms have
their magnetic dipole moments in the same direction. Then we have a {\em ferromagnet},
and examples of ferromagnets are iron (Fe), cobalt (Co) and nickel (Ni).

\vspace{0.2cm}
\noindent
$d)$ The maximum density of magnetic dipole moment in iron is equal to
the number of iron atoms pr unit volume times the magnetic dipole moment
pr iron atom:
$$\frac{m}{V}= 2\mu_B \cdot \frac{7.9\cdot10^{6}}{55.9}\cdot 6.02\cdot 10^{23} =
2 \cdot 9.27 \cdot 10^{-24} \cdot \frac{7.9\cdot10^{6}}{55.9}\cdot 6.02\cdot 10^{23} =
1.6 \cdot 10^6$$
The SI unit of $m$ is Am$^2$, the unit of $V$ is m$^3$. Hence, the unit of
magnetic dipole moment pr unit volume, or {\em magnetization}, is A/m.


\vspace{1.2cm}
\noindent
{\em Exercise 2}

\vspace{0.2cm}
\noindent
The direction of $\bold{B}$:
\begin{itemize}
\item
$B_y = 0$ because $d\bold{B} \sim \hat{y} \times \hat{r} \perp \hat{y}$ 
according to Biot-Savart's law. (All the contributions to the current run
in the $y$ direction.)
\item
$B_z=0$: Look at the figure below. Here, $d\bold{B}_{+}$ and $d\bold{B}_{-}$ are
contributions to the magnetic field above and below the $xy$ plane, respectively,
from "symmetrically located" infinitely long, thin current carrying wires in
position $\pm x$. The Biot-Savart law and inspection of the figure then yields
the result that $\bold{B}$ must point in the positive $x$ direction for $z>0$ 
and in the negative $x$ direction for $z<0$.
\end{itemize}

%%FIGURE
\vspace{0.5cm}
\centerline{\epsfxsize=2.5in \epsfbox{los13_fig5.eps}}
\vspace{0.5cm}
%%END FIGURE

\noindent
The absolute value of the magnetic field cannot depend upon $x$ or
$y$ when the current carrying plane is infinite. Further, $B$ must
have the same absolute value a distance $z$ above the $xy$ plane
as a distance $z$ below the same plane. ($B_{+} = B_{-} = B$, see figure
below) Then the best choice of amperian loop should be clear: A rectangle
with the surface normal in the positive $y$ direction, symmetrically
located with respect to the $xy$ plane:

%%FIGURE
\vspace{0.5cm}
\centerline{\epsfxsize=2.5in \epsfbox{los13_fig6.eps}}
\vspace{0.5cm}
%%END FIGURE
                                                                                                                         
\noindent
When the integration curve is chosen as shown in the figure, the current
enclosed by the amperian loop is {\em positive}, according to the
right hand rule. With a length $L$ in the $x$ direction, the enclosed current
is $I_{\rm in} = i \cdot L$. Then, Ampere's law yields:
$$\oint \bold{B}\cdot \; d\bold{l} = 2 \cdot B \cdot L = \mu_0 i \cdot L$$
or
$$B = \mu_0 i/2$$
(On the vertical pieces of the amperian loop, we have 
$\bold{B} \perp \;d\bold{l}$ so these give zero contribution to the integral.)

\noindent
Alternatively, we could initially have located the complete amperian
rectangle on one side of the $xy$ plane. Then we would have had zero enclosed
current, and thereby found that $B$ must be independent of $z$. Next, we
locate the amperian loop so that it encloses a part of the $xy$ plane, and
therefore a certain current, and finally find the same answer as above.
As far as I can see, it is sufficient to use the Ampere law {\em once}
when we choose the curve symmetric with respect to the $xy$ plane.

\vspace{1.3cm}
\noindent
{\em Exercise 3}

\vspace{0.2cm}
\noindent
$a)$ The centripetal acceleration is $v_0^2/R$ whereas the Coulomb
force is $e^2/4\pi\varepsilon_0 R^2$. Then, Newton's 2. law gives
$$m_e \frac{v_0^2}{R} = \frac{e^2}{4\pi\varepsilon_0 R^2} \;\;\; \Rightarrow \;\;\;
R = \frac{e^2}{4\pi\varepsilon_0 m_e v_0^2}$$
The orbital angular moment of the electron is
$$\bold{L}_0 = m_e\; \bold{r} \times \bold{v}_0 = m_e R\; v_0 \;\hat{z}$$
while its magnetic dipole moment is
$$\bold{m}_0 = I \bold{A} = - \frac{e}{2\pi R/v_0} \cdot \pi R^2 \;\hat{z} =
-\frac{1}{2}e v_0 R \;\hat{z} = - \frac{e}{2 m_e} \bold{L}_0$$


%%FIGURE
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%\centerline{\epsfxsize=2.8in \epsfbox{los14_fig1.eps}}
\vspace{0.5cm}
%%END FIGURE


\vspace{0.2cm}
\noindent
$b)$ The chosen direction of $\bold{B}$ implies that the magnetic force
$\bold{F}_m = -e\;\bold{v}\times\bold{B}$ is directed towards the nucleus,
i.e., in the same direction as the attractive Coulomb force. Assuming the
orbital radius $R$ does not change, the speed $v$ is determined by the
equation
$$m_e \frac{v^2}{R} = \frac{e^2}{4\pi\varepsilon_0 R^2} + evB$$
This is a 2. order equation for $v$,
$$v^2 - \frac{eBR}{m_e}\;v - \frac{e^2}{4\pi\varepsilon_0 m_e R} = 0,$$
with solution
$$v = \frac{eBR}{2m_e} + \sqrt{\frac{e^2}{4\pi\varepsilon_0 m_e R} + \left(\frac{eBR}{2m_e}\right)^2 }$$
(The solution with a negative sign in front of the square root is negative and
not applicable.)

\noindent
Let us go back and look at the speed without the magnetic field:
$$v_0 = \sqrt{\frac{e^2}{4\pi\varepsilon_0 m_e R}}$$
We see immediately that $v>v_0$. This means that the magnetic
dipole moment
$$\bold{m} = - \frac{1}{2}e v R \;\hat{z}$$
is larger than before we turned on the magnetic field. In other words,
the {\em change}
$$\Delta \bold{m} = \bold{m} - \bold{m}_0$$
is directed opposite to the external magnetic field.

\noindent
If the magnetic field instead was directed downwards,
$\bold{B} = -B\;\hat{z}$, the magnetic force would be directed radially
{\em outwards}, i.e., in the opposite direction of the Coulomb force,
so that the additional term $evB$ in the equation of motion would enter
with the opposite sign. Thus, the new speed $v$ would have become smaller
than the initial speed $v_0$, and the magnetic dipole moment
also smaller than before we turned on the magnetic field. Again: The {\em change}
in magnetic dipole moment would still be directed opposite to the external field.

\noindent
In conclusion: An external magnetic field influences the orbital motion in the
atom in such a way that the {\em induced} magnetic dipole moment, i.e., the
magnetic dipole moment associated with
the change in the orbital motion, will be opposite to the external field.
I.e., {\em diamagnetism}.

\end{document}
