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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 15
\end{large}
\end{center}

\noindent
Guidance: Week 17

\noindent
To be delivered by: Monday April 30 

\vspace{0.8cm}
\noindent
{\em Exercise 1}


%%FIGURE
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\noindent
A cylindrical iron rod with relative permeability $\mu_r = 2000$ is placed coaxially
inside a solenoid, but fills only partially the volume inside the solenoid. The
solenoid has a winding density (i.e., windings pr unit length) $n = 2000$ m$^{-1}$
and the current in the solenoid wire is $I = 3$ A. We assume that both the
solenoid and the iron rod are sufficiently long that we may neglect edge effects.

\noindent
Assume first that we have linear response in the iron rod, i.e.
$\bold{M} = \chi_m \bold{H}$, and determine $\bold{H}$, $\bold{B}$ and $\bold{M}$
inside the solenoid, both inside (index j) and outside (index 0) the iron rod.
(Remember that the $H$-field is determined by the "free" current, whereas $B$
is determined by the total current.)

\noindent
Discuss the calculated value of $M_j$ inside the iron rod, taking into account
the {\em saturation magnetization} in iron, i.e., the maximum possible magnetization,
which you calculated in exercise 1d in {\o}ving 14. Calculate next a corrected
(maximum) value of $B_j$.

\vspace{0.3cm}
\noindent
Given information
$$\bold{B} = \mu_0 \left( \bold{H} + \bold{M} \right) = \mu_r \mu_0 \bold{H} = \mu \bold{H}$$
$$\bold{M} = \chi_m \bold{H} = (\mu_r - 1) \bold{H}$$
(The last line is only valid when we have linear response.)

\vspace{0.3cm}
\noindent
A couple of answers: $B_j = 15$ T ("uncorrected"), $B_j = 2$ T ("corrected").


%\vspace{1.3cm}
\newpage
\noindent
{\em Exercise 2}

\vspace{0.5cm}
\noindent

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\noindent
The figure shows two solenoids 1 and 2 that are both wound onto
the same cylinder of radius $R$. We assume that the cylinder has
magnetic properties as vacuum, i.e., we assume there is no
magnetization of the cylinder. Solenoid 1 har $N_1$ windings,
solenoid 2 has $N_2$ windings. Both solenoids are wound onto
a length $d$ of the cylinder, which is (appoximately infinitely)
long compared to the radius of the cylinder. (So the figure is
only qualitatively correct...!) You may assume that both solenoids
are tightly wound, and that each winding of both solenoids enclose the
same amount of magnetic flux. (The wire of the solenoids is covered with
some kind of electrically insulating material, e.g. a layer of plastic,
so that an electric current is forced to follow the solenoid wire. This
assumption is by the way implicit in all such exercises with solenoids.)

\vspace{0.5cm}
\noindent
$a)$ Assume that solenoid 1 carries a current $I_1$. What is then the
strength of the magnetic field $B$ inside the solenoid? Next, what is
the {\em total} magnetic flux $\phi_1$ enclosed by the wire of
solenoid 1 (i.e., all the $N_1$ windings)? What is the total
magnetic flux $\phi_2$ enclosed by the wire of solenoid 2 (again: all
the $N_2$ windings)? (Note: There is no current in solenoid 2. The current
in solenoid 1 can be made e.g. by coupling it to a battery and a resistance.)

\vspace{0.5cm}
\noindent
$b)$ The ratio between the total enclosed magnetic flux $\phi_1$ and
the current $I_1$ in the current loop it{\em self} is, by definition,
a quantity which is called the {\em self inductance} $L$ of the loop:
$$L = \frac{\phi_1}{I_1}$$
Then, what is the self inductance $L$ of such a long cylindrical solenoid
with radius $R$, length $d$ and $N_1$ windings?

\vspace{0.5cm}
\noindent
$c)$ The ratio between the total enclosed magnetic flux $\phi_2$ that is
enclosed by solenoid 2 and the current $I_1$ in solenoid 1 is, by definition,
a quantity which is called the {\em mutual inductance} $M$ between
the two current loops:
$$M = \frac{\phi_2}{I_1}$$ 
Then, what is the mutual inductance $M$ between two such long cylindrical
solenoids, both being wound onto a cylinder of radius $R$ over a length
$d$, and with $N_1$ and $N_2$ windings, respectively?

\vspace{0.5cm}
\noindent
$d)$ Determine numerical values for $L$ and $M$ (in SI units) when
$R=1$ cm, $d=60$ cm, $N_1 = 1200$, and $N_2 = 600$. 

\noindent
(Answers: $L = 9.5 \cdot 10^{-4}$, $M = 4.7 \cdot 10^{-4}$)

\vspace{0.5cm}
\noindent
Comment: We will come back to mutual inductance and selfinductance in the
final lectures, and see why these are "useful" quantities in many connections.


\vspace{2cm}
\noindent
{\em Exercise 3}

\vspace{0.2cm}
\noindent
Boundary conditions for $\bold{E}$ and $\bold{B}$:

\vspace{0.2cm}
\noindent
Let us take a look at how the electric field and the magnetic field "behave" when we
cross a {\em boundary surface}. By "boundary surface", I simply mean a surface that
divides space into two regions, 1 "above" and 2 "below" the surface. Let's first look
at the electric field:

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\noindent
The electric field is {\em discontinuous} if such a boundary surface contains
electric charge $\sigma$ pr unit area:
$$\bold{E}_1 - \bold{E}_2 = \frac{\sigma}{\varepsilon_0} \;\hat{n} \;\;\;\;\;\;\; (*)$$
Here, $\bold{E}_1$ is the field in region 1 just above the surface,
$\bold{E}_2$  correspondingly in region 2 just below the surface, while
$\hat{n}$ is a unit normal vector directed upwards.

\noindent
You notice that the equation $(*)$ is a compact way of expressing that the {\em parallel
component} of $\bold{E}$ is continuous,
$$E_1^\parallel - E_2^\parallel = 0,$$
whereas the {\em normal component} is discontinuous,
$$E_1^\perp - E_2^\perp = \frac{\sigma}{\varepsilon_0},$$
when we cross the boundary.

\newpage
\noindent
Next, we look at the magnetic field:

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\noindent
Here, the boundary surface is oriented perpendicular to the paper plane. The magnetic
field is {\em discontinuous} if there runs a current $\bold{i}$ pr unit length in
the boundary surface:
$$\Delta \bold{B} = \mu_0 \; \bold{i} \times \hat{n}$$
This means that both $B_n$ and $B_{t\parallel}$ are continuous when crossing
the plane, while $B_{t\perp}$ is discontinuous with a disconinuity $\mu_0 i$.
Here, we have decomposed the tangential component $B_t$ of $\bold{B}$ into one
component that is parallel to the current direction, $B_{t\parallel}$, and one
component that is perpendicular to the current direction, $B_{t\perp}$.

\vspace{0.6cm}
\noindent
$a)$ Look at previous exercises and your lecture notes (or examples in your book)
and find a couple of examples where you can control that these boundary conditions
are fulfilled.

\vspace{1.0cm}
\noindent
If our "system" contains dielectric and/or magnetizable media, we may possibly have
interfaces where we know what the {\em free} charge $\sigma_f$ pr unit area is,
or what the {\em free} current $\bold{i}_f$ pr unit length is. (But perhaps we cannot
tell immediately what the {\em total} charge $\sigma$ pr unit area is, or what
the {\em total} current $\bold{i}$ pr unit length is.) In such circumstances, we must
in addition use the following boundary conditions for the normal component $D_n$ of
the electric displacement,
$$D_{1n} - D_{2n} = \sigma_f,$$
and the tangential component $\bold{H}_t$ of the $H$ field,
$$\Delta \bold{H}_t = \bold{i}_f \times \hat{n}$$

\newpage
\noindent
$b)$ Let us look at some examples, in which we must use the various boundary
conditions in order to determine the field strengths:

\noindent
Suppose we have a uniform electric field $\bold{E}_0 = E_0 \, \hat{z}$. In this field we put
a dielectric slab (overall electrically neutral) with approximately infinite extent in the
$x$ and $y$ directions, and thickness $h$ in the $z$ direction. In other words, the slab is
oriented perpendicular to the external field. The material in the slab has relative
permittivity $\varepsilon_r$.

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\noindent
What is the electric displacement $\bold{D}_1$ and the electric field $\bold{E}_1$ inside
the dielectric slab? Repeat with the slab oriented {\em along} the external field
direction! (I.e.: With infinite extent in the $y$ and $z$ directions, and thickness $h$ in
the $x$ direction.)

\noindent
Next, do the same things for an infinitely large {\em magnetizable} slab with thickness
$h$ and relative permeability $\mu_r$, oriented perpendicular to and along the direction
of a uniform external magnetic field $\bold{B}_0 = B_0 \, \hat{z}$, respectively. I.e.:
Determine $\bold{H}_1$ and $\bold{B}_1$ inside the slab.

\noindent
Did you get any surprising results? How do you explain that the electric field strength
is different inside the slab with the two orientations in the external field? And
correspondingly: How do you explain that the magnetic field strength is different inside
the magnetizable slab in the two cases?


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