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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}
Solution to {\o}ving 15
\end{large}
\end{center}

\noindent
Guidance week 17


\vspace{1.2cm}
\noindent
{\em Exercise 1}

\vspace{0.5cm}
\noindent
The {\em external} current $I$ generates an $H$ field $H = nI$ along
the solenoid everywhere inside the solenoid (because of Ampere's
law for $H$.) So, we can simply use that
$$\bold{B} = \mu_0 \left( \bold{H} + \bold{M} \right) =
\mu_r \mu_0 \bold{H}$$
to determine the various quantitites:

\vspace{0.3cm}
\noindent
Inside the iron:
$$H_j = nI = 2000 \;{\rm m}^{-1} \cdot 3\;{\rm A} = 6000 \;{\rm A/m}$$
$$B_j = \mu_r \mu_0 H_j = 2000 \cdot 4\pi \cdot 10^{-7} \;{\rm (Vs/Am)} \cdot
6000 \;{\rm A/m}\;= 15 \;{\rm T}$$
$$M_j = (\mu_r - 1)H_j = 1.2 \cdot 10^7 \;{\rm A/m}$$

\vspace{0.3cm}
\noindent
In the airfilled part inside the solenoid:
$$H_0=H_j=6000 \;{\rm A/m}$$
$$B_0=\mu_0 H_0=7.5\;{\rm mT}$$
$$M_0=0$$

\vspace{0.3cm}
\noindent
The calculated value of the magnetization inside the iron rod,
$M_j =1.2 \cdot 10^7 \;{\rm A/m}$, is larger than the saturation
magnetization $M_s = 1.6 \cdot 10^6 \;{\rm A/m}$, and therefore
not possible. The reason is that we have used the linear relation
$B = \mu_r \mu_0 H$ between the magnetic field $B$ and the field $H$
from the external current. However, here we have such a strong external
field $H$ that this linear relation is no longer valid. All magnetic
dipoles are already aligned with the external field when
$H \simeq M_s/\mu_r = 800$ A/m. An additional increase in $H$ cannot
raise the value of $M$ any further.

\noindent
Corrected, maximum value of $B_j$ becomes
$$B_j^{\rm korr} = \mu_0 \left( H_j + M_s \right) =
4\pi \cdot 10^{-7} \cdot \left(6000 + 1.6 \cdot 10^6 \right) = 2\;{\rm T}$$



\newpage
\noindent
{\em Exercise 2}

\vspace{0.2cm}
\noindent
$a)$ We have used Ampere's law in the lectures to calculate the
magnetic field inside a long solenoid:
$$B = \mu_0 n I_1 = \mu_0 \frac{N_1}{d} I_1$$
{\em One} winding of the solenoid wire encloses an area $A = \pi R^2$,
and therefore a magnetic flux
$$\phi = B A = \mu_0 \frac{N_1}{d} I_1 \pi R^2$$
Then, $N_1$ windings must enclose a flux which is $N_1$ times bigger,
because here, the magnetic field is constant everywhere inside the solenoid.
Hence:
$$\phi_1 = N_1 \phi = \mu_0 \frac{N_1^2}{d} I_1 \pi R^2$$
One winding of solenoid 2 encloses exactly the same area, and therefore
the same amount of flux $\phi$, so that $N_2$ windings of solenoid 2
must enclose a total magnetic flux equal to
$$\phi_2 = N_2 \phi = \mu_0 \frac{N_1 N_2}{d} I_1 \pi R^2$$
 
\vspace{0.2cm}
\noindent
$b)$ The self inductance $L$ becomes
$$L = \frac{\phi_1}{I_1} = \mu_0 \frac{N_1^2}{d}\pi R^2$$
 
\vspace{0.2cm}
\noindent
$c)$ Mutual inductance $M$ becomes
$$M = \frac{\phi_2}{I_1} = \mu_0 \frac{N_1 N_2}{d}\pi R^2$$
 
\vspace{0.2cm}
\noindent
$d)$ Numerical values:
$$L = 4\pi \cdot 10^{-7} \cdot \frac{1200^2}{0.6} \cdot \pi \cdot 0.01^2 =
9.5 \cdot 10^{-4}$$
$$M = 4\pi \cdot 10^{-7} \cdot \frac{1200 \cdot 600}{0.6} \cdot \pi \cdot 0.01^2 =
4.7 \cdot 10^{-4}$$
In the SI system, inductance has its own unit, the henry (H). So, here
the self inductance $L$ is 0.95 mH and the mutual inductance $M$ is 0.47 mH.
Alternatively, we may use the unit T m$^2$/A, since magnetic flux must have
the unit of magnetic field times area, i.e., T m$^2$.


\newpage
\noindent
{\em Exercise 3}

\vspace{0.2cm}
\noindent
$a)$ Examples:

\noindent
For an infinitely large charged plane with charge $\sigma$ pr unit area, the electric field
is oppositely directed on the two sides, and the field strength is $\sigma/2\varepsilon_0$.
Hence, a {\em discontinuity} of $\sigma/\varepsilon_0$ in the normal component of the field,
and no discontinuity in the tangential component of the field.

\noindent
Inside a metal sphere with charge $Q$, the electric field is zero. On the surface,
at $r=R$, the field is $Q/4\pi\varepsilon_0 R^2$, directed radially outwards
(if $Q>0$), i.e., normal to the surface. Again, a discontinuity of $\sigma/\varepsilon_0$ in
the normal component of the field, since $\sigma=Q/4\pi R^2$ is the charge pr unit area on
the surface of the sphere.

\noindent
An infinitely large plane carrying a uniform current $i$ pr unit length results
in a uniform magnetic field $\mu_0 i/2$, in opposite direction on the two sides of the plane,
see {\o}ving 13, exercise 2. I.e., a discontinuity of $\mu_0 i$, and if you take a look at
{\o}ving 13, you will find out that we're talking about the component of $\bold{B}$ that lies
in the plane of the current, and at the same time is normal to the current direction.

\noindent
Inside an infinitely long solenoid, the magnetic field is $\mu_0 n I$, outside it is zero. Hence,
a discontinuity of $\mu_0 n I$. The current is $I$ pr turn, while the number of turns pr unit
length is $n$, so $i = nI$ becomes the current pr unit length. Again, a discontinuity of $\mu_0 i$.

\noindent
Comment: No boundary surface is infinitely large, but if we come sufficiently close to the surface,
it will look as if it is infinite and flat. The total electric field on the surface must be equal to
the sum of the contributions form the "nearby region", i.e., the part of the surface that looks large
and flat, and the contribution from all the charges in "the rest of the world". The charges in the rest
of the world are all far away from the "crossing point", i.e., far away when compared to the charges
that are actually {\em in} the plane where we are crossing. "The rest of the world" therefore must
contribute with the same field just below and just above the boundary surface, i.e., with a contribution
that is {\em continuous}. In other words, the whole discontinuity in the electric field is due to the
charges in the plane that we cross. And correspondingly for the magnetic field: Total magnetic field is
the sum of the contributions from the current {\em in} the plane where we cross, and the contribution
from all other currents in the world. Only the current in the plane where we cross contribute to
the discontinuity.

\vspace{0.3cm}
\noindent
$b)$ Dielectric slab perpendicular to constant external electric field $\bold{E}_0$:

\noindent
Here we have boundary surfaces that are perpendicular to the fields. We cannot use the
boundary condition for $\bold{E}$ because we do not know how much charge we have in
the boundary surfaces between vacuum and the dielectric. We know that there is an induced
(bound) charge, positive in the upper surface and negative in the lower surface, but not
{\em how much}. However, we can use the boundary condition for $\bold{D}$ because we know
that there is zero {\em free} charge in the slab. Hence, we must have $D_1=D_0$, where
$D_0 = \varepsilon_0 E_0$ is the electric displacement outside the slab. In addition, we
have $D_1 = \varepsilon_1 E_1 = \varepsilon_r \varepsilon_0 E_1$. Thus:
\begin{eqnarray}
D_1 &=& \varepsilon_0 E_0 \nonumber \\
E_1 &=& \frac{1}{\varepsilon_r} E_0 \nonumber
\end{eqnarray}

\vspace{0.2cm}
\noindent
Dielectric slab parallel to constant external electric field $\bold{E}_0$:

\noindent
Now we have boundaries parallel to the field direction. Then we can use that the parallel
component of $\bold{E}$ is continuous, i.e., $E_1 = E_0$. The relation 
$D_1 = \varepsilon_r \varepsilon_0 E_1$ is of course still valid, so
\begin{eqnarray}
D_1 &=& \varepsilon_r \varepsilon_0 E_0 \nonumber \\
E_1 &=& E_0 \nonumber
\end{eqnarray}

\vspace{0.2cm}
\noindent
Magnetizable slab perpendicular to constant external magnetic field $\bold{B}_0$:

\noindent
Here we have again boundaries perpendicular to the fields. We may therefore use the
fact that $B_n$ is continuous, i.e., $B_1 = B_0$. In addition we have
$B_1 = \mu_1 H_1 = \mu_r \mu_0 H_1$. Hence
\begin{eqnarray}
H_1 &=& \frac{1}{\mu_r \mu_0} B_0 \nonumber \\
B_1 &=& B_0 \nonumber
\end{eqnarray}

\vspace{0.2cm}
\noindent
Magnetizable slab parallel to constant external magnetic field $\bold{B}_0$:

\noindent
Boundaries are now parallel to the fields. We know that there is an induced
magnetization current in the surface of the slab, but not how much. However, we may
use the boundary condition for $\bold{H}$, because we know that there is zero free
current in the slab. Hence we have $H_1=H_0$, where $H_0 = B_0/\mu_0$ is the $H$ field
outside the slab (vacuum). In addition we have $B_1 = \mu_r \mu_0 H_1$. Thus
\begin{eqnarray}
H_1 &=& \frac{1}{\mu_0} B_0 \nonumber \\
B_1 &=& \mu_r B_0 \nonumber
\end{eqnarray}

\vspace{0.2cm}
\noindent
Explanation of different $E_1$ and $B_1$ in the two situations:

\noindent
Dielectric slab perpendicular to external field results in polarization, and a corresponding
induced charge in the surface. The induced charge contributes with an electric field opposite
to the external field, so that $E_1$ becomes smaller than $E_0$. With the slab parallel to
the external field, the induced surface charge on the slab is localized infinitely far away
from "where we are". Therefore, it does not contribute anything to the field "where we are",
and $E_1 = E_0$.

\noindent
A magnetizable slab parallel to the external field results in magnetization, and a corresponding
induced current in the surface of the slab. The induced current contributes to the field in the
same direction as the external field, so that $B_1$ becomes larger than $B_0$. With the slab
perpendicular to the external field, the induced surface current is localized infinitely far away
from "where we are". Hence, it does not contribute anything to the field "where we are",
and $B_1 = B_0$.



\end{document}
