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Institutt for fysikk, NTNU

\noindent
TFY4155/FY1003: Elektrisitet og magnetisme

\noindent
V{\aa}r 2005

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{\O}ving 14
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\noindent
Guidance: Thursday April 21 and Friday April 22

\noindent
To be delivered by: Monday April 25

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\noindent
{\em Exercise 1}

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\noindent
$a)$ A bar magnet har its magnetization vector $\bold{M}$
in the direction from the south pole (S) towards the north pole (N).
Draw a bar magnet and sketch field lines for $\bold{B}$ inside and
around the bar magnet. Also include the bound surface current
$I_m$ (including its direction) associated with the magnetization
$\bold{M}$.

\vspace{0.5cm}
\noindent
$b)$ Explain, on the basis of $a)$, why two bar magnets placed
after each other with S agaist N attract each other and why they
repel each other when they are placed with S against S (or N against N).
(Hint: Use for example what you know about magnetic force on an
electric current in your argumentation.)

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\noindent
$c)$ Explain next why a sphere (or a needle or an object with
any shape) of unmagnetic iron is attracted both by the S-pole
and the N-pole of the bar magnet.


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\noindent
{\em Exercise 2}

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\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.)

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\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?

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

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\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.


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\noindent
{\em Exercise 3}

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\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...)

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\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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