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\noindent
Institutt for fysikk, NTNU

\noindent
TFY4155/FY1003: Elektrisitet og magnetisme

\noindent
V{\aa}r 2005

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{\O}ving 12
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\noindent
Guidance: Thursday April 7 and Friday April 8

\noindent
To be delivered by: Monday April 11

\vspace{0.7cm}
\noindent
{\em Exercise 1}

\vspace{0.2cm}
\noindent
Particles with mass $m$, charge $q$, and velocity $\bold{v}$ enter a region
with "crossed" electric and magnetic field, $\bold{E}$ and $\bold{B}$,
as shown in the figure. $\bold{E}$ is directed downwards, and
$\bold{B}$ is directed into the paper plane. In the region with extent
$L$ we assume that the fields are homogeneous. Outside this region,
$\bold{E}=\bold{B}=0$.

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\noindent
You keep the electric field strength $E$ constant throughout the experiment.
First, you set $B=0$ and register that the particles are deflected and hit
the detector in a distance $y$ above the center line (which is dashed). Next,
you repeat the experiment, but this time you adjust the value of $B$ until
the particles are no longer deflected.

\noindent
Show that the ratio between the charge and the mass of the particles is proportional
to the deflection (when $B=0$), i.e.,
$$\frac{q}{m} = - a y,$$
and find $a$ expressed in terms of $E$, $B$, $D$, and $L$.

\vspace{0.5cm}
\noindent
Given information: $\bold{F} = q \left(\bold{E} + \bold{v} \times \bold{B} \right)$ \hspace{0.3cm} 
(the Lorentz force)

\vspace{0.5cm}
\noindent
{\small In this manner, J. J. Thomson analyzed so called cathode rays in 1897, and
showed that these rays consisted of a particular type of particles with negative charge.
These were electrons that were emitted from the metal of the cathode. Thomson was the
first one to determine the ratio $e/m_e$. He found the same value for this ratio, independent
of what kind of metal he used in the cathode. Thus, he could conclude that the observed particles
had to be a {\em fundamental} ingredient of nature.}


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\noindent
{\em Exercise 2}

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\noindent
The figure shows a mass spectrometer. An ion source emits charged particles.
Two slits, $S_1$ and $S_2$, make sure that a well {\em collimated} beam of
particles enter into the region with the magnetic field $\bold{B}$ (which
is directed into the paper plane). Between $S_1$ and $S_2$, we have a voltage
difference $V$ which accelerates the ions. The particle velocity at $S_2$ is much larger
than at $S_1$, so we may put $v=0$ at $S_1$. The ions are deflected through
an angle 180$^\circ$ by the magnetic field and are detected on a photographic plate $P$.

\noindent
The spectrometer is supposed to be used to separate the bromine isotopes $^{79}$Br and $^{81}$Br.
The source emits these isotopes as ions with charge $-e$. You  may assume that
the isotopes have atomic masses 79$m_p$ and 81$m_p$, respectively.

\noindent
On the photographic plate, we want a separation of at least 1.0 cm between the points
where the two isotopes strike. At the same time, we must make sure that both isotopes
really do hit the plate, which has a width $L=250$ cm, measured from where the ions enter
into the magnetic field. With these two conditions, what is the upper and lower limit
for the strength of the magnetic field $B$, when the
accelerating voltage drop $V$ is 400 V?

\vspace{0.3cm}
\noindent
Given information: $e = 1.6 \cdot 10^{-19}$ C, $m_p = 1.67 \cdot 10^{-27}$ kg \hspace{1.5cm}

\noindent
{\small [One of the answers is: $B_{\rm min} = 21$ mT]}

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\vspace{0.7cm}
\noindent
{\em Exercise 3}

\vspace{0.2cm}
\noindent
A circular current loop with radius $R$ conducts an electric current $I$. The current
loop lies in the $xy$ plane, with its center at the origin. The direction of $I$ is
counterclockwise when we have the positive $z$ axis out of the paper plane. In this
exercise, we will determine the resulting magnetic field $\bold{B}(0,0,z) = \bold{B}(z)$ 
{\em on the symmetry axis} of the current loop (i.e., on the $z$ axis).

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\vspace{0.3cm}
\noindent
$a)$ Why are the $x$ and the $y$ components of $\bold{B}(z)$ equal to zero?
(Hint: Symmetry.)

\vspace{0.3cm}
\noindent
$b)$ What is the direction of $\bold{B}(z)$ for positive and negative values of $z$?

\vspace{0.3cm}
\noindent
$c)$ Use Biot--Savart's law,
$$d\bold{B}=\frac{\mu_0}{4\pi}\frac{Id\bold{l}\times\hat{r}}{r^2} \;\;(=\;\;
\frac{\mu_0}{4\pi}\frac{Id\bold{l}\times\bold{r}}{r^3}\;\;)$$ 
to show that
$$B(z) = \frac{\mu_0 I R^2}{2\left(z^2+R^2\right)^{3/2}}$$

\vspace{0.3cm}
\noindent
$d)$ Determine $B(z)$ far away from the current loop (i.e., to leading order
when $z \gg R$) and express your answer in terms of the {\em magnetic dipole moment} 
$m = |\bold{m}|$ of the current loop.

\vspace{0.1cm}
\noindent
Magnetic dipole moment $\bold{m}$ for a plane, closed current loop enclosing
an area $A$ is, by definition,
$$\bold{m} = I\bold{A} = I A \; \hat{n}$$
where $\hat{n}$ is the unit vector normal to the plane, enclosed surface.
Magnetic dipole moment is therefore a {\em vector} (just like electric dipole
moment $\bold{p}$ is a vector).
Positive direction of $\bold{m}$ is defined by the right hand rule: Four fingers
in the direction of the current makes the thumb point in the same direction as $\bold{m}$.


\vspace{0.5cm}
\noindent
{\small Comments:

\noindent
Note that different books use a bit different notation here:
Some call it "magnetic dipole moment", others simply "magnetic moment".
Some use the symbol $\bold{\mu}$, others use $\bold{m}$. 
Anyway, it is all the same physical quantity! We will use the symbol
$\bold{m}$ and call it magnetic dipole moment, like in e.g. the Norwegian book
(LHL) and in Griffiths. Fishbane, and Young and Freedman, use
$\bold{\mu}$.

\vspace{0.2cm}
\noindent
Perhaps this is also the right moment to mention that both electric and magnetic
dipole moment have a more {\em general} definition than what we use in this course.
(After all, the world around us does not only consist of pairwise point charges
of opposite sign and plane current loops...!) If we have a volume charge density
$\rho(\bold{r})$, the electric dipole moment $\bold{p}$ is, by definition,
$$\bold{p} = \int \bold{r} \rho(\bold{r})\; d^3 r$$
And, if we have some kind of current distribution, specified by the current density
$\bold{j}(\bold{r})$, the magnetic dipole moment $\bold{m}$ is, by definition,
$$\bold{m} = \frac{1}{2} \int \bold{r} \times \bold{j}(\bold{r})\; d^3 r$$
Here, the integral runs over "all space", i.e., wherever $\rho$ or $\bold{j}$
are different from zero.
For the {\em special cases} that we (mainly) concern ourselves with in this course, namely
pairwise point charges $\pm q$ separated by a distance described by the vector
$\bold{d}$, and plane current loops with stationary current $I$ enclosing an area
described by the vector (sometimes called the "vector area") $\bold{A} = A\;\hat{n}$,
these general definitions reduce exactly to
$$\bold{p} = q\bold{d}$$
and 
$$\bold{m} = I\bold{A}$$
The general definitions of $\bold{p}$ and $\bold{m}$ are {\em not} "pensum" in this
course.}

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