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\begin{center}
{\Large {Solution Derivations for Capa \#8\\[20pt]} }
\end{center}
1) A mass spectrometer applies a voltage of $2.00\ kV$ to accelerate a singly
charged ion ($+e$). A $0.400\ T$ field then bends the ion into a circular
path of radius $0.305\ m$. What is the mass of the ion?
\begin{quote}
$V$ = Given\\
$q$ = Given\\
$B$ = Given\\
$r$ = Given\\
A mass spectrometer will cause the particle to follow a circular pattern.
Thus, we know that
\[
F=ma=m\frac{v^{2}}{r}=qvB
\]
where the last step follows as a result of the magnetic field. To find the
mass of the ion, we can solve for $m$.
\begin{eqnarray*}
m\frac{v^{2}}{r} & = & qvB\\
m & = & \frac{qvBr}{v^{2}}=\frac{qBr}{v}
\end{eqnarray*}
Now the only problem is to find $r$. It is related to the kinetic energy
gained by accelerating though the electric field. Since it is initially at
rest, and exits the electric field before entering the mass spectrometer,
the final kinetic energy will be the potential energy gained by traveling
though the electric field. That is,
\begin{eqnarray*}
\frac{1}{2}mv^{2} & = & qV\\
v & = & \sqrt{\frac{2qV}{m}}
\end{eqnarray*}
Plugging $v$ into the above equation and squaring both sides,
\begin{eqnarray*}
m & = & \frac{qBr}{\sqrt{\frac{2qV}{m}}}\\
m^{2} & = & \frac{q^{2}B^{2}r^{2}}{\frac{2qV}{m}}=\frac{mqB^{2}r^{2}}{2V}\\
m & = & \frac{qB^{2}r^{2}}{2V}
\end{eqnarray*}
\end{quote}
2) A charged particle is moving perpendicularly to a magnetic field $B$.
Fill in the blank indicating the direction for the quantity missing in the
table. (If your three answers from top to bottom are +z, -x, -y, then
enter: +z-x-y in the computer). Use the diagram on the upper left for the
directions of the various axes.
\begin{quote}
Use the right hand rule. Remember that if the charge is negative, it will
move in the direction opposite that given by the right hand rule.
\end{quote}
3) The rectangular loop shown below is located in a uniform magnetic field
of $0.50\ T$ pointing in the positive x direction. What is the magnitude of
the magnetic dipole moment of the loop? DATA: $h=6.5\ cm$, $w=8.0\ cm$,
$\theta = 36.0^{\circ}$, $I = 5.0\ A$.
\begin{quote}
$h$ = Given\\
$w$ = Given\\
$\theta $ = Given\\
$I$ = Given\\
$B$ = Given\\
The magnetic dipole moment is defined by
\[
\mu =IA=I h w
\]
The direction is given by curling your fingers in the direction of $I$. It
is important for the next problem.
\end{quote}
4) Calculate the magnitude of the torque about the z-axis on the loop.
\begin{quote}
Torque is given by
\[
\tau =\mu \times B = |\mu |\, |B|\, \sin \phi
\]
Imagine looking at the problem from the top. In this case, $B$ is pointing
down ($+x$ direction), and $\mu $ points to the upper right. The total
angle between the two vectors is $180-\theta $ (if $\mu $ is pointing more
into the page than out). You can see this because if the wire loop were on
the y-axis, $\mu $ would be antiparallel to $B$. Since the loop is tilted
$\theta $ away from the y-axis, so is $\mu $. Thus, $\phi =180-\theta $.
\end{quote}
5) A proton, that is accelerated from rest through a potential of $7.0\ kV$
enters the velocity filter consisting of a parallel-plate capacitor and a
magnetic field, shown below. The E-field between the parallel capacitor
plates is $3.7 \times 10^5\ N/C$. What B-field is required so that the
protons are not deflected? (Ignore relativistic effects for high velocities.)
\begin{quote}
$V$ = Given\\
$E$ = Given\\
$q,m$ = Given (proton or electron)\\
$B$ = ?\\
We can find the force of the electric field and the force due to the
magnetic field. They must be equal for the particle to travel unaffected.
\begin{eqnarray*}
F & = & qE=qvB\\
B & = & \frac{E}{v}
\end{eqnarray*}
From problem 1, we know that $v=\sqrt{\frac{2qV}{m}}$. Thus,
\[
B=\frac{E}{\sqrt{\frac{2qV}{m}}}
\]
\end{quote}
6) A mass spectrometer, sketched below, is a device used to separate different
ions. Such ions with a well-defined velocity $v_0$ enter through a slit into
a region of uniform magnetic field B, where they follow a semicircular path
until they strike the detector slit D. The distance between the entry and the
detector slits is $d = 1.10\ m$.
Chlorine ions of mass $35\ amu$ ($1\ amu$ equals $1.66 \times 10^{-27}\ kg$),
carrying a charge of $+1e$, enter the spectrometer with initial speed of
$v_0 = 7.50 \times 10^5\ m/s$. What value of $B$ is required for these
ions to hit the detector?
\begin{quote}
$m$ = Given\\
$q$ = Given\\
$v$ = Given\\
$B$ = ?\\
\begin{eqnarray*}
F & = & \frac{mv^{2}}{r}=qvB\\
B & = & \frac{mv}{qr}
\end{eqnarray*}
Note that the mass will be some constant times one $amu$.
\end{quote}
7) A point charge moving in a magnetic field of $1.18\ Tesla$ experiences
a force of $0.744 \times 10^{-11}\ N$. The velocity of the charge is
perpendicular to the magnetic field.
In this problem, we use the points of the compass and `into' and `out of'
to indicate directions with respect to the page. If the magnetic field
points west and the force points out of the page, then give all possible
correct answers for the charge Q. (Give all correct statements in
alphabetical order: ACD, BE, etc.)
\begin{quote}
$B$ = Given\\
$F$ = Given\\
If the magnetic field points in one direction, the force perpendicular to
it, the particle must move in the other perpendicular direction. That is,
it cannot move parallel to one of the other forces. That being said, use
the right hand rule to find the direction of the force for a positive
particle moving in the direction said. If the direction of the force is
opposite to that given, the charge must be negative.
\end{quote}
8) The speed of the charge in the problem above is $0.410 \times 10^8\ m/s$.
Calculate the magnitude of the charge.
\begin{quote}
$v$ = Given\\
$q$ = ?\\
Using the equation for magnetic force,
\begin{eqnarray*}
F & = & qvB\\
q & = & \frac{F}{vB}
\end{eqnarray*}
\end{quote}
9) In the problem above, the path of the charge in the magnetic field is
a circle. Assume that the charge is positive and has a mass of
$1.673 \times 10^{-27}\ kg$, the mass of a proton. What is the radius of
the circle?
\begin{quote}
$m$ = Given\\
$r$ = ?\\
Using the equation for magnetic force and the charge calculated in the
previous problem
\begin{eqnarray*}
F & = & qvB=m\frac{v^{2}}{r}\\
r & = & \frac{mv^{2}}{qvB}=\frac{mv}{qB}
\end{eqnarray*}
\end{quote}
10) In the problem above, what is the orbital frequency, of the charge
in the previous problem?
\begin{quote}
The book gives the equation (and the derivation) for frequency on page
729-730 (eqn 29-5). Using previously calculated and given values,
\[
f=\frac{qB}{2\pi m}
\]
\end{quote}
11) Five particles follow the paths shown in the figure as they pass through
a magnetic field B, which is indicated by the dots in the figure (B out of
the page). What can you conclude about the charge of each particle? (If
your answers for charges of particles 1, 2, 3, 4, and 5, are, respectively,
$+$, $0$, $+$, $-$, $0$, then enter +0+-0 into the computer.)
\begin{quote}
Use the right hand rule to find the direction of the force. If
the particle moves in the direction of the force, it is a positively charged
particle. If it moves opposite of the force, it is negatively charged.
If it is unaffected by the force, then it is a neutral particle.
\end{quote}
12) A $7\ kg$ particle carrying a charge of $65\ \mu C$ enters a uniform
$4\ T$ magnetic field at a speed of $63\ m/s$ and with an angle of
$63^{\circ}$ with respect to the field lines, as shown in the figure.
Answer the following questions: (Select T-True, F-False, If the first is T
and the rest F, enter TFFFF).
\begin{quote}
QUESTION:\\
A) The y-component of the particle's velocity is unchanged as it passes
through the B-Field.\\
B) The Force on the particle is in the -z direction.\\
C) The particle's speed varies as it passes through the B-Field.\\
D) The work done by the field on the particle is zero as the force is normal
to the displacement.\\
E) The particle follows a helical path.\\
ANSWER:\\
A) False. Only the direction parallel to the magnetic field will be
unchanged. Another way to see this is that the force into the page will
cause the particle to move in a circular direction around the x-axis. This
must change the y-velocity (and the z-velocity).\\
B) True. Use the right hand rule.\\
C) False. The magnetic field cannot change the particles speed, only the
direction of the velocity.\\
D) True. The force of the magnetic field is always perpendicular to the
movement of the particle. Thus, the dot product involved in work equates
to zero.\\
E) True. As described in (A) and page 732 (fig. 29-15) of the book, a
particle moving in a magnetic field will travel a helical path.
\end{quote}
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