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\begin{bf}
\noindent
Institutt for fysikk, NTNU

\noindent
TFY4155/FY1003 Elektrisitet og magnetisme

\noindent
V{\aa}r 2005

\end{bf}

\begin{center}
\begin{large}
Solution to {\o}ving 12
\end{large}
\end{center}

\noindent
Guidance Thursday April 7 and Friday April 8


\vspace{0.5cm}
\noindent
{\em Exercise 1}

\vspace{0.2cm}
\noindent
In the first experiment, $B=0$. Then, Newton's second law is:
\begin{eqnarray}
\bold{F} &=& m\bold{a} = q\bold{E} \nonumber \\
\Rightarrow \frac{d\bold{v}}{dt} &=& \frac{q}{m} \bold{E} \nonumber \\
\Rightarrow \bold{v}(t) &=& \bold{v}(0) + \frac{q}{m} \bold{E} t = \frac{d\bold{r}}{dt} \nonumber \\
\Rightarrow \bold{r}(t) &=& \bold{r}(0) + \bold{v}(0)t + \frac{q}{2m} \bold{E} t^2 \nonumber
\end{eqnarray}
Here, it is natural to choose $t=0$ the moment the particle enters the region
with $E \neq 0$, and furthermore, to choose the origin in this position:
$$\bold{r}(0) = (x_0,y_0) = (0,0)$$
Here, the velocity is
$$\bold{v}(0) = v\;\hat{x}$$
when we orient the $x$ axis towards the right. The $y$ axis is oriented upwards,
so that
$$\bold{E} = -E \; \hat{y}$$
(i.e., with $E>0$) The particle trajectory thus becomes a parabola, just like when we
throw a mass in the field of gravity. The velocity in the $x$ direction is not affected
by the electric field, so
$$x(t) = vt$$
whereas the particle obtains a constant acceleration in the $y$ direction, i.e.,
the displacement in the $y$ direction, as a function of $t$, must be determined by
$$y(t) = - \frac{q}{2m} E t^2$$
The particle will leave the region where $E \neq 0$ at the moment
$$t_L = \frac{x(t_L)}{v} = \frac{L}{v}$$
The vertical position is then
$$y(t_L) = -\frac{q}{2m}E\frac{L^2}{v^2}$$
Already, we may conclude that $q < 0$ if $y(t_L) > 0$.

\noindent
The distance from $x=L$ to $x=L+D$ is then traveled without influence from any kind of forces,
and with a direction relative to the $x$ axis in terms of the angle $\alpha$, where
$$\tan \alpha = \frac{v_y(t_L)}{v_x(t_L)} = \frac{-\frac{q}{m}E\frac{L}{v}}{v} = -\frac{qEL}{mv^2}$$
Besides, we must have
$$\tan \alpha = \frac{y-y(t_L)}{D}$$
where $y$ is where the electron hits the detector, at $x=L+D$.

\vspace{0.5cm}
\noindent
The experiment is then repeated with the same $E$-field, but now we turn on a
magnetic field $B$ directed into the plane, so that the particles are no longer deflected
by the fields. This implies that the electric force (upwards) is exactly balanced by
the magnetic force (downwards). In other words:
\begin{eqnarray}
\bold{F} &=& q\bold{E} + q\bold{v}\times\bold{B} = 0 \nonumber \\
\Rightarrow E &=& vB \nonumber \\
\Rightarrow \frac{1}{v} &=& \frac{B}{E} \nonumber
\end{eqnarray}
Hence:
\begin{eqnarray}
\frac{y-y(t_L)}{D} &=& -\frac{qEL}{mv^2} = -\frac{qEL}{m} \cdot \frac{B^2}{E^2} \nonumber \\
\Rightarrow y + \frac{q}{2m}E L^2\frac{B^2}{E^2} &=& -\frac{qEL}{m} \cdot \frac{B^2}{E^2}D \nonumber \\
\Rightarrow yE &=& -\frac{q}{m} \cdot B^2 \left(DL+\frac{1}{2}L^2\right) \nonumber \\
\Rightarrow \frac{q}{m} &=& -\frac{yE}{B^2 \left(DL +\frac{1}{2}L^2\right)} \nonumber
\end{eqnarray}
I.e.,
$$a = \frac{E}{B^2 \left(DL + L^2/2 \right)}$$


\vspace{0.5cm}
\noindent
{\em Exercise 2}

\vspace{0.2cm}
\noindent
$a)$ The speed of the ions when they enter the region with magnetic field is
determined by the change in potential energy, going through the voltage difference
$V$, being equal to the change in the kinetic energy of the ions:
$$eV = \frac{1}{2}mv^2 \;\; \Rightarrow \;\; v = \sqrt{\frac{2eV}{m}}$$
The centripetal acceleration inside the magnetic field is
$$a = \frac{v^2}{r}$$
so that Newton's 2. law gives
$$F = m\frac{v^2}{r} = evB \;\;\; \Rightarrow \;\;\; r = \frac{mv}{eB}$$
Radius for the resulting circular path for a particle with mass $m$ becomes
$$r = \frac{1}{B} \sqrt{\frac{2Vm}{e}}$$
i.e., proportional with $\sqrt{m}$. Radii and masses for the different isotopes
must be related as follows:
$$\frac{r_i}{r_j} = \sqrt{\frac{m_i}{m_j}}$$
where $i,j = 79$ or 81.

\noindent
If the points where the ions hit the photographic plate are supposed to be separated
by a distance of (at least) $a=1.0$ cm, the {\em diameter} of the two circular paths must differ by 1.0 cm.
We obtain
$$a = 1.0 \;{\rm cm}\; = 2(r_{81} - r_{79}) = 2r_{79} \left(\sqrt{\frac{m_{81}}{m_{79}}} - 1 \right)$$
This gives
$$r_{79} = \frac{a}{2} \left(\sqrt{\frac{m_{81}}{m_{79}}} - 1 \right)^{-1} = 
0.5 \;{\rm cm}\; \cdot \left(\sqrt{\frac{81}{79}} -1 \right)^{-1} \simeq 39.7 \;{\rm cm}$$
and
$$r_{81} = r_{79} + \frac{a}{2} \simeq 40.2 \;{\rm cm}$$
Now, we can determine how strong magnetic field that can be used to achieve these radii:
$$B = \frac{1}{r_{81}} \sqrt{\frac{2Vm_{81}}{e}} = 
\frac{1}{0.402} \cdot \sqrt{\frac{2 \cdot 400 \cdot 81 \cdot 1.67 \cdot 10^{-27}}{1.6\cdot10^{-19}}} =
0.065 \; {\rm T}$$
This represents the upper limit of $B$: A stronger magnetic field will reduce both
$r_{79}$ and $r_{81}$, but $r_{81}$ the most, so that the "hit points" move closer to each other.
However, at the same time the diameter $d_{81} = 2 r_{81}$ must not be larger than the physical limit
of the instrument, given by $L = 250 cm$.
That corresponds to a minimum value of the magnetic field strength:
$$B_{\rm min} = \frac{1}{L/2} \sqrt{\frac{2Vm_{81}}{e}} =
\frac{1}{1.25} \cdot \sqrt{\frac{2 \cdot 400 \cdot 81 \cdot 1.67 \cdot 10^{-27}}{1.6\cdot10^{-19}}} =
0.021 \; {\rm T}$$
In other words, we may use a magnetic field between 21 and 65 mT.


\vspace{0.5cm}
\noindent
{\em Exercise 3}

\vspace{0.2cm}
\noindent
$a)$ In this exercise, we may argue somewhat along the same lines as we did when we
calculated the electric field on the symmetry axis of a uniformly charged ring. In that case,
we looked at the contributions to the total field from diametrically opposite charge
elements $dq$ and convinced ourselves that the total electric field had to be directed along
the symmetry axis.

\noindent
Here, we may e.g. look at the two current elements lying exactly on the positive and
the negative $y$ axis, respectively, and determine the direction of the contribution to
the magnetic field on the $z$ axis from these two. Let us look at positive values of $z$ first.
(See figure below.) The "current element" $I\;d\bold{l}$ which crosses the positive $y$ axis
has its direction along the negative $x$ axis. The cross product of this vector with $\bold{r}$ 
from the current element to the actual position on the positive $z$ axis becomes a vector lying
in the $yz$ plane, with positive $y$ and $z$ components. The diametrically opposite current
element, i.e., the one crossing the negative $y$ axis, is directed along the positive $x$ axis. The
cross product of this vector with $\bold{r}$ from the current element to the actual position
on the positive $z$ axis becomes a vector which also lies in the $yz$ plane, but this vector
will have a {\em negative} $y$ component and positive $z$ component. From symmetry reasons, these
two contributions to $\bold{B}$ must be equal in absolute value, have equal $z$ components (with
the same sign), and have equal $y$ components, but with the opposite sign.
Thus, the sum of these two contributions must point along the positive $z$ axis.

\noindent
In this way, we could argue for any pair of diametrically opposite current elements around
the ring. They will all have an equal $z$ component with the same sign, and equal $x$ and $y$
components with opposite sign.

\noindent
Conclusion: $\bold{B}$ on the positive $z$ axis is directed along the positive $z$ axis.

\vspace{0.2cm}
\noindent
$b)$ In $a)$, we convinced ourselves that $\bold{B}(z)$ is directed along the positive $z$
axis if $z>0$. What about $z<0$?

\noindent
Looking again at the figure below, we see that the current element that crosses the
positive $y$ axis will give a contribution to $\bold{B}(z)$ on the negative $z$ axis which
lies in the $yz$ plane, with positive $z$ component and negative $y$ component. For the
current element crossing the negative $y$ axis, we find a contribution with positive
$z$ component and positive $y$ component. Altogether, a magnetic field directed along
the positive $z$ axis.

\noindent
Conclusion: The magnetic field points along the postive $z$ axis everywhere on the $z$ axis.

\vspace{0.2cm}
\noindent
$c)$ The vectors $I\;d\bold{l}$ and $\hat{r}$ are perpendicular to each other. Thus
$$|I\;d\bold{l} \times \hat{r}| = I R\; d\phi \cdot 1$$
since a line element $dl$ along a circle equals the radius $R$ multiplied with
the "angle element" $d\phi$:

%%FIGURE
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\vspace{0.5cm}
%%END FIGURE

\noindent
The direction of $d\bold{B}$ must be as shown in the figure:

%%FIGURE
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\centerline{\epsfxsize=4.0in \epsfbox{los12_fig1.eps}}
\vspace{0.5cm}
%%END FIGURE

\noindent
From this figure we notice that
$$\frac{dB_z}{dB} = \sin\theta = \frac{R}{r} = \frac{R}{\sqrt{z^2+R^2}}$$
and this is precisely the $z$ component of the magnetic field that we are looking for.
The absolute value of $d\bold{B}$ becomes
$$dB = \frac{\mu_0}{4\pi}\cdot \frac{I R\; d\phi}{z^2+R^2}$$
so that
$$dB_z = dB\sin\theta = \frac{\mu_0}{4\pi}\cdot \frac{I R\; d\phi}{z^2+R^2}
\cdot \frac{R}{\sqrt{z^2+R^2}} = \frac{\mu_0 I R^2\; d\phi}{4\pi\left(z^2+R^2\right)^{3/2}}$$
The total $z$ component, and therefore the total magnetic field, is obtained by integrating
the contributions from all the current elements around the ring, i.e., by integrating this
expression over the angle $\phi$ from 0 to $2\pi$:
$$B(z) = \int dB_z = \frac{\mu_0 I R^2}{4\pi\left(z^2+R^2\right)^{3/2}}
\int_0^{2\pi}\; d\phi = \frac{\mu_0 I R^2}{2\left(z^2+R^2\right)^{3/2}}$$
which is what we were supposed to show.

\vspace{0.2cm}
\noindent
$d)$ Far away form the current loop, we may write
$$z^2+R^2 \simeq z^2$$
Thus, the magnetic field is approximately
$$B(z) \simeq \frac{\mu_0 I R^2}{2z^3}$$
The magnetic dipole moment of the current loop is
$$m = IA = I \cdot \pi R^2$$
so we may write this magnetic field as
$$B(z) = \frac{\mu_0 m}{2\pi z^3}$$
It is well worth comparing this result with the electric field on the
axis of an electric dipole, in large distance $z$ from the dipole. This is
something we already did in {\o}ving 5, where we found
$$E(z) = \frac{p}{2\pi\varepsilon_0 z^3}$$
Here, $p$ is the electric dipole moment of the dipole. In other words, exactly
the same result, with $m$ instead of $p$, and $\mu_0$ instead of $1/\varepsilon_0$.

\noindent
We will find more analogies between electrostatics and magnetostatics as we move along!


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