Subfield: Electromagnetism

$g=9.8 \mathrm{~N} / \mathrm{kg}$



$k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$



$c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$



$N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$



$\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$



$1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$



$m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$



$m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$



$\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$



$$

\begin{aligned}

& G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

& k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

& k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

& R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

& e=1.602 \times 10^{-19} \mathrm{C} \\

& h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\

& (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

& \ln (1+x) \approx x \text { for }|x| \ll 1 \\

& \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1

\end{aligned}

$$

$$

\begin{array}{lrl}

\text { velocity parameter } & \beta & =\frac{v}{c} \\

\text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\

\text { relativistic momentum } & p & =\gamma \beta m c \\

\text { relativistic energy } & E & =\gamma m c^{2} \\

\text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}}

\end{array}

$$

Extra Supplementary Reading Materials:

Beloit College has a "homemade" $500 \mathrm{kV}$ VanDeGraff proton accelerator, designed and constructed by the students and faculty.

<img_4547>



Accelerator dome (assume it is a sphere); accelerating column; bending electromagnet



The accelerator dome, an aluminum sphere of radius $a=0.50$ meters, is charged by a rubber belt with width $w=10 \mathrm{~cm}$ that moves with speed $v_{b}=20 \mathrm{~m} / \mathrm{s}$. The accelerating column consists of 20 metal rings separated by glass rings; the rings are connected in series with $500 \mathrm{M} \Omega$ resistors. The proton beam has a current of $25 \mu \mathrm{A}$ and is accelerated through $500 \mathrm{kV}$ and then passes through a tuning electromagnet. The electromagnet consists of wound copper pipe as a conductor. The electromagnet effectively creates a uniform field $B$ inside a circular region of radius $b=10 \mathrm{~cm}$ and zero outside that region.



<img_4492>



Only six of the 20 metals rings and resistors are shown in the figure. The fuzzy grey path is the path taken by the protons as they are accelerated from the dome, through the electromagnet, into the target.
Context question:
a. Assuming the dome is charged to $500 \mathrm{kV}$, determine the strength of the electric field at the surface of the dome.
Context answer:
\boxed{$10^{6}$}


Context question:
b. Assuming the proton beam is off, determine the time constant for the accelerating dome (the time it takes for the charge on the dome to decrease to $1 / e \approx 1 / 3$ of the initial value.
Context answer:
\boxed{0.556}


Question: c. Assuming the $25 \mu \mathrm{A}$ proton beam is on, determine the surface charge density that must be sprayed onto the charging belt in order to maintain a steady charge of $500 \mathrm{kV}$ on the dome.

Solution:
There are several ways the dome can discharge, two of which are along the resistors and the proton beam. We will ignore any other path.



At $500 \mathrm{kV}$, the current through the resistor chain is $50 \mu \mathrm{A}$, from $V=I R$. So the total current $I$ needed to be supplied to the dome is $75 \mu \mathrm{A}$. This is sprayed onto the belt, which moves at a rate of



$$

\frac{\delta A}{\Delta t}=v_{b} w

$$



so the necessary surface charge density is



$$

\sigma=\frac{I}{v_{b} w}=\frac{(75 \mu \mathrm{C} / \mathrm{s})}{(20 \mathrm{~m} / \mathrm{s})(0.10 \mathrm{~m})}=37.5 \mu \mathrm{C} / \mathrm{m}^{2}

$$

Final answer: $37.5$ $\mu \mathrm{C} / \mathrm{m}^{2}$