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}


Context 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.
Context answer:
\boxed{$37.5$}


Context question:
d. The proton beam enters the electromagnet and is deflected by an angle $\theta=10^{\circ}$. Determine the magnetic field strength.



<img_4438>
Context answer:
\boxed{0.0894}


Question: e. The electromagnet is composed of layers of spiral wound copper pipe; the pipe has inner diameter $d_{i}=0.40 \mathrm{~cm}$ and outer diameter $d_{o}=0.50 \mathrm{~cm}$. The copper pipe is wound into this flat spiral that has an inner diameter $D_{i}=20 \mathrm{~cm}$ and outer diameter $D_{o}=50 \mathrm{~cm}$. Assuming the pipe almost touches in the spiral winding, determine the length $L$ in one spiral.



Copyright (c)2017 American Association of Physics Teachers







<img_4372>

Solution:
Treat the problem as two dimensional. The area of the spiral is



$$

A=\frac{\pi}{4}\left(D_{o}^{2}-D_{i}^{2}\right)

$$



The area of the pipe is



$$

A=L d_{o}

$$



Equating and solving,



$$

L=\frac{\pi\left(D_{o}^{2}-D_{i}^{2}\right)}{4 d_{o}}=33 \mathrm{~m}

$$

Final answer: 33 m