In your laboratory, you set up an experiment with an electron gun that emits electrons with energy of \(7.50 \mathrm{keV}\) toward an atomic target. What deflection (magnitude and direction) would Earth's magnetic field \((0.300 \mathrm{G})\) produce in the beam of electrons if the beam is initially directed due east and covers a distance of \(1.00 \mathrm{~m}\) from the gun to the target? (Hint: First calculate the radius of curvature, and then determine how far away from a straight line the electron beam has deviated after \(1.00 \mathrm{~m}\).)

First, we will use the given energy of the electrons and convert it into Joules. Then, we will find the velocity using the kinetic energy formula. The given energy of the electrons is 7.50 keV. $$ \begin{aligned} E &= 7.50 \times 10^3 \mathrm{~eV} \end{aligned} $$ To convert the electron-volt to Joule, we will multiply by electron charge in Joule units \((1 \mathrm{~eV} = 1.6 \times 10^{-19}\mathrm{~J})\). $$ \begin{aligned} E_\text{Joule} &= E \times 1.6 \times 10^{-19}\mathrm{~J/eV} \end{aligned} $$ As we have the energy in Joule, the relationship between kinetic energy and velocity is given by $$ \begin{aligned} E_\text{Joule} &= \frac{1}{2}m_\text{electron}v^2 \end{aligned} $$ Where \(m_\text{electron} = 9.11 \times 10^{-31}\mathrm{~kg}\) is the mass of the electron. Solving for the velocity, we get $$ \begin{aligned} v &= \sqrt{\frac{2E_\text{Joule}}{m_\text{electron}}} \end{aligned} $$

The magnetic force acting on the electrons is given by $$ \begin{aligned} F_\text{mag} &= qvB \end{aligned} $$ Where \(q\) is the electron charge, \(v\) is the electron velocity, and \(B\) is the magnetic field strength. The strength of Earth's magnetic field is given as \(0.300\mathrm{~G}\), which we will need to convert to Tesla (1 Gauss = \(10^{-4}\) Tesla). $$ \begin{aligned} B_\text{Tesla} &= 0.300 \times 10^{-4}\mathrm{~T} \end{aligned} $$

The magnetic force acting on the electrons will cause them to move in a circular path. The centripetal force for the circular motion will be equal to the magnetic force. $$ \begin{aligned} F_\text{centripetal} = F_\text{mag} &= \frac{m_\text{electron}v^2}{r} \end{aligned} $$ Where \(r\) is the radius of the circular path. To find the radius, we can rearrange the equation as follows: $$ \begin{aligned} r &= \frac{m_\text{electron}v}{|q|B_\mathrm{Tesla}} \end{aligned} $$

To find the deflection (magnitude), we can use the equation for a right triangle, with its hypotenuse as the radius \(r\), and one side along the initially straight path as \(1.00\mathrm{~m}\). The angles between \(1.00\mathrm{~m}\) and the hypotenuse (at the target and electron gun) are the same, so let this angle be \(\theta\). Using the law of cosines, we can find the length of the other side, which is the deflection: $$ \begin{aligned} \text{Deflection}^2 &= r^2 + (1.00\mathrm{~m})^2 - 2 r (1.00\mathrm{~m})\cos(\theta) \end{aligned} $$ As the angle between the magnetic field and the velocity is \(90^\circ\), the magnetic field is pointing north. Since electrons are negatively charged, they will be deflected to the south. Thus, the direction of the deflection is south.