# Deflection of an Electron Beam by a Magentic Field

Topics: Electron, Cathode ray tube, Voltage Pages: 4 (1154 words) Published: November 6, 2011
Deflection of an Electron Beam by an Electric Field

Nicole N

Lab Problem 1.4 – February 3, 2011

Problem Statement:
We were asked to test the design of an electron microscope to determine how a change in the electric field affects the position of the beam spot. The goal is to find out how different variables, such as charge of the deflection plates providing a vertical electric field and initial velocity of the electron beam will affect the amount of deflection the electron beam experiences. We modeled the situation with a Cathode Ray Tube (CRT). Due to the known equation for the deflection of the beam, we predicted that the higher the voltage, the greater the deflection.

Prediction:
To begin making predictions for this experiment, we first made a diagram of what exactly our situation is, and how we can calculate the information we are looking to find, which is how the different variables will affect the total deflection of the electron beam passing through a Cathode Ray Tube with deflection plates emitting a vertical electric field. The CRT was placed in its holder at a 32.0˚ angle. To describe the total deflection, we first denoted that the change in the y-direction (coordinate system and labeled parts are indicated in the diagram below) as the electron passes through the electric field of the deflection plates is y1, and the deflection of the electron beam after it leaves the deflection plates until it hits the screen is y2. Therefore the total deflection in the y-direction = y1+ y2 . Also, L1 is the length of the region when the electron is passing between the deflection plates, and L2 is the distance from the end of the plates to the screen at the end of the CRT. Please see a diagram of the experiment below with the different variables labeled:

Next, we had to produce equations to determine the values of y1 and y2. The deflection of y1 is equal to: y1=12aT2
and using the equation Fe=qE=ma, we input qEm for the value...