SP211 Lab 6 e/me ratio of the electron We will measure e/me by accelerating electrons across a potential to give them an initial speed, then measuring properties of their uniform circular motion in a B-field. Put your work/notes for this lab in your course notebook in some organized fashion. 1. The magnetic field produced by a Helmholtz coil (along its axis of symmetry) with N turns, current i, radius R, a distance z from its center, is B= µ0 N iR2 3 . 2 (R2 + z 2 ) 2 Use this expression to calculate the B-field exactly half-way between the two Helmholtz coils, when z = R/2 (this is the setup we will use). 2. The electrons are accelerated across a potential difference Vacc within the apparatus. Using the conservation of mechanical energy, determine the speed of the electrons for both Vacc = 200 V and Vacc = 300 V. 3. An electron moving perpendicular to a uniform B-field feels a force FB = evB. The direction of the magnetic force is always ⊥ to the electron’s velocity, so this sets up uniform circular motion. The magnetic force is thus a centripetal force in this situation, so we have evB = me v2 r where r is the radius of the electron’s trajectory. Using this, derive an expression for the ratio e/me in terms of v, B, and r. 4. Turn on the apparatus; DO NOT INTRODUCE A VOLTAGE OR CURRENT FOR AT LEAST 5 MINUTES. After 5 minutes, set the apparatus to Vacc = 200 V. Record the currents necessary to bend the beam into diameters of 5 cm and 10 cm. For each measurement: i. determine the magnitude of the B-field produced by the coils (note that the apparatus has N = 132 turns), and ii. extract a value for e/me using your result from step 3. 5. Repeat step 4 for Vacc = 300 V. Do you get a consistent result?