Demonstration Description: The circular path taken by an electron beam in a magnetic field is used to determine the charge-to-mass ratio of an electron. Magnetic deflection of the circular beam can be altered by use of Hemholtz coils. One video camera displays the circular beam. The tube can also be rotated causing the beam to take a spiral path. This is displayed to the class by use of another video camera.

Demonstration Instructions: V7

Helmholtz Coils: 130 turns/coil Radius: .150m

Want e/m

mv2/2=eV .Conservation of Energy v=(2eV/m)1/2

evB=mv2/r Lorentz Force => Centripital Force eB=mv/r

Thus we get: e/m=2V/r2B2

Want to find B Intensity of magnetic field between two coils with radius R is given by: H=.7155I/R (A/m) With N turns/coil, intensity is multiplied N times Note: the permeability of a vacuum is 4p/107 So, the flux density of B field between coils is given by: B=.7155(4p/107)(NI/R) B=(8.99x10-7)NI/R (Wb/m2)

We know N=130 and R=.150m Thus B=7.79x10-4I (Wb/m2=Tesla)

To find e/m note coil current I B voltage V and beam radius r For example: V=300V (set) I=1.48A r=.05m (set either I or r and measure the other) e/m=2V/r2B2=2V/{(7.79x10-4)rI}2 =(2x300)/(7.79x10-4x.05x1.48)2 e/m=1.81x1011 C/kg