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Lab 2

by: Austin J Henry
Austin J Henry

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dr. jim
Class Notes
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This 3 page Class Notes was uploaded by Austin J Henry on Friday March 25, 2016. The Class Notes belongs to Earth 3 at University of California Santa Barbara taught by dr. jim in Winter 2016. Since its upload, it has received 27 views. For similar materials see msa in Earth Science at University of California Santa Barbara.


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Date Created: 03/25/16
Austin Henry 7874399 Wednesday 1—4 Eric Sandouk Charge to mass ratio of the Electron In this experiment we deduce the charge to mass ratio of the electron based on its deflection from a straight path under the influence of various magnetic fields. In this experiment we use a vacuum tube in the center of a pair of Helmholtz coils. Electrons are fired through a slit` parallel to the coils’ axis in low pressure to create a ribbon of electrons. This is made visible by a low concentration of mercury gas in the tube. This ribbon can then be deflected into an elliptical or circular path by running current through the coils, creating a magnetic field. The radii these paths form is measurable and the voltage controlled. Once the electron leaves the anode its energy will be entirely kinetic and 1 eq.1¿ mv =eV computable through the equation: 2 . V in this case is provided by the filament power supply and can be held constant. A magnetic field can then be introduced by the Helmholtz coils with its force acting perpendicular on the path of the electrons with the force computable by the equation: eq .2¿F=evB . For example, a charge of -2.6*10^ (-10) C moving at 10,000 m/s in the earth’s magnetic field experiences a force of 8.23*10^-11 N. Similarly, the force acting on the electron ribbon van be calculated. Using the two equations stated earlier, the charge to mass ratio of the electron can be found. The derivation of the equation used in this experiment is shown below mv 2 eq.3¿F= R Substituting F from eq. 3 into eq. 2 leads to: 2 mv R =evB mv =eB R eBR v= m e B R 2 v = 2 m Substituting this form of v^2 into eq. 1 leads to 2 2 2 1 m∗e B R =eV 2 m2 1 eB R2 =V 2 m Eq.3¿ e= 2V m B R 2 The value of B in tesla for the set of Helmholtz coils used in this experiment is: B=.00196 T ∗I ( A where I is the current measured by the coil power supply. The equipment used also includes a filament to supply a negatively charged electron cloud. An adjust-a-volt is attached to the filament power supply to ramp the current up from 0. The accelerating voltage, used to accelerate the electron cloud, is measured by an attached multimeter also used to adjust the accelerating voltage. This accelerating voltage is used to propel the electrons through a narrow slit into the vacuum chamber, forming a thin visible ribbon. The velocity of the electrons in this beam are computable from the known voltage measured by the multimeter in combination with Eq. 1). The Helmholtz coil circuit is then used to adjust the perpendicular magnetic force, used to bend the beams into a circular path. This measurable radius can be used with the other known or measurable elements of Eq. 3) to calculate the charge mass ratio of the electron. The results of our experiment are shown in the tables below: Helmholtz Accelerating B (Induced v (of Radius of e/m ratio, Coil Current, Voltage, V magnetic electron), Path (m) C/kg A field), T m/s 2.30 24.8 .032 238349313 0.004508 343833.2 3 1.97 “” .039 218729755 0.003861 329378.1 5 1.75 “” .045 208194095 0.00343 321347.6 7 1.45 “” .052 227105076 0.002842 335625 4 1.26 “” .057 250310306 0.00247 352354.8 6 2.77 30.1 .032 199445781 0.005429 346505.9 0 2.29 “” .039 196464537 0.004488 343906.5 3 2.95 “” 0.005782 231369.4 .045 889232422 1.68 “” .052 205333163 0.003293 351582.9 3 1.48 “” .057 220196903 0.002901 364085.9 0 3.01 35.0 .032 196405185 0.0059 370787.9 3 2.50 “” .039 191679940 0.0049 366300.4 5 2.08 “” .045 207986046 0.004077 381562.9 6 1.82 “” .052 203440040 0.003567 377369.9 7 1.62 “” .057 213700804 0.003175 386769.4 4 With my average e/m ratio equaling 2044176128 C/kg.


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