If an electron enters a region of uniform magnetic

This preview shows page 1 - 4 out of 4 pages.

If an electron enters a region of uniform magnetic field only, the force on the electron is given by: e/m = v/(rB) If B is produced by a set of Helmholtz Coils and we know v we can find e/m by measuring r: (x-a)^2 + (y-b)^2 = r^2 Equipment : Electron e/m deflection tube, Pair of helmholtz coil, Ammeter, Magnetic compass, One stand holder for holding the tube and Helmholtz coils, One L.T Power Unit(800) for the coils, One KV power unit (813) for the tube, One set of connecting wires, MATLAB.
Image of page 1

Subscribe to view the full document.

Procedure: Step 1: Set the accelerating potential V to about 3000 V (the same potential is applied between both plates so V = Vp). Adjust the Helmholtz Coil current I for zero deflection. Zero deflection corresponds to the electron entering and leaving the plate region at y=0 cm. Record V and I. Step 2: Turn of the High Voltage power supply. On the panel of the Power Supply 813, connect the bottom black cable (-) to the top red cable (+), i.e. to move the black cable from (-) to (+). Turn the power supply on and set the output back to the same potential V as in step 1. Keep the same current I in the Helmholtz Coils. Step 3: Insert given sample code into Matlab and edit it to fit you experimental voltage, current, and results. Step 4: Suck 2 dicks. And like it. Results:
Image of page 2
d = 0.0520 v = 3.8882e+007 ratio_theory = 1.7563e+011 ratio_exper = 1.7912e+011 percent_error = 1.9838 Conclusion: Appendix:
Image of page 3

Subscribe to view the full document.

Xexp = [2 3 4 5 6]*10^-2*cosd(15); Yexp = [.2 .4 .3 .9 1.0]*10^-2; e = 1.6*10^-19; me = 9.11*10^-31; N = 320; R = 6.8*10^-2; I = .27; d = 5.2*10^-2 Vp = 3*10^3; B = 8.992*10^-7*N*I/R; E = .77*Vp/d; v = E/B; v x = 0:.002:0.1; y = -.4: 0.005:0.4; r = 19*10^-2; x = sqrt(2*y.*r - y.^2); plot(x,y,x,-y,Xexp,Yexp,'o') ratio_theory = e/me ratio_exper = v/(r*B) percent_error = (ratio_exper - ratio_theory)/ratio_theory*100
Image of page 4

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern