you will see the equations in the pictures. e/me and Earth\'s Magnetic Field F.

Question

you will see the equations in the pictures.  

e/me and Earth's Magnetic Field F. Objective oserve the movement of electrons in electric E and magnetic B fields. The goal is to find the direction field and measure the mass to charge ratio of the electron of the Earth's magnetic F. Theory n electron beam is produced in a CRT by accelerating electrons emitted from a cathode through an electric field E= E, along the tube axis toward an anode grid within the CRT as illustrated within FIGURE 1 ine electrons gain an energy of evo mvo? by the time they pass the grid and the component of their velocity v, vo along the axis of the tube is (10.1 Where Vo Vonode -Veathode is the accelerating voltage and e and m are the charge and mass of the electron, respectively. Electrons in this focused beam travel in a vacuum toward a phosphor coated screen at the end of the tube which glows where and when they hit it. Our interest is in how the electrons are traveling at a velocity vo are moved by E and B fields. The force vector is both the electrostatic and magnetic forces F = e (E + v × B) [the x indicates the cross product, the use of the right hand rule] The path in three dimensions is actually a spiral, the detailed explanation is given in Appendix T10. A magnetic field B will turn electrons in a semicircle of radius r in the x,y plane and for each 1809 hit the screen 900 from where it would have with zero magnetic B field. It will have a speed v given by eBr (10.2) The time it takes the electrons to move from the center point between the Y plates along the axis of the tube a distance D to the face of the tube is the same as the time it takes to complete the semicircle. In both cases the time is given by distance divided by speed: (time along axis) (ime along semicircle) (10.3) Substituting vo from Equation (10.1) and v, from Equation (10.2) into Equation (10.3) and solving for m/e yields.

Explanation / Answer

A. eq 10.1

vo = sqrt(eVo/m)

KE of electron moving with speed vo = 0.5mvo^2

this KE is due to acceleration of the electron by voltage Vo

hence form conservation of energy

eVo = 0.5mvo^2

vo = sqrt(eVo/m)

10.2

v' = eBr/m

for an electon in electric field B, speed v' ( tangential speed)

then

ev'B = mv'^2/r ( r is radius of the circle it is rotating in)

hence

v' = eBr/m

B. if B is doubled

V is the same

equation 10.4 becomes

e/m = 2*pi^2*V/B^2D^2

D = pi*r*vo/v'

hence

e/m = 2*V*e^2/vo^2*m^2

now, vo remains the same

v' doubles

hence

D halves

e/m remains the same

C. this needs experimental data, but the direction of the magnetic field of the earth's srface is directed towards north at approximate angle of 23 deg with magnitude of 5*10^-5 T

D. Va = 250 V

Vo = sqrt(eVo/m) = 6629935.4413179 m/s

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