Considers Thompson’s 2nd experimental measurement of e/m. Thankfully, legions of

Question

Considers Thompson’s 2nd experimental measurement of e/m. Thankfully, legions of physicists have already figured out for you the exact value of e/m.
a) Knowing that, if I were to give you a cathode-ray-tube, where the length of the deflection plates were 1 cm, and the distance from the end of the deflection plates to the phosphorescent screen was 20 cm, and your electric field strength was E=10 kV/cm (1 MV/m), and the strength of your magnetic field was B=0.01 T, what is the magnitude of the deflection you could expect at the screen?
b) If you have a good month and don’t spend too much money on ITunes or at Starbucks, you could go to Magnets R’ Us and buy a better magnet, say one that has a strength of B=1 T. If you kept everything else the same with your cathode-ray-tube and used the new magnet, what would the deflection of the electron beam be?

Explanation / Answer

length of the plates   L = 1 cm = 0.01 m distance between the plates and screen D = 20 cm = 0.2 m electric field strength   E = 10 kV /cm   = 1 M V /m strength of magnetic field   B = 0.01 T initial speed of electron                      B q v = E q                            v = E / B              time taken by the electron to travel distance d = L +D                     t = L +D / v        magnitude of the deflection you could expect at the screen                    y =    (L +D )^2 g / 2 v ^2                        = ( 0.21^2 )(9.8 )(B^2 ) / ( 2 * E^2 )                          =   2.1609*10^ -17 m at B = 1 T                y =    (L +D )^2 g / 2 v ^2                        = ( 0.21^2 )(9.8 )(B^2 ) / ( 2 * E^2 )                          =   2.1609*10^ -11 m                        = ( 0.21^2 )(9.8 )(B^2 ) / ( 2 * E^2 )                          =   2.1609*10^ -11 m                        

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