Considers Thompson’s 2nd experimental measurement of e/m. Thankfully, legions of
ID: 1693086 • Letter: C
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 mRelated Questions
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