a) Estimate the approximate maximum deflection of the electron beamnear the cent
ID: 1670943 • Letter: A
Question
a) Estimate the approximate maximum deflection of the electron beamnear the center of a CRT television screen due to Earth's 5.0*10^-5T field. Assume the screen is 18 cm from the electron gun, wherethe electrons are accelerated by 2.0 kV. b) Same problem except 28 kV.Note that in color TV sets, the beam must be directedaccurately to within less than 1mm in order to strike the correctphosphor. Because Earth's field is significant here, mu-metalshields are used to reduce the Earth's field in the CRT.
I am having trouble setting up the formula for the deflection.Any help you can give me will be greatly appreciated!! b) Same problem except 28 kV.
Note that in color TV sets, the beam must be directedaccurately to within less than 1mm in order to strike the correctphosphor. Because Earth's field is significant here, mu-metalshields are used to reduce the Earth's field in the CRT.
I am having trouble setting up the formula for the deflection.Any help you can give me will be greatly appreciated!!
Explanation / Answer
Potential difference V = 2 kV = 2000 V
Speed of the eeltron after passing through V is v= [ 2Vq / m
Where m = mass = 9.11 * 10 ^ -31 kg
q =charge = 1.6 * 10 ^ -19 C
Plug the values we get v = 26.505 * 10 ^ 6 m / s
Horizontal distance X = 18 cm =0.18 m
Time t= X / v
= 6.79 * 10 ^ -9 s
In vertical direction :
Force acting on electron F = Bvq
= ( 5 * 10 ^ -5 T * 26.505 * 10^ 6 * 1.6 * 10 ^ -19 )
= 2.1204 * 10 ^ -16 N
Accleration a = F / m
= 2.32 * 10 ^ 14 m / s ^ 2
Initial celocity u = 0
From the relation S = ut + ( 1/ 2) at^2
S = 0 + 0.00536 m
= 5.36 * 10 ^ -3 mm
Therefore required deflection = 5.36 * 10 ^ -3 m
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