An electron in the beam of a TV tube is accelerated (from the rest) by a potenti

Question

An electron in the beam of a TV tube is accelerated (from the rest) by a potential difference of VAB=20 kV between points A and B, Figure 2. Then it passes through a region of transverse magnetic field, where it moves first straight between parallel plates with electric field established between them (BC is a straight line) and then along a circular arc with diameter CD. Report your results into a Table on the front page

a) Is VAB >0?

b) What is the speed of the electron at point B?

c) What is direction of the magnetic field, toward the page or out of page? Explain your choice.

d) What is orientation of the electric field between the plates: toward G, F, B or C? Explain your choice.

e) What is the polarity of the voltage (show “+” and “–“ signs) established between the plates. Report on front page like this .

f) If B=0.01 T, what is the distance CD?

g) What is the magnitude of the electric field between the parallel plates?

h) If instead of the electron, a proton is accelerated between points A and B that have the same magnitude of VAB =20 kV but of opposite polarity, what magnitudes of the electric field between parallel plates and magnetic field are required so that the proton also arrives to point D?

for picture of fig2

https://www.chegg.com/homework-help/questions-and-answers/2-electron-beam-tv-tube-accelerated-rest-potential-difference-vab-20-kv-points-b-figure-2--q11521535

or use

http://i65.tinypic.com/29vkwau.png

please answer all the questions

Explanation / Answer

a) No Vab must be negative to acclerate electron.

b) KE gain = q deltaV

m v^2 /2 = q (20 kV)

9.109 x 10^-31 x v^2 /2 = 1.6 x 10^-19 x 20 x 10^3

v = 8.38 x 10^7 m/s

c) magnetic force = q ( v X B)   {this is cross product)
at the instance C,

force is toward C to D

velocity to the right

q is negative

hence B will be out of the page.

d) magnetic force is upwards hence electric force on elctron must be downwards

Fe = q E

Q is negative

hence E will be upwards/

e) field is from high potential to low potentail.

G to F.

f) Fm = m v^2 / r

q v B = m v^2 / r

r = m v / qB

r = (9.109 x 10^-31 x 8.38 x 10^7 ) / (1.6 x 10^-19 x 0.01)

r = 0.0477 m

CD = 2r = 0.095 m

g) E = vB = 8.38 x 10^7 x 0.01 = 838000 N/C

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