Early black-and-white television sets used an electron beam to draw a picture on

Question

Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were accelerated by a voltage of 3.0 kV; the beam was then steered to different points on the screen by coils of wire that produced a magnetic field of up to 0.59T .

A) What is the speed of electrons in the beam? using KE= PE, I got (1/2)mV^2= e*Voltage, the answer is suppose to be 3.2*10^7 but I can't get that answer, can you please elaborate?

B) What acceleration do they experience due to the magnetic field, assuming that it is perpendicular to their path? What is this acceleration in units of g? This one I can't even if I use F=qvB, and F=ma, where a= qvB/m, I thought the mass of the electron was

9.109 x10^-31 kg, but I could be wrong?

c)If the electrons were to complete a full circular orbit, what would be the radius?

this one I used a=v^2/r, and plugged in acceleration equation from be to use r= v^2/(qvB/m)

I can't seem to get the correct digits... can someone please check my work? and the math?

Explanation / Answer

yes. apply KE = PE

0.5 mv^2 = Vq

so velocoty v = sqrt (2eV/m)

v = sqrt ( 2*1 .6*10^-19 * 3000/9.11*10^-31)

v = 1053.787 *10^12

v= 3.26 *10^7 m/s

b. yes again ma = qvB

a = 1.6*10^-19 * 3.26*10^7 * 0.59/9.11*10^-31

a = 3.37*10^18 m/s^2

and in terns of g it is ( as g = 9.8 m/s^2)

a = 3.438 *10^17 g

c. apply mv^2/r = qvB

r = mv/qB

r = 9.11*10^-31* 3.26*10^7/(1.6*10^-19*0.59)

r = 0.31 mm or 3.14 *10^-4 m

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