Question

Older televisions display a picture using a device called a cathode ray tube, where electrons are emitted at high speed and collide with a phosphorescent surface, causing light to be emitted. The paths of the electrons are altered by magnetic fields. Consider one such electron that is emitted with an initial velocity of 2.00 x10^7 m/s in the horizontal direction when magnetic forces deflect the electron with a vertically upward acceleration of 5.25 x 10^15 m/s^2. The phosphorescent screen is a horizontal distance of 6 cm away from the point where the electron is emitted.
Required:
a. How much time does the electron take to travel from the emission point to the screen?
b. How far does the electron travel vertically before it hits the screen?

Answer 1

Answer:

a) $3*10^{-9}\ s$

b) 2.36 cm

Explanation:

a) The horizontal distance = x = 6 cm

1 cm = 0.01 m

6 cm = 6 cm * 0.01 m/cm = 0.6 m

Therefore the time taken (t) by the electron  to travel from the emission point to the screen can be gotten from:

x = t * $V_i$

$V_i=initial\ velocity=2*10^7\ m/s$

$x=tV_i\\\\t=\frac{x}{V_i}\\ \\t=\frac{0.06}{2*10^7}\\ \\t=3*10^{-9}\ m/s$

b) The vertical distance (y) traveled by the electon before it hits the screen is given by:

$y=\frac{1}{2}at^2\\ \\a=upward\ acceleration=5.25*10^{15}\ m/s^2\\\\Substituting:\\\\y=\frac{1}{2}*5.25*10^{15}* (3*10^{-9})\\\\y=0.0236\ m\\y=2.36\ cm$