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 1.85 107 m/s in the horizontal direction when magnetic forces deflect the electron with a vertically upward acceleration of 5.30 1015 m/s2. The phosphorescent screen is a horizontal distance of 6.9 cm away from the point where the electron is emitted. (a) How much time does the electron take to travel from the emission point to the screen? Incorrect: Your answer is incorrect. Think about the horizontal component of velocity—does it change? How can you use it and the horizontal distance to find the time? s (b) How far does the electron travel vertically before it hits the screen?

Answer 1

The first thing to realize is that all the things like velocity and acceleration are just vectors, so, you can treat those on the x axis and those in the y axis seperately.

So, since the electron is moving in the horizontal direction, and the acceleration due to mag. field is perpendicular to it, only the vertical motion is affected.
so, a)--> t = d/v = 6.9*10^-2/1.85*10^7 = 3.729*10^-9 sec = 3.72 nano seconds
b) --> (considering only vertical components), vertical distance = 1/2*vertical acceleration*t^2

[since s=ut + 1/2 at^2 and u=0]

vertical distance = 1/2*vertical acceleration*t^2

= 1/2 * 5.30*10^15*3.797*10^-9 = 0.03820m = 3.82cm

= 42.114 * 10^-18 m = 4.21*10^-19 m