Question

You can use either electric or magnetic fields to steer moving charges. Suppose you have a beam of electrons moving in the xdirection with speed 1.0×107 m/s, and you want them to turn 30° to their right in a distance of 1.00 cm.(a) [2 pts.] One way to turn the electrons is to run them between two charged plates that are 1.00 cm wide. There will be a uniform electric field between the plates. What direction does the electric field have to have to get the electrons to turn to their right?(b) [2 pts.] What will be the shape of the path the electrons follow?(c) [10 pts.] What magnitude of electric field is needed to get the electrons to turn by 30°? (d) [1 pts.] What is the difference in potential that you need to put across the plates, and which plate should be at the higher potential?(e) [4 pts.] Another way to turn the electrons is to apply a magnetic field (instead of the electric field!). What direction must the magnetic field have in order for the electrons to turn to their right?

Answer 1

given
vx = 1*10^7 m/s

a) The direction of electric field must be towards +y axis

b) the electrons follow parabola path.

c)

let v is the velocity of electron when it is deflected by 30 degrees.

we know, v*cos(30) = vx ----(1)

and v*sin(30) = vy ---(2)

take equation(2)/equation(1)

tan(30) = vy/vx

==> vy = vx*tan(30)

= 1*10^7*tan(30)

= 5.77*10^6 m/s

time taken for the electrons to cross to travel 1 cm distance,

t = x/vx

= 0.01/(1*10^7)

= 1*10^-9 s

let ay is the acceleration in y-direction.

use, ay = (vf - vi)/a

= (5.77*10^6 - 0)/(1*10^-9)

= 5.77*10^15 m/s^2

now use, Fe = q*E

m*a = q*E

E = m*a/q

= 9.1*10^-31*5.77*10^15/(1.6*10^-19)

= 3.28*10^4 N/C <<<<<<----------Answer

d) the distance bwteeen the plates, d = 1 cm = 0.01 m

delta_V = E*d

= 3.28*10^4*0.01

= 328 V <<<<<<----------Answer

e) The direction of magnetic field must be towards -z axis
(into the page)