Question

An accelerating voltage of 2.37 103 V is applied to an electron gun, producing a beam of electrons originally traveling horizontally north in vacuum toward the center of a viewing screen 36.3 cm away. What is the magnitude of the deflection on the screen caused by the vertical component of the Earth's magnetic field, taken as 20.0 µT down?

Answer 1

I have tried using \[r = \frac{ mv }{ qB }\] but it's not giving me the correct answer.

Answer 2

I don't have the final answer and am heading to bed, but this might help.

Force = q v B, and given the directions, I think the particle starts by going east and the deflection might be small enough not considerate part of a circle.

acceleration = a = F/m = q v B/m

deflection distance = (1/2) a t^2, with t given by the beam velocity and the distance to the screen. Get the beam velocity from the energy in electron-volts converted to Joules and then equated to the electron kinetic energy (1/2) m v^2.

Not easy, but doable.

Answer 3

Well, I've come to doubt my "solution."

Try this: m v^2 / r = e v B.

r = m v^2 / e v B

Centripetal acceleration of the electron beam, toward the west (?).

r is radius of big (?) circle centered in the west.

Have the circle intersect the screen to get your drflection.

I think this is right and earlier suggestion wrong.
Can't stop to try the calculation snow. Good luck.

Answer 4

You are trying r =m v / eB right?
All in m k s units ? (meters kilograms seconds tesla)
Then seeing how this produces an arc of a circle and intersects the screen?

Well, I just plugged in the numbers and came up with r=0.827 x 10^19 m,
which seems ridiculously large, so this seems wrong. I guess it is possible that the earth's field is so weak the beam hardly deviates, but I am puzzled. Sorry!

Answer 5

the answer is close to 6.2 x 10^-3
I appreciated the effort @douglaswinslowcooper

Answer 6

How was it obtained?