An electron shot from the back of a T.V. help?

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In summary, the conversation discusses the calculation of the magnetic field required for an electron to make a circular path with a radius of 30m when shot from the back of a TV towards the front at half the speed of light. The correct calculation for the magnetic field is B = 3.06x10^-5 and the conversation also mentions taking "baby steps" to arrive at the correct calculation.
  • #1
acsis
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Homework Statement



If I shoot an electron from the back of a TV at a screen towards the front at .5c (e= speed of light) What magnetic field must be in the coils around the cathode ray tube (picture tube) if the electron would have made a circle radius 30m if it hadn't hit the screen and made it glow. (your TV show)

Homework Equations



r= (mv)/qB

The Attempt at a Solution



Using the equation rearranged to solve for B:

B=mv/qr

mass of an electron = 9.1 x 10^-31
charge of an electron = 1.602 x 10^-19
.5c=...well it's half of the spd of light so I assume : 1.5 x 10^8 since the spd of light is 3x10^8 if I remember correctly.
and radius = 30m

so:

B= (9.1x10^-31)(1.5x10^8)/(1.602x10^-19)(30)
B= .026?

Could someone check if I did this correctly, did I forget something?
 
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  • #2
Your calculation is wrong. Check it.
 
  • #3
Hm.

Now I'm getting

B = 3.06x10^-5

Is that correct?

I tried taking baby steps to see what I did incorrectly but I'm not sure if it's helping.
 
  • #4
Yes. It is correct.
 
  • #5
Thanks, now I know those baby steps helped.
 
  • #6
acsis said:
Hm.

Now I'm getting

B = 3.06x10^-5

Is that correct?

I tried taking baby steps to see what I did incorrectly but I'm not sure if it's helping.

Have checked those calculations and got 2.8402x10^-5... :rolleyes:
 

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