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PatchworkTeapot
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#1
Report Thread starter 6 years ago
#1
An electron in a cathode-ray tube is accelerated through a potential difference of 10 kV, then passes through the d = 1.9-cm-wide region of uniform magnetic field in the figure below. Theta=12 degrees.

How do I find the magnetic field in region d?

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#2
Report 6 years ago
#2
(Original post by PatchworkTeapot)
An electron in a cathode-ray tube is accelerated through a potential difference of 10 kV, then passes through the d = 1.9-cm-wide region of uniform magnetic field in the figure below. Theta=12 degrees.

How do I find the magnetic field in region d?

Image
what have you tried so far?

what can you work out with the information you've been given? - how about a velocity for the electron? what do you know about charged particles moving through constant magnetic fields.
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PatchworkTeapot
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#3
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#3
I converted 10kV to 1.6x10^-15 eV in energy, and used kinetic energy to find the velocity (v= 5.93x10^7/s). I assume that you would also need acceleration, so you can find the force on the electron, but I don;t know how to find that. Since F=qvB, once you find acceleration, would you just multiply it by mass of electron and divide by velocity and charge? Do you also just multiply by sin(12)?
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Jpw1097
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#4
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#4
(Original post by PatchworkTeapot)
I converted 10kV to 1.6x10^-15 eV in energy, and used kinetic energy to find the velocity (v= 5.93x10^7/s). I assume that you would also need acceleration, so you can find the force on the electron, but I don;t know how to find that. Since F=qvB, once you find acceleration, would you just multiply it by mass of electron and divide by velocity and charge? Do you also just multiply by sin(12)?
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I know this is an oversimplification. I assumed that the acceleration was always vertically downwards since the direction which the electron moves in does not change much as the angle is 12°. However, any moving charged particle in a magnetic field will experience a force that is perpendicular to the direction of motion and so the charged particle will move in a circular orbit, where F is the centripetal force. As such, the force will act towards the centre of the circle and so the direction of the force in constantly changing. Since the angle is so small though, I have assumed the difference in force to be negligible (so my answer may be slightly inaccurate).

However, have a read and see what you think. Hopefully it helps. If there is a way to calculate the centripetal force, then that would be your answer.
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#5
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#5
Unit problem.
the energy of electrons accelerated through 10kV is 10 keV or 1.60x10^-15 J

since the path of a moving electron while in a perpendicular and uniform magnetic field is circular, I was thinking of finding the radius of the arc and using the circular motion formulae.
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PatchworkTeapot
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#6
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#6
How would you find the radius? Do you use angle and something to do with angular momentum maybe?
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#7
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#7
(Original post by PatchworkTeapot)
How would you find the radius? Do you use angle and something to do with angular momentum maybe?
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find the hypotenuse
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PatchworkTeapot
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#8
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#8
Let me go through the steps to make sure I am doing this correctly.
K=1/2mv^2
v=sqrt(2K/m)
K=10kV x 1.6x10^-19C= 1.60x10^-15 J

v=sqrt(2 x 1.60x10^-15 J/9.11x10^-31kg)
v=5.93x10^7m/s

a=v^2/r
r=d/cos(theta)=0.019m/cos(12)=0.0194m
a=(5.93x10^7m/s)^2/(0.0194m)=1.809x10^17m/s^2

B=ma/qvsin(theta)=(9.11x10^-31 x 1.809x10^17)/(1.6x10^-19 x 5.93x10^7)sin(12)
B=83.7mT

Is this correct? Can you please tell me what was done wrong if there are mistakes?
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#9
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#9
(Original post by PatchworkTeapot)
Let me go through the steps to make sure I am doing this correctly.
K=1/2mv^2
v=sqrt(2K/m)
K=10kV x 1.6x10^-19C= 1.60x10^-15 J

v=sqrt(2 x 1.60x10^-15 J/9.11x10^-31kg)
v=5.93x10^7m/s

a=v^2/r
r=d/cos(theta)=0.019m/cos(12)=0.0194m
a=(5.93x10^7m/s)^2/(0.0194m)=1.809x10^17m/s^2

B=ma/qvsin(theta)=(9.11x10^-31 x 1.809x10^17)/(1.6x10^-19 x 5.93x10^7)sin(12)
B=83.7mT

Is this correct? Can you please tell me what was done wrong if there are mistakes?
I think you should have used sin(12). 1.9cm is the length of the opposite side to the 12 degree angle and you're interested in the length of the hypotenuse. http://mathworld.wolfram.com/SOHCAHTOA.html

if you've been taught any relativistic formula yet... 10keV is getting into the territory where starts diverging from classical by a significant amount. if you haven't then don't worry about it.
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