Find Speed of Particles in B-Field

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Homework Help Overview

The discussion revolves around calculating the speed of particles entering a magnetic field (B-field), with a specific focus on kinetic energy provided in electronvolts (eV).

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster seeks to determine the speed of particles based on their kinetic energy. Some participants attempt to apply the kinetic energy formula, questioning the correctness of their calculations.

Discussion Status

Participants are actively engaging with the problem, with one providing a calculation attempt and seeking validation. There is a focus on ensuring the correct application of the kinetic energy formula, but no consensus has been reached regarding the correctness of the calculations.

Contextual Notes

There is mention of a specific kinetic energy value (100 keV) and the mass of a proton, which may influence the calculations. The discussion highlights the need for clarity on units and conversions in the context of the problem.

Bert
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given is:
http://img340.imageshack.us/img340/6684/oplosqa0.jpg

So if you have a question with a B-field you need any speed given from the particles who entering in this field.
Who can I find this speed? Thanks.
 
Last edited by a moderator:
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Bert said:
given is:
http://img340.imageshack.us/img340/6684/oplosqa0.jpg

So if you have a question with a B-field you need any speed given from the particles who entering in this field.
Who can I find this speed? Thanks.

Get the speed from the kinetic energy 100keV.
 
Last edited by a moderator:
I try this but I fail so:

[tex]\frac{1}{2}m . v^2=U[/tex]
then we have
[tex]1,602 \ . \ 10^{-19} \ . \ 1 . \ 10^5=1,602 \ . \ 10^{-14}= \frac{1}{2} 10^{-27}.v^2[/tex]

so [tex]v[/tex] must be [tex]\sqrt{ \frac{2 \ . \ 1,602 . 10^{-14} } {10 ^{-27} } }[/tex]

is this correct?
 
Bert said:
I try this but I fail so:

[tex]\frac{1}{2}m . v^2=U[/tex]
then we have
[tex]1,602 \ . \ 10^{-19} \ . \ 1 . \ 10^5=1,602 \ . \ 10^{-14}= \frac{1}{2} 10^{-27}.v^2[/tex]

so [tex]v[/tex] must be [tex]\sqrt{ \frac{2 \ . \ 1,602 . 10^{-14} } {10 ^{-27} } }[/tex]

is this correct?

mass of a proton = [tex]1.67*10^{-27}kg[/tex]. Other than that everything looks good.
 
Thanks.
 

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