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NotTim

  • 4 years ago

A proton of mass 1.67x10^27 kg, moves in a circle in the plane perpendicular to a uniform magnetic field of magnitude 1.8T. The radius of curvature is 3.0cm. what is the speed of the proton?

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  1. NotTim
    • 4 years ago
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    To speed things up, the equation is \[v=qBr/m\]

  2. NotTim
    • 4 years ago
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    My only difficulty with this situation is the matter of "q". I don't know where the value comes from.

  3. jim_thompson5910
    • 4 years ago
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    q looks like the charge (in Coulombs perhaps?) of the proton, but i'm not entirely sure

  4. NotTim
    • 4 years ago
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    I think you are correct. Q is the charge.

  5. NotTim
    • 4 years ago
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    IF anyone ever comes: value of q still undetermined.

  6. jim_thompson5910
    • 4 years ago
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    If q is the charge, then in coulombs, the charge of a proton is q = 1.602 * 10^(-19) C where C is the abbreviation of a coulomb

  7. JamesJ
    • 4 years ago
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    Yes, that's the charge of a proton. You can solve it now. Just make sure you use SI units throughout.

  8. NotTim
    • 4 years ago
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    wait, so Q is a constant?

  9. JamesJ
    • 4 years ago
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    q is the charge of object you're interested in, which is a proton. And a proton most definitely has a fixed charge.

  10. NotTim
    • 4 years ago
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    So this is knowledge I should have know already?

  11. JamesJ
    • 4 years ago
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    it doesn't seem like an unreasonable thing that you should.

  12. jim_thompson5910
    • 4 years ago
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    yeah it seems like they were assuming that you would have or should have known that

  13. jim_thompson5910
    • 4 years ago
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    but if you didn't, you can easily look it up

  14. NotTim
    • 4 years ago
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    Ok. Thank you very much.

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