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NotTim Group Title

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?

  • 2 years ago
  • 2 years ago

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  1. NotTim Group Title
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    To speed things up, the equation is \[v=qBr/m\]

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

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

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

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

    • 2 years ago
  6. jim_thompson5910 Group Title
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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

    • 2 years ago
  7. JamesJ Group Title
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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.

    • 2 years ago
  8. NotTim Group Title
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    wait, so Q is a constant?

    • 2 years ago
  9. JamesJ Group Title
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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.

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

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

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

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

    • 2 years ago
  14. NotTim Group Title
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    Ok. Thank you very much.

    • 2 years ago
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