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Magnetic Fields, Deuterium and curved tracks

  1. May 20, 2008 #1

    TFM

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    [SOLVED] Magnetic Fields, Deuterium and curved tracks

    1. The problem statement, all variables and given/known data

    A deuteron (the nucleus of an isotope of hydrogen) has a mass of [tex] m_D [/tex] and a charge of e. The deuteron travels in a circular path with a radius of r in a magnetic field with a magnitude of B.

    Find the time required for it to make 1/2 of a revolution.

    2. Relevant equations

    Cyclotron frequency: [tex] \omega = \frac{v}{R} [/tex]

    3. The attempt at a solution

    IO have already calculated the velocity in the previous part to be:

    [tex] \frac{reB}{m_D} [/tex]

    Frequncy is 1/Period, so I get

    [tex] period = \frac{1}{\frac{v}{R}} = \frac{R}{v} [/tex]

    [tex] = \frac{r}{\frac{reB}{m_D}} [/tex]

    Which I have rearranged to get:

    [tex]\frac{m_D}{eB} [/tex]

    Sincve this is for one whol revolution, I multiplied it by a half to get:

    [tex]\frac{m_D}{2eB} [/tex]

    Which Mastering Physics says is wrong, but also says:

    Your answer is off by a multiplicative factor.

    Any Ideas?

    TFM
     
  2. jcsd
  3. May 20, 2008 #2
    T = 2*pi/omega
     
  4. May 20, 2008 #3

    TFM

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    So that would make:

    [tex] period = \frac{2\pi}{\frac{v}{R}} = \frac{2\piR}{v} [/tex]

    and this the fraction would be:

    [tex]\frac{2\pi m_D}{2eB} [/tex]

    Does this look more right?

    TFM
     
  5. May 20, 2008 #4

    TFM

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    Does the above look correct now?

    TFM
     
  6. May 21, 2008 #5
    Yes.
     
  7. May 21, 2008 #6

    TFM

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    Thanks,

    The next part of the question asks:

    Through what potential difference would the deuteron have to be accelerated to acquire this speed?

    But I am niot sure what formula to use.

    Any suggestions,

    TFM
     
  8. May 21, 2008 #7
    Use equation of equivalence of potential and kinetic energy (potential energy turns to kinetic)
     
  9. May 21, 2008 #8

    TFM

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    Kinetic Energy:

    [tex] K.E. = \frac{1}{2}mv^2 [/tex]

    Electric Potential:

    [tex] U= q_0V [/tex]

    Equate:

    [tex] \frac{1}{2}mv^2 = q_0V [/tex]

    [tex] V = \frac{mv^2}{2q_0} [/tex]

    Is this correct?

    TFM
     
  10. May 21, 2008 #9

    TFM

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    Putting in my values I get

    [tex] \frac{4 \Pi^2m_D^3}{\frac{4e^2B^2}{2e}} [/tex]

    Does this look correct?

    TFM
     
  11. May 21, 2008 #10
    Nope, there is no place for pi here v=omega*r.

    using omega=eB/m you should get it
     
    Last edited: May 21, 2008
  12. May 21, 2008 #11

    TFM

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    I looked at the wrong part :redface: - i had already found the speed to be:

    [tex] v = \frac{reB}{m_D} [/tex]

    so putting in this v, I get:

    [tex] V = \frac{m_D(\frac{r^2e^2B^2}{m_D^2})}{2e} [/tex]

    is this better?

    TFM
     
  13. May 21, 2008 #12
    This should be correct.
     
  14. May 21, 2008 #13

    TFM

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    Indeed it is correct.

    Thanks, michalll :smile:

    TFM
     
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