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Finding Velocity of Positive ion with Volts given.

  1. Feb 24, 2008 #1
    1. The problem statement, all variables and given/known data

    An ion is accelerated through potential diff. of 250 V, directly perpendicular to mag. field.
    Find radius of the path of the ion in the field.

    Given:
    mass (m)=2.5 x 10^-26
    charge(q)= 1.6 x 10^-19
    Volts= 250 V
    Mag. Field (B) = 0.50 T

    2. Relevant equations

    radius= (mass x VELOCITY)/ (charge x B)

    3. The attempt at a solution

    I know the answer is 1.77 cm for the radius, but i need to prove how, but i cant seem to find an equation or reason to use Volts to get Velocity in the equation.
     
  2. jcsd
  3. Feb 24, 2008 #2

    Fredrik

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    Staff Emeritus
    Science Advisor
    Gold Member

    What's the definition of "potential difference"? Thinking about that should help you figure out how to use the number of volts. Hint: energy.
     
  4. Feb 25, 2008 #3
    Pe

    First i find PE with:

    PE=Volts x charge

    plug in to:

    velocity=[(-2/mass) x PE]^(1/2)

    seems right...

    PE= 250V / 1.6 x 10^-19 C = 1.56 x 10^21 J

    plug in:
    and you get velocity at 3.54 x 10^23 cm/s

    but when you use that number in the final radius equation, its not = to 1.77 cm (and i know thats right)
     
  5. Feb 25, 2008 #4
    First i find PE with:

    PE=Volts x charge

    plug in to:

    velocity=[(-2/mass) x PE]^(1/2)

    seems right...

    PE= 250V / 1.6 x 10^-19 C = 1.56 x 10^21 J

    plug in:
    and you get velocity at 3.54 x 10^23 cm/s

    but when you use that number in the final radius equation, its not = to 1.77 cm (and i know thats right)
     
  6. Feb 25, 2008 #5
    nvm... its good now, thanks for the help
     
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