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Homework Help: Current Through a Beam of Protons

  1. Oct 5, 2009 #1
    A cylindrically shaped beam of protons has a diameter of 1.20 mm and has 1.40e6 protons per cubic centimeter. The kinetic energy of each proton is 1.15 keV.
    What is the beam current in microamperes?

    I know that:
    I am 99% sure this is the equation that you use.
    q is the charge of the proton or q=1.602e-19C

    n is the charge density of the protons which is 1.4e6 protons/cubic centimeter. in order to get this into cubic meters I should divide by 1e6. Then n= 1.4 protons/cubic meter

    A is the cross sectional area of the beam or. (1.2/1000)= .012/2= [tex]\pi[/tex](6e-4)^2= 1.3097e-6

    using 1/2mv^2=ke I found that v=469352.4916m/s

    I plugged all these values into the above equation to get the current in amperes, then I multiplied this by 1e6 to get it into microamperes. I am still getting it wrong, and I don't know why. The homework program only lets me get point if I am 1% within the scientifically notated answer. My answer was 1.1905e-13 microamperes...

    This is so frustrating, and I am about to pull my hair out...
  2. jcsd
  3. Oct 5, 2009 #2


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    I agree with your velocity.
    is difficult to understand. Why divide the radius by 2? I think that is your only mistake - I've got twice your answer.
    I found the volume of the cylinder of electrons that passes by any given point in a second using d = vt, then used V = pi*r^2*L to get the the volume. Finally, calculated the number of protons and charge in that volume.
  4. Oct 6, 2009 #3
    isn't the variable n the cross sectional area of the beam. since the beam is a cylinder, the cross sectional area is a circle. the area of a circle is [tex]\pi[/tex]r^2. the problem gives us the diameter in mm. i converted this to m and then divided it by two to get the radius...
    n should be the cross sectional area because I is in amperes (which has a unit of C/s). when you multiply charge by charge density by A by velocity you should get C/s. Then (C)(1/m^3)(A)(m/s)= C/s, so A must have a value of m^2.
    Last edited: Oct 6, 2009
  5. Oct 6, 2009 #4
    I don't understand this. Why are there less protons in a cubic meter (1.4) than there would be in a cubic centimeter (1.4e6)?
  6. Oct 6, 2009 #5


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    Oops - sorry! I wrote down 1.2 mm as the radius and never checked it!

    With that change I agree with your answer of 1.19 x 10-7 A.

    I guess you have a typo in this bit: "pi (6e-4)^2= 1.3097e-6"
    I get 1.13E-6.
  7. Oct 6, 2009 #6
    The answer of 1.19 x 10-7 A was wrong. The computer won't let me guess anymore, but I tried calculating it out again just to see. This time I said that n should be 1.4 x 10^6 x 1x10^6. Then I=.119 A. Does this seem right?
  8. Oct 6, 2009 #7


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    Using meters throughout, I had
    I = qnAv = 1.6E-19*1.4E12*1.13E-6*4.69E5 = 1.19E-7 Amps.
    Oh, that is 0.119 microamps not the 1.1905e-13 microamperes you had in the first post. What a shame if that is all that was wrong!
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