- #1
cwatki14
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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=qnA<v>
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 let's 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...
What is the beam current in microamperes?
I know that:
I=qnA<v>
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 let's 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...