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Homework Help: Proton in Magnetic field - am I correct?

  1. Jan 19, 2004 #1
    Proton in Magnetic field -- am I correct?

    One of my Physics ISU questions is as follows:
    - A proton moves in a circular path perpendicular to a 1.10 T magnetic field. The radius of the path is 4.5 cm. calculate energy of the proton in eV

    Well i had hard time doing it but here we go: my solution

    mV^2/r=QrB V=QrB/m

    V= (1.6x10^-16) x o.o45 m x 1.10 T) / 1.67x10^-27 kg

    V= 4.7x10^6 m/s

    Ek= 1/2mV^2 = 1/2 x 1.67x10^-27 kg x (4.7x10^6 m/s)^2
    = 1.8x10^-14 J

    Ek= 1.87x10^-14J x eV/1.6x10^-19J = (1.87x10^-14J)/ (1.6x10^-19J)

    eV=1.3x10^5 eV=1.3x10^5



    Fellow physisists is my solution correct?

    Thank you
     
  2. jcsd
  3. Jan 19, 2004 #2
    Hmm...I got close to 1.17*10^5 eV

    Here's how I did it:

    We want to know the kinetic energy of the proton, so:

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

    We know the mass of a proton is 1.67*10^-27, so the variable we need is the velocity at which it is travelling.

    As this proton is moving in a circle in a magnetic field,

    [tex]F_m=F_c[/tex] (where [tex]F_m[/tex] is the magnetic force and [tex]F_c[/tex] is the centripetal force

    representing these with equations, we have

    [tex]qvB=\frac{mv^2}{r}[/tex]

    now, we isolate v, since that is what we are trying to get

    [tex]v=\frac{qBr}{m}[/tex]

    substituting into the kinetic energy equation, we get

    [tex]E_{k}=\frac{1}{2}m(\frac{qBr}{m})^2[/tex]

    which reduces to

    [tex]E_{k}=\frac{(qBr)^2}{2m}[/tex]

    substituting in,

    [tex]E_{k}=\frac{((1.6*10^-19)(1.10)(0.045))^2}{2(1.67*10^-27)}[/tex]

    which gets 1.878*10^-14 J

    since 1 eV = 1.6*10^-19 J, the answer in eV is

    (1.878*10^-14)/(1.6*10^-19) = 117,375 eV (1.17*10^5)

    There are two things about yours that confuse me. First off, the magnitude of the charge on a proton is the elementary charge, which is 1.6*10^-19, whereas you have it as 1.6*10^-16...and yet you get the same energy in joules. Then at the end when you do your conversion from joules to electron volts, there seems to be a discrepency of some kind, because (1.87*10^-14)/(1.6*10^-19) does not equal 1.3*10^5. Also, I would have rounded to 1.88*10^5 if I were rounding. Apart from that the solution the same as mine.
     
    Last edited: Jan 19, 2004
  4. Jan 19, 2004 #3
    Ok I'll look more into it

    thanks a bunch, help is always appreciated

    BTW that -16... I just copied it wrong from my notes:smile:

    Hmm also you must've read my first post wrong since

    Ek= 1.87x10^-14J x eV/1.6x10^-19J = (1.87x10^-14J)/ (1.6x10^-19J)

    eV=1.3x10^5 eV=1.3x10^5
     
    Last edited: Jan 19, 2004
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