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Radiation, Magnetic fields and Linear Acceleration?

  1. Jun 9, 2009 #1
    Hi I've been revising and came across some questions which I do not understand how they obtained the answer could you guys help?

    1.

    Show that the momentum, p, of a particle of mass m is related to its kinetic energy, KE by the relationship p^2 = 2m KE

    Use the above relationship to calculate the accelerating potential required for
    electrons to have a de Broglie wavelength of 4·5 × 10^(–11) m.

    And the answer turns out to be 645V.

    2.

    A magnet moves towards a coil as shown (solenoid circuit). Use Lenz’s law to explain in which direction the current will flow through the turns of the coil. The coil is now situated in a uniform magnetic field changing at a rate of 0·43Ts^–1.

    r = 0.046m
    B = 0.43Ts^-1
    N = 1300
    Total Resistance = 6.6 ohms
    A = 2 *(pi)* r^2 = 0.0133m^2

    Calculate the current flowing in the ammeter.

    Formulas I've thought of is R = V/I, E = NBA / t and I set V = 1

    However the formula turned out to be

    I = (1300 * 0.43 * (pi) * ((0.046)^2)) / 6.8 = 0.55A

    Problem I got is how did R become the demominator (6.8)?

    3.

    In the LINAC, the protons are accelerated from rest through 3 gaps each with an accelerating p.d. of 36kV.

    KE in J is 1.7*10^(-14)J
    Speed of protons 4.5 * 10^6 ms^-1

    The radius of the synchrotron is 12·5 m. Calculate the value of the magnetic flux
    density in the synchrotron when the speed of the protons is 9·4 × 10^6 ms^–1.

    Equation I could think of is

    F = (mv^2) / r where do I go from here?

    Any help would be gr8

    Cheers
     
  2. jcsd
  3. Jun 9, 2009 #2

    Astronuc

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    Staff: Mentor

    In 3. what is the Lorentz force?

    Also please show one's work.
     
  4. Jun 9, 2009 #3
    1.##Attempt##

    KE by the relationship p^2 = 2m KE

    Use the above relationship to calculate the accelerating potential required for
    electrons to have a de Broglie wavelength of 4·5 × 10^(–11) m.

    [tex]\lambda[/tex] = h/mv

    since p^2 = 2m KE
    [tex]\lambda[/tex]^2 = h^2/(mv)^2
    (mv)^2 = h^2 / [tex]\lambda[/tex]^2 = 2m KE

    This is the point I couldnt continue...


    2. #Attempt#

    r = 0.046m
    B = 0.43Ts^-1
    N = 1300
    Total Resistance = 6.6 ohms
    A = 2 *(pi)* r^2 = 0.0133m^2

    Calculate the current flowing in the ammeter.

    Formulas I've thought of is R = V/I,
    E = NBA / t
    and I set V = 1

    Rearranging would give me

    E = NB2(pi)r^2/t
    E = IVt = NB2(pi)r^2/t

    Rearranging and cancellation provides

    I = NBA/Vt^2

    However the formula turned out to be

    I = NBA/R
    I = (1300 * 0.43 * (pi) * ((0.046)^2)) / 6.8 = 0.55A

    3. #Attempt#

    Equation I could think of is

    F = (mv^2) / r = BQv

    F = (1.67 x 10^-27 x 9·4 × 10^6) / (12.5 x Q ) = B

    Im sure that F= BQv but prob is I cannot find Q...
     
  5. Jun 9, 2009 #4

    dx

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    Homework Helper
    Gold Member

    Now write KE in terms of V. If an electron at rest is accelerated through a potential V, what is it's KE?
     
  6. Jun 9, 2009 #5
    The Kinetic Energy I worked out for the proton accelerated through 3 gaps with pd of 36kV is 108kEv which is 1.7 x 10^-14 J
     
  7. Jun 9, 2009 #6

    dx

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    Homework Helper
    Gold Member

    I was talking about your attempt at question 1.
     
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