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Homework Help: Magnetic field and momentum

  1. Nov 25, 2004 #1
    here is the question:

    Protons having a kinetic energy of 8.6eV are moving in the positive x direction and enter a magnetic field B=0.0390 (k hat)T directed out of the plane of the page and extending from x=0 to x=1.08. Ignore relativistic effects

    a) calculate the Y component of the protons momentum as they leave the magnetic field

    Well i can find the velocity from the kinetic energy, but how can i find the momentum of the Y component when it leaves the magnetic field??

    b) find the angle (alpha) b/w the initial velocity vector of the protom beam and the velocity vectr after the beam emerges from the field...

    How could u find this angle? i hvae no idea
  2. jcsd
  3. Nov 25, 2004 #2
    Here are some tips : suppose x axis is horizontal, y axis vertical and z axis points out of our paper towards us...Use the Lorentz force knowing that B = 0.039e_z and v is in the positive x direction. So F = qvB and think of what the position of F is with respect to the Y-axis. Calculate this force...this number (let's call it A) is equal to m*a and use this trick in order to calculate the velocity at x = 1.08 : a = dv/dt so you have that
    F = mdv/dt = A or m*dv =A*dt and dt = dt*dx/dx and put dt*dx at the right hand side of the equation in order to come to : m*dv*dx/dt = A*dx and dx/dt = v. Finally start integrating and your problem is solved...good luck

  4. Nov 25, 2004 #3
    The angle of the velocity vector after leaving the B-field can be calculating by using the Lorentz-force, calculate the velocity from this formula (as shown above) and projecting it out onto the x-axis...

  5. Nov 25, 2004 #4
    so i understand how u derived this formula: m*dv*dx/dt = A*dx and dx/dt = v

    so im solving for dx/dt to find the y component right?

    and A=qvB (velocity can be solved from the given kinetic energy)

    and dx is just 1.08, since we are starting from an initial point of 0 to a final point of 1.08 right?

    but how can i solve for dv?
  6. Nov 25, 2004 #5
    well on the left hand side you will have A * m*v*dv, if you integrate this you will have

    [tex]A*m*\frac{v^2}{2}[/tex]...then fill in the beginvalue of v (v_i)(calculated from the given kinetic energy) and the endvalue is just v (v_f),which is asked...

    so you have then [tex]A*m*\frac{v_f ^2- v_i^2}{2}[/tex]

  7. Nov 25, 2004 #6
    isnt A suppose to be on the right side? hmm..this is what i got

    m*dv*v= A*dx and dx/dt = v

    m*((vf^2)-(vi^2))/2)* = A*dx

    A*dx integrating u just get A*x and x=1.08..soo

    m*((vf^2)-(vi^2))/2)* = A*x

    i solve for vf, and it asks for momentum, so dont i just multiply the final velocity by the mass of the proton to find the momentum?

    but i keep getting the wrong answer?
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