1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Motion in a Magnetic Field and TV

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data

    The Picture tube in an old black and white TV uses magnetic deflection coils rather than electric plates. Suppose an electron beam is accelerated from rest through a 50kV potential difference and then through a region with uniform magnetic field 1cm wide. The screen is located 10cm from the coils and is 50cm wide. When the field is off, the electron beam hits the centre of the screen. Ignoring relativistic corrections, what field magnitude is needed to deflect the beam to the side of the screen.

    m_e = 9.11E-31 kg;
    q_e = 1.6E-19 C;
    delta V = 50E3 V;
    d_1 = 0.01m
    d_2 = 0.1m
    d_3 = 0.5m

    2. Relevant equations

    F = qv x B = qvBsin(theta) = qvB (assuming the magnetic field is at right angles);
    1/2mv^2 = q*deltaV;
    F = ma;
    v_f = v_i + at;
    v=d/t

    3. The attempt at a solution

    I will say the velocity from the potential difference is positive x and screen to be in the yz plane. I also assume the magnetic field is pointed in the positive z direction, giving the velocity from magnetic field in the y direction.


    First I got the velocity from potential difference.

    1/2mv^2=qV
    => v = sqrt(2q_eV/m_e)

    Then I worked out the velocity in y from magnetic field, I'll call this v'.

    F = qvB = ma
    => a=qvB/m.
    v'_f = v'_i + at v'_i = 0;
    v'_f = at =qvBd_1/mv = qBd_1/m_e (using t = d_1/v)

    Now, time taken for beam to travel 0.1m in x is

    t = d/v = 0.1/sqrt(2q_eV/m_e) = 7.5E-10

    Beam must travel d_3/2 = 0.25m in this time, so v' = d/t becomes

    0.25/7.5E-10 = 3.3E8.

    We have v' = qBd_1/m_e.

    Rearranging I got

    B = m_e v/q_e d_1.

    Plugging in numbers I get B =0.18 T.

    The answer the book gives is 70 mT.





    In the solutions the book works out R as a radius of curvature from Pythagorus and uses qvB=mv^2/R. My problem with this is that the magnetic field only exists for 0.01m, not 0.1m which is what they use.


    Any help/pointing out the obvious and making me feel like an idiot will be greatly appreciated.


    Thanks,
    Michael.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 2, 2012 #2
    Ok, I see what they do in the solutions now.

    They use trig and get R for the magnetic field (giving it value 1.08cm) the use qvB=mv^2/R, solve for B and get the answer. I know my way is a lot messier than that, but why doesn't it work?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Motion in a Magnetic Field and TV
Loading...