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Homework Help: Electric and magnetic fields with electron

  1. Apr 7, 2008 #1
    1. The problem statement, all variables and given/known data
    at time t1, an electron is sent along the positive direction of an x axis, through both an electric field E and a magnetic field B, with E directed parallel to the y axis. Figure 28-34 gives the y component Fnet, y of the net force of the electron due to the two fields, as a function of the electron's speed v at time t1. the x and z components of the net force are zero at t1. Assuming Bx=0, find the magnitude E and B in unit-vector notion.

    the graph has the y axis set as the net force in the y direction and the x axis as the velocity of the electron. the y axis goes from -2 through 2 and the x axis goes from 0 to 100. the x axis is zero at -2 on the y axis, and 100 at 2 on the y axis. the graph is a straight line through those two points.

    2. Relevant equations
    im not really sure

    3. The attempt at a solution

    to be honest i have no idea where to start on this one, any help would be appreciated.
     
  2. jcsd
  3. Apr 13, 2008 #2
    I think I'm in your class. I don't know if this is right but I think I have the idea.

    Ok, becuase B is always perpendicular to v and E, B should be pointing in the page in the k direction. The formla you should use is Fnet=q(E+VxB) E,V,B are vectors. Becuase V and B are parallel their cross product is vB so now Fnet=q(E+vB). Your vectors will be stright lines when Fnet=0 so 0=q(E+vB). Obviously the q goes away and you are left with B=E/v, where E is the only vector. From here it gets a little trick. You have to use the KE formula for velocity. This is as far as I have gotten becuase you have to use voltage in the KE formula and I can't find a relation in the problem to get velocity for anything. If you figure the rest out could you post it so I can finish? I couldn't find the problm in the soultions manual unless it is number 7. 7 doesn't have a part b and it doesn't tell you how to find E. It uses V/m to find E, and again I have to clue to to find V.
     
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