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Electric and magnetic fields

  1. Sep 26, 2007 #1
    A charged particle of mass m and positive charge q moves in uniform electric and magnetic fields, E and B, both pointing in the z direction. The net force on the particle is F=q(E+v x B)*. Write down the equation of motion for the particle and resolve it into its 3 components. Solve the equations and describe the particles motion.

    Since there are both electric and magnetic fields, should I consider the motion with respect to each individually?
    Considering the Electric field, the trajectory is a straight line along z, since its parallel to E, and a parabola along x,y? Yes?

    The Magnetic field will have a force perpendicular to both v and B right?

    Now, how does one put the two together?

    *Is v (cross) B, x is cross product not a variable.
  2. jcsd
  3. Sep 26, 2007 #2
    I need help on this physics question:

    An electron is shot at 10^6 ms-1 between two parallel charged plates. if (e) = 1 KN/C where will the electron strike the upper plate? Assume vacuum conditions.
  4. Sep 26, 2007 #3
    Rather than conceptually deducing the particle's motion, it'd probably be better to write down what F_x, F_y, and F_z are (the three components of F) with respect to E_x, E_y, E_z, B_x, B_y, B_z, v_x, v_y, and v_z. At this point a lot should cancel since only E_z and B_z are nonzero.
  5. Sep 26, 2007 #4
    Use the force, Luke. Ignoring relativistic effects,

    [tex]m \mathbf{\ddot{x}} = \mathbf{F} = q\left(\mathbf{E} + \mathbf{v}\times\mathbf{B} \right)[/tex]

    Resolving the components, [tex]\mathbf{x} = (x,y,z)[/tex], and using the fact that only the z-components of E and B exist:

    m \ddot{x} &= q \left( v_y B_z \right) \\
    m \ddot{y} &= q \left( -v_x B_z \right) \\
    m \ddot{z} &= q E_z

    You should now be able to solve for z, at least. Combine the equations for x and y to get a 3nd order equation for x, or 2nd order equation for [tex]\dot{x}[/tex].
    Last edited: Sep 26, 2007
  6. Sep 26, 2007 #5
    lol :)

    Can't you just integrate for each seperately? I don't know you would put x and y together.
    After you have each, what is 'solving' them, as asked in the question?
  7. Sep 26, 2007 #6
    Just posting to bring this back up on the forum. Still not sure why I'd combine x and y
  8. Sep 26, 2007 #7
    is my v(z) still constant with the electric field present?
  9. Sep 26, 2007 #8
    The idea is certainly to just integrate them. However, quite often that's not directly possible. For this case, you can do so for z, as it does not involve any other function of t. To be very explicit, what we're really doing is trying to find three function x(t), y(t) and z(t). So we begin with z(t):

    m \frac{d^2 z(t)}{dt^2} &= q E_z \\
    v_z(t) = \frac{d z(t)}{dt} &= \frac{q E_z}{m} t + v_z(0) \\
    z(t) &= \frac{q E_z}{2 m} t^2 + v_z(0) t + z(0)

    So [tex]v_z[/tex] is actually a linear function of t.

    Now we could try the same with x(t). However, you notice that when trying to integrate, you've got [tex]v_y = \dot{y}[/tex] on the right hand side, which is a function of t. So you can't do the integral unless you've already got [tex]v_y[/tex]. Thus you get around this by differentiating the first equation wrt t, so that you've got [tex]\frac{d^3 x}{dt^2}[/tex] on the left and [tex]\ddot{y}[/tex] on the right. Then you can substitute the 2nd equation in, and treat it as a 2nd order differential equation in [tex]\dot{x}[/tex].
  10. Sep 26, 2007 #9
    Physics Question about an electron

    I need help on this physics question:

    An electron is shot at 10^6 ms-1 between two parallel charged plates. if (e) = 1 KN/C where will the electron strike the upper plate? Assume vacuum conditions.
  11. Sep 26, 2007 #10
  12. Sep 27, 2007 #11
    Aslam O Alaikum.

    I am a new member here an dthis is my first post.
    I want to know that What is difference between Magnetic field and Electric field?
    Because If we read their general definations use in general physics Defination of both is same??
    So whats their main difference ?


    M.Ahsen tahir

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