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    Motion of charged particle in uniform electric & magnetic fields

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    In this question we will consider the motion of a charged particle in uniform electric and magnetic fields that are perpendicular. You may ignore the effects of gravity throughout this question.

    (a)(i) Consider a charged particle of mass m and charge q that is moving in a uniform electric field E = Ee_y and a perpendicular uniform magnetic field B = Be_z. Show that the equations of motion for the particle are:
    (see the attached file for the equations)

    (ii) If the particle is at rest at time t = 0, verify that the velocity components (see the attached file for the equations) satisfy the differential equations given in part (i) and the initial conditions.

    (iii) Hence determine the position of the particle at time t, assuming that it was located at the origin at t = 0.

    b) An infinite metal plate occupies the xz-plane (y=0). The plate is kept at zero potential, V=0 (see Figure 2). Photoelectrons are liberated from the plate at y=0 by ultraviolet radiation. The initial velocity of the photoelectrons is negligible. A uniform magnetic field B is maintained parallel to the plate in the positive z-direction and a uniform electric field E is maintained perpendicular to the plate in a negative y-direction. (Neither field is shown in Figure 2). The electric field is produced by a second infinite plate parallel to the first plate, maintained at a constant positive voltage V_0 with respect to the first plate. The separation of the plates is d.

    Using the results of part (a), show that the electrons will fail to reach the second plate if (see the attached file for the equation) where -e and m are the charge and mass of an electron.

    I am having difficulty with my maths and not arriving at all equations of motion/mass particularly mass in part a and I do not know what to chose at time t=0.

    I am trying to use the Lorenz force.

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    Solution Preview

    The force on a charged particle (charge q) moving with velocity v in the presence of magnetic field B and electric field E is:
    In our case:
    For a general velocity vector
    We get:
    Together, the total force is:

    The acceleration is defined as the derivative with respect to time of the velocity:
    Newton's second law states that .

    Comparing (1.6) to (1.5) component by component we get:
    The particle start from rest, hence the initial conditions are:
    The equation in the z direction is pretty simple. The derivative of with respect to time is zero. Therefore, is constant. Since it is constant at all times, it is equal
    Taking the derivative of (1.7b) ...

    Solution Summary

    The motion of charged particles in uniform electric and magnetic fields are determined.