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Thread: Lorentz force law

  1. #1
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    Lorentz force law

    According to the Lorentz force law, an electron moving in a constant magnetic field follows a path $\displaystyle c(t) $ which satisfies $\displaystyle \ddot c = \dot c \times B $, where $\displaystyle B $ is a constant vector on $\displaystyle \mathbb{R}^3 $. Prove that this path is a helix.
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    Quote Originally Posted by mathman88 View Post
    According to the Lorentz force law, an electron moving in a constant magnetic field follows a path $\displaystyle c(t) $ which satisfies $\displaystyle \ddot c = \dot c \times B $, where $\displaystyle B $ is a constant vector on $\displaystyle \mathbb{R}^3 $. Prove that this path is a helix.
    We can choose to align the z axis with the magnetic field B. Then:

    $\displaystyle B=B_z$, $\displaystyle B_x=0$, $\displaystyle B_y=0$

    So:

    $\displaystyle \ddot c_x = \dot c_y B_z $

    $\displaystyle \ddot c_y = -\dot c_x B_z $

    $\displaystyle \ddot c_z = 0 $

    Solving the third equation by integrating twice with respect to t: $\displaystyle c_z = kt $ where k is a constant.

    Now if you differentiate the second equation and then substitute for $\displaystyle \ddot c_x$ from the first then you will get:
    $\displaystyle \ddot \dot c_y=-\dot c_yB_z^2$

    which has the general solution:
    $\displaystyle \dot c_y=-a sin(B_zt)+b cos(B_zt) = csin(B_zt+\phi)$

    So
    $\displaystyle c_y=\frac{-c cos(B_zt + \phi)}{B_z}$

    Now using equation 1
    $\displaystyle \ddot c_x=\dot c_y B_z= cB_zsin(B_zt+\phi)$

    So
    $\displaystyle c_x= \frac {-csin(B_zt+\phi)}{B_z}$

    Now using $\displaystyle c_x$, $\displaystyle c_y$, $\displaystyle c_z$ we can write:

    $\displaystyle \bar c = \frac {-csin(B_zt+\phi)}{B_z} \hat e_x-\frac {ccos(B_zt+\phi)}{B_z} \hat e_y+kt\hat e_z$

    This is the equation of a helix.
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