an-electron-moves-in-a-helical-trajectory-within-a-uniform-magnetic-field-orient

Question

An electron moves in a helical trajectory within a uniform magnetic field oriented along the y-axis. The position of the electron is described by the vector

$\begin{array}{l}\vec{r}=b\cos \left(\frac{2\pi y}{c}\right)\hat{\mathbf{i}}+y\ \hat{\mathbf{j}}+b\sin (\frac{2\pi y}{c})\hat{\mathbf{\ k}}\end{array}$

where $b$ represents the radius of the helical path, $c$ denotes the pitch of the helix, and $y$ varies as $y=v_{y}t$ , where $v_{y}$ is the constant component of velocity along the $y-axis$ . Determine the time-dependent angular momentum $\vec{L}$ of the electron about the origin.

Accepted Answer

$mbv_y\left[\frac{2\pi y}{c}\cos (\frac{2\pi y}{c})-\sin (\frac{2\pi y}{c})\hat{\mathbf{i}}-\frac{2\pi }{c}b\ \hat{\mathbf{j}}+\left(\cos (\frac{2\pi y}{c})+\frac{2\pi y}{c}\sin (\frac{2\pi y}{c})\right]\hat{\mathbf{k}}\right]$

Answer

$mbv_y\left[\frac{2\pi y}{c}\cos (\frac{2\pi y}{c})-\sin (\frac{2\pi y}{c})\hat{\mathbf{i}}-\frac{2\pi }{c}b\ \hat{\mathbf{j}}+\left(\cos (\frac{2\pi y}{c})+\frac{2\pi y}{c}\sin (\frac{2\pi y}{c})\right]\hat{\mathbf{k}}\right]$

Answer

$mbv_y\left[\frac{2\pi y}{c}\cos (\frac{2\pi y}{c})-\sin (\frac{2\pi y}{c})\hat{\mathbf{i}}-\frac{2\pi }{c}b\ \hat{\mathbf{j}}+\left(\cos (\frac{2\pi y}{c})+\frac{2\pi y}{c}\sin (\frac{2\pi y}{c})\right]\hat{\mathbf{k}}\right]$

Answer

$mbv_y\left[\frac{2\pi y}{c}\cos (\frac{2\pi y}{c})-\sin (\frac{2\pi y}{c})\hat{\mathbf{i}}-\frac{2\pi }{c}b\ \hat{\mathbf{j}}+\left(\cos (\frac{2\pi y}{c})+\frac{2\pi y}{c}\sin (\frac{2\pi y}{c})\right]\hat{\mathbf{k}}\right]$

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