MEC BYU 19

Analysis of Motion:

The particle experiences a magnetic force (perpendicular to velocity) and a viscous drag force (opposite to velocity). The magnetic force changes the direction, while the viscous force reduces the speed.

Equation of motion: $m\frac{d\vec{v}}{dt} = q(\vec{v} \times \vec{B}) – b\vec{v}$.

In complex notation $\eta = v_x + i v_y$, this becomes:

$$ \frac{d\eta}{dt} = – \left( \frac{b}{m} – i\frac{qB}{m} \right) \eta $$

The solution is $\eta(t) = v_0 e^{-\frac{b}{m}t} e^{i\omega t}$ where $\omega = \frac{qB}{m}$.

1. Time Calculation:

The velocity vector turns by $2\pi$ radians when the phase $\omega t = 2\pi$.

$$ t = \frac{2\pi}{\omega} = \frac{2\pi m}{qB} $$

2. Distance Calculation:

The speed of the particle decays exponentially: $v(t) = v_0 e^{-\frac{b}{m}t}$.

Distance $S = \int_0^t v(t’) dt’$.

$$ S = \int_0^{\frac{2\pi m}{qB}} v_0 e^{-\frac{b}{m}t’} dt’ $$ $$ S = v_0 \left[ -\frac{m}{b} e^{-\frac{b}{m}t’} \right]_0^{\frac{2\pi m}{qB}} $$