Question 5.2: Find and sketch the trajectory of the particle in Ex. 5.2, i...

The general solution is (Eq. 5.6):

\left.\begin{array}{l}y(t)=C_{1} \cos \omega t+C_{2} \sin \omega t+(E / B) t+C_{3} \\z(t)=C_{2} \cos \omega t-C_{1} \sin \omega t+C_{4}\end{array}\right\}                                  (5.6)

y(t)=C_{1} \cos (\omega t)+C_{2} \sin (\omega t)+\frac{E}{B} t+C_{3} ; \quad z(t)=C_{2} \cos (\omega t)-C_{1} \sin (\omega t)+C_{4} .

\text { (a) } y(0)=z(0)=0 ; \dot{y}(0)=E / B ; \dot{z}(0)=0 \text {. Use these to determine } C_{1}, C_{2}, C_{3}, \text { and } C_{4} .

y(0)=0 \Rightarrow C_{1}+C_{3}=0 ; \dot{y}(0)=\omega C_{2}+E / B=E / B \Rightarrow C_{2}=0 ; z(0)=0 \Rightarrow C_{2}+C_{4}=0 \Rightarrow C_{4}=0 ;

\dot{z}(0)=0 \Rightarrow C_{1}=0, \text { and hence also } C_{3}=0 . \text { So } y(t)=E t / B ; z(t)=0 . Does this make sense? The magnetic
force is q( v \times B )=-q(E / B) B \hat{ z }=-q E which exactly cancels the electric force; since there is no net force, the particle moves in a straight line at constant speed. 

(b) Assuming it starts from the origin, so C_{3}=-C_{1}, C_{4}=-C_{2}, \text { we have } \dot{z}(0)=0 \Rightarrow C_{1}=0 \Rightarrow C_{3}=0 ;

\dot{y}(0)=\frac{E}{2 B} \Rightarrow C_{2} \omega+\frac{E}{B}=\frac{E}{2 B} \Rightarrow C_{2}=-\frac{E}{2 \omega B}=-C_{4} ; y(t)=-\frac{E}{2 \omega B} \sin (\omega t)+\frac{E}{B} t ;

z(t)=-\frac{E}{2 \omega B} \cos (\omega t)+\frac{E}{2 \omega B} , or  y(t)=\frac{E}{2 \omega B}[2 \omega t-\sin (\omega t)] ; \quad z(t)=\frac{E}{2 \omega B}[1-\cos (\omega t)] . Let  \beta \equiv E / 2 \omega B .

Then y(t)=\beta[2 \omega t-\sin (\omega t)] ; \quad z(t)=\beta[1-\cos (\omega t)] ;(y-2 \beta \omega t)=-\beta \sin (\omega t),(z-\beta)=-\beta \cos (\omega t) \Rightarrow (y-2 \beta \omega t)^{2}+(z-\beta)^{2}=\beta^{2} . This is a circle of radius  \beta whose center moves to the right at constant speed:

y_{0}=2 \beta \omega t ; \quad z_{0}=\beta .

\text { (c) } \dot{z}(0)=\dot{y}(0)=\frac{E}{B} \Rightarrow-C_{1} \omega=\frac{E}{B} \Rightarrow C_{1}=-C_{3}=-\frac{E}{\omega B} ; C_{2} \omega+\frac{E}{B}=\frac{E}{B} \Rightarrow C_{2}=C_{4}=0 .

y(t)=-\frac{E}{\omega B} \cos (\omega t)+\frac{E}{B} t+\frac{E}{\omega B} ; z(t)=\frac{E}{\omega B} \sin (\omega t) \cdot \quad y(t)=\frac{E}{\omega B}[1+\omega t-\cos (\omega t)] ; z(t)=\frac{E}{\omega B} \sin (\omega t) .

\text { Let } \beta \equiv E / \omega B \text {; then }[y-\beta(1+\omega t)]=-\beta \cos (\omega t), z=\beta \sin (\omega t) ;[y-\beta(1+\omega t)]^{2}+z^{2}=\beta^{2} \text {. This is a circle } of radius \beta whose center is at   y_{0}=\beta(1+\omega t), z_{0}=0 .