Question

Find and sketch the trajectory of the particle in Ex. 5.2, if it starts at

the origin with velocity

$$\begin{gathered} \mathbf{\left(}\mathbf{a}\mathbf{\right)}\overrightarrow{\mathbf{v}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{E}}{\mathbf{B}}\stackrel{\mathbf{⏜}}{\mathbf{y}} \\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\overrightarrow{\mathbf{v}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{E}}{\mathbf{2}\mathbf{B}}\stackrel{\mathbf{⏜}}{\mathbf{y}} \\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}\overrightarrow{\mathbf{v}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{E}}{\mathbf{B}}\mathbf{(}\stackrel{\mathbf{⏜}}{\mathbf{y}}\mathbf{+}\stackrel{\mathbf{⏜}}{\mathbf{z}}\mathbf{)}\mathbf{.} \end{gathered}$$

Short answer

(a) The trajectory for $\overrightarrow{v}\left(0\right)=\frac{E}{B}\stackrel{⏜}{y}$is

$$\begin{gathered} y\left(t\right)=\frac{E}{B}t \\ z\left(t\right)=0 \end{gathered}$$

(b) The trajectory for $\overrightarrow{v}\left(0\right)=\frac{E}{2B}\stackrel{⏜}{y}$is

$$\begin{gathered} y\left(t\right)=-\frac{E}{2\omega B}\left(\omega t\right)+\frac{E}{B}t \\ z\left(t\right)=-\frac{E}{2\omega B}\cos \left(\omega t\right)+\frac{E}{2\omega B} \end{gathered}$$

(c) The trajectory for $\overrightarrow{v}\left(0\right)=\frac{E}{B}\left(\stackrel{⏜}{y}+\stackrel{⏜}{z}\right)$is

$$\begin{gathered} y\left(t\right)=-\frac{E}{\omega B}\cos \left(\omega t\right)+\frac{E}{B}t+\frac{E}{\omega B} \\ z\left(t\right)=\frac{E}{\omega B}\sin \left(\omega t\right) \end{gathered}$$

Step-by-step solution

Given data

Initial velocity of the particle is

$$\begin{gathered} \left(\mathrm{a}\right)\overrightarrow{v}\left(0\right)=\frac{E}{B}\stackrel{⏜}{y}, \\ \left(\mathrm{b}\right)\overrightarrow{v}\left(0\right)=\frac{E}{2B}\stackrel{⏜}{y}, \\ \left(\mathrm{c}\right)\overrightarrow{v}\left(0\right)=\frac{E}{B}\left(\stackrel{⏜}{y}+\stackrel{⏜}{z}\right) \end{gathered}$$

General trajectory of a particle in crossed electric and magnetic field

The general trajectory of a charge in the given electric field $\overrightarrow{E}$and magnetic field $\overrightarrow{B}$is

$$\begin{gathered} \mathbf{y}\left(t\right)\mathbf{=}{\mathbf{C}}_{\mathbf{1}}\mathbf{cos}\left(\mathrm{\omega t}\right)\mathbf{+}{\mathbf{C}}_{\mathbf{2}}\mathbf{sin}\left(\mathrm{\omega t}\right)\mathbf{+}\frac{\mathbf{E}}{\mathbf{B}}\mathbf{t}\mathbf{+}{\mathbf{C}}_{\mathbf{3}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right) \\ \mathbf{z}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{C}}_{\mathbf{2}}\mathbf{cos}\mathbf{\left(}\mathbf{\omega t}\mathbf{\right)}\mathbf{+}{\mathbf{C}}_{\mathbf{1}}\mathbf{sin}\mathbf{\left(}\mathbf{\omega t}\mathbf{\right)}\mathbf{+}{\mathbf{C}}_{\mathbf{4}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

Here, $\omega$is the frequency and $C_1$are constants to be set from initial conditions.

Trajectory for the first case

The initial conditions are

(i)$y\left(0\right)=0$

Apply this to equation (1) .

${\mathrm{C}}_1+{\mathrm{C}}_3=0$

(ii) $z\left(0\right)=0$

Apply this to equation (2) .

$C_2+C_4=0$

(iii) $\stackrel{.}{y}\left(0\right)=\frac{E}{B}$

Applied this to equation (1) .

$C_2=0$

Thus,$C_4=0$

(iv) $\stackrel{.}{z}\left(0\right)=0$

Apply this to equation (2) .

$C_1=0$

Thus,$C_3=0$

Hence, the trajectory is

$$\begin{gathered} y\left(t\right)=\frac{E}{B}t \\ z\left(t\right)=0 \end{gathered}$$

Trajectory for the second case

The initial conditions are

(i)$y\left(0\right)=0$

Apply this to equation (1) .

$C_1+C_3=0$

(ii) $z\left(0\right)=0$

Apply this to equation (2) .

$C_2+C_4=0$

(iii) $\stackrel{.}{y}\left(0\right)=\frac{E}{2B}$

Apply this to equation (1) .

$C_2=-\frac{E}{2\omega B}$

Thus, $C_4=\frac{E}{2\omega B}$

(iv) $\stackrel{.}{z}\left(0\right)=0$

Apply this to equation (2) .

$C_1=0$

Thus,$C_3=0$

Hence, the trajectory is

$$\begin{gathered} y\left(t\right)=-\frac{E}{2\omega B}\sin \left(\omega t\right)+\frac{E}{B}t \\ z\left(t\right)=-\frac{E}{2\omega B}\cos \left(\omega t\right)+\frac{E}{2\omega B} \end{gathered}$$

Trajectory for the third case

The initial conditions are

(i)$y\left(0\right)=0$

Apply this to equation (1) .

$C_1+C_3=0$

(ii) $z\left(0\right)=0$

Apply this to equation (2) .

$C_2+C_4=0$

(iii) $\stackrel{.}{y}\left(0\right)=\frac{E}{B}$

Apply this to equation (1) .

$C_2=0$

Thus, $C_4=0$

(iv) $\stackrel{.}{z}\left(0\right)=\frac{E}{B}$

Apply this to equation (2) .

$C_1=-\frac{E}{\omega B}$

Thus, $C_4=\frac{E}{\omega B}$

Hence, the trajectory is

$$\begin{gathered} y\left(t\right)=-\frac{E}{\omega B}\cos \left(\omega t\right)+\frac{E}{B}t+\frac{E}{\omega B} \\ z\left(t\right)=\frac{E}{\omega B}\sin \left(\omega t\right) \end{gathered}$$