91Ó°ÊÓ

Initial velocity of the particle is

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

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

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

Here, $\mathrm{Ó\neg }$is the frequency and $C_1$are constants to be set from initial conditions.

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

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

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) .

$C_2=-\frac{E}{2\mathrm{Ó\neg }B}$

Thus, $C_4=\frac{E}{2\mathrm{Ó\neg }B}$

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

Apply this to equation (2) .

$C_1=0$

Thus,$C_3=0$

Hence, the trajectory is

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

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) .

$C_1=-\frac{E}{\mathrm{Ó\neg }B}$

Thus, $C_4=\frac{E}{\mathrm{Ó\neg }B}$

Hence, the trajectory is