An electron accelerated by a potential difference $\text{V=3.6V}$ first enters a region of uniform electric field of
a parallel-plate capacitor whose plates extend over a length $\text{l=6\hspace{0.17em}cm}$ in the direction of initial velocity. The electric field is normal to the direction of initial velocity and its strength varies with time as $\text{E=a}\times \text{t}$ where ${\text{a=3200\hspace{0.17em}Vm}}^{\text{-1}}{\text{s}}^{\text{-1}}$. Then the electron enters a region of uniform magnetic field of induction $\text{B=}\pi \times {\text{10}}^{-9}T$. Direction of magnetic field is same as that of the electric field. Calculate the pitch (in mm) of helical path traced by the electron in the magnetic field. (Mass of electron,  $m=9\times {10}^{-31}kg,e=1.6\times {10}^{-19}C$)
[Neglect the effect of induced magnetic field].
 

Since, electron is accelerated through a potential difference V, its initial velocity $v_0$ is given by  $\frac{1}{2}m{v}_0^2=eV$
Or  $v_0=\sqrt{\frac{2eV}{m}}---------\left(1\right)$
Since, initial velocity is parallel to plates or normal to the direction of electric field, component of velocity parallel to plates remains constant as  $v_0$.
Hence, time taken by the electron to cross electric field is $t_0=\frac{l}{v_0}$
Now consider motion of electron, normal to plates.
At some instant t, its acceleration $=\frac{eE}{m}=\frac{eat}{m}$
Let velocity component normal to plates be $v_y$
$\therefore \frac{d}{dt}{v}_y=\frac{eat}{m}$
$\therefore {\int }_0^{v_y}dv=\frac{ea}{m}{\int }_0^{t_0}tdt$
Or  $v_y=\frac{ea}{2m}{t}_0^2=\frac{a{l}^2}{4V}-------\left(2\right)$
If is angular deviation of electron from its initial direction of motion, then pitch of its helical path.
$p=\frac{2\pi m}{eB}v\cos (90-\theta )$
$\therefore p=\frac{2\pi m}{eB}v\sin \theta =\frac{2\pi m}{eB}{v}_y$
Or  $p=\frac{\pi ma{l}^2}{2eBV}$
p =9mm