Question

An electron accelerated by a potential difference $V=3$ volt first enters into a uniform electric field of a parallel plate capacitor whose plates extend over a length $l=6cm$ in the direction of initial velocity.
The electric field is normal to the direction of initial velocity and its strength varies with times as $E=\alpha t$, where $\alpha =3600V{m}^{-1}{s}^{-1}$.
Then the electron enters into a uniform magnetic field of induction $B=\pi \times {10}^{-9}T$. Direction of magnetic field is same as that of the electric field.
The pitch of helical path traced by the electron in the magnetic field is $1215\times {10}^{-x}cm$. Find x.

(Mass of electron, $m=9\times {10}^{-31}kg)$

Answer 1

Since, electron is accelerated through a potential difference V, therefore, its initial velocity $v_0$ is given by:

$\frac{1}{2}m{v}_0^2=eV$
$v_0=\sqrt{\frac{2eV}{m}}$ .... (i)
Since, initial velocity is parallel to plates or normal to the direction of electric field, therefore, component of velocity parallel to plates remains constant as $v_0$.
Hence, time taken by the electron to cross the electric field is, $t_0=\frac{l}{v_0}$
Now consider motion of electron, normal to plates.
At some instant t, its acceleration $=\frac{eE}{m}=\frac{e\alpha t}{m}$
Let velocity component normal to plates to $v_y$.
Then this acceleration is equal to $\frac{d}{dt}{v}_y$.
$\frac{d}{dt}{v}_y=\frac{e\alpha t}{m}$ or $d{v}_y=\frac{e\alpha t}{m}dt$

But at initial moment $t=0,v_y=0\Rightarrow {\int }_0^{v_y}d{v}_y=\frac{e\alpha }{m}{\int }_0^{l_0}tdt$

$v_y=\frac{e\alpha }{2m}{t}_0^2=\frac{\alpha l^2}{4V}$ (ii)

Angular deviation $\theta$ of electron from its initial direction of motion is shown in fig.
Now electron enters into magnetic field.

Pitch of its helical path:
$P=\frac{2\pi m}{eB}vcos(90-\theta )$
$=\frac{2\pi m}{eB}vsin\theta =\frac{2\pi m}{eB}{v}_y$
$=\frac{\pi m\alpha I^2}{2eBV}=1.215cm$