An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30° with the initial velocity.

Potential difference, V = 2 KV = 2000 volt
Uniform magnetic field, B = 0.15 T 

(a) If magnetic field is transverse to initial velocity of electron then, the velocity vector has no component in the direction of magnetic field. 
∴ Force on electron = Bev sin 90° = Bev 

This force acts as the centripetal force 

$\therefore$               $\boxed{\mathrm{Bev} = \frac{{\mathrm{mv}}^2}{\mathrm{r}}}$
             
                      $\mathrm{r} = \frac{\mathrm{mv}}{\mathrm{eB}}$                       ...(1)

But,         $\boxed{\frac{1}{2}{\mathrm{mv}}^2 =\mathrm{eV}}$                         ...(2) 

 $\Rightarrow$             $\mathrm{v}=\sqrt{\frac{2\mathrm{eV}}{\mathrm{m}}}$
$\therefore$        $\mathrm{r} = \frac{\mathrm{m}}{\mathrm{eB}}\sqrt{\frac{2\mathrm{eV}}{\mathrm{m}}}$               [From 1, 2]

            $$\begin{gathered} \mathrm{v} = \frac{1}{\mathrm{B}}\sqrt{\frac{2\mathrm{mV}}{\mathrm{e}}} \\ = \frac{1}{0.15}\sqrt{\frac{2\times 9.1\times {10}^{-31}\times 2000}{1.6 \times {10}^{-19}}}\mathrm{m} \\ = {10}^{-3}\mathrm{m} = 1.0 \mathrm{mm}. \end{gathered}$$ 

The electron would move in a circular trajectory of radius 1.0 mm. The plane of the trajectory is normal to B. 

(b) If velocity makes an angle 30° with the direction of magnetic field, the velocity can be resolved into v_⊥ and v_||

i.e., v cos 30° and v sin 30° respectively.

Due to v_⊥ the electron will move on a circular path. The resultant path will be a combination of straight line motion and circular motion which is called helical. 

Thus,        $\mathrm{evB} \sin \mathrm{\theta } = \frac{\mathrm{m}(\mathrm{v} \sin \mathrm{\theta }{)}^2}{{\mathrm{r}}_{\mathrm{n}}}$ 

for circular motion of radius ${\mathrm{r}}_{\mathrm{n}}$ 

                 ${\mathrm{ev}}_{\perp } \times \mathrm{B} = \frac{{\mathrm{mv}}_{\perp }^2}{{\mathrm{r}}_{\mathrm{n}}}$
                        ${\mathrm{v}}_{\perp } = \mathrm{v} \sin \mathrm{\theta }$
   
                         ${\mathrm{r}}_{\mathrm{n}} = \frac{\mathrm{mv} \sin \mathrm{\theta }}{\mathrm{eB}}$

 ${\mathrm{r}}_{\mathrm{n}} = \frac{9.1 \times {10}^{-31} \times 2.65 \times {10}^7 \times \sin 30}{1.6 \times {10}^{-19} \times 0.15}$

   $$\begin{gathered} = 0.49 \times {10}^{-3}\mathrm{m} \\ = 0.49 \mathrm{mm} \simeq 0.5 \mathrm{mm} \end{gathered}$$ 

The linear velocity is   = $\mathrm{v} \cos \mathrm{\theta }$
                              $$\begin{gathered} = 2.65 \times {10}^7 \times \cos 30° \\ = 2.65 \times {10}^7\times \frac{\sqrt{3}}{2} \\ = 2.3 \times {10}^7 {\mathrm{ms}}^{-1} \end{gathered}$$ 

Thus, the electron moves in a helical path of radius 0.49 mm with a velocity component of 2.3 x 10⁷ ms^–1 in the direction of magnetic field.