A proton accelerated by a potential difference $V = 500 kV$ flies through a uniform transverse magnetic field with induction$B = 0.51 T$. The field occupies a region of space $d = 10 cm$thickness (see fig). Find the angle through which the proton deviates from the initial direction of motion.

If $v$ is the speed of the proton, then

$$\begin{gathered} \frac{1}{2}m{v}^2 = qV \\ \Rightarrow v=\sqrt{\frac{2qV}{m}} \end{gathered}$$

Here $q$ and $m$ are the charge and mass of the proton respectively. The proton traverses circular path of radius $R$ in the magnetic field, where

$R = \frac{mv}{qB}$

From the figure 

$$\begin{gathered} \sin \alpha = \frac{d}{ R}=\frac{d}{{\displaystyle \frac{m\sqrt{\frac{2qV}{m}}}{qB}}} \\ =Bd\sqrt{\frac{q}{2mV}} \end{gathered}$$

On substituting the values, we get $\alpha = {30}^o$