A charged particle projected in a magnetic field $\overrightarrow{\mathrm{B}}=10\stackrel{\wedge }{\mathrm{k}}\mathrm{T}$ from the origin in x – y plane. The particle moves in a circle and just touches a straight line $\mathrm{y}=5\left(\mathrm{m}\right)\text{ at }\mathrm{x}=5\sqrt{3}\left(\mathrm{m}\right)$ . The mass of particle = 5 × 10^-5 kg charge = 1$\mathrm{\mu }$C.

The trajectory of the particle will be a circle with its centre in the fourth quadrant , it touches the line y = 5 and passes through the origin .

We can write , $R^2 = {\left(5\sqrt{3}\right)}^2 +{\left(R - 5\right)}^2 \Rightarrow R = 10 m.$

Angular frequency $\omega = \frac{qB}{m }=0.2 rad/s$

 Therefore v = R$\omega =2 m/s .$

The arc of the circle starting from the origin upto the point of tangency with the line y = 5, subtends an angle of ${\tan }^{-1}\left(\frac{10-5}{5\sqrt{3}}\right) ={30}^0$. 

So angle between the velocity vector and x axis = ${90}^0-{30}^0 ={60}^0$