A particle of mass $m=1.6\times {10}^{-27 }kg$ and charge $q=1.6\times {10}^{19} C$ enters a region of uniform magnetic field of strength $1T$ along the direction shown in the figure. The speed of the particel is ${10}^7 m/s$.

(a) The magnetic field is directed along the inward normal to the plane of the paper. If particle leaves the region of the field at the point $F$. Find the distance $EF$ and the angle $\theta$.

(b) If the direction of the field is along the outward normal the plane of the paper, find the time spent by the particle in the region of the magnetic field after entering it at $F$. 

Inside a magnetic field, speed of the charged particle does not change. Further, velocity is perpendicular magnetic field in both the cases hence path of the particle in the magnetic field will be circular. Centre of circle can be obtained by drawing perpendicular to velocity at $E$and $F$. Radius and angular speed of circular path would be $r=\frac{mv}{Bq}$ and $\omega =\frac{Bq}{m}$.

(a) Refer to figure (i)

$\angle CFG=90°-\theta$and $\angle CEG=90°-45°=45°$

Since, $CF=CE$

$\therefore \angle CFG= \angle CEG$

or $90°-\theta =45° \theta =45°$

Further, $FG=GE=r\cos 45°$

$$\begin{gathered} \therefore EF=2FG=2r \cos 45°=\frac{2mv\cos 45°}{Bq} \\ =\frac{2\left(1.6\times {10}^{-27}\right)\left({10}^7\right)\left({\displaystyle \frac{1}{\sqrt{2}}}\right)}{\left(1\right)\left(1.6\times {10}^{-19}\right)} \\ =0.14 m \end{gathered}$$

(b) Refer to figure (ii), In this case particle will complete $\frac{3}{4}th$ of the circle in the magnetic field. Hence, the time spent in the magnetic field:

$t=\frac{3}{4}$(time period of circular motion)

$$\begin{gathered} t=\frac{3}{4}\left(\frac{2\mathrm{\pi m}}{Bq}\right)=\frac{3\mathrm{\pi m}}{2Bq} \\ =\frac{\left(3\mathrm{\pi }\right)\left(1.6\times {10}^{-27}\right)}{\left(2\right)\left(1\right)\left(1.6\times {10}^{-19}\right)} \\ =4.712\times {10}^{-8} s \end{gathered}$$