A particle of mass `m=1.6xx10^-27` kg and charge `q=1.6xx10^-19C` enters a region of uniform magnetic field of stregth `1 T` along the direction shown in figure. The speed of the particle is `10^7 m//s`

a. The magnetic field is directed along the inward normal to the plane of the paper. The particle leaves the region of the fiedl at the point `F`. Find the distasnce `EF` and the angle theta.
b. If the direction of the field is along the outward normal to the plane of the paper find the time spent by the particle in the regin of the magnetic field after entering it at `E`.

Explanation

First of all let us see how the path look like. At least for the part when the particle is inside the magnetic field, the path will be part of a circle. To locate the center, let us find the direction of `q(vec(v) xx vec(B))`.
As can be seen in the figure, it is directed perpendicular to velocity and is toward O. Since this force is centripetal in nature, the force is directed toward the center. The coordinates of center can be easily found out. The charge particle will move in anticlockwise direction and exit the magnetic field at the boundary.
When a charged particle moves perpendicular to a magnetic field the path is a circle with magnetic force qvB directed radially toward the center of the path. So, if we draw normal at E and F they will meet at O which is the center of the circle. By geometry of Fig.
`theta = phi_(2) = phi_(1) = 45^(@)`

Calculation : As `phi_(1) = phi_(2) = theta`, therefore
`EF = 2EG = 2 rsin theta`.
Since
`r = (mv)/(qB) = (1.6 xx 10^(-27) xx 10^(7))/(1.6 xx 10^(-19) xx 1) = 0.1 m`,
therefore
`EF - (2 xx 0.1 xx sin 45^(@)) m = 10sqrt(2) cm`.
Now, as in a magnetic field the speed of a particle remains constant, so if t is the time taken by the particle in the field, then
`S = v xx t = r theta`
`t = (r theta)/(v)`.
Hence,
`t = (0.1 xx pi//2)/(10^(7)) = 5pi xx 10^(-9) = 1.57 xx 10^(-8)s`.