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

Inside a magnetic field speed of charged particle does not change. Further, velocity is perpendicular to 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 perpendiculars to velocity ( or tangent to the circular path) at E and F. Radius and angular speed of circular path would be
`r=(mv)/(Bq)` and `omega=(Bq)/m`

A) Refer figure (a) : `angleCFG=90^(@)-theta` and `angle=90^(@)-45^(@)=45^(@)`, Since CF=CE
`thereforeangleCFG=angleCEG` or `90^(@)-theta=45^(@)` or `theta=45^(@)`
Further, `FG = GE= R cos 45° therefore EF = 2FG = 2r cos 45°`
`=(2mvcoss45^(@))/(Bq)=(2(1.6xx10^(-27))(1/(sqrt2)))/((1)(1.6xx10^(-19)))=0.14m`
In this case particle completes `1/4`th of circle in the magnetic field
B) Refer figure (b) : In this case particle will complete `3/4` th of circle in the magnetic field. Hence, the time spent in the magnetic field.
`t=3/4` (time period of circular motion)
`=3/4((2pim)/(Bq))=(3pim)/(2Bq)=((3pi)(1.6xx10^(27)))/((2)(1)(1.6xx10^(-19)))=4.712xx10^(-8)s`