Saved Bookmarks
| 1. |
In the figure shown a positively charged particle of charge q and mass m enters into a uniform magnetic field of strength B as shown in the figure.The magnetic field points inwards and is present only within a region of width d.The initial velocity of the particle is perpendicular to the magnetic field and phi=30^(@).Find the time spent by the particle inside the magnetic field if d=0.2 m,B=1T,q=1C,m=1kg and v=1m//s.Use sin 45^(@)=0.7 if required . |
|
Answer» we assumed `d` to be sufficiently large so that the particle emerges out of region of magnetic field at `Q` figure `-(a)`. `therefore x=R-R COS 60^(@) =0.5 gt d` `therefore` The charge will CROSS the field and emerge from the right side. `therefore` The trajectory of the particle in the region of magnetic field is as shownin figure `-b` In the figure (b)`PQ` is the chord and `OC` is `_|_` bisector of line `PQ`.`Q` is the point from where the particle emerges out.We can see from the geometry that `/_APQ=phi+theta/2` `PQ=d sec (phi +theta/2)=2R sin theta/2` `rArr d=2R sin theta/2cos (phi+theta/2)=R[sin (theta/2+phi+theta/2)+sin (-phi)]` `rArr d=R [sin (theta+phi)-sin phi] rArr sin (theta+phi)=d/15+sin theta=0.7` `rArr theta+phi=45^(@) rArr t=pi12sec`. 
|
|