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

`pi/12`sec.

Explanation

As the particle enters the region of magnetic field, it moves in a circular path of radius `R=(mv)/(qB)=1 m` whose centre is at `O`.and `omega=(qB)/m=1 rad//sec`.
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 shown in 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`.