Question

A particle of mass m and charge q is projected into a region that has a perpendicular magnetic field B. Find the angle of deviation (figure) of the particle as it comes out of the magnetic field if the width d of the region is very slightly smaller than 

(a) `(mv)/(qB)`  (b)`(mv)/(2qB)` (c)`(2mv)/(qB)`

Answer 1

Given:
Mass of the particle = m
Charge of the particle = q
Magnetic field = B
As per the question, the particle is projected into a perpendicular magnetic field.

(a) When the width, d = `(mv)/(qB)`
d is equal to the radius and θ is the angle between the radius and tangent, which is equal to `pi/2`.

(b) When the width, d = `(mv)/(2qB)`
Width of the region in which a magnetic field is applied is half of the radius of the circular path described by the particle.
As the magnetic force is acting only along the y direction, the velocity of the particle will remain constant along the x direction. So, if d is the distance travelled along the x axis, then
d = v_xt 
`t = (d)/(V_x)`.........(i))
(i)


(ii)

The acceleration along the x direction is zero. The force will act only along the y direction.
Using the equation of motion for motion along the y axis:
vy = uy + a_yt 
`vy = 0 + (qu_xBt)/(m)`
`= (qu_xBt)/m`
Putting the value of t from equation (i), we get:

`(qu_xBd)/(mv_x)`
we know
`tantheta = (vy)/(vx)`

`(qBd)/(mv_x) = (qBmv_x)/(2qBmvx) = 1/2`

`⇒theta = tan^-1(1/2)`
`= 26.4 = 30^circ = pi/6`
(c) When the width, d = (c) When the width, d = `(2mv)/(qB)`