A charged partide enters a magnetic field at right angles to the magnetic field. The field exists for a ler
equal to 1.5 times the radius of circular path of the particle. The particle will be devated from its path by
(1) 90°
(2)sin-1(2/3)
(3)30°
(4) 180°
Answers
Let magnetic field strength be B
Charge q
Mass m and
Radius of the circular path be r
B=B
k
^
v
=vcosθ
i
^
+vsinθ
j
^
F
=(vcosθ
i
^
+vsinθ
j
^
)×Bq
k
^
⇒
F
=Bqv(cosθ
i
^
+sinθ
j
^
)
θ=ωt
a
x
=
m
Bqv
sinωt
dt
2
d
2
x
=
m
Bqv
sinωt
Integrating
⇒x=
mω
2
Bqv
sin(ωt+θ)
So magnitude of maximum horizontal displacement suffered by the particle in the field will be
x
max
=
mω
2
Bqv
Again,
x
max
=
mω
2
Bqv
Bqv=mω
2
r
ω=
m
Bq
x
max
=
ω
v
v=rω
⇒x
max
=r
This means for a particle to reach maximum displacement the field must have a minimum length of r
since length of field is 1.5r the particle cannot come out of field horizontally so to come out of it must reverse its direction and thus displacement.
Thus the particle will deflect by 180°
Answer:
Please see diagram.
As I understand the question, let the uniform magnetic field B be into the plane of the diagram. Let the charged particle +q enter from one side of magnetic field with a velocity v.
The force on q is F = q v x B.
It is a vector cross product. The force is in the direction towards the center of circle and supplies the centripetal force. Since force and acceleration on q are in perpendicular direction to velocity v, velocity does not change.
The particle q moves in a circle and exits the magnetic field at the other end of the diameter at the entry point. Hence the angle turned by the particle is 180 degrees.
