Download BilaksPhysiks

A proton is shot with a velocity v at an angle 
to a uniform magnetic field B, which is directed
along the x-axis as shown below. Assume that
the only force on the proton is the magnetic
force. Explain why the proton will move in a
helical path. Find an expression for the pitch of
the helix.
v
+

B
y
z
x
Bold type denotes
vector quantities
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
Question 7
Question 8
Question 9
Question 10
Question 11
1. Why does the proton feel a
magnetic force?
i) Because it has a charge
ii) Because it is moving with a velocity which is not parallel to the
magnetic field
iii) Because it has a mass
A : i only
B : ii only
C : iii only
D : i and ii only
E : ii and iii only
Choice A
Incorrect
In order to feel a magnetic force, a particle
must be charged. There is another condition
that must be met for the particle to feel a
magnetic force.
Think about the following equation:
Fmag  q(v  B)  qvB sin n
Unit vector
perpendicular to
v and B
Choice B
Incorrect
This is true, because if a particle is
moving with a velocity that is parallel to the
magnetic field (=0), the particle will not feel a
Unit vector
force:
Fmag  q(v  B)  qvB sin( 0)n  0
perpendicular to
v and B
BUT, this is not the only reason that the
proton will feel a magnetic force.
Choice C
Incorrect
Mass has nothing to do with a particle
feeling a magnetic force.

Choice D
Correct
Unit vector in
the direction of
the force
Fmag  q(v  B)  qvB sin n
Notice that if the charge q of a particle or the angle 
between the magnetic field and the velocity of the
particle is zero, the magnetic force on the particle will be
zero.
Choice E
Incorrect
One reason that the proton feels a
magnetic force is because it is moving
with a velocity that is not parallel to the
magnetic field, but mass has nothing to
do with a particle feeling a magnetic
force. Try again
• We will break the motion of the proton into
components to see the contributions of each.
• Let’s first take a look at the motion of the
proton along the x-axis. Notice that the axes
are defined on the first slide.
2. Which of the following choices correctly
describes the x-component of the proton’s
motion?
A: The proton is accelerating in the xdirection (ax>0)
B: There is no motion in the x-direction.
C: The x-component of the proton’s velocity vx
is constant.
Choice A
Incorrect
The proton is not accelerating in the x-direction,
because the x-component of the magnetic force Fx is
zero.
The x-component of the proton’s velocity is parallel to
the magnetic field, therefore a force will not be felt by
the proton due to the horizontal component of motion.
v
vy

vx
B
Fx  qvBsin   0
Choice B
Incorrect
Initially, there is motion in the x-direction
and the y-direction, because it is stated
that the proton is shot with a velocity at
an angle to the x-axis. It is not shot
directly upwards in the y-direction.
Choice C
Correct
This is true because there is no x-component of
magnetic force on the proton.
The x-component of the proton’s velocity is parallel to
the magnetic field, therefore a force will not be felt by
the proton due to the horizontal component of motion.
Fx  qvBsin   0
3. What type of path will the proton follow if
only the motion in the x-y plane is
considered?
A: Straight line (Linear)
B: Parabolic
C: Circular
This is due to the horizontal
component of the initial
velocity.
Choice A
Correct
If we only consider the motion in the x-y
plane, the particle will move with constant
velocity in the positive x-direction and will not
change directions. This describes a linear
path.
Choice B
Incorrect
If we only consider the motion in the x-y
plane, the particle will move with constant
velocity in the positive x-direction and will not
change directions. This describes a linear
path.
Choice C
Incorrect
If we only consider the motion in the x-y
plane, the particle will move with constant
velocity in the positive x-direction and will not
change directions. This describes a linear
path.
4. Which of the following choices correctly
describes the y-component of the proton’s
motion as it moves through the magnetic field?
A: The proton is accelerating in the ydirection (ay≠0)
B: There is no motion in the y-direction.
C: The y-component of the proton’s velocity
vy is constant.
Choice A
Correct
The proton would accelerate in the y-direction
because the y-component of velocity causes
a magnetic force to be felt by the proton.
F  qvB sin 
F  qv yB

Where: ≠0
Choice B
Incorrect
There is motion in the x-direction and
the y-direction because it is stated that
the proton is shot with a velocity at an
angle  to the x-axis. It is not shot
horizontally.
Choice C
Incorrect
The y-component of the proton’s velocity vy
is not constant, because this motion causes a
magnetic force to be felt by the proton.
F  qvBsin 
Where: ≠0

5. What type of path will the proton follow if only the
motion of the proton in the y-z plane is considered?
A: Straight line (Linear)
B: Parabolic
C: Circular
This is due to the vertical
component of the initial
velocity.
Choice A
Incorrect
Using the right hand rule we see that the magnetic
force on the proton will cause it to travel on a circular
path (counterclockwise if looking down the x-axis
from the origin).
At any instant during
F
the proton’s motion,
v
your middle finger
B
The force must be centripetal and
have constant magnitude to
obtain uniform circular motion.
should point down the
positive x-axis. and
your pointer finger
should point in the
direction of the ycomponent of the
velocity. This will direct
your thumb in the
direction of the force
that pushes the proton.
Choice B
Incorrect
Using the right hand rule we see that the magnetic
force on the proton will cause it to travel on a circular
path (counterclockwise if looking down the x-axis
from the origin).
At any instant during
v
F
B
The force must be centripetal and
have constant magnitude to
obtain uniform circular motion.
the proton’s motion,
your middle finger
should point down the
positive x-axis. and
your pointer finger
should point in the
direction of the ycomponent of the
velocity. This will direct
your thumb in the
direction of the force
that pushes the proton.
Choice C
Correct
If we only consider the motion of the
proton in the y-z plane due to its initial
vertical motion, the proton’s path would
be circular.
6. What type of path do you get if you
combine the contributions of each
component of the proton’s motion ?
A: Linear
B: Helical
C: Circular
Choice A
Incorrect
This is the result if only the component of
the motion along the magnetic field (xcomponent) is considered.
Choice B
Correct
If you trace out a circle while moving
down an axis, the resulting path is
helical.
In this case, the proton’s motion along
the field is at a constant velocity
Choice C
Incorrect
This is only the effect of the proton’s
motion in the y-z plane.
7. To find the radius of the projected motion of the proton in
the y-z plane, choose all of the statements that are relevant.
(i) the magnetic force provides the centripetal acceleration
(ii) the magnetic force should be equated to the mass of
the
proton times centripetal acceleration
(iii) the centripetal acceleration is:
where R is the radius of the circle
A: i & ii only 
B: i & iii only
C: i, ii, and iii
v2
a
R
Choice A
Incorrect
2
v
a
R
This equation is the correct expression
for centripetal acceleration and will be
 find the radius of the proton’s
used to
path.
Choice B
Incorrect
As usual, we set F=ma. In this case
the mass is the mass of a proton, and a
is centripetal acceleration expressed as:
v2
a
R

Choice C
Incorrect
All of these statements are relevant to
finding the radius of the proton’s path.
8. Which of the following expressions is correct
for the magnitude of the component of the
velocity perpendicular to the magnetic field?
A:
qBm
v
R
B:
qBR
v
m
C:
qBRsin 
v
m


Choice A
Incorrect
Check your algebra
ma  Fmag
mv2
 qvB sin 
R
qBRsin 
v
m

Choice B
Incorrect
Check your algebra
ma  Fmag
mv2
 qvB sin 
R
qBRsin 
v
m

Choice C
Correct
ma  Fmag
mv2
 qvB sin 
R
qBRsin 
v
m
9. Which of the following
expressions is correct for the
period T?
A:
qBsin 
2m
B:
T
C:
2Rm
T
qB


2m
T
qBsin 
Choice A
Correct
2R
2Rm
2m
Period: T 


v
qBRsin  qBsin 

Choice B
Incorrect
2R
2Rm
2m
Period: T 


v
qBRsin  qBsin 

Choice C
Incorrect
2R
2Rm
2m
Period: T 


v
qBRsin  qBsin 



10. Which one of the following expressions
gives the pitch P of the helix (the distance
traveled in the x-direction during one
revolution) ?
A: P  v y T
vx
B: P 
T
C:
P  vxT
Choice A
Incorrect
The pitch of a helix is the distance
traveled along the helical axis during
one revolution. The helical axis is the xaxis in this case. We are not concerned
with the y-component of motion here.
Choice B
Incorrect
The pitch of a helix is given by the
following expression:
P  vxT

Choice C
Correct
This is the correct expression for the
pitch of the helix since the axis of the
helix is the x-axis.
11. Find the expression for the pitch
of the helical path of the proton
described in the original problem.
Answer
Since the x-component of the
proton’s velocity is constant:
From question 9 we have:

vx  vcos
2m
T
qBsin 
P  vxT  vcosT

vcos2m 2mv
P

cot 
 
qBsin
qB