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.

+ B

v

x

y

zBold type denotes vector quantities

Question 1Question 2Question 3Question 4Question 5Question 6Question 7Question 8

Question 9Question 10Question 11

1. Why does the proton feel a magnetic force?

i) Because it has a chargeii) Because it is moving with a velocity which is not parallel to the

magnetic fieldiii) Because it has a mass

A : i onlyB : ii onlyC : iii onlyD : i and ii onlyE : ii and iii only

Choice AIncorrect

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:

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Fmag=q(v ×B) = qvBsinθn

Unit vector perpendicular to v and B

Unit vector perpendicular to v and B

Choice BIncorrect

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 force:

BUT, this is not the only reason that the proton will feel a magnetic force.

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Fmag=q(v ×B) = qvBsin(0)n = 0Unit vector perpendicular to v and B

Unit vector perpendicular to v and B

Choice CIncorrect

Mass has nothing to do with a particle feeling a magnetic force.

Choice DCorrect

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Fmag=q(v ×B) = qvBsinθn

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.

Unit vector in the direction of the force

Unit vector in the direction of the force

Choice EIncorrect

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 x-direction   (ax>0)

B: There is no motion in the x-direction.

C: The x-component of the proton’s velocity vx is   constant.

Choice AIncorrect

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.

vy

vx

v

B

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Fx =qvBsinθ = 0

Choice BIncorrect

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 CCorrect

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.

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Fx =qvBsinθ = 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 ACorrect

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 BIncorrect

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 CIncorrect

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 y-direction   (ay≠0)

B: There is no motion in the y-direction.

C: The y-component of the proton’s velocity vy    is constant.

Choice ACorrect

The proton would accelerate in the y-direction because the y-component of velocity causes a magnetic force to be felt by the proton.

Where: ≠0Where: ≠0

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F =qvBsinθ

F = qv yB

Choice BIncorrect

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 CIncorrect

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.

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F =qvBsinθWhere: ≠0Where: ≠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 AIncorrect

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).

F

B

v

The force must be centripetal and have constant magnitude to obtain uniform circular motion.

At any instant during 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 y-component of the velocity. This will direct your thumb in the direction of the force that pushes the proton.

At any instant during 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 y-component of the velocity. This will direct your thumb in the direction of the force that pushes the proton.

Choice BIncorrect

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).

F

B

v

At any instant during 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 y-component of the velocity. This will direct your thumb in the direction of the force that pushes the proton.

At any instant during 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 y-component of the velocity. This will direct your thumb in the direction of the force that pushes the proton.

The force must be centripetal and have constant magnitude to obtain uniform circular motion.

Choice CCorrect

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 AIncorrect

This is the result if only the component of the motion along the magnetic field (x-component) is considered.

Choice BCorrect

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 CIncorrect

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

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a=v2

R

Choice AIncorrect

This equation is the correct expression for centripetal acceleration and will be used to find the radius of the proton’s path.

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a=v2

R

Choice BIncorrect

As usual, we set F=ma. In this case the mass is the mass of a proton, and a is centripetal acceleration expressed as:

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a=v2

R

Choice CIncorrect

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:

B:

C:

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v=qBmR

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v=qBRm

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v=qBRsinθm

Choice AIncorrect

Check your algebra

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ma=Fmag

mv2

R=qvBsinθ

v =qBR sinθ

m

Choice BIncorrect

Check your algebra

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ma=Fmag

mv2

R=qvBsinθ

v =qBR sinθ

m

Choice CCorrect

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ma=Fmag

mv2

R=qvBsinθ

v =qBR sinθ

m

9. Which of the following expressions is correct for the

period T?

A:

B:

C:

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T =2πRm

qB

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T =2πm

qBsinθ

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T =qBsinθ2πm

Choice ACorrect

Period:

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T =2πR

v=

2πRmqBRsinθ

=2πm

qBsinθ

Choice BIncorrect

Period:

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T =2πR

v=

2πRmqBRsinθ

=2πm

qBsinθ

Choice CIncorrect

Period:

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T =2πR

v=

2πRmqBRsinθ

=2πm

qBsinθ

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:

B:

C:

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P =vxT

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P =vyT

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P =vx

T

Choice AIncorrect

The pitch of a helix is the distance traveled along the helical axis during one revolution. The helical axis is the x-axis in this case. We are not concerned with the y-component of motion here.

Choice BIncorrect

The pitch of a helix is given by the following expression:

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P =vxT

Choice CCorrect

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

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vx =vcosθ

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P =vcosθ2πmqBsinθ

=2πmv

qBcot θ

Since the x-component of the proton’s velocity is constant:

Since the x-component of the proton’s velocity is constant:

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P =vxT =vcosθT

From question 9 we have:

From question 9 we have:

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T =2πm

qBsinθ