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5a. Magnetism Due: 11:00pm on Sunday, February 7, 2010

Note: To understand how points are awarded, read your instructor's Grading Policy.

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A Wire and a Compass

Description: Short conceptual problem on the deflection of a compass needle placed in proximity to a wire through which a current flows. Based on Young/Geller Conceptual Analysis 20.7.

You are looking straight down on a magnetic compass that is lying flat on a table. A wire is stretched horizontally under the table, parallel to and a short distance below the compass needle. The wire is then connected to a battery so that a current flows through the wire. This current causes the north pole of the compass needle to deflect to the left.

The questions that follow ask you to compare the effects of different actions on this initial deflection.

Part A

If the wire is lowered farther from the compass, how does the new angle of deflection of the north pole of the compass needle compare to its initial deflection?

As the distance between the wire and the compass increases, the magnetic field produced by the wire weakens. Therefore, the north pole of the compass needle will deflect away from its original direction (determined by

Hint A.1 How to approach the problem

When the switch is closed, current passes through the wire and a magnetic field is created near the wire. This causes the north pole of the compass needle to deflect. The magnitude of the magnetic field produced by the conductor changes as you move farther away from the conductor. Therefore, when the wire is lowered, the compass needle will detect a different magnetic field. To determine the effect on the angle of deflection, find how the magnetic field's magnitude varies with the distance from the conductor.

Hint A.2 Find an expression for the magnetic field's magnitude produced by a conductor

Which of the following expressions gives the magnitude of the magnetic field , produced by a conductor

carrying a current , at a distance from the axis of the conductor? In the equations below, let denote the

permeability of free space.

As you discovered, the magnitude of the magnetic field produced by a straight conductor carrying a current decreases with the distance from the conductor. Therefore, what will be the effect on the north pole of the compass needle as the wire is lowered farther away from the compass?

ANSWER:

ANSWER: It is smaller.

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the magnetic field of the earth) by a smaller angle.

Part B

With the wire back at its initial location, you connect a second identical battery in series with the first one. When you close the switch, how does the new angle of deflection of the north pole of the compass needle compare to its initial deflection?

Hint B.1 How to approach the problem

When a second battery is added in series with the first one, a different emf is delivered to the wire, resulting in a different current passing through the wire. How does the magnitude of the magnetic field produced by a conductor carrying a current change as a function of the current?

Hint B.2 Find how the emf delivered to the circuit changes

When a second battery is added in series with the first one, how does the potential difference across the wire change?

What is the effect of a larger potential difference between the ends of the wire on the current passing through it?

Hint B.2.1 Batteries in series

When two or more batteries are connected in series to form a source of emf, the potential difference between the ends of the emf source is equal to the sum of the voltages of each battery.

ANSWER: It increases.

Hint B.3 Find how the current in the wire changes

When a second battery is added in series with the first one, how does the current in the wire change?

Because the batteries are connected in series, a larger emf will be delivered to the circuit, causing the wire to draw a larger current. How does the magnitude of the magnetic field produced by a conductor carrying a current change as a function of the current?

Hint B.3.1 How to approach the problem

Because the batteries are connected in series, a larger emf will be delivered to the circuit.

Hint B.3.2 Ohm's law

Recall that the potential difference across a resistor is equal to the product of the current passing through

the resistor and the resistor's resistance :

.

ANSWER: It increases.

Hint B.4 Magnetic field produced by a conductor

The magnitude of the magnetic field produced by a conductor carrying a current at a distance from the

axis of the conductor is given by



Since the magnitude of the magnetic field produced by the wire increases due to the larger current flowing through the wire, the compass needle will detect a stronger magnetic field, and the north pole of the needle will deflect away from its original direction by a larger angle.

,

where denotes the permeability of free space.Therefore, the magnetic field's magnitude is directly proportional

to the current in the conductor.

ANSWER: It is larger.

Magnetic Force Vector Drawing

Description: Simple conceptual question about the magnetic force on charges moving through a magnetic field. (vector applet)

For each of the situations below, a charged particle enters a region of uniform magnetic field. Draw a vector to represent the direction of the magnetic force on the particle.

Part A

Hint A.1 The right-hand rule for magnetic force

A charged particle moving through a region of magnetic field experiences a magnetic force. This force is directed perpendicular to both the velocity vector and the magnetic field vector at the point of interaction. The requirement that the force be perpendicular to both of the other vectors specifies the direction of the force to within an algebraic sign. This algebraic sign is determined by the right-hand rule. To employ the right-hand rule:

1. Spread your right thumb and index finger apart by 90 degrees. 2. Orient your hand so that your index finger points in the direction of the velocity. 3. Curl your fingers toward the direction of the magnetic field.

If the charge is positive, your thumb is now pointing in the direction of the force as shown.



If the charge is negative, the force is in the direction opposite your thumb.

Hint A.2 Apply the right-hand rule


Your hand should be shaped as seen in the figure.

At this point, is your thumb pointing up or down?

The direction of the magnetic force on a positive charge is the direction in which your middle finger is pointing.

ANSWER: up

down



Draw the vector starting at the location of the charge. The location and orientation of the vector will be graded. The length of the vector will not be graded.

ANSWER:

View

Part B


Hint B.1 Apply the right-hand rule


At this point, is your thumb pointing up or down?


ANSWER: up

down

ANSWER:

View

Part C

Hint C.1 Apply the right-hand rule


At this point, what direction is your thumb pointing?





ANSWER: up and to the right

up and to the left

down and to the right

down and to the left

ANSWER:

View

Magnetic Force on Charged Particles Conceptual Question

Description: Simple conceptual questions about the direction of the magnetic force vector acting on various charges moving through a magnetic field.

For each of the situations below, a charged particle enters a region of uniform magnetic field. Determine the direction of the force on each charge due to the magnetic field.

Part A

Determine the direction of the force on the charge due to the magnetic field.

Hint A.1 Determining the direction of a magnetic force



A charged particle moving through a region of magnetic field experiences a magnetic force, unless the velocity and magnetic field are parallel. If the velocity is parallel to the magnetic field, then the force is zero. Otherwise, the direction of the force can be found by using the right-hand rule.

To employ the right hand rule:

1. Open your hand so that it is completely flat, and point the fingers of your right hand in the direction of the velocity vector.

2. Rotate your wrist until you can bend your fingers to point in the direction of the magnetic field. 3. The direction of your outstretched thumb is the direction of the magnetic force on a positive charge.

ANSWER:

points into the page.

points out of the page.

points neither into nor out of the page and .

.

Part B

Determine the direction of the force on the charge due to the magnetic field.

Hint B.1 Determining the direction of a magnetic force

A charged particle moving through a region of magnetic field experiences a magnetic force, unless the velocity and magnetic field are parallel. If the velocity is parallel to the magnetic field, then the force is zero. Otherwise, the direction of the force can be found by using the right-hand rule.

To employ the right hand rule:

1. Open your hand so that it is completely flat, and point the fingers of your right hand in the direction of the velocity vector.

2. Rotate your wrist until you can bend your fingers to point in the direction of the magnetic field. 3. The direction of your outstretched thumb is the direction of the magnetic force on a positive charge.

ANSWER:






.

Part C

Determine the direction of the force on the charge due to the magnetic field. Note that the charge is negative.

Hint C.1 Effect of a magnetic field on a negative charge

You can use the right-hand rule to determine the direction of the force exerted on a positive charge. Once you find the direction of the force that would be exerted on a positive charge, the force on a negative charge will point in the opposite direction.

ANSWER:




.

Charged Particles Moving in a Magnetic Field Ranking Task

Description: Short conceptual problem to derive charge and relative speed of particles from their trajectories in a uniform magnetic field. (ranking task)

Five equal-mass particles (A–E) enter a region of uniform magnetic field directed into the page. They follow the trajectories illustrated in the figure.



Part A

Which particle (if any) is neutral?

Hint A.1 Neutral particles

Since the magnitude of the magnetic force acting on a particle is given by

,

a neutral particle (with ) will not experience a magnetic force.

ANSWER: particle A

particle B

particle C

particle D

particle E

none

Part B

Which particle (if any) is negatively charged?

Hint B.1 Find the direction of the magnetic force

The direction of the magnetic force is determined by the right-hand rule. With the given directions for velocity and magnetic field, what is the direction of the magnetic force on a positively charged particle?

ANSWER: left

right

ANSWER: particle A

particle B

particle C

particle D

particle E

none

Part C

Rank the particles on the basis of their speed.

Hint C.1 Determining velocity based on particle trajectories

A charged particle moving in a uniform magnetic field follows a circular trajectory. By Newton's second law, the magnetic force acting on the particle must be equal to the product of its mass and acceleration:

.



Rank from largest to smallest. To rank items as equivalent, overlap them.

In our scenario, the velocity and field vectors are perpendicular, so . Also, since the particle

moves along a circular path, the acceleration must equal the expression for centripetal acceleration:

.

This can be solved for velocity to yield

.

Thus, the speed of a particle can be determined by measuring the radius of its circular path in a known magnetic field, assuming that you also know the charge and mass of the particle.

ANSWER:

View

Part D

Rank the particles A, B, C, and E on the basis of their speed.


ANSWER:

View

Part E

Now assume that particles A, B, C, and E all have the same magnitude of electric charge. Rank the particles A, B, C, and E on the basis of their speed.

Hint E.1 Charged particle trajectories in magnetic fields

Particles A, B, C, and E are charged. A charged particle moving in a uniform magnetic field follows a circular




trajectory. The speed of the particle has two distinct effects on the radius of its circular path. First, the faster the particle moves, the larger the magnetic force acting on it, by

.

However, the faster it moves, the larger its centripetal acceleration, by

,

and therefore the larger the force needed to keep it in its circular path.

ANSWER:

View

The Cyclotron

Description: Several qualitative and quantitative questions: radius and frequency of revolution, energy of the particles. Focuses on classical model but last follow-up statement makes reference to relativistic model.

Learning Goal: To learn the basic physics and applications of cyclotrons.

Particle accelerators are used to create well-controlled beams of high-energy particles. Such beams have many uses, both in research and industry.

One common type of accelerator is the cyclotron, as shown in the figure. In a cyclotron, a magnetic field confines charged particles to circular paths while an oscillating electric field accelerates them. It is useful to understand the details of this process.

Consider a cyclotron in which a beam of particles of positive charge and mass is moving along a

circular path restricted by the magnetic field (which

is perpendicular to the velocity of the particles).



Part A

Before entering the cyclotron, the particles are accelerated by a potential difference . Find the speed with

which the particles enter the cyclotron.

Express your answer in terms of , , and .

ANSWER:

=

Part B

Find the radius of the circular path followed by the particles. The magnitude of the magnetic field is .

Express your answer in terms of , , , and .

Hint B.1 Find the force

Find the magnitude of the force acting on the particles.

Express your answer in terms of , , , and . You may or may not use all these variables.

ANSWER: =

Hint B.2 Find the acceleration

Find the magnitude of the acceleration of the particles.

Express your answer in terms of and the radius of revolution .

ANSWER:

=

Hint B.3 Solving for

Use Newton's second law to construct an equation to be solved for .

ANSWER: =

Part C

Find the period of revolution for the particles.

Hint C.1 Relationship between and




Note that the period does not depend on the particle's speed (nor, therefore, on its kinetic energy).

The period is the amount of time it takes a particle to make one complete orbit. Since the speed of the particle is constant, the period will be equal to the distance the particle travels in one orbit divided by the particle's speed:

.

ANSWER:

=

Part D

Find the angular frequency of the particles.


Hint D.1 Relationship between and

Recall that the frequency of revolution is equal to ; it represents the number of revolutions a particle

makes per second. The angular frequency is equal to ; it is the number of radians the particle traverses per

second.

ANSWER:

=

Part E

Your goal is to accelerate the particles to kinetic energy . What minimum radius of the cyclotron is required?


Hint E.1 Find in terms of

Find an expression for the speed of a particle in terms of its kinetic energy .

Express your answer in terms of and .

You can now substitute this value for into your expression for from Part B.

ANSWER:

=

ANSWER:



=

Part F

If you can build a cyclotron with twice the radius, by what factor would the allowed maximum particle energy increase? Assume that the magnetic field remains the same.

Hint F.1 Find in terms of

Using your result from Part E, solve for in terms of .


In other words, the kinetic energy is proportional to .

ANSWER:

=

ANSWER: 2

4

8

16

Part G

If you can build a cyclotron with twice the radius and with the magnetic field twice as strong, by what factor would the allowed maximum particle energy increase?

Hint G.1 Find in terms of and

Using your result from Part E, solve for in terms of and .


It's now clear that the kinetic energy is proportional to .

ANSWER:

=

ANSWER: 2



Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 5 points.

One limitation of the cyclotron has to do with the failure of the laws of classical mechanics to accurately predict the behavior of high-energy particles. When their speeds become comparable to the speed of light ,

the angular frequency is no longer what you determined in Part D. Using special relativity, one can show that the angular frequency is actually given by the formula

.

As you can see, the frequency drops as the energy (and speed) increases; the particles' motion falls out of phase with the pulsating voltage, restricting the cyclotron's ability to accelerate the particles further

4

8

16



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