Solutions] Mastering Physics HW20

HW20 Due: 11:59pm on Tuesday, October 20, 2009 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View]

Force on Moving Charges in a Magnetic Field Learning Goal: To understand the force on a charge moving in a magnetic field. Magnets exert forces on other magnets even though they are separated by some distance. Usually the force on a magnet (or piece of magnetized matter) is pictured as the interaction of that magnet with the magnetic field at its location (the field being generated by other magnets or currents). More fundamentally, the force arises from the interaction of individual moving charges within a magnet with the local magnetic field. This force is written , where is the force, is the individual charge (which can be negative),

is its velocity, and

is the local magnetic field.

This force is nonintuitive, as it involves the vector product (or cross product) of the vectors

and

. In the following questions

we assume that the coordinate system being used has the conventional arrangement of the axes, such that it satisfies where

, , and

,

are the unit vectors along the respective axes.

Let's go through the right-hand rule. Starting with the generic vector cross-product equation forefinger of your right hand in the direction of be pointing in the direction of

point your

, and point your middle finger in the direction of

. Your thumb will then

.

Part A Consider the specific example of a positive charge moving in the +x direction with the local magnetic field in the +y direction. In which direction is the magnetic force acting on the particle? Express your answer using unit vectors (e.g., ANSWER:

Direction of

=

- ). (Recall that

is written x_unit.)

Correct

Part B Now consider the example of a positive charge moving in the +x direction with the local magnetic field in the +z direction. In which direction is the magnetic force acting on the particle? Express your answer using unit vectors. ANSWER:

Direction of

= Correct

Part C Now consider the example of a positive charge magnitude

at angle

moving in the xy plane with velocity

(i.e., with

with respect to the x axis). If the local magnetic field is in the +z direction, what is the direction of the

magnetic force acting on the particle? Hint C.1

Finding the cross product

The direction can be found by any of the usual means of finding the cross product: 1. Use the determinant expression for the cross product. (See your math or physics text.) 2. Use the general definition , where any term with the three directions in the normal order of xyz or any cyclical permutation (e.g., yzx or zxy) has a positive sign, and terms with the other order (xzy, zyx, or yxz) have a negative sign. Express the direction of the force in terms of , as a linear combination of unit vectors, ANSWER:

Direction of

= Correct

, , and .

Part D First find the magnitude of the force magnetic field

(of magnitude

on a positive charge

Express your answer in terms of , , ANSWER:

in the case that the velocity

(of magnitude ) and the

) are perpendicular. , and other quantities given in the problem statement.

= Correct

Part E Now consider the example of a positive charge magnitude

in the +z direction. Find

Express your answer in terms of , , ANSWER:

=0

moving in the -z direction with speed

with the local magnetic field of

, the magnitude of the magnetic force acting on the particle. , and other quantities given in the problem statement.

Correct

There is no magnetic force on a charge moving parallel or antiparallel to the magnetic field. Equivalently, the magnetic force is proportional to the component of velocity perpendicular to the magnetic field. Part F Now consider the case in which the positive charge

is moving in the yz plane with a speed

at an angle

with the z axis as

shown (with the magnetic field still in the +z direction with magnitude ). Find the magnetic force on the charge.

Hint F.1

Direction of force

Find the direction of the force

. Note that it must be perpendicular to both

and

.

Express the direction of the force using unit vectors. ANSWER:

Hint F.2

Direction of

=

Correct

Relevant component of velocity

Of course, this problem can be solved by simply applying a rule for finding the vector product. However, another useful way to think about it is to realize that only the component of velocity perpendicular to the field generates any force. Find , the component of velocity perpendicular to

.

Express your answer in terms of , , and unit vectors. ANSWER:

= Answer not displayed

Express the magnetic force in terms of given variables like , , ANSWER:

, , and unit vectors.

= Correct

Charged Particles Moving in a 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 ANSWER:

) will not experience a magnetic force.

particle A particle B particle C particle D particle E none Correct

Part B Which particle (if any) is negatively charged? Hint B.1

Find the direction of the magnetic force Hint not displayed

ANSWER:

particle A particle B particle C particle D particle E none Correct

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

View Correct

Part D Rank the particles A, B, C, and E on the basis of their speed. Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER:

ANSWER:

View Correct

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 Hint not displayed

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

View Correct

For Part D of "Determining the Velocity of a Charged Particle", if you do it in the most straightforward manner, it is essential that you keep all numbers to full precision in your calculator until you plug into your final expression for the dot product of force and velocity. GBA

Determining the Velocity of a Charged Particle A particle with a charge of

5.50

is moving in a uniform magnetic field of 3.30×10 −7

particle is measured to be

7.60×10 −7

.

Part A Are there components of the velocity that cannot be determined by measuring the force? Hint A.1

Magnetic force on a moving charged particle

Recall the following formula: . If you know

, does

ANSWER:

yes no

uniquely define

?

Correct

Part B Calculate the x component of the velocity of the particle. Hint B.1

Relation between

and

Which component of the force depends on the x component of the velocity? ANSWER:

x Answer not displayed y

Express your answer in meters per second to three significant figures. ANSWER:

= -106 Correct

Part C Calculate the y component of the velocity of the particle. Hint C.1

Relation between

and Hint not displayed

1.30

) . The magnetic force on the

Express your answer in meters per second to three significant figures. ANSWER:

= -46.2 Correct

Part D Calculate the scalar product

. Work the problem out symbolically first, then plug in numbers after you've simplified the

symbolic expression. Hint D.1

Formula for dot product

The dot product of two vectors

and

is given by .

Express your answer in watts to three significant figures. ANSWER:

0 Correct

Part E What is the angle between Hint E.1

and

?

Another dot product formula

Recall that , where

is the angle between

and

.

Express your answer in degrees to three significant figures. ANSWER:

90 Correct

Notice that the dot product of the velocity and the force is zero. This will always be the case. Since be perpendicular to both

and

,

must

. This result is important because it implies that magnetic fields can only change the

direction of a charged particle's velocity, not its speed.

Charge Moving in a Cyclotron Orbit Learning Goal: To understand why charged particles move in circles perpendicular to a magnetic field and why the frequency is an invariant. A particle of charge

and mass

magnetic field of magnitude

moves in a region of space where there is a uniform magnetic field

(i.e., a

in the +z direction). In this

problem, neglect any forces on the particle other than the magnetic force.

Part A At a given moment the particle is moving in the +x direction (and the magnetic field is always in the +z direction). If positive, what is the direction of the force on the particle due to the magnetic field? Hint A.1

is

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. Bend your middle finger so that it is perpendicular to your thumb and index finger. 3. Orient your hand so that your thumb points in the direction of the velocity and your index finger in the direction of the magnetic field. If the charge is positive, your middle finger is now pointing in the direction of the force as shown in the figure.

If the charge is negative, the force is in the direction opposite your middle finger. ANSWER:

x direction x direction y direction y direction z direction z direction Correct

Part B This force will cause the path of the particle to curve. Therefore, at a later time, the direction of the force will ____________. ANSWER:

have a component along the direction of motion remain perpendicular to the direction of motion have a component against the direction of motion first have a component along the direction of motion; then against it; then along it; etc. Correct

Part C The fact that the magnetic field generates a force perpendicular to the instantaneous velocity of the particle has implications for the work that the field does on the particle. As a consequence, if only the magnetic field acts on the particle, its kinetic energy will ____________. ANSWER:

increase over time decrease over time remain constant oscillate Correct

Part D The particle moves in a plane perpendicular to the magnetic field direction as shown in the figure. What is frequency of the circular motion?

Hint D.1

How to approach the problem Hint not displayed

Hint D.2

Determine the magnetic force Hint not displayed

Hint D.3

Determine the acceleration of the particle Hint not displayed

Hint D.4

Express the angular speed in terms of the linear speed Hint not displayed

, the angular

Express

in terms of ,

, and

.

ANSWER: = Correct Note that this result for the frequency does not depend on the radius

of the circle. Although it appeared in the

equations of force and motion, it canceled out. This implies that the frequency (but not the linear speed) of the particle is invariant with orbit size. The first particle accelerator built, the cyclotron, was based on the fact that the frequency of a charged particle orbiting in a uniform field is independent of the radius. In the cyclotron, radio frequency voltage is applied across a gap between the two sides of the conducting vacuum chamber in which the protons circulate owing to an external magnetic field. Particles in phase with this voltage are accelerated each time they cross the gap (because the field reverses while they make half a circle) and reach energies of millions of electron volts after several thousand round trips.

Mass Spectrometer J. J. Thomson is best known for his discoveries about the nature of cathode rays. Another important contribution of his was the invention, together with one of his students, of the mass spectrometer. The ratio of mass to (positive) charge of an ion may be accurately determined in a mass spectrometer. In essence, the spectrometer consists of two regions: one that accelerates the ion through a potential and a second that measures its radius of curvature in a perpendicular magnetic field. The ion begins at potential

and is accelerated toward zero

potential. When the particle exits the region with the electric field it will have obtained a speed .

Part A With what speed Hint A.1

does the ion exit the acceleration region?

Suggested general method Hint not displayed

Hint A.2

Initial energy Hint not displayed

Hint A.3

Final energy Hint not displayed

Find the speed in terms of

, ,

, and any constants.

ANSWER: = Correct

Part B After being accelerated, the particle enters a uniform magnetic field of strength

and travels in a circle of radius

(determined by observing where it hits on a screen--as shown in the figure). The results of this experiment allow one to find in terms of the experimentally measured quantities such as the particle radius, the magnetic field, and the applied voltage. What is Hint B.1

? Cyclotron frequency Hint not displayed

Hint B.2

Relationship of

and Hint not displayed

Hint B.3

Putting it all together Hint not displayed

Express

in terms of

,

,

ANSWER: = Correct

, and any necessary constants.

By sending atoms of various elements through a mass spectrometer, Thomson's student, Francis Aston, discovered that some elements actually contained atoms with several different masses. Atoms of the same element with different masses can only be explained by the existence of a third subatomic particle in addition to protons and electrons: the neutron.

Score Summary: Your score on this assignment is 98.9%. You received 39.54 out of a possible total of 40 points.