Force on Moving Charges in a Magnetic Field

MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assig...

1 of 13 17/4/07 16:06

[ Assignment View ]

Eðlisfræði 2, vor 2007

27. Magnetic Field and Magnetic Forces

Assignment is due at 2:00am on Wednesday, February 28, 2007

Credit for problems submitted late will decrease to 0% after the deadline has passed.The wrong answer penalty is 2% per part. Multiple choice questions are penalized as described in the online help.The unopened hint bonus is 2% per part.You are allowed 4 attempts per answer.

Forces on moving charges and currents in Magnetic Field

Force on Moving Charges in a Magnetic FieldLearning 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 point your forefinger of your right hand in the direction of , and point your middle finger in the direction of . Your thumb will then be pointing in the direction of .

Part AConsider 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., - ). (Recall that is written x_uni t . )

ANSWER: Direction of =

Part BNow 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.


Part C

[


2 of 13 17/4/07 16:06

Now consider the example of a positive charge moving in the xy plane with velocity (i.e., with magnitude at angle 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 productThe direction can be found by any of the usual means of finding the cross product:

Use the determinant expression for the cross product. (See your math or physics text.)1.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.

2.

Express the direction of the force in terms of , as a linear combination of unit vectors, , , and .


Part DFirst find the magnitude of the force on a positive charge in the case that the velocity (of magnitude ) and the magnetic field (of magnitude ) are perpendicular.

Express your answer in terms of , , , and other quantities given in the problem statement.

ANSWER: =

Part ENow consider the example of a positive charge moving in the -z direction with speed with the local magnetic field of magnitude in the +z direction. Find , the magnitude of the magnetic force acting on the particle.

Express your answer in terms of , , , and other quantities given in the problem statement.

ANSWER: = 0

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 FNow 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.


3 of 13 17/4/07 16:06

Part F.1 Direction of forcePart not displayed

Part F.2 Relevant component of velocityPart not displayed

Express the magnetic force in terms of given variables like , , , , and unit vectors.

ANSWER: =

Electromagnetic Velocity Filter

When a particle with charge moves across a magnetic field of magnitude , it experiences a force to the side. If the proper electric field is simultaneously applied, the electric force on the charge will be in such a direction as to cancel the magnetic force with the result that the particle will travel in a straight line. The balancing condition provides a relationship involving the velocity of the particle. In this problem you will figure out how to arrange the fields to create this balance and then determine this relationship.

Part AConsider the arrangement of ion source and electric field plates shown in the figure. The ion source sends particles with velocity along the positive x axis. They encounter electric field plates spaced a distance apart that generate a uniform electric field of magnitude in the +y direction. To cancel the resulting electric force with a magnetic force, a magnetic field (not shown) must be added in which direction? Using the right-hand rule, you can see that the positive z axis is directed out of the screen.

Hint A.1 Method for determining directionHint not displayed

Hint A.2 Right-hand ruleHint not displayed

Choose the direction of .

ANSWER: Answer not displayed

Part BPart not displayed

Part CPart not displayed

A Conductor Moving in a Magnetic Field

A metal cube with sides of length is moving at velocity across a uniform magnetic field . The cube is oriented so that four of its edges are parallel to its direction of motion (i.e., the normal vector of two faces are parallel to the direction of motion).


4 of 13 17/4/07 16:06

Part AFind , the electric field inside the cube.

Hint A.1 Net force on charges in a conductorElectrons in a conductor are more or less free to move within the conductor. As a result electrons in a conductor placed in an electric field (with no magnetic field present) will move until the electic field that they generate inside the conductor cancels the applied field (i.e., until the net internal electric field equals zero). In general, the charges in a conductor move until the net force on them is zero. This happens almost instantaneously in a good conductor.

In this problem, there is no external electric field but there is a force on the electrons due to the applied magnetic field. The electrons will thus move in such a way as to create an electric field that will cancel the magnetic force on them (leaving a net force of zero on the electrons).

Part A.2 Find the magnetic force magnitudeWhat is , the magnitude of the force due to the magnetic field exerted on an electron with charge inside the cube?

Express your answer in terms of some or all of the variables , , and .

ANSWER: =

Part A.3 Find the magnetic force directionUse the vector equation and the right-hand rule to determine the direction of the force from the magnetic field. Remember to use for the electron charge.

Answer in terms of the unit vectors , , and .

ANSWER:

Part A.4 Determine the force due to the electric fieldWhat is , the force on an electron inside the cube with charge due to an induced electric field ?

Express your answer in terms of one or both of the variables and .

ANSWER: =

Express the electric field in terms of , , and unit vectors ( , , and/or ) .

ANSWER: =

Now, instead of electrons, suppose that the free charges have positive charge . Examples include "holes" in semiconductors and positive ions in liquids, each of which act as "conductors" for their free charges.

Part BIf one replaces the conducting cube with one that has positive charge carriers, in what direction does the induced electric field point?


5 of 13 17/4/07 16:06

ANSWER:

The direction of the electric field stays the same regardless of the sign of the charges that are free to move in the conductor.

Mathematically, you can see that this must be true since the expression you derived for the electric field is independent of .

Physically, this is because the force due to the magnetic field changes sign as well and causes positive charges to move in the direction (as opposed to pushing negative charges in the direction). Therefore the result is always the same: positive charges on the side and negative charges on the side. Because the electric field goes from positive to negative charges will always point in the direction (given the original directions of and ).

Rail Gun

A Rail Gun uses electromagnetic forces to accelerate a projectile to very high velocities. The basic mechanism of acceleration is relatively simple and can be illustrated in the following example. A metal rod of mass and electrical resistance rests on parallel horizontal rails (that have negligible electric resistance), which are a distance apart. The rails are also connected to a voltage source , so a current loop is formed.

The rod begins to move if the externally applied vertical magnetic field in which the rod is located reaches the value . Assume that the rod has a slightly flattened bottom so that it slides instead of rolling. Use for the magnitude of the acceleration due to gravity.

Part AFind , the coefficient of static friction between the rod and the rails.

Hint A.1 How to approach this problemHint not displayed

Part A.2 Force due to the magnetic fieldPart not displayed

Part A.3 Frictional forcePart not displayed

Express the coefficient of static friction in terms of variables given in the introduction.

ANSWER: =

Charge moving in Cyclotron Orbits


6 of 13 17/4/07 16:06

Charge Moving in a Cyclotron OrbitLearning 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 moves in a region of space where there is a uniform magnetic field (i.e., a magnetic field of magnitude in the +z direction). In this problem, neglect any forces on the particle other than the magnetic force.

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

Hint A.1 The right-hand rule for magnetic force Hint not displayed


Part BPart not displayed


Part DPart not displayed

Motion of Electrons in a Magnetic Field

An electron of mass and charge is moving through a uniform magnetic field in vacuum. At the origin, it has velocity , where and . A screen is mounted perpendicular to the x axis at a distance from the origin.

Throughout, you can assume that the effect of gravity is negligible.

Part AFirst, suppose . Find the y coordinate of the point at which the electron strikes the screen.

Hint A.1 Forces acting on electronHint not displayed

Hint A.2 Two-dimensional kinematicsHint not displayed


7 of 13 17/4/07 16:06

Part BNow suppose , and another electron is projected in the same manner. Which of the following is the most accurate qualitative description of the electron's motion once it enters the region of nonzero magnetic field?



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 AWith what speed does the ion exit the acceleration region?

Hint A.1 Suggested general methodPerhaps the easiest method to use for solving this problem is conservation of energy.

Part A.2 Initial energyFind the initial total mechanical energy (which includes electric potential energy) of the particle.

Express in terms of , , , and any constants.

ANSWER: =

Part A.3 Final energyFind , the total mechanical energy of the ion as it exits the region with the electric field (i.e., when it reaches the region of zero electric potential and enters the magnet).Express in terms of , , and any needed constants.

ANSWER: =

Find the speed in terms of , , , and any constants.

ANSWER: =

Part B


8 of 13 17/4/07 16:06

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 ?

Part B.1 Cyclotron frequencyFind the cyclotron frequency , which is the (angular) frequency of the orbital motion of the ion in the magnetic field. There is a skill builder problem on this if you need help.

Hint B.1.a General method to find Hint not displayed

Express the cyclotron frequency in terms of , , and .

ANSWER: =

Part B.2 Relationship of and

What is the relationship between the cyclotron frequency and the kinematic variables and ? (This is the definition of angular speed.)

Express in terms of and .

ANSWER: =

Hint B.3 Putting it all togetherEliminate from the two equations from the previous hints, and then eliminate using the result of Part A. The result should give in terms of the experimentally measured parameters.

Express in terms of , , , and any necessary constants.

ANSWER: =

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.

Torques on Magnetic Dipoles in Uniform Field

Torque on a Current Loop in a Magnetic FieldLearning Goal: To understand the origin of the torque on a current loop due to the magnetic forces on the current-carrying wires and to introduce the magnetic moment of the loop .

This problem will show you how to calculate the torgue on a magnetic dipole in a uniform magnetic field. We start with a rectangular current loop, the shape of which allows us to calculate the Lorentz forces explicitly. Then we generalize our result. Even if you already know the general formula to solve this problem, you might find it instructive to discover where it comes from.

Part AA current flows in a plane rectangular current loop with height and horizontal sides . The loop is placed into a uniform magnetic field in such a way that the sides of length are perpendicular to , and there is an angle between the sides of length and .


9 of 13 17/4/07 16:06

Calculate , the magnitude of the torque about the vertical axis of the current loop due to the interaction of the current through the loop with the magnetic field.

Hint A.1 How to approach the problemFirst find the forces (direction and magnitude) on each segment of the loop. Then find the torque due to each force and add them up.

Part A.2 Forces on the parts of the loop that have length As current is moving through the loop, forces act on its different parts. These result from Lorentz forces on the charges that move through the wire. What is the direction of the forces on the two pieces of wire of length ?

Hint A.2.a Force on a straight current-carrying wire in a magnetic fieldThe force on a straight wire of length carrying current in a magnetic field is given by .

Select the correct diagram from the four options. The forces are symbolized by the red arrows.

ANSWER: A B C D

The forces on these parts of the loop don't cause a motion of the center of mass, since they are equal and opposite, but they do produce a net torque for general , since their lines of action do not pass through the center of mass of the loop.

Part A.3 Force on a current-carrying wire in a magnetic fieldFind , the magnitude of the force on each of the vertical wires (those with length ).

Hint A.3.a Relevant equationHint not displayed

Express in terms of , , , and . Note that not all of these may appear in the answer.

ANSWER: =

Part A.4 Torque on a loopFind , the component of the torque around the vertical axis of the loop (i.e., out of the plane of the top-view figure).

Hint A.4.a Definition of torqueHint not displayed

Express in terms of , , and .


10 of 13 17/4/07 16:06

ANSWER: =

Part A.5 Forces on the parts of the loop that have length

As current is moving through the loop, forces act on its different parts. These result from Lorentz forces on the charges that move through the wire. What can you say about the forces on the two pieces of wire of length ?

Select the most accurate qualitative description from the following list.

ANSWER: The forces point away from the center of the loop and cancel each other.The forces act as a torque about an axis through the midpoints of the two two pieces of length

b and make the loop turn in the direction of positive .The forces point toward the center of the loop and cancel each other.Both forces point upward perpendicular to and .

Since the forces on these two parts of the loop cancel out, and their line of action goes through the center of mass of the loop, we need not consider them in the calculation of the torque on the loop. The key is that the forces are in line, so their torque contributions about any point cancel.

Express the magnitude of the torque in terms of the given variables. You will need a trigonomeric function [e.g., or ]. Use for the magnitude of the magnetic field.

ANSWER: =

Part BGive a more general expression for the magnitude of the torque . Rewrite the answer found in Part A in terms of the magnitude of the magnetic dipole moment of the current loop . Define the angle between the vector perpendicular to the plane of the coil and the magnetic field to be , noting that this angle is the complement of angle in Part A.

Hint B.1 Definition of the magnetic dipole momentHint not displayed

Give your answer in terms of the magnetic moment , magnetic field , and .

ANSWER: =

The more general vector form of this expression is

.

Part CA current flows around a plane circular loop of radius , giving the loop a magnetic dipole moment of magnitude

. The loop is placed in a uniform magnetic field , with an angle between the direction of the field lines and the magnetic dipole moment as shown in the figure. Find an expression for the magnitude of the torque on the current loop.


11 of 13 17/4/07 16:06

Hint C.1 Formula for the area of a circleHint not displayed

Express the torque explicitly in terms of , , , , and (where and are the magnitudes of the respective vector quantities). Do not use . You will need a trigonomeric function [e.g.. or ] .

ANSWER: =

A DC Motor

A shunt-wound DC motor with the field coils and rotor connected in parallel (see the figure) operates from a 115 DC power line. The resistance of the field windings, , is 235 . The resistance of the rotor, , is 6.00 . When the motor is running, the rotor develops an emf . The motor draws a current of 4.08 from the line. Friction losses amount to 41.0 .

Part ACompute the field current .

Hint A.1 Voltages in a parallel circuitRecall that for each element in a parallel circuit there is the same voltage difference between the individual elements. As a result, the same voltage difference of = 115 lies across both the field windings with resistance 235 and the rotor with resistance 6.00 .

Hint A.2 Current through a resistorRecall that the current passing through an element with resistance is given by

,

where is the voltage across the resistor.

Express your answer in amperes.

ANSWER: = 0.489

Part BCompute the rotor current .

Hint B.1 Currents in a parallel circuitRecall that for a circuit in parallel, there are multiple paths for the total current to take in getting from the positive electrode of the power supply to the negative electrode. As a result, different amounts of current pass through the elements in a parallel circuit according to the resistance and the voltage differences on each element. The current passing through each of the sections must add up to the total current being supplied by the power supply.

ANSWER: = 3.59


12 of 13 17/4/07 16:06

Part CCompute the emf .

Hint C.1 How to approach the problemThink of the back emf from the rotor as reducing the voltage difference from the power supply across the rotor. You can picture the system diagrammatically by putting the resistance part of the rotor and the back emf part of the rotor in series. This gives you the equation for total voltage across the rotor, where is the voltage from the power supply, is the voltage across the resistance element of the rotor, and is the back emf from the rotor. We already know that , and we can find by using the current calculated in Part B.

Hint C.2 Relation between voltage and currentHint not displayed

ANSWER: = 93.5

Part DCompute the rate of development of thermal energy in the field windings.

Part D.1 Equation for thermal energyPart not displayed

Express your answer in watts.

ANSWER: = 56.3

Part ECompute the rate of development of thermal energy in the rotor.

Part E.1 Equation for thermal energyWhat is the power dissipated via heat through a resistor of resistance that has a current of passing through it?

ANSWER:


ANSWER: = 77.4

Part FCompute the power input to the motor .

Part F.1 Equation for total power inputWhat is the power through an electric circuit of voltage and current ?

ANSWER:


13 of 13 17/4/07 16:06


ANSWER: = 469

Part GCompute the efficiency of the motor.

Hint G.1 Definition of efficiencyThe efficiency of an electrical motor (or any motor for that matter) is given by the ratio of the power coming out of the circuit, after all losses are taken into account, and the total power initially supplied to the circuit. This efficiency must be a value less than or equal to one, since the power coming out of the circuit must be less than or equal to the power going in (i.e., the circuit cannot generate its own energy, but can only consume it).

ANSWER: 0.628

Changing Energy of a Magnetic Coil

A coil with magnetic moment 1.49 is oriented initially with its magnetic moment antiparallel to a uniform magnetic field of magnitude 0.810 .

Part AWhat is the change in potential energy of the coil when it is rotated 180 degrees, so that its magnetic moment is parallel to the field?

Hint A.1 Potential energy for a magnetic dipoleHint not displayed

Hint A.2 Potential energy differenceHint not displayed

Hint A.3 Definition of antiparallelHint not displayed

Express your answer in joules.

ANSWER: -2.41

Summary 7 of 10 problems complete (67.63% avg. score)33.82 of 35 points

Documents

Force on Moving Charges in a Magnetic Field