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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 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
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 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?
Direction of
- ). (Recall that
is written x _ u n i t . )
=
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?
Direction of
=
Part C
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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 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,
, and .
Direction of
,
=
Part D
First find the magnitude of the force on a positive charge in the case that the velocity
the magnetic field (of magnitude ) are perpendicular.
statement.
(of magnitude ) and
, and other quantities given in the problem
=
Part E
Now 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.
statement.
, and other quantities given in the problem
= 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 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.
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Part F.1 Direction of force
Part not displayed
Part F.2 Relevant component of velocity
Part not displayed
Express the magnetic force in terms of given variables like , ,
, , and unit vectors.
=
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 A
Consider 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 direction
Hint not displayed
Hint A.2 Right-hand rule
Hint not displayed
Choose the direction of
.
Part B
Part not displayed
Part C
Part 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).
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Part A
Find , the electric field inside the cube.
Hint A.1 Net force on charges in a conductor
Electrons 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 magnitude
What is
cube?
, the magnitude of the force due to the magnetic field exerted on an electron with charge
Express your answer in terms of some or all of the variables ,
, and
inside the
.
=
Part A.3 Find the magnetic force direction
Use the vector equation
field. Remember to use
and the right-hand rule to determine the direction of the force from the magnetic
for the electron charge.
Answer in terms of the unit vectors
, , and .
Part A.4 Determine the force due to the electric field
What 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
=
Express the electric field in terms of
and .
,
, and unit vectors ( , , and/or ) .
=
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 B
If one replaces the conducting cube with one that has positive charge carriers, in what direction does the induced
electric field point?
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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 A
Find , the coefficient of static friction between the rod and the rails.
Hint A.1 How to approach this problem
Hint not displayed
Part A.2 Force due to the magnetic field
Part not displayed
Part A.3 Frictional force
Part not displayed
Express the coefficient of static friction in terms of variables given in the introduction.
=
Charge moving in Cyclotron Orbits
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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 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 A
At a given moment the particle is moving in the +x direction (and the magnetic field is always in the +z
direction). 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 B
Part not displayed
Part C
Part not displayed
Part D
Part not displayed
Motion of Electrons in a Magnetic Field
An electron of mass
and charge
is moving through a uniform magnetic field
origin, it has velocity
, where
perpendicular to the x axis at a distance
in vacuum. At the
and
. A screen is mounted
from the origin.
Throughout, you can assume that the effect of gravity is negligible.
Part A
First, suppose
. Find the y coordinate of the point at which the electron strikes the screen.
Hint A.1 Forces acting on electron
Hint not displayed
Hint A.2 Two-dimensional kinematics
Hint not displayed
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Part B
Now 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?
Part C
Part not displayed
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 does the ion exit the acceleration region?
Hint A.1 Suggested general method
Perhaps the easiest method to use for solving this problem is conservation of energy.
Part A.2 Initial energy
Find the initial total mechanical energy
Express
in terms of
, ,
(which includes electric potential energy) of the particle.
, and any constants.
=
Part A.3 Final energy
Find , 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.
=
Find the speed in terms of
, ,
, and any constants.
=
Part B
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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 frequency
Find 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
What is the relationship between the cyclotron frequency
definition of angular speed.)
.
=
Part B.2 Relationship of
Express
, , and
in terms of
and the kinematic variables
and ? (This is the
and .
=
Hint B.3 Putting it all together
Eliminate from the two equations from the previous hints, and then eliminate
result should give
in terms of the experimentally measured parameters.
Express
in terms of
,
,
using the result of Part A. The
, 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.
Torques on Magnetic Dipoles in Uniform Field
Torque on a Current Loop in a Magnetic Field
Learning 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 A
A 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 .
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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 problem
First find the forces (direction and magnitude) on each segment of the loop. Then find the torque due to each force
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 field
The 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.
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 field
Find , the magnitude of the force on each of the vertical wires (those with length ).
Hint A.3.a Relevant equation
Hint not displayed
Express
in terms of , , , and
. Note that not all of these may appear in the answer.
=
Part A.4 Torque on a loop
Find , 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 torque
Hint not displayed
Express
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in terms of
, , and .
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=
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.
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.
=
Part B
Give 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 moment
Hint not displayed
, magnetic field
, and .
=
The more general vector form of this expression is
.
Part C
A 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.
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Hint C.1 Formula for the area of a circle
Hint 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
].
=
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 A
Compute the field current .
Hint A.1 Voltages in a parallel circuit
Recall 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 resistor
Recall that the current passing through an element with resistance
is given by
,
where
is the voltage across the resistor.
= 0.489
Part B
Compute the rotor current .
Hint B.1 Currents in a parallel circuit
Recall 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.
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= 3.59
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Part C
Compute the emf .
Hint C.1 How to approach the problem
Think 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
, and we can find by using the current calculated in Part B.
Hint C.2 Relation between voltage and current
Hint not displayed
= 93.5
Part D
Compute the rate of development of thermal energy in the field windings.
Part D.1 Equation for thermal energy
Part not displayed
= 56.3
Part E
Compute the rate
of development of thermal energy in the rotor.
Part E.1 Equation for thermal energy
What is the power dissipated via heat through a resistor of resistance
that has a current of passing through it?
= 77.4
Part F
Compute the power input to the motor
.
Part F.1 Equation for total power input
What is the power through an electric circuit of voltage
and current ?
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= 469
Part G
Compute the efficiency of the motor.
Hint G.1 Definition of efficiency
The 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).
0.628
Changing Energy of a Magnetic Coil
A coil with magnetic moment 1.49
magnetic field of magnitude 0.810 .
is oriented initially with its magnetic moment antiparallel to a uniform
Part A
What 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 dipole
Hint not displayed
Hint A.2 Potential energy difference
Hint not displayed
Hint A.3 Definition of antiparallel
Hint not displayed
Summary
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-2.41
7 of 10 problems complete (67.63% avg. score)
33.82 of 35 points
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