[ Problem View ]
Force on Moving Charges in a Magnetic Field
Description: Student goes through right-hand rule questions and then looks at force on a
charge moving at particular velocity through uniform 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
direction of
. Your thumb will then be pointing in the
.
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., - ). (Recall that is written x_unit.)
ANSWER:
Direction of
= z_unit
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
= -y_unit
Part C
Now consider the example of a positive charge moving in the xy plane with velocity
(i.e., 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 .
ANSWER:
Direction of
= - cos(theta)*y_unit + sin(theta)*x_unit
Part D
First find the magnitude of the force
field are perpendicular.
in the case that the velocity and the magnetic
Express your answer in terms of , ,
statement.
, and other quantities given in the problem
ANSWER:
q*v*B
= abs(q)*v*B
Part E
Now consider the example of a positive charge moving in the -z direction with the local
magnetic field in the +z direction. Find
on the particle.
Express your answer in terms of , ,
statement.
ANSWER:
, the magnitude of the magnetic force acting
, 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 charge is moving in the yz plane at an angle with
the z axis as shown
(with the
magnetic field still in the +z direction). Find the magnetic force
on the charge.
Part F.1
Part not displayed
Part F.2
Part not displayed
Express the magnetic force in terms of given variables like , ,
ANSWER:
, , and unit vectors.
= q*v*B*sin(theta)*x_unit
[ Print ]
[ Problem View ]
Electromagnetic Velocity Filter
Description: Find the velocity of a charged particle that is undeflected in crossed electric
and magnetic fields. Look at relation between mass, charge, and acceleration as charged
particle traverses the fields.
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
.
ANSWER:
Part B
Now find the magnitude of the magnetic field that will cause the charge to travel in a
straight line under the combined action of electric and magnetic fields.
Part B.1
Part not displayed
Part B.2
Part not displayed
Express the magnetic field
that will just balance the applied electric field in terms
of some or all of the variables , , and
ANSWER:
.
= E/v
Part C
It may seem strange that the selected velocity does not depend on either the mass or the
charge of the particle. (For example, would the velocity of a neutral particle be selected
by passage through this device?) The explanation of this is that the mass and the charge
control the resolution of the device--particles with the wrong velocity will be
accelerated away from the straight line and will not pass through the exit slit. If the
acceleration depends strongly on the velocity, then particles with just slightly wrong
velocities will feel a substantial transverse acceleration and will not exit the selector.
Because the acceleration depends on the mass and charge, these influence the sharpness
(resolution) of the transmitted particles.
Assume that you want a velocity selector that will allow particles of velocity to pass
straight through without deflection while also providing the best possible velocity
resolution. You set the electric and magnetic fields to select the velocity . To obtain
the best possible velocity resolution (the narrowest distribution of velocities of the
transmitted particles) you would want to use particles with __________.
Hint C.1 Use Newton's law
If the velocity is "wrong" the forces won't balance and the resulting transverse force
will cause a transverse acceleration. Use
to determine how this acceleration
will depend on and . You want particles with the incorrect velocity to have the
maximum possible deviation in the y direction so that they will not go through a slit
placed at the right end. This means that the acceleration should be maximum.
Assume that the selector is short enough so that particles that move away from the axis
do not have time to come back to it.
ANSWER:
both and
large
large and
small
small and
large
both and
small
You want particles with the incorrect velocity to have the maximum possible deviation
in the y direction so that they will not go through a slit placed at the right end. The
deviation will be maximum when the acceleration is maximum. The acceleration is
directly proportional to and inversely proportional to :
.
So for maximum deviation, should be large and
small.
[ Print ]
Rail Gun
=(V*B*L)/(R*m*g)
[ Problem View ]
Charge Moving in a Cyclotron Orbit
Description: General problem which goes through charged particle motion perpendicular
to magnetic field; reviews cyclotron frequency derivation. Goes through kinematics and
frequency invariance.
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
.
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 direction (and the magnetic field is
always in the direction ). If is positive, what is the direction of the force on the
particle due to the magnetic field?
Hint A.1 Finding direction using the cross product
Hint not displayed
Answer in terms of unit vectors , , and or linear combinations of them.
ANSWER:
Direction of force = -y_unit
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;
then along; etc.
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
Part D
The particle moves in a plane perpendicular to the magnetic field direction. What is ,
the angular frequency of the circular motion?
Hint D.1 Broad approach to this problem
This is a circular dynamics problem. Set
to solve the problem. Note that
angular velocity and angular frequency are the same physical quantity.
Part D.2 Determine the magnetic force
Part not displayed
Part D.3
Part not displayed
Part D.4 Find the acceleration from kinematics
Part not displayed
Part D.5 Describe the motion of a particle in a magnetic field
Part not displayed
Express in terms of ,
ANSWER:
, and
.
= q*B_0/m
Part E
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. What can
you conclude from this?
Hint E.1 Relation between linear speed and angular frequency (angular
velocity)
The linear speed relates to the angular frequency as
ANSWER:
.
The frequency and the linear speed of the particle are invariant
with orbit size.
The linear speed (but not the frequency of the particle) is invariant
with orbit size.
The frequency (but not the linear speed) of the particle is invariant
with orbit size.
Both the frequency and the linear speed of the particle depend on
the orbit size.
[ Print ]
Mass Spectrometer
=sqrt(2*(q/m)*V)
=(R^2*B_0^2)/(2*V)
Torgue on a Current Loop
=I*a*b*B*cos(theta)
=m*B*sin(phi)
=I*B*pi*r^2*sin(phi)