VI-1
Unit VI – MAGNETIC FORCES & FIELDS
References:
PHYSICS FOR SCIENTISTS AND ENGINEERS, Serway & Beichner, 5th ed., Ch. 29
FUNDAMENTALS OF PHYSICS, Halliday, Resnick, & Walker, 6th ed., Ch. 29
Unit Objectives
When you have completed Unit VI, you should be able to:
1. State the conditions necessary for a particle to experience a magnetic force when it is in a
magnetic field.
2. Describe qualitatively the motion of a charged particle moving through a magnetic field
that is constant, is changing with time, or is changing with position.
3. Given any three of the following quantities: a particle's charge, its velocity, the magnetic
force it experiences, or the magnetic field through which it is moving, determine the
magnitude and sign or direction of the fourth quantity.
4. State the conditions necessary for a charged particle to move with uniform circular motion
in the presence of a magnetic field and, beginning with Newton's second law derive the
expression for the radius of its circular path.
5. Describe qualitatively and quantitatively the motion of a charged particle moving in a
region containing BOTH electric and magnetic fields.
6. Calculate the magnitude and direction of the force on a current-carrying wire in a uniform
magnetic field.
7. Calculate the magnitude and direction of the torque due to a uniform magnetic field, on a
rectangular loop of wire carrying a current.
VI-2
Unit VI – MAGNETIC FORCES & FIELDS
Before beginning the “real stuff” on magnetism let’s point out some similarities between past
diagrams and ones we will use in this unit and agree on some new notation for diagrams.
Figure VI –1 shows a diagram of the -field in the
region around a (+) and (–) point charge. The diagram
meant that the -field had a greater magnitude at point
A than at point B because the field lines were closer
together. It also told what the direction of the total field was at a given point, that is, the was tangent to
the field line at the point and in the direction of the
arrow on the field line as illustrated at points C and D.
Figure VI – 2 is the diagram for the magnetic field
in the region around a bar magnet. This diagram
has a meaning similar to the -field diagram
above only in this case we refer to the magnetic
field, . The -field has a greater magnitude at
point A than at point B because the field lines in
the diagram are closer together at point A than at
point B. The direction of the -field at a given
point is tangent to the field line at the point and in
the direction of the arrow on the field line as
illustrated by the vectors representing the
magnetic field at points C and D.
In dealing with magnetic fields forces, velocities, current elements, etc., we will usually be
considering sets of three vector quantities, two of which can be drawn in the plane of the page.
The other, however, is usually perpendicular to the page, so we need to agree on a notation
for vector quantities directed into-the page (N2pg) and vector quantities directed out of-the
page (OOpg). We will take a dot “ i” to mean that the vector quantity is toward the reader or
out of the page (like the “stabbing” end of an arrow coming toward you). An “X“ we will take
to mean the vector quantity is directed away from the reader or into the page (like the feathers
of an arrow moving away from you). For example, in Figure VI – 3(a)
and are vectors in
the plane of the page and in the case on the left
is a vector out of the page and on the right
is into the page. In Figure VI-3(b) the situation on the left shows the velocity and force on
a particle moving through a uniform -field directed into the page while the diagram on the
right shows the velocity and force on a particle moving through a -field that is directed out of
the page and whose magnitude is increasing as we move from left to right in the diagram as
shown by the dots being closer together.
VI-3
Unit VI – MAGNETIC FORCES & FIELDS
In previous sections of this course we defined different types of “fields” in terms of the force
experienced by some type of particle or test object. For example, the gravitational field is the
force per unit mass in some region of space, the electric field is the force per unit charge, etc.
In a similar way we will define the magnetic field in terms of the force on a moving charged
particle.
Imagine a charged particle being held at rest in a region of space devoid of a gravitational field.
If the particle is released and it moves off in some direction, we say “AHA!” there must be an
-field in this region and it must be directed in the direction of the acceleration of the charged
particle (assuming a + charge). Now suppose in the same region we release the particle from
rest and it just sits there. We take this as evidence of no -field in our region of space.
Suppose we now do a little work on our particle, get it moving and notice the particle is
somehow deflected from the direction it was initially moving. This being so, we conclude that
the particle experiences an acceleration and thus a force. We conclude then that our particle
must be moving through some new type of force field where the force only exists if our charged
particle is moving in the field. To properly define the field we must investigate the
characteristics of the force which we will call the magnetic force, FM , experienced by the
particle. We find the force has the following properties:
I. When a charged particle experiences a magnetic force, the direction of FM is always
perpendicular to the velocity of the particle:
FM ^ v
II. In a given region containing our field, there is one particular direction such that if is in
this direction or in exactly the opposite direction, FM will be zero. Since this direction is
unique let’s define the direction of our field in this direction. Thus, if
magnetic field then
represents our
III. No matter what the direction of v the magnitude of FM is directly
proportional to the component of v perpendicular to
the right,
is the component of v perpendicular to
between vectors v and
then
. In the sketch at
. If  is the angle
and thus
IV. FM is always at right angles to the plane containing both v and
, that is:
FM ^ v and FM ^ B
V. FM is directly proportional to the magnitude of the charge on the particle:
FM µ q
Reversing the sign of the charge reverses the direction of FM .
VI-4
Unit VI – MAGNETIC FORCES & FIELDS
The force law that incorporates I – V above is:
FM = qv ´ B
Note that since FM , q and v are experimentally measurable quantities, this force law defines
the magnitude and direction of . For example, suppose we determine, for a particle of known
q and v , the magnitude and direction of the maximum FM . From above,
Therefore, the magnitude of
is defined by this relation to be
, or if the angle () between v and
The cross product uniquely defines the direction of
is known:
.
(
)
. The choice of v ´ B rather than
(B ´ v ) was an arbitrary choice since both will give a F
M
perpendicular to
. It is an arbitrary
choice just like choosing the direction of E to be in the direction of the force on a (+) charge.
Thus the direction of is defined as in a direction such that for a known v and FM , the
(
)
product v ´ B gives a vector in the direction of FM . A little less jazzy definition is: “
is
defined as in the direction of the north seeking end of a compass needle.” No matter, the
result is the same. The v ´ B choice also has the advantage that it obeys our time honored
(
)
right-hand-rule (RHR) and if q is (–) we can employ a similar left-hand-rule (LHR) by pushing v
into with the left hand and your left thumb will point in the direction of the magnetic force,
FM , on the (–) charge.
Let’s show that the relation FM = qv ´ B correctly describes results I through V on page VI-3.
(
)
I & IV. Due to how the cross product is defined, a vector ( FM in this case) given by v ´ B is
perpendicular both v and
.
II. Since the magnitude of FM = qvBsin, where  is the angle between v and
and
, when v
are parallel ( = 0o) or anti-parallel ( = 180o), FM is zero.
III. Since the magnitude of FM = qvBsin, and the magnitude of the component of v
, then for a given q, v , and
perpendicular to is
proportional to v ^ .
the magnetic force is
V. Since the magnitude of FM = qvBsin, then FM is proportional to the charge q. If q is (+)
(
)
(
)
FM is in the direction of the v ´ B vector and if q is negative FM = - q v ´ B , i.e., in the
(
)
direction of – v ´ B .
VI-5
Unit VI – MAGNETIC FORCES & FIELDS
Units of magnetic field – the units of
are defined by the expression
.
In MKSA units:
As usual the
is given a special name: 1
is defined as 1 tesla = 1 T.
Another common magnetic field unit is the gauss (G) where 104 G = 1 T. The earth’s -field is
about 0.5 G = 5 x 10-5 T. For a small magnet: B ~ 100 G = 10-2 T. A big lab electromagnet
can produce a field of about 20kG = 2 T.
Now for a little practice using FM = qv ´ B .
For answers & solutions see “In-Text Fillin Answers” beginning on page VI-18
I. Given the sign of the charge q and two of the three vectors v ,
instant, state the direction of the third vector.
, or FM at a particular
II. Sketch in the resulting motion of the charged particle in the following situations.
(VI-5 #1)
(VI-5 #2)
VI-6
Unit VI – MAGNETIC FORCES & FIELDS
For review: The potential difference
between plates = V, the distance
between them = d.
i. Draw in the E -between the
plates.
ii. The force on q (in terms of q,
V, and d is given by:
i. Let q be (+)
ii. Let q be (–)
iii. Let q = 0
In i & ii what is the shape
of the paths? (Hint: FM is
always ^ v )
____________________
(VI-6 #1)
magnitude
direction
iii. What is the shape of the path of
q? ______________________
(VI-6 #2)
III. A uniform -field points horizontally northward and has a magnitude of 1.5 T. A 5 MeV
proton moves vertically downward through the -field. Find:
a. the magnitude and direction of FM acting on the proton
b. the ratio of FM on the proton to the weight of the proton
c. the resulting acceleration of the proton
d. the change in the kinetic energy of the proton after 1 second
Solution:
a. The magnitude of using FM = qv ´ B is given by
FM = ____________________ , where in this case  = _______.
(VI-6 #3)
(equation)
First we must find the proton’s speed, v. We are given that it is a 5 MeV proton. Since
an eV is a unit of ______________ , 5 MeV must be its __________________. (VI-6 #4)
Now,
thus,
(VI-6 #5)
VI-7
Unit VI – MAGNETIC FORCES & FIELDS
So,
(VI-7 #1)
Direction of FM : Since qp is (+), the _____________ hand rule gives
(
)
(right, left)
the direction of v ´ B and therefore, FM to be _____________ .
(N, S, E, W, in, out)
Thus,
FM = ___________________ , ______________
(magnitude)
(VI-7 #2)
(direction)
b.
and
(VI-7 #3)
This illustrates why we can ignore gravitational forces when working with magnetic
forces on charged particles: the gravitational forces are teensy weensy in comparison
to the magnetic forces.
c. If FM is the only significant force on the proton, then from Newton’s Second Law:
(VI-7 #4)
d. K = 0. If the net force is just FM and FM is always perpendicular to v (and
the
displacement) then the net work done is zero and thus K = 0. This illustrates that the
magnetic force never does any work on a charged particle moving in a -field.
IV. Suppose, in each of the drawings below, the -field is zero and at the instant the charged
particle is at the position shown is turned on and its magnitude (the direction is given)
increases linearly with time. If the particles have a v as indicated, sketch the paths of the
particles from t = 0 to t = later.
(VI-7 #5)
VI-8
Unit VI – MAGNETIC FORCES & FIELDS
Charged Particles In Uniform Circular Motion (UCM)
Recall that if a body is moving with UCM, the net force on it (AKA the centripetal force) is
1. Constant in magnitude.
2. Always directed perpendicular to the velocity of the body (thus, its speed is constant).
3. Always directed toward the center of the object’s circular path.
Notice then, that with a charged particle moving through a -field we have the possibility that
the particle will move with UCM provided certain conditions are met. Hints as to what these
conditions are were provided in the problems on pages VI-5 thru 7. In problem II (c) i and ii, for
example, the particles moved with UCM along the arc of a circle, whereas in problems II (a),
(b) and (c) iii and IV (a) and (b) the particles did not. Take another look at each situation and
compare the situation with the conditions for UCM listed above and see if you can discover the
conditions that must be met if the particle is to move with UCM. In problems II (a) and (c) iii
the particle does not move with UCM because in both cases FM = 0. In problem IV (a) and (b)
the particle does not move with UCM because in both cases FM violates condition #_____
above. (VI-8 #1)
What is the difference in the relationship between the directions of v and in II (b) where the
particle is not moving with UCM and in II (c) i and ii where the particles are moving with UCM?
(Hint: in II (b), what is the angle  between v and and in II (c) I and ii what is ?)
Conclusion, for a charged particle to move with UCM in a -field  must be _______. (VI-8 #2)
Summary: For a charged particle to move with UCM in a -field the following conditions must
be met:
1.
must be constant in magnitude and direction throughout the path of the particle.
2. v must be perpendicular to (i.e., ( = 90o)
In the situation at the right the conditions above are satisfied so
that particle is moving with UCM. For q to be moving clockwise
as shown, the sign of q must be _____ (VI-8 #3). Let’s derive the
expression for the radius of its path. The magnitude of the FM
acting on q is :
.
(VI-8 #4)
Applying Newton’s second law, the net force on q equals its
mass m, times its acceleration. In this situation the net force is the centripetal force which in
this case is FM . Thus, FM equals the mass of the particle times its centripetal acceleration.
Writing the centripetal acceleration in terms of the particle’s speed v, and the radius of its path
R:
(VI-8 #5)
VI-9
Unit VI – MAGNETIC FORCES & FIELDS
Setting the expression of the magnitude of FM (in terms of q, v, & B) equal to the mass of the
particle times is ac (see answer to VI-8 #5):
(VI-9 #1)
Solving for R we get:
R = _______________
(VI-9 #2)
Sample Exercise VI – 1: A bubble chamber photograph shows a proton moving in a circular
arc with a radius of 20 cm at right angles to a -field having a magnitude of 0.3 T. Find:
a. the magnitude of the momentum of the proton
b. its speed
c. its kinetic energy in MeV
Solutions:
a. Notice that in the expression above for the radius of the circular path the magnitude of
the momentum of the charged particle appears on the right-hand-side. Solving it for the
momentum and substituting for R, q, and B:
p = mv = __________ = (__________)(__________)(__________)
equation
substitution
= ______________
(VI-9 #3)
b. Solving (a) for v and grinding out the answer:
= (__________) / (__________) = _______________
(VI-9 #4)
c. Using the definition of kinetic energy,
K=
(____)(____) =
equation
(__________)(__________)2 = __________ J
substitution
= __________ MeV
At this point try problems 1 through 5 at the back of this unit.
(VI-9 #5)
VI-10
Unit VI – MAGNETIC FORCES & FIELDS
There are several simple but very practical devices that utilize electric and magnetic forces to
measure some of the physical properties of charged particles. Here we will describe two of
them: the velocity selector (or velocity filter) and the mass spectrometer. The Hall effect, which
is used in many -field measuring devices, and the cyclotron particle accelerator also use the
magnetic force in their operation. You should read about these in your text and/or
supplemental references.
The Velocity Selector
Suppose we have parallel plates, charged oppositely, separated
by a distance d, with the voltage between them being V as
shown. A particle having a charge q and velocity v enters the
region between the plates. Between the plates the E -field is
directed _________ (VI-10 #1) and is uniform (neglecting fringing
(up?, down?)
effects), therefore, if q is (+), it will experience an electric force given by
= ______________ , ____________
equation
Or since
(VI-10 #2)
direction
can be written in terms of V and d,
can also be written in terms of q, V and
d as:
= _____________, ____________
equation
(VI-10 #3)
direction
Also, from previous work we know that the shape of its path in the E -field will be
____________.
If q were (–), the magnitude of
would be the same but it would be directed
____________.
(VI-10 #4)
(VI-10 #5)
Let q be (–) and suppose we want it to pass between the plates undeflected. That is, we want
its v to be constant. For this to be so, the net force on q must be ________. (VI-10 #6) One way
to achieve this is to set up a -field between the plates, orienting it in such a direction that the
magnetic force, FM . will be in the opposite direction to . Then, if we make their magnitudes
equal, the net force on q will be zero and it will pass between the plates undeflected.
On the drawing at the right, indicate the directions of E and for
the situation where – q will travel with constant v . Now since FM
depends on v , then there must be a unique value of v (for a
given E and ) where the charge will travel in a straight line.
Let’s find it. The requirement is:
Writing FM in terms of q, v, and B and
in terms of q and E
_______________ = _______________
(VI-10 #7)
VI-11
Unit VI – MAGNETIC FORCES & FIELDS
Solving for v we get
(VI-11 #1)
Or putting E in terms of easily measurable quantities like V and d,
(VI-11 #2)
Notice then, if we shoot a whole mess of particles
having different charges and velocities into our crossed
E and fields, only particles having a velocity given
by the relation above will pass through the exit
aperture, all others will be deflected. Knowing E (or V
and d) and we can determine this v . Notice that all
particles no matter what their charge, mass or
whatever, will pass through undeflected provided they
have this particular velocity. This device sorts out or selects particles that have a particular
velocity and allows them to pass through so we call it a velocity selector.
At this point try problems 6 and 7 at the back of this unit.
The Mass Spectrometer
The mass spectrometer uses the magnetic force to determine the
relative masses of the isotopes of an element.
Consider a bunch of particles having different masses and charges but
all having the same velocity (we could use a v selector to achieve this)
moving through aperture A into a uniform -field as shown. The
-field first sorts the particles between (+) and (–) with the (+) charges
moving along paths _____ and _____ and the (–) charges along paths
_____ and _____. (VI-11 #3) Recall that the radius of the circular path of a
charged particle moving in a uniform -field is given by
R = mv / qB
If two of the particles have the same mass but different (+) charges, the one with the bigger (+)
charge would most likely follow path _____ and the one with the smaller (+) charge path
_____. (VI-11 #4)
If two of the particles have the same (+) charge but different masses, the more massive
particle would most likely follow path _____ and the less massive path _____. (VI-11 #5)
The mass spectrometer uses this concept that the radius of the path is proportional to the
mass of the particle to measure the masses of atomic particles.
VI-12
Unit VI – MAGNETIC FORCES & FIELDS
The diagram at the right illustrates the essential features
of a mass spectrometer. Singly charged ions of a
particular element are emitted from a source S, with
widely varying velocities into a velocity selector. Let the
potential difference across the plates of the velocity
selector be V and the separation of the plates be d.
The magnetic field between the plates is , and is
directed into the page as shown. As we found
previously, the velocity selector allows all ions to pass
through it undeflected if they have a speed given by
v = ____________ .
(VI-12 #1)
These ions pass through aperture A into another region where there is a uniform magnetic
field,
directed out of the page. Since
is perpendicular to the velocity of the ions, a
particular ion will move in a circular path and strike a strip of photographic film placed along the
inner wall of the spectrometer exposing the film a teensy bit. If all the ions entering through A
have the same mass, all of them will hit roughly at the same spot exposing the film a lot. The
distance between A and an exposed spot on the film gives he diameter, D, of the circular path
of those ions. Let’s derive the expression for the mass of an ion in terms of the known
quantities. Previously we found that the radius of the path of a particle of mass m, charge q,
and speed v traveling through a magnetic field that is perpendicular to the velocity was given
by
(VI-12 #2)
Writing R in terms of D, replacing B with
, and solving for m we get
(VI-12 #3)
But v can be written in terms of the parameters of the velocity selector (V, d, and B1).
Substituting this for v in the expression for m results in the expression
m = ___________________ .
(VI-12 #4)
Thus, if all our ions have the same charge and B1, B2, V, and d are known, a measurement
of D will give their mass. If the beam of ions contain different isotopes of the element having
slightly different masses, we will get a bunch of exposed spots on the film each with a slightly
differing value of D. A careful measurement of these diameters will give the relative masses of
the isotopes.
At this point try problems 8 through 10 at the back of this unit.
VI-13
Unit VI – MAGNETIC FORCES & FIELDS
The Magnetic Force on a Current-Carrying Wire
Consider a hunk of wire lying in a uniform magnetic field. If a
difference in potential exists across the ends of the wire the
conduction electrons will drift with some velocity v d toward the
positive end of the wire. Since the electrons are moving in
-field they each feel a magnetic force given by
. The electrons, however, are constricted to move within the wire so this magnetic force is
transmitted to the wire itself. Hence, the net force on the wire due to the -field is just the sum
of the forces on each electron. That is,
Where n is the total number of conduction electrons in the part of the wire that is in the
-field.
In the drawing at the right let :
= the -field that the wire is lying in
= a vector whose magnitude equals the length of the wire
that is actually in the -field. Its direction is defined as
being in the direction of conventional current flow. (to
the right in this case)
I = the conventional current in the wire
n = the number of conventional (+) charges in length of the wire (equal to the number of
electrons in )
t = the time required for a charge to move distance
q = charge on our conventional (+) charge
v = average speed of a charge along the wire (equal to vd of the conduction electrons)
Note:
.
The force on one individual charge is
Since our (+) charge q is moving in the same direction as , then
The total force on all n charges in length
and thus
is
This total force as we stated before, is the total force transmitted to the wire by the charges,
therefore,
,
VI-14
Unit VI – MAGNETIC FORCES & FIELDS
and since I = (amount of charge passing) / time, then I = nq/t. Hence, if there is a hunk of
wire of length carrying a current I in a magnetic field
wire is given by
The magnitude of the force is
, the magnetic force on the
where  is the angle between
and . The
direction is given by the right-hand-rule if the current is conventional or (+) current and by the
left-hand-rule if the current is electron or (–) current.
For practice consider the following wires in
-fields.
(VI-14 #1)
What if the wire is not straight? Check out Example 29.2 on page 913 of your text.
VI-15
Unit VI – MAGNETIC FORCES & FIELDS
The Magnetic Torque on a Current-Carrying Loop
Consider the loop pictured in Figure A. The loop lies in the y – z
plane and is in a uniform -field directed in the +x direction.
Application of
to each side of the loop gives the
magnetic forces in the directions shown and since
| F1 | = | F3 | = IaB and | F2 | = | F4 | = IbB , the net force on
the loop is zero. Also, since all of the forces are coplanar the net
torque (recall
) is zero.
Now let’s rotate the loop through and angle about the z axis as
shown in Figure B. The net force is still zero since the
magnitudes and directions of all the forces are the same but now
there is a net torque due to the -field. Viewing downward in the
+z direction as in Figure C it can be seen that the forces
and
exert torques that, if the loop is free to rotate, will rotate about
an axis parallel to the z axis at a distance a/2 from the z axis.
The net torque due to is given by
(a/2)(IbB)sin and hence
Now
the magnitude of the net torque is
(IbB)sin = IabBsin
with its direction in the +z direction as given by applying the RHR.
If there are N loops of wire, this is the same as having one loop carrying a current of NI.
Writing the expression for a rectangular loop having N turns of wire:
= NIabBsin
The area of the loop is ab and it can be shown that for any arbitrarily shaped loop of area A,
the net torque is
= NIABsin
Earlier when we discussed the flux of we defined area to be a vector where its direction was
perpendicular to the surface under discussion. Here we will use that idea to define the
magnetic dipole moment , m . By definition,
where A is the area of the loop made up of N turns of wire carrying current I. The direction of
VI-16
Unit VI – MAGNETIC FORCES & FIELDS
A is illustrated at the right. Wrapping the fingers of your right hand
around the loop in the direction of conventional current flow causes your
right thumb to point in the direction of A and therefore, m . This permits
us to simplify the expression for the torque, namely,
Since  is the angle between and A (see Figure B, p. VI-15), let’s apply the above stuff to
a simple problem to get a feel for the relationships.
Sample Exercise VI–2: A thin, circular coil of wire containing 5 loops
and having a radius of 10 cm carries a current of 5 amps. Calculate:
a. the magnetic dipole moment of the coil
b. the torque on the coil if it is in a uniform -field having a
strength of 0.2 T and making an angle of 45o down toward the
bottom of the page relative to the perpendicular to the plane of
the loop.
Solution:
a. By definition the magnetic dipole moment is
m
= (_____)(_____)(_____)
and plugging in the numbers:
m
= ( _______ )( _______ )( _______ ), ____________
(direction)
= ______________________, ____________
(direction)
(VI-16 #1)
b. The magnitude of the torque is given by
and plugging in the numbers:
Finding the direction of the
t by using the RHR with
t = ____________ , __________
we get:
(VI-16 #2)
Note that t will always be in a direction that will tend to make
of the coil.
perpendicular to the plane
VI-17
Unit VI – MAGNETIC FORCES & FIELDS
The torque on a current carrying coil of wire in a magnetic field is put to practical use in all
types of meters that use a moving coil galvanometer (e.g. voltmeters, ammeters, etc) and in
electric motors. You should read about these simple applications in your text and/or the
supplemental references.
At this point try problems 11 through 16 at the back of this unit.
– End Unit VI –
VI-18
Unit VI – MAGNETIC FORCES & FIELDS
In-Text Fill-In Answers
Page
Fill-In #
VI-5
1
Fill-In Answer
I (a) q is (+) thus the RHR and
gives
into the page (N2Pg)
I (b) q is (–) thus the LHR and
requires that
toward the top of the page (TOPg), to produce a
I (c) q is (+) thus the RHR and
requires that
have a component
N2Pg.
have a component
between out of the page (OOPg) and N2Pg (i.e., not parallel or antiparallel to ) will give a
in the direction shown.
I (d) Since and
results in
are in opposite directions ( = 180o), the LHR and
= 0.
I (e) q is (+) thus the RHR and
requires that
perpendicular to plane containing
and
roughly 45o toward the upper left to give a
I (f) q is (–) thus the LHR and
OOPg, to produce a
2
, i.e., a component directed
in the direction shown.
requires that
toward the right.
II (a) Since the angle between
then
= qvBsin = 0.
and is 180o
\ the charged
particle will continue moving in a straight
line.
II (b)
VI-6
1
II (c) Since
is always ^ to
in i & ii we have the condition
necessary for UCM, \ the
paths are circular.
have a component
have a component
VI-19
Unit VI – MAGNETIC FORCES & FIELDS
In-Text Fill-In Answers
Page
Fill-In #
VI-6
2
Fill-In Answer
II (d)
, down
Shape of path: Parabolic
VI-7
= qvBsin , where in this case  = 90o.
3
III (a)
4
III (a) We are given that it is a 5 MeV proton. Since an eV is a unit of
energy, 5 MeV must be its kinetic energy .
5
III (a)
1
III (a)
2
III (a) Since qp is (+), the right hand rule gives the
and therefore,
to be
direction of
east.
Thus,
3
III (b)
4
III (c) acceleration of the proton = 4.4 x 1015 m/s2, east
5
IV.
VI-20
Unit VI – MAGNETIC FORCES & FIELDS
In-Text Fill-In Answers
Page
Fill-In #
Fill-In Answer
VI-8
1
Condition #1 -
2
For a charged particle to move with UCM in a -field the angle
between the -field and the velocity of the particle must be 90o.
3
q must be (+) for
of the circle.
4
5
VI-9
1
2
3
p = mv = 9.6 x 10-31
4
v = p/m = 5.6 x 106 m/s
5
VI-10
1
down
2
3
4
parabolic
5
upward
6
7
qvB = qE
to be directed toward the center
VI-21
Unit VI – MAGNETIC FORCES & FIELDS
In-Text Fill-In Answers
Page
Fill-In #
VI-11
1
Fill-In Answer
2
VI-12
3
The (+) charges move along paths c and d and the (–) charges along
paths a and b.
4
The bigger (+) charge would follow path d and the one with the smaller
(+) charge path c.
5
The more massive particle would most likely follow path c and the less
massive path d.
1
2
3
4
VI-14
1
a. The RHR and
gives
out of the page
b. To give
downward the LHR requires that and \ the electron
current flow to the right
c. If the direction of I+ is 0o must be directed in quadrant I & II at an
angle > 0o but < 180o.
d. Since  = 180o,
=0
VI-16
1
a.
2
b.
, OOpg
VI-22
Unit VI – MAGNETIC FORCES & FIELDS
End of Unit Problems
1. Four particles follow the paths shown at the right as they travel
through a magnetic field. If the particles have identical speeds,
a. What can you conclude about the charge of each particle if
their masses are the same?
b. What can you conclude about the mass of particles 1, 2 and
4 if the magnitude of particle’s charge is the same?
2. The earth is under continual bombardment by
energetic charged particles. The magnetic field of
the earth shields us from most of these particles by
deflecting them back out into space.
a. Will the particles be more strongly deflected if
they approach the earth along its polar axis or
along the magnetic equator?
b. The intensity of the bombarding particles has
been found to be the greatest from the west.
What then, must be the sign of the charge of the
majority of the particles that approach the earth?
Explain your answer.
3. The uniform magnetic field of a laboratory magnet has a magnitude of 1.5 Tesla and is in a
horizontal, northward direction. Find the magnitude and direction of the magnetic force on
an electron moving with the speed of 3 x 107 m/s at the instant the electron is moving:
a. South
b. West
c. Vertically up
d. In a horizontal plane 53 o N of E
[Ans. a. zero b. 7.2 pN, up c. 7.2 pN, east d. 4.3 pN, down]
4. At the equator, near the surface of the earth, the magnetic field is approximately 50 T
northward, and the electric field due to an excess (–) charge on the earth is about 100 N/C.
Find the magnitude and direction of the gravitational, electric, and magnetic forces on a 100
eV electron moving eastward in this environment.
[Ans. FG = 9 x 10-30 N, down; FE = 1.6 x 10-17 N, up; FM = 4.8 x 10-17 N, down]
VI-23
Unit VI – MAGNETIC FORCES & FIELDS
End of Unit Problems
5. An -particle is a particle having twice the charge of a proton and approximately four times
its mass. Suppose an -particle moves with constant speed in a horizontal circle through a
uniform 1.2 Tesla magnetic field. As viewed from above, the particle moves clockwise in a
circle with a 4.5 cm radius. Find
it’s speed.
[Ans. 2.6 x 106 m/s]
its period of revolution. (Recall that one expression for ac is ac = 2v/T) [Ans: 0.1 s]
its kinetic energy in units of eV.
[Ans: 142 keV]
the difference in potential through which it would have to be accelerated to achieve its
energy.
[Ans: 72 kV]
e. the direction of the magnetic field.
[Ans: out of the page]
a.
b.
c.
d.
6. Electron moving with a constant horizontal velocity enters a
region of uniform E -field of 2 x 103 N/C directed upward
between two parallel plates.
a. If the E -field is in the direction stated, what must be the sign of the charge on each
plate?
[Ans: upper plate (–) & lower plate (+)]
b. What is the magnitude and direction of the force on the electron due to the E -field?
[Ans: 0.32 fN, down]
c. What is the shape of the electron’s path as it moves through the E -field? (Hint: Is it’s
acceleration uniform?)
[Ans: It’s path is a parabola]
d. Suppose we set up a magnetic field in the region between the plates so that we now
have both E and B fields in the region through which the electron travels. We also find
that if the magnitude of the B -field is 0.2 Tesla the electron passes between the plates
undeflected. What must be the direction of the B -field for this to occur?
[Ans: Out of the page]
e. Under the conditions stated in (d), what is the speed of the electron? [Ans: 10 km/s]
7. A velocity selector consists of a region where the electric and magnetic fields are described
by
and
, respectively. If B = 0.015 T, find the value of E such that a 750 eV
electron moving away from the origin along the x-axis will travel with constant velocity.
[Ans: 0.24 MV/m]
VI-24
Unit VI – MAGNETIC FORCES & FIELDS
End of Unit Problems
8. As shown in the drawing at the right, two particles, a
proton and an -particle, each having with the same
velocity, move from Region I where there is no magnetic
field into Region II where there is a magnetic field.
a. What must be the characteristics of the B -field in
Region II if the paths of the p + and -particle are to
be circular?
[Ans: B must be uniform & into the page]
b. What is the ratio of the radius of the proton’s path to the radius of the -particle’s path?
[Ans: 1:2]
5
c. If the radius of Path II is 10 cm and the speed of the particle is 10 m/s, what is the
magnitude of the B -field in Region II?
[Ans: 0.02 T]
9. The diagram at the right shows a slightly different form of
mass spectrometer. In this one an ion of mass m and
charge +q is produced essentially at rest within source S.
The ion is accelerated through a difference in potential
V and enters through aperture A into a region
containing magnetic field B . In the B -field it moves with
constant speed along a circular path striking the
photographic film at a distance x from the aperture.
Show that the mass of the ion is given by:
10. Two types of singly charged ions having charge +q and masses differing buy a small
amount m are introduced into the mass spectrometer described in problem 9.
a. Show that the difference in mass between the two ions is given by:
where m is the mass of either ion in x is the distance between the spots made by each
ion on the photographic plate.
b. Calculate x for a beam of singly ionized chlorine atoms having masses of 35 and 37
proton masses if V equals 7.3 kV and B equals 0.5 T.
[Ans: 16 mm]
VI-25
Unit VI – MAGNETIC FORCES & FIELDS
End of Unit Problems
11. A rigid rectangular loop of wire hangs with it's lower end
between the poles of the magnet as shown in the sketch of the
right. The magnetic field of the magnet is directed perpendicular
to the plane of the loop and into the page. The loop is
suspended from an equal arm balance and a pan having the
same mass as the loop things from the other end. The lower
end of the loop is 5 cm wide.
a. If the equal arm balance is to be “balanced” by placing various numbers of teensy tiny
but very testy toads (AKA: T4) on the pan when there is a current flowing in the loop,
should the current in the loop be flowing clockwise (CW) or counter-clockwise (CCW)?
[Ans: CW]
b. When the current is adjusted to 4 amps, it is found that 0.1 kg of T 4’s are needed to
balance the balance. What is the magnitude of the magnetic field? [Ans: 5 T]
12. A conducting rod having a mass of 0.1 kg slides without
friction on two conducting rails 50 cm apart and inclined
at an angle of 10° with the horizontal. A uniform 0.1 T
magnetic field acts vertically upward. A battery attached
to the rails causes a current to flow up one rail rail
through the rod to the other rail and back to the battery.
What must be the current and it’s direction of flow in the rod that will cause the rod to slide
up the incline at constant speed?
[Ans: 3.5 A, right to left rail]
13. What must be the direction of a magnetic field if it is to cause the loop
shown at the right to rotate counterclockwise as viewed along the z-axis
from above if I is conventional current? . . . . if I is electron current?
[Ans:
]
14. A circular coil 2 cm in diameter contains 300 turns of wire. Find:
a. the maximum torque on this coil when it carries a current of 10 mA while in a magnetic
field of 0.5 Gauss.
[ Ans:
]
b. the angle between the perpendicular to the plane containing the coil and the magnetic
field when the torque is a maximum. . . . when the torque is a minimum. [Ans: 90 o, 0o]
VI-26
Unit VI – MAGNETIC FORCES & FIELDS
End of Unit Problems
15. An N-turn circular coil of radius R is suspended in a uniform magnetic field B directed
vertically upward. The coil can rotate about a horizontal axis through its center. A mass
hangs by a string from the bottom of the coil. When a current I is put through the coil it
eventually assumes an equilibrium position, in which the perpendicular to the plane of the
coil makes an angle  relative to the direction of B . Find  and draw a sketch of the coil in
its equilibrium position. Let B = 0.5 T, R = 10 cm, N = 10 turns, m = 0.5 kg, and I = 1 A.
[Ans:  = 72o]
16. In the Bohr model of the hydrogen atom the electron moves in a circular about the proton,
with the Coulomb force being the centripetal force. In its lowest orbit the radius of the
electron’s path is 52.9 pm.
a. Find the equivalent current the moving electron generates.
b. Find the magnetic dipole moment of this current loop.
[Ans: ~1 mA]
[Ans: 8.8 x 10-24
]
If the earth’s magnetic field at the equator is 0.5 Gauss directed horizontally northward, find
the torque on a hydrogen atom if the magnetic dipole moment vector of the atom is directed
c. due west.
d. due north.
[Ans: 4.4 x 10-28
, down toward the center of the earth]
[Ans: zero]