Download I. Magnetic forces

In Unit 401 we turn our attention to the last of the major topics, magnetism. In this unit
we focus on steady-state magnetic fields, a.k.a. magnetostatics. The final unit after this
one will look at magnetic fields that change with time or through space. In this unit we
consider how single charges move through steady magnetic fields as well as uniform
currents flowing across magnetic fields. We also consider how magnetic fields are
produced. In other words, this unit considers magnetic forces and magnetic sources.
Before taking a test on this unit students should be familiar with the material outlined
below.
I. Magnetic forces
A. Charged Particles
1. Magnitude and direction of force
2. Velocity selector
3. Mass spectrometer
B. Uniform Currents
1. Forces, magnitude and direction
2. Torques and magnetic dipole moments
II. Magnetic Sources
A. Fields of long, current carrying wires
1. Field of a single current
2. Field due to multiple currents
3. Force between two parallel currents
B. Biot-Savart Law
1. Field of a long, straight current
2. Field along the axis of a circular current loop
3. Field at the center of a circular loop
C. Ampere’s Law
1. Magnetic field due to a single uniform current
2. Magnetic field due to a hollow, uniform current
3. Magnetic field due to current in a coaxial cable
4. Magnetic field due to a single, non-uniform current
You should expect about one problem for each of the above lettered sections on the next
test.
Lesson 4-01
Introduction
Read 29. 1
Before we start using the text there are a few questions that should be answered. Most
importantly, “What is the difference between an electric field and a magnetic field?” In
this introductory lesson you will see a comparison and a contrast between the two fields.
Electric Field Sources
Magnetic Field Sources
 The symbol for the magnetic field is B.
 The symbol for the electric field is E.
 Units are Newton/(Coulombm/sec).
 Units are Newtons/Coulomb or
This set of units is defined to be Tesla.
Volts/meter.
The units differ from the E field by a
factor of velocity units. (A hint!)

A B field can be generated by passing
 An E field is generated by creating a
current through a coil of wire.
potential difference across parallel
L
plates.
E
B
I

d
The value of E is found by dividing a
voltmeter reading by the gap distance.

B = o N I / L
L is length of coil. N is number of
turns around coil. o = 4 E - 7 N/A2
E =  V / d

The value of B is determined by
multiplying an ammeter reading by the
geometrical properties of the coil.

Magnetic fields come from electric
charges moving through space. This
explains why units of “m/s” appear in
the definition of Tesla. Magnetic field
lines do not begin or end.
You can determine field strengths of either kind by simple measurements using a meter
stick, voltmeter and ammeter. The aim of this section is entirely qualitative in nature. At
this time do not be concerned about the math. The biggest contrasts seems to be
a) Sources of E are charges at rest while sources of B are charges in motion (current).
b) Units are similar with the exception of “m/s” in the definition of Tesla.
c) Use a voltmeter to determine E; use an ammeter to calculate B.
d) Field lines can stop and start for E but not for B.
In the next section forget about where the field comes from. Assuming that the field is
there consider how electric charges move through a uniform E or B field.
Electric fields come from charges at
rest. The electric field lines begin on
positive charges and end on negative
charges.
Charge in a Pure, Uniform Electric Field
Consider four, positively charged particles
in a pure E field as shown below.
qA
qC
qB
Charge in a Pure, Uniform Magnetic Field
Consider four, positively charged particles
in a pure B field as shown below.
qA
qD
The above charges have the following
initial conditions:
a) qA is at rest.
b) qB is initially moving to the right.
c) qC is initially moving up the page.
d) qD is initially moving up and to the
right at some angle from the vertical.
The resulting descriptions of motion that
will be observed are outlined below.
a) The first charge will accelerate from
rest to the right. You can use the
kinematics equations with initial
velocities of zero.
b) The second charge will also accelerate
to the right but the initial velocity
terms in the kinematics will no longer
be zero.
c) The third charge will move along a
parabolic path. For constant speed in
the “Y” direction use d = vYt. For
constant acceleration in the “X”
direction use the kinematics as
described in part (a).
d) The fourth charge will move along a
parabolic path. For constant speed in
the “Y” direction use d = vYt. For
constant acceleration in the “X”
direction use the kinematics as
described in part (b).
qC
qB
qD
The above charges have the following
initial conditions:
a) qA is at rest.
b) qB is initially moving to the right.
c) qC is initially moving up the page.
d) qD is initially moving up and to the
right at some angle from the vertical.
The resulting descriptions of motion that
will be observed are outlined below.
a) Nothing happens and the charge will
remain at rest in the B field. Boring!
b) The charge will continue to move to
the right at a constant speed according
to d = vt. Boring, part II!
c) The charge will move in a circular
motion wrapping around the magnetic
field lines. To determine which way
the charge circulates use your right
hand. Point your index finger in the
direction of the charge’s motion. Point
your middle finger in the direction of
the magnetic field. Your thumb will
point to the direction of the center of
the curve as long as it is perpendicular
to the direction of index and middle
finger directions. Try this out on
figures 29.6, 14, 37 and 29.33.
d) Since the velocity has perpendicular
and parallel parts the resulting motion
is a combination of (b) and (c). A
coiled or helical path as shown in
figure 19.11 will result.
We see that not only does an electric charge have to move as suggested by the units but
now it must move perpendicular to the B field lines. Why? Also recognize the resulting
path is circular rather than parabolic.
Why Does a Charge Curve When Crossing B Field Lines?
Before finishing this lesson this question must be addressed. The answer is the key to
understanding over half of this unit. As a charged particle moves through space it must
drag its E field lines along. The motion of the E field lines being pulled through space
causes the lines to wrap around the path. The resulting wrapped lines are what we
perceive as magnetism.
As a charged particle moves through space it will create a tubular, magnetic vortex
that curls around its path. Picture the charge leaving a magnetic tunnel in its wake. In
figure 30.3 the magnetic tunnel is mapped out using iron filings as charges flow up the
wire. To determine the direction of the magnetic tube use the “right-hand curl” rule
demonstrated in figure 30.4a&b. Both two and three-dimensional representations are
shown below. Note that direction of the magnetic curl is reversed for negative charge.
v
v
B
· ·
⊗⊗ B
· ·
⊗⊗
· ·
⊗⊗
+q
+q
Consider a charged particle moving across a magnetic field from some other source.
As the charge moves there is an action-reaction between the magnetic tube of the moving
charge and the external magnetic field. Suppose that there is a magnetic field pointing
into the page as shown in the left-hand figure below. Now imagine a proton attempting
to move up the page through that field. The proton’s tubular field would have the form
shown in the second figure below. A combination of the two fields is shown below right.
⊗ ⊗ ⊗ ⊗ ⊗
· ·
⊗⊗
· ·
⊗⊗
· ·
⊗⊗
⊗ ⊗ ⊗ ⊗ ⊗
⊗ ⊗ ⊗ ⊗ ⊗
External B field
+q
B field from moving q
⊗ ⊗ ⊗ ⊗ ⊗
· ·
⊗⊗
⊗ ⊗ ⊗ ⊗ ⊗
· ·
⊗⊗
⊗ ⊗ ⊗ ⊗ ⊗
· ·
⊗⊗
Superposition of both
magnetic fields
The combined magnetic fields will weaken on the left side of the path since the two fields
are anti-parallel. The two fields strengthen on the right side of the path since the fields
are parallel. The greater magnetic pressure from the right side of the path will curve the
charge to the left. As the path curves the fields continue to add on the outside and
partially cancel on the inside. It is the superposition of the external magnetic field and
the charged particle’s field that makes any charge curve when crossing another B field.
Lesson 4-02
Read 29. 3-6
Magnetic Force on a charge
The previous lesson looked at the qualitative considerations of magnetic force acting on a
charged particle moving through a magnetic field. In today’s lesson we consider more
quantitative questions of the same material.
The culminating application of this lesson is the ability of the mass spectrometer
to determine the mass of any subatomic particle. The mass spectrometer is actually a
combination of three unique parts: (1) a particle accelerator, (2) a velocity selector and
(3) analyzing chamber.
Magnetic Force
 If a particle enters a uniform magnetic field from the outside it will follow a semicircular path before once again emerging from the B field. See figure 29.14.
 If a particle is created in a uniform magnetic field it will travel along a circular path
of constant radius. See figure 29.10. In real lab situations the circle decays creating
an inward spiral. See figure 29.3 on page 703.
Magnetic Force
When a particle of charge “q” moves at
speed “v” through a magnetic field of strength
FB = q v x B = qvB
B the field will exert a force on the charge as
where FB is  to the v-B plane
shown in the equation box to the right.
The directions of v, B and F were described yesterday. Note that the magnetic force is
velocity dependent. The faster a charge moves through a magnetic field the greater the
force. Also, the magnetic force is a centripetal force. No work is done to change the
speed. The force merely changes direction. By combining the magnetic force in today’s
lesson with the centripetal force equation from mechanics you can get an equation for the
radius of the motion.
Magnetic & Centripetal F’s
Radius of Curvature
Equation of Motion
qvB = m(v2) / R
R = mv / qB
where a = v2 / R
2R = v  (complete circle)
or
R = v t (semicircle)
Note: Work done by a centripetal force is zero since kinetic energy is constant.
See Sample Problem 29-3 on pages 710 and 711. The above theory can be used to
analyze the mass of subatomic particles in the mass spectrometer.
The Mass Spectrometer
As mentioned above the mass spectrometer has three
parts. The maximum speed of a particle upon exiting
vMAX  (2qV / m)
the particle accelerator can be found using the potential
difference, V, of the accelerator. The maximum speed is
found using the formula in the box to the right. Note that a “singly ionized atom” has a
net charge of one unpaired proton or +1.6E-19 C.
The problem with the particle accelerator is that not all particles fall through the entire
potential difference. This means that the particles in the beam have a distribution of
speeds. This is not good because it will cause the particle beam to spread or diverge. In
order to combat this problem a velocity selector is used to clean up the beam.
Velocity Selector
The velocity selector works off the principle
x
x
x
that the electric field acts with the same force
on all particles independent of how fast they
q+ 
x
x
x
x
move. The magnetic field on the other hand
is velocity dependent. The idea is to shoot a
x
x
x
charged particle through crossed electric and
magnetic fields. See the diagram to the right.
The end of the coil is shown as a black circle with a current flowing in a clockwise
manner. The clockwise current produces a magnetic field into the page shown in blue.
Also shown in black are parallel plates with voltage differences so that the resulting
electric field lines are shown in red.
As a positively charged particle passes from left to right through the crossed fields the
electric force is down while the magnetic force is up (or vice-versa for negative charge).
 FY : FB – FE = qvB – qE = ma
A particle moving too fast will experience a larger magnetic force and be pulled up out of
the beam. A particle moving too slowly will experience a weak magnetic force causing
the electric field to pull the particle downward out of the beam. There is exactly one
speed however, that will perfectly balance the upward magnetic force and the downward
electric force. Particles at this speed will pass
through the selector without deflection. To
vEq = E / B
determine the speed that gets though without
deflection merely set the acceleration in the y
direction equal to zero and solve for v. By adjusting either the voltage across the parallel
plates to change E or the current passing through the coil to change B you can dial up any
desired speed for particles before they enter the pure magnetic field of the mass
spectrometer. By knowing the radius, length and number of turns in the coil a reading
from an ammeter is needed to calculate B. By knowing the gap distance between parallel
plates a reading from a voltmeter is all that is necessary to get E. Since the q’s cancel out
in the derivation of the equilibrium speed the velocity selector will not discriminate
between doubly or singly charged particles. Be sure that you can do the following:
Given the direction of E, B, FE or FB for a charged particle moving in a known direction
in a velocity selector you should be able to determine the direction of the other vectors.
Now that the speed of the particle is known the analysis suggested at the
beginning of the lesson can be implemented in the analyzing chamber.
Homework Problems Ch 29: 5, 9, 13, 23, 25, 31, 37
Lesson 4-03
Read 29. 7-9
Magnetic Force on current
Consider a conductor that is crossing a magnetic field. What happens when a current is
forced to flow through the wire? As each charge moves along the wire it will experience
a magnetic force. Since the charges cannot leave the wire the result is that the entire wire
experiences a force. Today’s lesson concerns the resulting force on a length of current
flowing across a magnetic field.
Force on a Length of Current
The amount of force acting on a length of wire is


FB  i L  B
given to the right. Note the cross product that
demands that you use the part of the B field
perpendicular to the current. Sample problems 29-6,7&8 demonstrate the use of this
equation. Pay careful attention to the latter examples that break the wire into segments of
length idL (a point current) producing a small force dF. The net force is a sum of dF’s.
Magnetic Torques
Consider the rectangular loop of current in an external magnetic field shown in figure 2923. Two sides have current running parallel to the field so that they experience no force.
The other two sides experience a force as shown in the figure. These forces tend to spin
the current loop. The authors stress the fact that the forces attempt to align the normal
vector to the loop area with the external magnetic field. The direction of the normal
vector to the loop is determined using the right-hand curl rule for the current flowing
through the loop as shown in figure 29-24b.
From the mechanical definition of torque you know that torque () is defined as
the cross product of the moment length (r = b/2) and force (2F= iaB). The two is inserted
since two sides experience a magnetic force that produces torque. As a result the net
torque on the loop is  = iabBsin. It so happens that there is a second way to define the
torque on a current loop aligning with an external B field.
We begin the second way of determining torque by defining

  NiA
something known as the magnetic dipole moment, . The dipole
moment is calculated as the product of the number of current loops,
N, and the amount of current, i, and the area of the loop. The direction of the dipole
moment is determined as shown in figure 29-24b. If you compare our definition of the
dipole moment to the first three variables of torque in the previous paragraph you will
note that they are the same assuming that N = 1. This allows us to
  
define the magnetic torque on multiple current loops as the cross
  B
product of the dipole moment and the external magnetic field. Review
Sample problems 29-9 and 29-10 at this time.
Homework Problems Ch 29: 46, 49, 53(dF/d=0): 55, 62 (2.45A), 63, 65
Lesson 4-04
Read 30. 1-2
Magnetic Fields due to
Long, Straight Currents
In Chapter 29 you considered the impact of an external magnetic field on a moving
charge or current. But where do these magnetic fields originate? What is their source?
In chapter 30 we will address the sources of magnetic fields. In this initial lesson we
review basics that you learned in your previous physics course. The three lessons that
follow we will get down to some advanced calculations of magnetic field sources.
Long Straight Currents
When a current flows through a long, straight conductor, a magnetic field will wrap
concentrically around the wire as shown in figures 30-2 and 30-3. In order to determine
the direction of the magnetic field lines use your right-hand curl rule as demonstrated in
figure 30-4. The strength of the magnetic field depends
 i
on the perpendicular distance from the wire to any point
B O
in space. The magnitude of the field strength is found
2r
using the equation shown in the above box where
o = 4E-7 Tm/A.
If there are multiple currents then you use the above equation to get the magnetic
field due to each individual current at a specified point. The total magnetic field at the
specific point can be found by invoking the principle of superposition. Review Sample
Problems 30-2 and 30-3.
Parallel Currents
When parallel currents occur, each current will create a magnetic field that pulls
or pushes on the other current. Consider figure 30-8 on page 733. The magnetic field
from current ia will exert a force on current ib. At the same time the magnetic field from
current ib exerts a force on current ia. First, realize that parallel currents will attract while
anti-parallel currents repel. Second, recognize that the force that each current exerts on
the other must be equal in magnitude and opposite in direction according to Newton’s
Third Law of Motion. Finally, the magnitude of the force of either current on the other is
found by inserting the above expression into F = iLB. The result is equation (30-14).
Homework Problems Ch 30: 26[dB = o(i/w)dx/(2x)], 27, 28, 29, 32, 33, 34, 37, 38
Lesson 4-05
Read 30. 1-2, 5
Biot-Savart Law
Recall how we went through electricity starting with point charges, then through
continuous charges and finally into Gauss’s Law of Electricity when high degrees of
symmetry were present. We are taking a similar track in finding magnetic fields. In the
last lesson we looked at straight parallel currents. In this lesson we consider the
equivalent of summing magnetic fields due to “point currents” using Biot-Savart Law.
In order to use the Biot-Savart Law you must consider any current to be made of
small increments of current “i” that are of length “dL”. Consider figure 30-1 where the
authors construct an analogy between dq’s producing dE’s and idL’s producing dB’s.
The authors use a notation of idS where “dS” is an increment of arc-length. Notice that
the dE’s are always either parallel or anti-parallel to the radius vector. In figure 30-1b the
dB’s are directed perpendicular to the plane defined by the direction of “idL” and the
radius vector. More specifically, the direction of dB is found using the cross product of
idL cross r. Different books write the Biot-Savart Law in two distinct ways. Some books
use the cross product of idLr where r is the radius vector

  O id L  rˆ
having magnitude of r and direction of r. Other books prefer
dB 
to use a unit vector for r known as “r hat”. This will insert
4 r 2
the direction of r into the cross product but leave out the
magnitude of r. We will use the latter notation. Written in this form one sees something
similar to E = kq/r2 where the constant k is replaced with 2E-7, the point charge is
replaced with a point current and the function is still over r2.
According to The College Board each student should know how to apply the BiotSavart Law to find the magnetic field under the given conditions:
1. Find B at a perpendicular distance r from a very long, straight current source. See
section 30-1.
2. Find B at a perpendicular distance r along the perpendicular bisector from a straight
current of length 2L.
3. Find B at the center of a circular loop of current with radius R. See section 30-1.
4. Find B along the axis of a ring of current. See section 30-5.
Students should commit results of the above list to memory as well as be able to
derive them.
Homework Problems Ch 30: 11, 13, 17, 19, 49, 68, 69
Lesson 4-06
Read 30. 3-4
Ampere’s Law
In the previous lesson we saw a very complicated integration technique for finding the
magnetic field strength at a specified point in space. The results of the previous method
determined B(r) for a long, straight current, B at the center of any current arc and B at a
given distance along the axis of a current ring. There is another method that does not
work for many of these examples but is fast when a high degree of symmetry is present.
The method was developed by Ampere' and is known as “Ampere’s Law”. This
method does for magnetism what Gauss’s Law of Electricity did for static charge
distributions. Recall from Gauss’s Law how the charge distribution had to be completely
surrounded by a closed surface. In Ampere’s Law the current must be completely
surrounded by a closed loop or closed path. The summation for this method will be along
a path instead of over a surface. This summation is known as a path integral. See figures
30:12-16, 30:20 and figure 30:21 for examples of this summation.
Ampere’s Law


The law is written in the text box to the right and
 B  dL   o i
stated without proof. The circle on the integral
sign is to remind us that we are summing along a
closed path. The end of the path must return to the beginning of the path forming a
circle, square, triangle, hexagon and so forth. I replaced the “dS” used by the book with a
“dL” to farther remind you that we are summing along a length with the unit of meter.
The dot product is a reminder that we sum only the parts of B that are parallel to “dL”. If
you can choose the shape of the path so that B is always parallel or perpendicular to dL
then the integration will be simplified.
This method is first demonstrated in figure 30-14 where we select the shape of the
orange path as our closed loop. This orange path is everywhere parallel to B while
encompassing the long, straight current. Since B has the same value everywhere along
the path it is a constant of the integration and can factor outside of the integral. This
leaves B*(integral of dL) on the left-hand side of the equation which quickly is replaced
by B*(2r). The right-hand side of the equation is o times the current flowing through
the area enclosed by the orange path. As a result, the magnetic field at a distance of “r”
from the current is found using the equation from lesson 4-04. The equation should be
very familiar to you by now. Until now, we have only mentioned the magnetic field
outside of the wire. How does the magnetic field vary within the current flow?
Ampere’s Law can be used to find B(r) inside of the wire also. As long as the
current density, J, is uniform or only a function of r, you can take the same approach
outlined in the previous paragraph. The left-hand side is done the same way as before.
The only difference is that only the portion of the current flowing inside of the path of
integration is used. See figure 30-15. The current flowing inside of the path is I(r/R)2
where “r” is the radius of the path and R is the radius of the wire. The resulting
expression for the magnetic field is linear. You should not be surprised to find this result.
There are only three major variations on this theme. What if the current is nonuniform? What if the current flow is both in and out at the same time such as in the case
of a coaxial cable? Finally, consider breaking the path integral into several parts where B
is sometimes parallel and sometime perpendicular to the path. This is the case of finding
the magnetic field inside of a long, tightly wound solenoid (coil).
To get an idea of the types of variations, look at the figures associated with
Sample Problem 30-4 and homework problems 30: 43, 46, 47 and 50. You should be
able to submit rough sketches of B(r) for each of these cases by inspection. For each of
these problems a simple ratio of the partial current to the total current will do. The
method suggested two paragraphs above would do for these problems.
If the current is non-uniform but is a function of distance from the center then you
can modify Ampere’s Law by replacing the current inside the closed path by another
integral expression. To be honest, the left-hand side is treated as in the previous
problems and you merely integrate the current density in
order to get the total current inside. Usually, dA is broken


 
B

d
L


J
 dA
into concentric rings where each ring has a differential area
O

of 2r*dr. But it looks more complicated and everybody
will think you are a genius if you write the right-hand side in integral form. See problem
30:48 for an example of this type.
The final variation with Ampere’s Law has to do with solenoids and torroidal
solenoids. For the solenoid, the magnetic field near the axis is found by choosing an
Amperian loop that is rectangular with long sides being down the axis of the loop and just
outside of the surface of the coil. The two short sides of the rectangle are perpendicular
to the field and contribute zero to the summation due to the dot product being zero. The
portion of the path outside of the coil is assumed to be zero since B is very weak there
approaching zero in value. This leaves the path along the axis of the coil as the only
contributing part of the integral. But the current has looped back through the rectangular
path N times. The result for the magnetic field near the
central axis of the tightly wound coil is shown in the box B = Oni where n = N/L
to the right. Be sure that you know this equation from
memory as well as being able to derive it from Ampere’s Law. Many students are able to
remember this as the “money or moni” equation by letting mu have an “m” phonetic
sound as it does in the Greek language. Also be able to get B inside of a torroidal
solenoid. For the torroidal solenoid, or coil bent into a donut shape, the magnetic field
inside is not uniform but varies slightly being stronger near
the hole of the donut. The magnetic field strength for the
B = ONi / (2r)
torroidal solenoid is determined in the box to the right.
Recommended homework problems:
Homework Problems Ch 30: 43, 46, 47, 48, 50, 53, 55, 56, 61
Expect a test within about two days.
You should be able to start with the Biot-Savart Law or Amepre’s Law and derive each
of the following. Each derivation should take less than 5 minutes and including
explanation of each step. You should also be able to state the final result without proof.
1.
2.
3.
4.
5.
6.
Find B(r) inside and outside of a very long, straight wire using Amp’s Law.
Find B along a perpendicular bisector of a short straight wire using the BS Law.
Find B at the center of a single loop of current of radius R using the BS Law.
Find B along the axis of a current loop or radius R using the BS Law.
Find B along the axis of a tightly wound solenoid using Amp’s Law.
Find B(r) inside of a torroidal solenoid using Amp’s Law.