Physics 122B
Electricity and Magnetism
Lecture 19 (Knight: 32.4-32.6)
Biot-Savart and Ampere’s Laws
Martin Savage
Lecture 19 Announcements
 Lecture HW has been posted on Tycho
and is due at 10 PM Wednesday,
The deadline for requests for regrades
of Exam 2 (see Syllabus for procedure) is
noon Monday.
7/19/2016
Physics 122B - Lecture 19
2
Magnetic Force
A current consists of moving charges. Ampere’s experiment
implies that a magnetic field exerts a force on a moving charge.
This is true, although the exact form of the force relation was not
discovered until later in the 19th century. The force depends on the
relative directions of the magnetic field and the velocity of the
moving charge, and is perpendicular to both..

F  q vB
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
Physics 122B - Lecture 19
3
Cyclotron Motion
Consider a positive charged particle
with mass m and charge q moving at
velocity v perpendicular to a uniform
magnetic field B. The particle will
move in a circular path of radius rcyc
because of the force F on the particle,
which is:
mv 2
F  qvB 
rcyc
f cyc 
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rcyc 
mv
qB
v
q B

(independent of rcyc and v)
2 rcyc m 2
Physics 122B - Lecture 19
4
Example: The Radius of
Cyclotron Motion
An electron is accelerated from rest
through a potential of 500 V, then
injected into a uniform magnetic field B.
Once in the magnetic field, it completes
a half revolution in 2.0 ns.
What is the radius of the orbit?
1
2
v
mv2  (e)V  0  0
f cyc 
B
2eV
 1.33 107 m/s
m
1
1

 2.5 108 Hz
9
T 2(2.0 10 s)
2 mf cyc
e
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3
 8.94 10 T
rcyc 
mv
 8.47 103 m  8.47 mm
qB
Physics 122B - Lecture 19
5
The Cyclotron Accelerator
When a charged particle moves
in a uniform field, fcyc is independent
of both radius and energy.
One can “pump” energy into the
particle as it cycles, using an
electric field that varies with the
frequency fcyc.
This is the basic principle of the
cyclotron, a particle accelerator
that can increase the energies of
charged particles.
The ultimate energy of a
cyclotron is determined by the
radius R of the magnet:
Emax = (RqB)2/2m
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Physics 122B - Lecture 19
6
Measuring Momentum with
Magnetic Deflection*
rorbit
mv
p

=
so p  qBrorbit
qB qB
In other words, by measuring the orbit
radius of curvature rorbit of a particle with
charge q that is deflected by a uniform
perpendicular magnetic field B into a
circular orbit, one is measuring the
momentum p of the particle.
The figure shows a “split-pole magnetic
spectrograph”, a magnetic analyzer used in
nuclear physics research to analyze
particles from nuclear reactions into
groups with the same momentum.
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Physics 122B - Lecture 19
7
The Earth’s Van Allen Belt
Charged particles tend to spiral in a magnetic
field, moving along field lines in a spiral path. When
the field lines “pinch” together, the particles
“bounce” and reverse the direction of their motion
along the field lines. This forms a “trap” that can
catch and hold charged particles.
The Earth’s magnetic field forms such a trap for
electrons and protons emitted by the Sun. These
trapped charged particles form the Van Allen
radiation belts in the space around the Earth.
Energetic electrons and protons leaking from the trap
near the Earth’s magnetic poles, ionize the air in the
upper atmosphere and produce the Aurora Borealis in
the northern sky.
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Physics 122B - Lecture 19
p
p
e
8
The Hall Effect
When a charged particle moves in
a vacuum, it experiences a force
that is perpendicular to its
velocity in a magnetic field. In
1879, Edwin Hall, a graduate
student at Johns Hopkins Univ.,
discovered that the same
behavior is true of charged
particles moving in a conductor.
VH
Fm  evd B  Fe  eE  e
w
VH  wvd B
vd 
J I/A
I


ne
ne
wtne
Edwin Herbert Hall
(1855 – 1938)
VH  wB
I
IB

wtne tne
The sign of the mobile charges matters !!!
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Physics 122B - Lecture 19
9
Example: Hall Probe
Measurement of Magnetic Field
A Hall probe consists of a strip of metallic bismuth that is 0.15 mm
thick and 5.0 mm wide. Bismuth is a poor conductor with a charge
carrier density of n = 1.35x1025 m-3. The Hall voltage on the probe is
measured to be 2.5 mV when the current is 1.5 A.
What is the magnetic field, and what is the electric field inside
the bismuth?
tne
B
VH
I
(1.5 104 m)(1.35 1025 m-3 )(1.6 1019 C)

(0.0025 V)
(1.5 A)
 0.54 T
E
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VH (0.0025 V)

 0.50 V/m
w
(0.005 m)
Physics 122B - Lecture 19
10
The Cross Product Form
of the Biot-Savart Law
The Biot-Savart Law can be represented
more compactly using a vector cross
product. This automatically gives a B field
that is perpendicular to the plane of the
charge velocity and radius vector to the
point at which the field is being evaluated.
0 q  v  rˆ 
B
4
r2
[Note that the r in the numerator is
^
r (unit vector), not r (vector)!]
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Physics 122B - Lecture 19
11
The Magnetic Field of a Current
Consider the charge
Q moving at speed v
through a short
segment of wire s.
Q v = I t v = I t
s
= I s
t
Therefore, we can use the Biot-Savart law
to find the magnetic field B produced by the
wire segment:
0 q v  rˆ 0 I s  rˆ
B

2
4 r
4 r 2
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Physics 122B - Lecture 19
0 ds  rˆ
dB 
I
4
r2
The Biot-Savart Law
for current elements
12
Example: The Magnetic Field
of a Long Straight Wire
A long straight wire carries current I
in the x direction.
Find the magnetic field a distance d
from the wire.
0 ds  rˆ
dB 
I 2
4
r

B 0
4
I
 0
4


dx
 x2  d 2

0 I 
x


4  d x 2  d 2
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
I dx  rˆ 0
 r 2  4
I dx sin 
 r2

I
 0
x 2  d 2 4
d



dB all in same direction



dx d
 x2  d 2 
3/ 2
0 2 I
4 d
Physics 122B - Lecture 19
B is
constant
along the wire
and falls off as 1/d.
13
Example: The Magnetic Field
near a Heater Wire
A 1.0 m long 1.0 mm diameter nichrome heater wire is connected
to a 12 V battery.
What is the magnetic field, B, at 1.0 cm from the wire?
(rnichrome = 1.5 x 10-6 W m.)
R  r L / A  (1.5 106 Wm)(1.0 m) /  (0.0005 m) 2  1.91 W
I  V / R  (12.0 V) /(1.91 W)  6.28 A
B
0 2 I
 (107 Tm/A)2(6.28 A)/(0.01 m)  1.26 10-4 T
4 d
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Physics 122B - Lecture 19
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Example: The Magnetic Field
of a Current Loop
A loop of radius R
carries a current I.
Find the magnetic
field at a distance z
from the center of
the loop along the z
axis.
0 Ids 0 I cos  ds 0
IRds
dBz  dB


z
4 r 2 4 z 2  R 2
4  z 2  R 2 3/ 2
2 R
0
IR
Bz 
ds
3/ 2 
2
2
4  z  R 
0
0 I 2 R
0
IR


4  z 2  R 2 3/ 2
2  z 2  R 2 3/ 2
2
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2
Bloop 
Physics 122B - Lecture 19
0
I
2 R 1  ( z / R) 2  3/ 2


15
Example: Matching
the Earth’s Magnetic Field
What current is needed in a 5 turn coil 10 cm
in diameter to cancel the Earth’s magnetic field
at the center of the coil? (BEarth = 5 x 10-5 T)
Bloop 
0
I
2 R 1  ( z / R) 2  3/ 2


 NI
BN loops ( z  0)  0
2 R
2 RB 2(0.05 m)(5 105 T)
I

 0.80 A
7
N 0
5(4 10 T m/A)
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Right-Hand Rule for Loops:
Fingers follow current and
thumb points along B.
Physics 122B - Lecture 19
16
Loops as Magnetic Dipoles
S
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N
A current loop is a magnet.
It has two distinct sides, that
can be identified as its north
and south poles.
Magnetic lines of flux come
out of the north pole and go into
the south pole.
A loop suspended from a
thread aligns itself with the
Earth’s field, with the loop’s
north pole pointing north.
The side of a loop from
which lines of flux come out (N
pole) is repelled by the north
pole of a bar magnet and
attracted by the south pole of a
bar magnet.
Physics 122B - Lecture 19
17
The Magnetic Dipole Moment
Remember that the on-axis
electric field produced by an
electric dipole of electric dipole
moment p has the form:
Edipole 
1
2p
4 0 z 3
The on-axis magnetic field
of a current loop has a similar
form, where its magnetic dipole
moment  is equal to AI:
Bdipole 
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0
IR 2
2  z 2  R 2  3/ 2


0 2( R 2 ) I 0 2 AI


3
z  R 4
z
4 z 3
Physics 122B - Lecture 19
Bdipole 
0 2 AI 0 2

3
4 z
4 z 3
18
Example: The Field of a
Magnetic Dipole
a. The on-axis magnetic field 10 cm from a magnetic dipole is 1.0 x 10-5 T.
What is the magnetic moment of the dipole?
b. If this magnetic dipole is created by a single current loop 4.0 mm in
diameter, what is the current I the loop?
Bdipole
1
3
B
z



dipole
so    0 
2
 4 
 2
 0 3
4 z
  10 Tm/A 
7
1
(1.0 10-5 T)(0.10 m)3
 0.050 Am 2
2

0.050 Am2
  AI so I  
 4,000 A
2
A  (0.002 m)
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Physics 122B - Lecture 19
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Question 1
What is the current direction of the loop
when looking down on the loop, and which
side is the north magnetic pole?
(a) Current cw, N pole on top; (b) Current cw, N pole on bottom;
(c) Current ccw, N pole on top; (d) Current ccw, N pole on bottom;
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Physics 122B - Lecture 19
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Line Integrals (1)
A line integral is a special kind of vector integral in which the projection
of some vector quantity is projected on a straight or curved linear path
connecting two points, and the product of vector’s projection times
infinitesimal path distance is summed. The simplest line integral is just the
sum over the path length L:
f
L   sk   ds
k
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i
We have previously seen this kind of integration in
our discussion of work:
f
f
i
i
W   Fs ds   F  ds
Physics 122B - Lecture 19
21
Line Integrals (2)
Now consider a line integral in the
presence of a magnetic field. Divide
the path up into line segments of length
s. At the kth segment the magnetic
field is Bk.
B
k
k
f
 sk   B  ds
i
If B is always in the same direction as ds
and constant over the path, then:
f
f
f
i
i
i
 B  ds   Bds  B ds  BL
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Physics 122B - Lecture 19
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Lecture 19 Announcements
 Lecture HW has been posted on Tycho
and is due at 10 PM Wednesday,
The deadline for requests for regrades
of Exam 2 (see Syllabus for procedure) is
noon Monday.
7/19/2016
Physics 122B - Lecture 19
23