Download Biot-Savart Law

22.7 Source of magnetic field
due to current
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The Oersted’s discovery in 1819 indicates an
electric current can act as a source of
magnetic field.
Biot and Savart investigated the force
exerted by an electric current on a nearby
magnet in the 19th century.
They arrived at a mathematical expression for
the magnetic field at some point in space due
to an electric current.
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Biot-Savart Law
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The magnetic field is dB at some
point P
An infinitesimal length element is ds
The wire is carrying a steady
current of I
The vector dB is perpendicular to
both ds and the unit vector r̂
directed from the element toward
P
The magnitude of dB is inversely
proportional to r2, where r is the
distance from the element to P
2
Observations

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The magnitude of dB is proportional to
the current and to the magnitude ds of
the length element ds
The magnitude of dB is proportional to
sin q, where q is the angle between the
vectors ds and r̂
3
Biot-Savart Law, Equation

The observations are summarized in the
mathematical equation called BiotSavart Law:
I ds  rˆ
dB  km
2
r

The Biot-Savart law gives the magnetic
field only for a small length of the
conductor
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Permeability of Free Space
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mo
7 T  m
km 
 10
4p
A
The constant mo is called the
permeability of free space
mo = 4 p x 10-7 T. m / A
The Biot-Savart Law can be written as
mo I ds  rˆ
dB 
2
4p r
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Total Magnetic Field due a wire
of current

To find the total field, you need to sum
up the contributions from all the current
elements

You need to evaluate the field by
integrating over the entire current
distribution
6
Exercise 52
The magnetic field at the point P is
m0 I
B
(cosq1  cosq2 )
4p a
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B for an Infinite-Long, Straight
Conductor, Direction
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The magnetic field lines are
circles concentric with the wire
The field lines lie in planes
perpendicular to to wire
The magnitude of the field is
constant on any circle of
radius a
The right hand rule for
determining the direction of
the field is shown
moI
B
2p r
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9
10
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B for a Circular Current Loop
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The loop has a radius of
R and carries a steady
current of I
Find B at point P
Bx 

moIR 2

2 x R
2
2

3
2
The field at the center of
the loop
m0 I
Bx 
2R
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22.8 Magnetic Force Between
Two Parallel Conductors
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Two parallel wires each carry a steady
current
The field B2 due to the current in wire
2 exerts a force on wire 1 of F1 = I1l
B2
Substituting the equation for B2 gives
m II
F1  o 1 2
2p a
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Parallel conductors carrying currents in the
same direction attract each other
Parallel conductors carrying current in
opposite directions repel each other
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Magnetic Force Between
Two Parallel Conductors
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The result is often expressed as the
magnetic force between the two wires, FB
This can also be given as the force per
unit length,
FB m oI1I2

 2pa
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22.9 Definition of the Ampere

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The force between two parallel wires
can be used to define the ampere
When the magnitude of the force per
unit length between two long parallel
wires that carry identical currents and
are separated by 1 m is 2 x 10-7 N/m,
the current in each wire is defined to
be 1 A
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Magnetic Field of a Wire Without
Current
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A compass can be used to
detect the magnetic field
When there is no current in the
wire, there is no field due to the
current
The compass needles all point
toward the earth’s north pole
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Due to the earth’s magnetic field
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Magnetic Field of a Wire With a
Current
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The wire carries a
strong current
The compass
needles deflect in a
direction tangent to
the circle
This shows the
direction of the
magnetic field
produced by the wire
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André-Marie Ampère
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1775 –1836
Credited with the
discovery of
electromagnetism

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The relationship
between electric
currents and
magnetic fields
Died of pneumonia
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Ampere’s Law
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The product of B  ds can be evaluated for
small length elements ds on the circular path
defined by the compass needles for the long
straight wire
Ampere’s Law states that the line integral of
B  ds around any closed path equals moI
where I is the total steady current passing
through any surface bounded by the closed
path
 B  ds  m o I
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Field Due to a Long Straight Wire
with a finite radius
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Want to calculate the
magnetic field at a
distance r from the
center of a wire carrying
a steady current I
The current is uniformly
distributed through the
cross section of the wire
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Field Due to a Long Straight Wire
- From Ampere’s law

Outside of the wire, r > R
 B  ds  B(2p r )  m I
o
mo I
B
2p r

Inside the wire, we need I’, the current
inside the amperian circle
 B  ds  B(2p r )  m I '
o
 mI 
B   o 2 r
 2p R 
r2
I'  2 I
R
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Field Due to a Long Straight
Wire – Summary
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The field is
proportional to r
inside the wire
The field varies as
1/r outside the wire
Both equations are
equal at r = R
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Magnetic Field of a Toroid
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Find the field at a
point at distance r
from the center of
the toroid
The toroid has N
turns of wire
 B  ds  B(2p r )  m NI
o
mo NI
B
2p r
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22.10 Magnetic Field of a
Solenoid
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A solenoid is a long wire wound in the form
of a helix
A reasonably uniform magnetic field can be
produced in the space surrounded by the
turns of the wire
Each of the turns can be modeled as a
circular loop
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The net magnetic field is the vector sum of all the
fields due to all the turns
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Ideal Solenoid –
Characteristics
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An ideal solenoid is
approached when
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The turns are closely
spaced
The length is much
greater than the radius of
the turns
For an ideal solenoid,
the field outside of
solenoid is negligible
The field inside is
uniform
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Ampere’s Law Applied
to a Solenoid, cont

Applying Ampere’s Law gives
 B  ds  
B  ds  B
path1


ds  B
path1
The total current through the
rectangular path equals the current
through each turn multiplied by the
number of turns
 B  ds  B
 mo NI
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Magnetic Field of
a Solenoid, final
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Solving Ampere’s Law for the magnetic
field is
N
B  mo I  mo nI
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n = N / l is the number of turns per unit
length
This is valid only at points near the
center of a very long solenoid
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Magnetic Field of a
Solenoid with a finite length
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The field distribution is similar to that
of a bar magnet
As the length of the solenoid
increases
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The field lines in the interior are
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The interior field becomes more uniform
The exterior field becomes weaker
Approximately parallel to each other
Uniformly distributed
Close together
This indicates the field is strong and
almost uniform
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Magnetic field lines of a solenoid
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22.11 Magnetic Moment –
Bohr Atom
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The electrons move in
circular orbits
The orbiting electron
constitutes a tiny current loop
The magnetic moment of the
electron is associated with
this orbital motion
The angular momentum and
magnetic moment are in
opposite directions due to the
electron’s negative charge
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Magnetic Moments
of Multiple Electrons
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In most substances, the magnetic
moment of one electron is canceled by
that of another electron orbiting in the
opposite direction
The net result is that the magnetic effect
produced by the orbital motion of the
electrons is either zero or very small
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Electron Spin
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Electrons (and other particles) have an
intrinsic property called spin that also
contributes to its magnetic moment
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The electron is not physically spinning
It has an intrinsic angular momentum as if
it were spinning
Spin angular momentum is actually a
relativistic effect
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Electron Magnetic Moment
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In atoms with multiple
electrons, many electrons
are paired up with their spins
in opposite directions
 The spin magnetic
moments cancel
Those with an “odd” electron
will have a net moment
Some moments are given in
the table
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Ferromagnetic Materials
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Some examples of ferromagnetic materials are
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Iron
Cobalt
Nickel
Gadolinium
Dysprosium
They contain permanent atomic magnetic
moments that tend to align parallel to each
other even in a weak external magnetic field
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Domains
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All ferromagnetic materials are made up
of microscopic regions called domains
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The domain is an area within which all
magnetic moments are aligned
The boundaries between various
domains having different orientations
are called domain walls
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Domains,
Unmagnetized Material
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The magnetic moments
in the domains are
randomly aligned
The net magnetic
moment is zero
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Domains,
External Field Applied
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A sample is placed in an
external magnetic field
The size of the domains
with magnetic moments
aligned with the field
grows
The sample is
magnetized
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Domains,
External Field Applied, cont
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The material is placed in
a stronger field
The domains not aligned
with the field become
very small
When the external field
is removed, the material
may retain most of its
magnetism
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22.12 Magnetic Levitation
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The Electromagnetic System (EMS) is
one design model for magnetic
levitation
The magnets supporting the vehicle are
located below the track because the
attractive force between these magnets
and those in the track lift the vehicle
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EMS
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German Transrapid –
EMS Example
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Close surface and line integrals
of Electric and Magnetic fields

Electric field, E
Close Surface
Integral
Magnetic field,
  qin
S E  dA  0

B
？
(Gauss Law)
Close Line
Integral
？
 
 B  dl  μ0 I tot
C
(Ampere’s Law)
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Line integral of electric field along
a close path
Potential difference between a and b points,
Vb  Va  
b
a
 
E  dl
For a close path, a=b and Va=Vb.
 
 E  dl  0
c
The electric force is
conservative.
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Magnetic Flux
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The magnetic flux
is expressed as
The flux depends on
the magnetic field
and the area:
 B   B  dA
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Surface integral of magnetic field
around any close surfaces
 
 B  dA  0
S
No magnetic monopole is found so far!
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Close surface and line integrals
of Electric and Magnetic fields
Steady state only

Electric field, E
Close Surface
Integral
  qin
S E  dA  0
(Gauss Law)
Close Line
Integral
 
 E  dl  0
c
The electric force
is conservative.
Magnetic field,

B
 
 B  dA  0
S
No magnetic monopole!
 
 B  dl  μ0 I tot
C
(Ampere’s Law)
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Exercises of Chapter 22
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5, 8, 14, 18, 20, 22, 23, 29, 35, 38, 44,
49, 55, 68, 69
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