Chapter 30
Sources of the Magnetic Field
Amperes and Biot-Savart Laws
FB on a Charge Moving
in a Magnetic Field


Magnitude proportional to charge and speed
of the particle
Direction depends on the velocity of the
particle and the direction of the magnetic field


It is perpendicular to both
FB = q v x B




FB is the magnetic force
q is the charge
v is the velocity of the moving charge
B is the magnetic field
Direction given by
Right-Hand Rule

Fingers point in the
direction of v



(for positive charge;
opposite direction if
negative)
Curl fingers in the
direction of B
Then thumb points in the
direction of v x B; i.e. the
direction of FB
Hall Effect



When the charge carriers are negative, the upper
edge of the conductor becomes negatively charged
When the charge carriers are positive, the upper edge
becomes positively charged
Sign of Hall voltage, VH, gives the sign of the
charges
Biot-Savart Law – Introduction


Biot and Savart conducted experiments
on the force exerted by an electric
current on a nearby magnet
They arrived at a mathematical
expression that gives the magnetic field
at some point in space due to a current
Biot-Savart Law – Set-Up


Measured magnetic field
dB at some point P,
distance r from wire
Wire carrying a steady
current of I


Length element is ds
Biot & Savart found that dB


is proportional to current, I,
length, ds and sin()
is inversely proportional to r2
Biot-Savart Law – Equation

Observations are summarized in the
Biot-Savart law:
dB 


2
4
r
Constant o called the permeability of free
space


 0 I d s  rˆ
o = 4 x 10-7 T. m / A
The magnetic field is due to the currentcarrying conductor
Total Magnetic Field


dB is the field created by the current in the
length segment ds
To find total field, B, sum up contributions
from all current elements I.ds
0
I d s  rˆ
B 

2
4
r



The integral is over the entire current distribution
B-S Law is also valid for a current consisting
of charges flowing through space
Quick Quiz 30.1
Consider the current in the length of wire shown in the figure
below. Rank the points A, B, and C in terms of magnitude of
the magnetic field due to the current in the length element
shown, from greatest to least.
(a) A, B, C
(b) B, C, A
(c) C, B, A
(d) C, A, B
(e) An equal field applies at all these points.
Quick Quiz 30.1
Answer: (b). Point B is closest to the current element. Point
C is farther away and the field is further reduced by the sin θ
factor in the cross product d s x rˆ . The field at A is zero
because θ = 0.

B Compared to E, 1

Distance


Magnitude of magnetic field varies as
inverse square of distance from source
Magnitude of electric field due to point
charge also varies as inverse square of
distance from the charge
B Compared to E, 2

Direction


Electric field created by a point charge is
radial in direction
Magnetic field created by a current element
is perpendicular to both the length element
ds and the unit vector r̂
B Compared to E, 3

Source


Electric field is established by an isolated
electric charge
Current element producing a magnetic field
must be part of an extended current
distribution

Need to integrate over entire current
distribution
B Compared to E, 4

Ends of Field Lines

Magnetic field lines have no beginning and
no end


They form continuous circles
Electric field lines begin on positive
charges and end on negative charges
B for Long, Straight Conductor



Thin, straight wire carrying
constant current I
d s  rˆ  dx sin(  ) kˆ
Integrating over all the current
elements gives (see ex 30.1)
B 

 0I
4a
 0I
4a

2

sin(  ) d 
1
cos(  1 ) 
cos(  2 ) 
B for a Long, Straight
Conductor, Special Case

For infinitely long,
straight wire



1= 0° and 2= 1
cos(11cos(21
The field becomes
B 
 0I
2a
B for a Long, Straight
Conductor, Direction




Magnetic field lines are circles
concentric with the wire
Field lines lie in planes
perpendicular to to wire
Magnitude of B is constant on
any circle of radius a
The right-hand rule for
determining the direction of B
is shown

Grasp wire with thumb in
direction of current. Fingers
wrap in direction of B.
B for a Curved Wire Segment




What is the field at point O
due to the wire segment?
I and R are constants
ds is parallel to r along AA’
and CC’ so we only need to
calculate the field from AC.
We find that
B 


 0I

4R
 in radians
See Example 30.2

B for a Circular Loop of Wire


Put  = 2in previous result
 0I
 0I

B 
2 
4R
2R

This is the field at the centre of the loop
Demo Ec10:
Forces between parallel wires

Wires can be made to
attract and repel one
another depending on
the direction of the
current flow through
each wire.
Magnetic Force Between Two
Parallel Conductors



Two parallel wires carrying
steady currents I1 and I2
Field B2 due to the current in
wire 2 exerts a force on wire 1
of F1 = B2 I1 ℓ
However B2 = 0 I2 / 2a, so
F1 
 0 I 1I 2
2a
l

Magnetic Force Between Two
Parallel Conductors, cont.
F1 


 0 I 1I 2
2a
l
Parallel conductors carrying currents in
the same direction attract each other
Parallel conductors carrying current in
opposite directions repel each other
Quick Quiz 30.2
For I1 = 2 A and I2 = 6 A in the figure below, which is true:
(a) F1 = 3F2
(b) F1 = F2/3
(c) F1 = F2
Quick Quiz 30.2
Answer: (c). F1 = F2 as required by Newton’s third law.
Another way to arrive at this answer is to realize that
F1 
 0 I 1I 2
2a
l
gives the same result whether the multiplication of currents
is (2 A)(6 A) or (6 A)(2 A).

Quick Quiz 30.3
A loose spiral spring is hung from the ceiling, and a large
current is sent through it. The coils move
(a) closer together
(b) farther apart
(c) they do not move at all
Quick Quiz 30.3
Answer: (a). The coils act like wires carrying parallel
currents in the same direction and hence attract one another.
Definition of the Ampere

F1
l

The force between two parallel wires can be
used to define the ampere

 0 I1 I 2
2a
with
 0  4   10
7
T mA
-1
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
Definition of the Coulomb


The SI unit of charge, the coulomb, is
defined in terms of the ampere
When a conductor carries a steady
current of 1 A, the quantity of charge
that flows through a cross section of the
conductor in 1 s is 1 C
Magnetic Field of a Wire


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

Needle points towards the
Earth’s north pole
Magnetic Field of a Wire, 2

When wire carries a
current a compass
needle deflects in a
direction tangent to
the circle

Shows the direction
of the magnetic field
produced by the wire
Ampere’s Law

The line integral of B . ds around any
closed path equals µoI, where I is the
total steady current passing through any
surface bounded by the closed path.

B  ds   0 I
Ampere’s Law, cont.


Ampere’s law describes the creation of
magnetic fields by all continuous current
configurations
Direction: put thumb of right hand in the
direction of the current through loop.
Your fingers curl in the direction you
integrate around the loop.
Quick Quiz 30.4
Rank the magnitudes of  B .d s for the closed paths in the
figure below, from least to greatest.
(a) a, b, c, d
(b) b, d, a, c
(c) c, d, b, a
(d) c, b, a, d
(e) d, c, a, b

Quick Quiz 30.4

B  ds   0 I
Answer: (b). Ampere’s law (above) indicates that the value
of the line integral depends only on the net current through
each closed path. Path b encloses 1 A, path d encloses 3 A,
path a encloses 4 A, and path c encloses 6 A.

Quick Quiz 30.5
Rank the magnitudes of  B .d s for the closed paths in the
figure below, from least to greatest.
(a) a, b, c, d

(b) b, c, d, a
(c) b, d, a, c
(d) d, c, a, b
(e) The criteria are
badly chosen in this case.
Quick Quiz 30.5
Answer: (e). Ranked from least to greatest, first comes b,
then a = c = d. Paths a, c and d all give the same nonzero
value μ0I because the size and shape of the paths do not
matter. Path b does not enclose the current, and hence its line
integral is zero.
MFM04AN1: Ampere’s law for
magnetic field near a currentcarrying wire
Field Due to a Long Straight
Wire – from Ampere’s Law


Calculate the magnetic
field at a distance r from
the centre of a wire
carrying a steady
current I
The current is uniformly
distributed through the
cross section of the wire
Field Due to Long
Straight Wire
1. Outside of the wire, r > R

B  ds  B  2  r   0 I
 B 
 0I
2r
2. Inside the wire, current I' at radius r:
I  


r
2
R
2
I , so that
 r 2 
B  ds  B  2  r   0 I    0  2 I 
 R

 B 
 0I
2R
2
r
Field Due to a Long Straight Wire
– Results Summary




Field proportional to r
inside the wire
Field varies as 1/r
outside the wire
Both equations are
equal at r = R
Note: much easier to
calculate using
Ampere’s rather than
Biot-Savart law!
Magnetic Field of a Toroid


Find field at distance
r from the centre of
the toroid
Toroid has N turns
of wire

B  ds  B  2  r   0 NI
 B 
 0 NI
2r
Magnetic Field of a Solenoid



A solenoid is a long wire
wound in the form of a
helix
Reasonably uniform
magnetic field can be
produced in its interior.
Here the field lines are



approximately parallel
uniformly distributed
close together
Magnetic Field of a Tightly
Wound Solenoid


Field distribution is
similar to a bar magnet
As the length of the
solenoid increases



the interior field becomes
more uniform
the exterior field
becomes weaker
i.e. an “ideal” solenoid
Ideal Solenoid –
Characteristics

An ideal solenoid is
approached when:


turns closely spaced
length much greater
than radius of turns
Field in interior
of a Solenoid



Apply Ampere’s law
Consider rectangle with side ℓ
parallel to the interior field and
side w perpendicular to the field
The side of length ℓ inside the
solenoid contributes to the field

This is path 1 in the diagram
B  ds 
 B  d s  B  ds  B 
path 1
path 1
Magnetic Field of Solenoid

Total current through the rectangular path
equals the current through each turn
multiplied by the number of turns
 B  d s  B    o NI

Ampere’s law then gives
B  0


N
l
I   0 nI
n = N / ℓ is the number of turns per unit length
valid only at points near centre of long solenoid
Quick Quiz 30.6
Consider a solenoid that is very long compared to the radius.
Of the following choices, the most effective way to increase
the magnetic field in the interior of the solenoid is:
(a) double its length, keeping the number of turns per unit
length constant
(b) reduce its radius by half, keeping the number of turns per
unit length constant
(c) overwrapping the entire solenoid with an additional layer
of current-carrying wire
Quick Quiz 30.6
Answer: (c). The magnetic field in a very long solenoid is
independent of its length or radius. Overwrapping with an
additional layer of wire increases the number of turns per
unit length.
Ampere’s vs. Gauss’s Law




E  dA 
0
Integrals around closed path vs. closed surface.


B  ds   0 I
q
i.e. 2D vs. 3D geometrical figures
Integrals related to fundamental constant x source
of the field.

Concept of “Flux” – the flow of field lines through
a surface.
MFM05AN1: Magnetic Flux
Magnetic Flux



Magnetic flux defined in a
similar way to electric flux
Consider an area element dA
on an arbitrarily shaped
surface, and magnetic field B
The magnetic flux FB is
F

B


B .d A
Units T.m2 = Wb

Wb is a weber
Magnetic Flux
through a Plane
Suppose a plane, of area A,
makes an angle  with dA

Magnetic flux FB = BA cos 
a) When field parallel to plane,
 and FB = 0
b) When field perpendicular to
plane,  and FB = BA


maximum value
Gauss’ Law in Magnetism

Magnetic fields do not begin or end at any
point


i.e. they form closed loops, with the number of
lines entering a surface equaling the number of
lines leaving that surface
Gauss’ law in magnetism says:
F

B


B .d A  0
Would not be true if magnetic monopoles
were found!
MFA05AN2: Gauss’s law for
magnetism, FB=0
Earth’s Magnetic Field

The Earth’s magnetic
field resembles that of a
huge bar magnet buried
deep in its interior


South magnetic pole is
located near the north
geographic pole
North magnetic pole is
located near the south
geographic pole
Dip Angle of
Earth’s Magnetic Field


A compass needle free to rotate vertically as
well as horizontally will point towards the
Earth’s surface
Angle between horizontal and direction of the
magnetic field is called the dip angle

The closer the compass is moved to the magnetic
poles, the farther from horizontal its needle will be


Needle horizontal at equator, with dip angle = 0°
Needle points straight down at south magnetic pole, with dip
angle = 90°
Earth’s Magnetic Poles

Dip angle of 90° near Hudson Bay in Canada


Dip angle of -90° near Dumont Durville, Antarctica


Earth’s south magnetic pole
Earth’s north magnetic pole
The magnetic and geographic poles are not at the
same location

The difference between true north and magnetic north is
called the magnetic declination

Declination varies by location on the Earth’s surface
Source of the Earth’s
Magnetic Field


There cannot be large masses of
permanently magnetized materials
in Earth’s core since the high
temperatures there prevent
materials from retaining
permanent magnetization
Most likely source of the Earth’s
magnetic field is believed to be
convection currents in the liquid
part of the core

It is also likely related to rate of the
Earth’s rotation: dynamo action
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
Reversals of the Earth’s
Magnetic Field

The direction of the Earth’s magnetic
field reverses every few million years


Evidence of these reversals are found in
basalts resulting from volcanic activity
The origin of the reversals is not
understood
Demo Ec19: Magnetic Flux


An Earth inductor compass.
Coil is rotated at constant
speed, with its axis orientated
either parallel or anti-parallel
to the Earth’s magnetic field.
The galvanometer will show
a different deflection,
depending on the direction
the axis is pointing.
Puzzle: try to work out the
direction of the Earth’s
magnetic field?
End of Chapter
Quick Quiz 30.7
In an RC circuit, the capacitor begins to discharge. During
the discharge, in the region of space between the plates of
the capacitor, there is
(a) conduction current but no displacement current
(b) displacement current but no conduction current
(c) both conduction and displacement current
(d) no current of any type
Quick Quiz 30.7
Answer: (b). There can be no conduction current because
there is no conductor between the plates. There is a timevarying electric field because of the decreasing charge on
the plates, and the time-varying electric flux represents a
displacement current.
Quick Quiz 30.8
The capacitor in an RC circuit begins to discharge. During
the discharge, in the region of space between the plates of
the capacitor, there is
(a) an electric field but no magnetic field
(b) a magnetic field but no electric field
(c) both electric and magnetic fields
(d) no fields of any type
Quick Quiz 30.8
Answer: (c). There is a time-varying electric field because of
the decreasing charge on the plates. This time-varying
electric field produces a magnetic field.
Quick Quiz 30.9
Which material would make a better permanent magnet?
-one whose hysteresis loop looks like Figure (a)
-one whose hysteresis loop looks like Figure (b)
Quick Quiz 30.9
Answer: (a). The loop that looks like Figure 30.32a is better
because the remnant magnetization at the point
corresponding to point b in Figure 30.31 is greater.
Quick Quiz 30.10
If we wanted to cancel the Earth’s magnetic field by running
an enormous current loop around the equator, the current
loop would be directed:
(a) east to west
(b) west to east
Quick Quiz 30.10
Answer: (b). The lines of the Earth’s magnetic field enter the
planet in Hudson Bay and emerge from Antarctica; thus, the
field lines resulting from the current would have to go in the
opposite direction. Compare Figure 30.7a with Figure 30.36.