Sources of the Magnetic Field
Physics 231
Lecture 8-1
Fall 2008
Magnetic Field of a Point Charge
Given a point charge, q, we know that it generates an
electric field regardless of whether it is moving or not
If the charge is in fact moving then the charge also
generates a magnetic field
The magnetic field generated by the charge however
does not behave in the same way the electric field of
the charge does
Physics 231
Lecture 8-2
Fall 2008
Magnetic Field of a Point Charge
It is found that the magnetic field is perpendicular to
both the velocity of the charge and the unit vector
from the charge to the point in question
The magnitude of the field
is given by
µ0 q v sin φ
B=
4π
r2
where φ is the angle between
the velocity and unit vector
The magnetic field lines form concentric circles about
the velocity vector
Physics 231
Lecture 8-3
Fall 2008
Magnetic Field of a Point Charge
The direction of the magnetic field is
given by one of the right-hand rules
Point your thumb, of your right hand,
in the direction of the velocity and curl
your fingers. The direction of the field
is then the direction your fingers curl.
The magnetic field for a negative
charge will be in the opposite direction.
The vector equation for this is
r µ0 q vr × rˆ
B=
4π r 2
Physics 231
Lecture 8-4
Fall 2008
Magnetic Field of a Current Element
If instead of a single charge, we have a current what is the
magnetic field then given by
We start with the principle of superposition – we add the
magnetic fields of the individual charges that make up the
current
Given n moving charges per unit
volume each carrying a charge q,
then the total charge in this
element is
dQ = n q A dl
where A is the cross sectional area
Physics 231
Lecture 8-5
Fall 2008
Magnetic Field of a Current Element
We then have for dB
µ0 dQ vd sin φ µ0 n q vd A dl sin φ
=
dB =
2
4π
4π
r
r2
But from our work on currents we also have that I = n q vd A
Therefore
Or in vector form
Physics 231
µ0 I dl sin φ
dB =
4π
r2
r
r µ0 I dl × rˆ
dB =
4π r 2
Lecture 8-6
Fall 2008
Biot-Savart Law
For a complete circuit we then use the integral form of the
previous equation
r
r µ0 I dl × rˆ
B=
4π ∫ r 2
This can be a very difficult integral to evaluate
The level of difficulty depends upon the variable chosen to
integrate over and the presence of any symmetries that can be
exploited
We will develop another, much simpler method, shortly
Physics 231
Lecture 8-7
Fall 2008
Field of a Straight Wire
We set up the problem as shown
The integral becomes
µ0 I ⎛
x dy
⎜
B=
∫
4π −a ⎜⎝ x 2 + y 2
a
(
⎞
⎟
3/ 2 ⎟
⎠
)
The result of this integral is
µ0 I
2a
B=
4π x x 2 + a 2
In the limit that a >> x
Physics 231
Lecture 8-8
µ0 I
B=
2π x
Fall 2008
Magnetic Force Between Two Current
Carrying Wires
We have two wires carrying currents I
and I’ separated by a distance r
The first wire carrying current I sets up a magnetic field
at the second wire given by
µ0 I
B=
2π r
Physics 231
Lecture 8-9
Fall 2008
Magnetic Force Between Two Current
Carrying Wires
From before we know that a wire
carrying a current that is placed in a
magnetic field will experience a force per
unit length that is given by
F
I I'
= µ0
L
2π r
This is in fact what the second
wire will experience
And from
r
r r
F = I'L× B
This force is directed towards the
other wire – the force is attractive
Physics 231
Lecture 8-10
Fall 2008
Magnetic Force Between Two Current
Carrying Wires
If we had started with the wire carrying the current I’, we
would have ended up with the same result
If the wires were carrying currents in opposite directions, then
the force between the wires would be repulsive
Physics 231
Lecture 8-11
Fall 2008
Circular Current Loop
Another circuit for which it is
worthwhile to calculate the
magnetic field is the circular
current loop
We start by utilizing the
Biot-Savart law
r
r µ0 I dl × rˆ
dB =
4π r 2
with r 2 = x 2 + a 2
Physics 231
a being the radius of the loop
Lecture 8-12
Fall 2008
Circular Current Loop
A current element at the top
will yield the magnetic field as
shown
A current element at the bottom
will yield a magnetic field of the
same magnitude, but directed
downward
The vertical components will cancel and all that will be left is a
horizontal magnetic field of incremental value
µ0 I dl
µ0 I
a dl
dB x =
cosθ =
2
2
4π x + a
4π x 2 + a 2
(
Physics 231
Lecture 8-13
)
3/ 2
Fall 2008
Circular Current Loop
The integral is over just the circumference of the loop as all
other variables are constant
Bx=
(
µ0 I a
4π x + a
2
)
2 3/ 2
∫ dl =
(
µ0 I a 2
2 x +a
2
)
2 3/ 2
If there are N loops, turns, then the result is just N times
above results
Bx=
Physics 231
µ0 N I a 2
(
2 x +a
2
)
2 3/ 2
Lecture 8-14
Fall 2008
Circular Current Loop
An important point for the field of a current loop is at the
center of the loop where x = 0
Bx =
µ0 N I
2a
A current loop consisting of N turns is often referred to as a
Solenoid
A final caution: These results for a current loop only apply on
the axis of the loop
Physics 231
Lecture 8-15
Fall 2008
Ampere’s Law
Just as we developed an easy way to calculate the electric
field in symmetric situations, Gauss’ Law, there is a similar
technique we can use for calculating magnetic fields
This is known as Ampere’s Law
r r
∫ B ⋅ dl = µ 0 I
where the integral is around a closed path and I is the
total current passing through the area bounded by
this path
Physics 231
Lecture 8-16
Fall 2008
Ampere’s Law
There is one big difference to remember between the two
methods
Gauss’ Law is an integral over a closed surface
r r Q
∫ E ⋅ dA =
ε0
Whereas Ampere’s Law is an integral over a closed path
r r
∫ B ⋅ dl = µ 0 I
Physics 231
Lecture 8-17
Fall 2008
Ampere’s Law
Given a path around which the integration is to be done,
there is a sign convention for the current
Determine the direction of the normal, as defined by the
sense of movement around the integration path
If the current is going in the same direction as the normal,
the current is considered to be positive otherwise it is
negative
Physics 231
Lecture 8-18
Fall 2008
Long Straight Wire
Current in same direction as normal of integration path
r r
∫ B ⋅ dl = µ 0 I
B ∫ dl =µ0 I
B 2π r = µ0 I
µ0 I
B=
2π r
This is the same as we derived from the Biot-Savart Law
Physics 231
Lecture 8-19
Fall 2008
Long Straight Wire
Current in opposite direction to normal of integration path
r r
∫ B ⋅ dl = µ 0 I
− B ∫ dl = − µ0 I
− B 2π r = − µ0 I
µ0 I
B=
2π r
Again this is the same as we derived from the BiotSavart Law
Physics 231
Lecture 8-20
Fall 2008
Caution
Ampere’s Law gives you only the magnetic field due to any
currents that cut through the area bounded by the integration
path
For the indicated integration
path, the result of Ampere’s Law
is zero
Physics 231
Lecture 8-21
Fall 2008
Example
Given three currents and
their associated loops
A
→
B
C
→
1) For which loop is ∫ B • dl the smallest?
A
B
C
Same
The currents that are used in Ampere’s Law are the currents
that are within the loop that is used for integration
Loop A has no current going through it whereas the other two
loops do have currents going through them, the integral for
Loop A is zero
Physics 231
Lecture 8-22
Fall 2008
Example
Given three currents and
their associated loops
A
C
B
→
→
2) Now compare loops B and C. For which loop is ∫ B • dl the
greatest?
B
C
Same
The right hand side of Ampere’s law is the same for both loops,
therefore the integral for both loops is the same
Physics 231
Lecture 8-23
Fall 2008
Example
y
A current I flows in an infinite straight
wire in the +z direction as shown. A
concentric infinite cylinder of radius R
carries current 2I in the -z direction.
a
x
x
x
x
b
x 2I
I
x
x
x
x
1) What is the magnetic field Bx(a) at point a, just outside
the cylinder as shown?
(a) Bx(a) < 0
(b) Bx(a) = 0
(c) Bx(a) > 0
The loop used for application of Ampere’s Law will be outside
the cylinder
The total current going through this loop is
the sum of the two current yielding a
current of –I. This gives a magnetic that
points in the positive x direction at a
Physics 231
Lecture 8-24
B
B
x
I
B
B
Fall 2008
Example
A current I flows in an infinite straight
wire in the +z direction as shown. A
concentric infinite cylinder of radius R
carries current 2I in the -z direction.
y
a
x
x
x
b
x
x 2I
I
x
x
x
x
2) What is the magnetic field Bx(b) at point b, just inside
the cylinder as shown?
(a) Bx(b) < 0
(b) Bx(b) = 0
(c) Bx(b) > 0
This time, the Ampere loop only encloses current I which is
in the +z direction — the loop is inside the cylinder!
The current in the cylindrical shell does not contribute to at
point b.
Physics 231
Lecture 8-25
Fall 2008
Long Cylindrical Conductor
The problem can be divided into
two regions
Outside the conductor
Inside the conductor
Physics 231
Lecture 8-26
Fall 2008
Long Cylindrical Conductor
Outside the conductor
We start with an integration path around
the conductor at an arbitrary distance r
from the central axis of the conductor
Then using Ampere’s Law
we then have
or
Physics 231
µ0 I
B=
2π r
r r
∫ B ⋅ dl = µ 0 I
B ∫ dl = B 2 π r = µ0 I
Lecture 8-27
Fall 2008
Long Cylindrical Conductor
Inside the conductor
Since we have not been told otherwise, we
assume that the current is uniformly
distributed across the cross section of the
conductor
I0
I0
We define a current density J =
=
A π R2
We now apply Ampere’s Law around an integration path
centered on the cylindrical axis and radius r < R
r r
∫ B ⋅ dl = µ 0 I
Physics 231
with I being the current going through
the area defined by the integration path
Lecture 8-28
Fall 2008
Long Cylindrical Conductor
Inside the conductor
2
2
π
r
r
I = Jπ r 2 = I 0
= I0 2
2
πR
R
r r
∫ B ⋅ dl = B 2 π r
r2
and µ0 I = µ0 I 0
R2
r
So after equating, we then have B = µ0 I 0
2π R2
The magnetic field increases linearly with the distance from
the axis of the cylinder
Physics 231
Lecture 8-29
Fall 2008
Long Cylindrical Conductor
Inside the wire: (r < a)
a
r
B = µ0 I 0
2π R2
B
Outside the wire: ( r > a )
µ0 I
B=
2π r
Physics 231
r
Lecture 8-30
Fall 2008
Solenoid
We choose a path of integration
as shown with one part of the
path inside the solenoid and one
part of the path outside
We assume that the path on the outside, cd, is
sufficiently far from the solenoid, compared to the
diameter of the solenoid, that the field is approximately
equal to zero
Physics 231
Lecture 8-31
Fall 2008
Solenoid
We use Ampere’s Law
r r
∫ B ⋅ dl = µ 0 I
First the Left-Hand-Side
r r br r c r r
∫ B ⋅ dl = ∫ B ⋅ dl + ∫ B ⋅ dl
a
d
b
a
c
d
r r
r r
+ ∫ B ⋅ dl + ∫ B ⋅ dl
aÆb: Using the right hand rule for
currents, the B field is in the same
direction as the path and it is
constant, we then have
Physics 231
Lecture 8-32
b
r r
∫ B ⋅ dl = BL
a
Fall 2008
Solenoid
bÆc: Along the part of the path
that is inside the solenoid, B is
perpendicular to dl, so that
contribution is zero, while for the
part of the path that is outside of
the solenoid, B ~0, so we then
c r
have
r
∫ B ⋅ dl = 0
a
The same is true for dÆa
b
r r
∫ B ⋅ dl = 0
d
cÆd: Here B is zero so we again have
d
r r
∫ B ⋅ dl = 0
c
Physics 231
Lecture 8-33
Fall 2008
Solenoid
Now for the Right-Hand-Side
The total current passing through
the area defined by the integration
path is NI where N is the number
of turns and I is the current in the
loop, so the right-hand-side is
µ0 N I
Now equating the left and right-hand-sides we have
B L = µ0 N I
Or solving for B
Physics 231
N
B = µ0 I
L
Lecture 8-34
Fall 2008
Solenoid
The magnetic field of a solenoid looks like the following with
the exact shape of the field lines being dependent upon the
the number of turns, the length, and the diameter of the
solenoid
Physics 231
Lecture 8-35
Fall 2008