Example 1!
Ch 30 - Sources of Magnetic
Field!
Determine the magnitude
and direction of the
magnetic field at the point O
in the diagram. (Current I
flows from top to bottom,
radius of curvature = r.)!
1.)
The Biot-Savart Law!
!
!
I d s ! rˆ
dB = km
r2
magnetic field at a point
P perpendicular to
segment of wire ds &
perpendicular to unit
vector r from ds to P.!
steady
current in
direction
ds!
3.)
determines
direction to
point!
element
of wire!
Determine the direction and
total magnitude of the
magnetic field at the point P
shown here, near a long,
thin, wire.!
P
r
a
ds
I
P
dB out
The magnetic field for an infinitely long straight wire is!
B=
!
ˆr
For straight segments, ds ! r = 0.
For curved segment, ds ! r = ds, because s and r are ".
µ I ds
B= o $ 2
4# r
µ I
B = o 2 $ rd%
4#r
µo I%
B=
4#r
Example 2!
r
km= 10-7 T•m/A !
km= µo /4π , where !
µo = permeability of free space!
µo =4π x10-7 !
ds
!
µo I
2!a
ds
2.)
4.)
Magnetic Force between wires!
F = I! ! B = I!B sin "
µI
B (due to wire) = o
2#a
$µ I '
F = I1!& o 2 ), or
% 2#a (
F µo I1I2
=
!
2#a
Ampère’s Law!
! !
B • ds
If we evaluate the dot-product #
in a circle around this wire, and sum this
product over the entire path of the circle,
we get!
! !
#µ I&
B
! • d s = B ! ds = %$ 2"o r ('(2"r) = µoI
B
ds
This result is valid for any closed path that
encloses a wire conducting a steady current,
and is known as Ampère’s Law:!
Line integral,
taken around a
closed path that
surrounds the
current.!
OR, if you find that the direction of the magnetic field lines in between the two
sources are parallel to one another, it means the sources will repulse, whereas if
the direction of the magnetic field lines between the two sources are anit-parallel
to one another, it means the sources will attract one another.
! !
B
! • d s = µo I
5.)
Oersted (1820)!
7.)
Gauss & Ampère!
If the wire is grasped with the
right hand, with the thumb in the
direction of current flow, the
fingers curl around the wire in
the direction of the magnetic
field. !
The magnitude of B is the same
everywhere on a circular path
perpendicular to the wire and
centered on it. Experiments
reveal that B is proportional to I,
and inversely proportional to the
distance from the wire.!
!
! qin
" E • dA = !
o
! !
"! B • ds = µ o Ithru
6.)
8.)
Example 3!
Example 3!
Use Ampère’s Law to calculate the
magnetic field at a distance r away from
the central axis of a current-carrying wire
of cross-sectional area R and current I.
Do this for:!
i.) r>R!
i.) r<R!
The current through the face of the Amperian path will be
the fraction of current through the whole face. That is:
" !r 2 %
r2
ithru = $
I= 2I
2'
R
# !R &
Amperian
path
R
R
So Ampere’s Law yields:
r
! B • ds = µ i
B! ds = µ i
o
thru
o thru
ii.) r<R!
# r2 &
B(2"r) = µ o % 2 I (
$R '
) B=
µoI
r
2"R 2
9.)
Example 3!
11.)
Example 4!
Use Ampère’s Law to
calculate to calculate
the magnetic field
inside a toroid of N
loops, at a distance r
from the center.!
i.) r>R!
! B • ds = µ i
B! ds = µ i
o
thru
Amperian
path
o thru
B(2"r) = µ o I
µI
B= o
2"r
R
r
10.)
side view:!
I
edge view:!
12.)
side view:!
As the wire passes through that
Amperian path N times, the
current through the face will be
Ni. With that, we can write:!
! B • ds = µ I
B ! ds = µ I
Example 5!
Find the magnetic field B for an
thin, infinite sheet of current,
carrying current of linear density
J in the z direction.!
I
o
imaginary
Amperian path!
edge view:!
! B • ds = µ I
B ! ds = µ (J!)
o
o
B(2"r) = µo NI
µ NI
B= o
2"r
L
o
B(2!) = µo (J!)
µJ
B= o
2
w
13.)
current out of page!
current into page:!
Note that if the path had
been outside the toroid,
there would have been as
much current moving
inward through the face as
outward through the face. In
other words, the net current
through the face would have
been zero and the B field
outside zero.!
I
imaginary Amperian !
path!
15.)
Electromagnets & Solenoids!
A solenoid is a long wire
wound in the form of a
helix. Typically, you are
given the number of winds
per unit length, or “n.”!
I
If the turns are closely
space and the solenoid is
finite in length, the field
lines from a solenoid look
very similar to those of a
bar magnet.!
14.)
I
16.)
B Field for a Solenoid!
Inside = strong B field!
Outside = weak B field!
Between coils = 0 field!
! B • ds = µ i
B! ds = µ ( nLI )
o thru
o
BL = µ o ( nLI )
" B = µ o nI
This is the magnetic field function for the magnetic field down the central
axis of a very long coil.
17.)
Electric Flux!
L
Classically, the Amperian path for a very
long coil, used to determine the effective
magnetic field in the central region of the
coil, is as shown to the right. Below is a
“side” view.
Note that the net magnetic field
components perpendicular to the coil’s
axis are zero, and the magnetic field
strength far away from the coil is also
approximately zero, so the only
“circulation” that occurs is down the axis
of the coil. With the path length of “L”
and the amount of current passing
through the Amperian face being (nL)i,
where “n” is the number of winds-perunit-length and “nL” identifies how many
wires there are inside the Amperian
surface, we can write:
19.)
Brief review of Electric Flux through an
open surface is:!
! !
! = E•A
! = EAcos "
from side
#
!
"
!
“The net electric flux through any closed surface is
equal to the net charge inside the surface divided
by ! o .” As such, we can write:!
• • • • • • • • • • •
L
18.)
!
! q
!c = " E • dA = in
#o
20.)
Example 7!
Magnetic Flux!
#
By the same token, through an open surface we can
also write:
! !
!B = B • A
!B = BAcos "
Find the total magnetic flux
through the loop shown here.!
!
B"
! B = " B • dA
!
$ µ I'
= " & o ) ( b dx )
% 2#x (
µ Ib a + c 1
= o "
dx
2# c x
µ Ib $ a + c '
= o ln &
)
2# % c (
“The net magnetic flux through any closed surface is
equal to zero as there are no magnetic monopoles
(i.e., situations in which you can find a North Pole
that isn’t attached to a corresponding South Pole.
That is:!
!
!
!B = " B • d A = 0
dx
x
b
I
c
a
Note that being able to do magnetic flux derivations will be needed in the next
chapter on Faraday’s Law where we will be determine the time rate of change of
magnetic flux.
21.)
23.)
Example 7!
Gauss’s Laws!
Find the total magnetic flux
through the loop shown here.!
Electric field intensity depends only on the net internal charge. It is
possible to have a single charge whose field lines produce a net
electric flux through a closed surface. As such, we can write:!
!
! q
!c = " E • dA = in
#o
It isn’t possible to have a single North Pole
without a South Pole attached, which means
that magnetic field lines are continuous and
form closed loops (and magnetic field lines
created by currents don’t begin or end at any
point. Conclusion: the net magnetic flux
through any closed surface is always zero,
which means: !
b
I
c
a
!B =
22.)
!
!
" B • dA = 0
N
S
24.)
Problems w/Ampère’s Law!
d! E
=
dt
! !
! B • d s = µo I
=
(dq dt)
"o
idisplacement
"o
# id = " o
Two problems: !
1.) What if current is changing?!
2.)What if Amperian path doesn’t
enclose current?!
d! E
dt
If we include this displacement current in Ampere’s Law, we
take care of the pesky problems. Doing so yields a more
complete version of Ampere’s Law which is:
! !
B
"! • d s = µ o (I + I d )
open end
= µ o (I + " o
25.)
There is an electric field between the plates, and according
to Gauss’s Law, there is an electric flux:
!E =
!
!
q
" E • dA = #
S
o
Furthermore, the electric flux is changing in time. As such,
we can write:
d! E d
=
dt
( " E • dA) = (dq dt )
!
!
S
dt
#o
Ignoring the middle expression and noting that dq/dt is
current I (referred to as the displacement current), we can
write:
26.)
d# E
)
dt
27.)