Download Electricity and Magnetism Force on Parallel Wires Gauss`s Law

Electricity and Magnetism
Force on Parallel Wires
Gauss’s Law
Ampère’s Law
Lana Sheridan
De Anza College
Nov 12, 2015
Last time
• magnetic field of a moving charge
• magnetic field of a current
• the Biot-Savart law
• magnetic field around a straight wire
• magnetic field around a loop
a current loop; one side of the loop acts as a north pole (in the direction of !)
and the other side as a south pole, as suggested by the lightly drawn magnet in
the figure. If we were to place a current-carrying coil in an external magnetic
field, it would
rotate just like a bar
would.
Consider
thetend
fourto arrangements
of magnet
circular
loops of radius r or 2r ,
Warm Up Question
centered on vertical axes (perpendicular to the loops) and carrying
CHECKPOINT
3 in the directions indicated. Rank the
identical
currents
The figure here shows four arrangements of circular loops of radius r or 2r, centered on
arrangements
according to the magnitude of the net magnetic field
vertical axes (perpendicular to the loops) and carrying identical currents in the direcRank the arrangements
according
of the net
magnetic
at tions
theindicated.
dot, midway
between the
loopsto the
onmagnitude
the central
axis,
greatest
field at the dot, midway between the loops on the central axis, greatest first.
first.
(a)
(b)
A a, b, c, d
B b, c, d, a
C d, a, (b and c)
D (b and c), d, a
1
Halliday, Resnick, Walker, pg 779.
(c)
(d)
a current loop; one side of the loop acts as a north pole (in the direction of !)
and the other side as a south pole, as suggested by the lightly drawn magnet in
the figure. If we were to place a current-carrying coil in an external magnetic
field, it would
rotate just like a bar
would.
Consider
thetend
fourto arrangements
of magnet
circular
loops of radius r or 2r ,
Warm Up Question
centered on vertical axes (perpendicular to the loops) and carrying
CHECKPOINT
3 in the directions indicated. Rank the
identical
currents
The figure here shows four arrangements of circular loops of radius r or 2r, centered on
arrangements
according to the magnitude of the net magnetic field
vertical axes (perpendicular to the loops) and carrying identical currents in the direcRank the arrangements
according
of the net
magnetic
at tions
theindicated.
dot, midway
between the
loopsto the
onmagnitude
the central
axis,
greatest
field at the dot, midway between the loops on the central axis, greatest first.
first.
(a)
(b)
A a, b, c, d
B b, c, d, a
C d, a, (b and c)
←
D (b and c), d, a
1
Halliday, Resnick, Walker, pg 779.
(c)
(d)
Overview
• Forces between current-carrying wires
• Gauss’s Law for magnetism
• Ampère’s Law
• B-field outside and inside a wire
• (Solenoids)
Two Wires carrying Current
With two long, straight current carrying wires, each creates its own
magnetic field:
µ0 I
B=
2πa
The result is that the wires interact, much like two bar magnets
producing magnetic fields would.
Forces on parallel wires
Currents in opposite directions repel, currents in the same direction
attract.
1
Figure from salisbury.edu.
Forces on parallel wires
To find the magnitude of the force, we need to recall how force
relates to the magnetic field for a wire.
The force on a long, straight current carrying wire is:
F = IL × B
Forces on parallel
wires
770
CHAPTER 29 MAGNETIC FIELDS DUE TO C
Suppose that wire a produces a field at b: B = µ0 Ia
a
The field due to a
at the position of b
creates a force on b.
2πd
29-3 Force
a
d
Fba
b
L L
ia
ib
Ba (due to ia )
The force on wire
b is: Two parallel wires carrying
Fig. 29-9
currents in the same µ
direction
attract
each
0 Ia
◦
:
Fb =
Ib LB
×a isBthe
Ib L
sin(90
) [toward
a =magnetic
other.
field
at
wire
b
pro- a]
2πd
:
duced by the current in wire a. Fba is the resulting force acting on wire b because it
:
carries current in Ba. µ0 Ia Ib
FB =
2πd
L
Two long parall
shows two such
Let us analyze th
We seek firs
current produce
causes the force
direction of the
wire b is, from E
The curled – stra
down, as Fig. 29Now that w
Equation 28-26
Forces on parallel wires
It is a bit more intuitive to think about the force per unit length on
the wires (since longer wires will experience larger forces).
The force per unit length on a wire due to another parallel wire at
a distance d:
FB
µ0 I1 I2
=
L
2πd
Forces on parallel wires
1
Figure from Stonebrook Physics ic.sunysb.edu.
! 10 6g, and then launches it with a speed of 10 km/s, all within 1 ms. S
Question
rail
guns may be used to launch materials into space from mining opera
he Moon or an asteroid.
The figure here shows three long, straight, parallel, equally spaced
CHECKPOINT 1
wires with identical currents either into or out of the page. Rank
he figure
threeto
long,
parallel,
spaced
thehere
wiresshows
according
thestraight,
magnitude
of theequally
force on
each wires
due towith iden
Rank
the
wires
according
to
the magnitud
urrentsthe
either
into
or
out
of
the
page.
currents in the other two wires, greatest first.
he force on each due to the currents in the other two wires, greatest first.
a
b
c
-4 Ampere’s
Law
B b, c, a
A a, b, c
C c,
a electric field due to any distribution of charges by first w
can find
theb,net
:
differential electric field dE due to a charge element and then summin
:
tributions of dE from all the elements. However, if the distribution is co
ed, we may have to use a computer. Recall, however, that if the distrib
1
Halliday, Resnick, Walker, pg 771.
! 10 6g, and then launches it with a speed of 10 km/s, all within 1 ms. S
Question
rail
guns may be used to launch materials into space from mining opera
he Moon or an asteroid.
The figure here shows three long, straight, parallel, equally spaced
CHECKPOINT 1
wires with identical currents either into or out of the page. Rank
he figure
threeto
long,
parallel,
spaced
thehere
wiresshows
according
thestraight,
magnitude
of theequally
force on
each wires
due towith iden
Rank
the
wires
according
to
the magnitud
urrentsthe
either
into
or
out
of
the
page.
currents in the other two wires, greatest first.
he force on each due to the currents in the other two wires, greatest first.
a
b
c
-4 Ampere’s
B b, c, a ←Law
A a, b, c
C c,
a electric field due to any distribution of charges by first w
can find
theb,net
:
differential electric field dE due to a charge element and then summin
:
tributions of dE from all the elements. However, if the distribution is co
ed, we may have to use a computer. Recall, however, that if the distrib
1
Halliday, Resnick, Walker, pg 771.
Definition of the Ampère (Amp)
This relation:
FB
µ0 I1 I2
=
L
2πd
gives us the formal definition of the Ampère.
Ampère Unit
Two long parallel wires separated by 1 m are said to each carry 1 A
of current when the force per unit length on each wire is
2 × 10−7 N/m.
B-Field around a wire revisited
Using the Biot-Savart law we found that the field around an
infinitely long straight wire, carrying a current I was:
B=
µ0 I
2πa
at a distance a from the wire.
Is the another way we could have solved this problem?
When we had an infinite line of charge, there was a law we could
use to find the E-field...
B-Field around a wire revisited
Using the Biot-Savart law we found that the field around an
infinitely long straight wire, carrying a current I was:
B=
µ0 I
2πa
at a distance a from the wire.
Is the another way we could have solved this problem?
When we had an infinite line of charge, there was a law we could
use to find the E-field... Gauss’s law.
Could we use something similar here?
Gauss’s Law for Magnetic Fields
Gauss’s Law for Electric fields:
I
qenc
ΦE = E · dA =
0
The electric flux through a closed surface is equal to the charge
enclosed by the surface, divided by 0 .
There is a similar expression for magnetic flux!
First we must define magnetic flux, ΦB .
Magnetic(30.18)
Flux
Definition of magnetic flux
S
field B that makes an
s case is
(30.19)
a, then u 5 908 and the
the plane as in Figure
maximum value).
a weber Magnetic
(Wb); 1 Wb
5
flux
The magnetic flux
e plane is a
magnetic
r to the plane.
S
dA
S
B
u
S
dA
Figure 30.19 The magnetic
flux through an area element dA
S
S
is B ? d A 5 B dA cos u, where
Sa magnetic field through
of
d A is a vector perpendicular to
Z
the surface.
a surface A is
ΦB = B · dA
Units: Tm2
If the surface is a flat plane and B is uniform, that just reduces to:
ΦB = B · A
Gauss’s Law for Magnetic Fields
Gauss’s Law for magnetic fields.:
I
B · dA = 0
Where the integral is taken over a closed surface A.
We can interpret it as an assertion that magnetic monopoles do
not exist.
In differential form:
∇·B=0
The magnetic field has no sources or sinks.
Gauss’s Law for Magnetic Fields
I
32-3 INDUCED MAGNETIC
FIELDS
B · dA = 0
ore complicated than
e does not enclose the
f Fig. 32-4 encloses no
ux through it is zero.
only the north pole of
el S. However, a south
ace because magnetic
like one piece of the
I encloses a magnetic
ttom faces and curved
s B of the uniform and
A and B are arbitrary
s of the magnetic flux
B
Surface II
N
S
Surface I
PA R T 3
863
B-Field around a wire revisited
Gauss’s law will not help us find the strength of the B-field around
the wire: the flux through any closed surface will be zero.
Another law can.
Ampère’s
Law
distribution
of currents
due to aFor
current-length
constant currents
l the elements.Again we
However, if the distribre’s law to find the magan be derived from the
André-Marie Ampère
owever, the law actually
(magnetostatics):
I
B · ds = µ0 Ienc
The line integral of the magnetic field around a closed loop is
proportional to the current that flows through the loop.1
Only the currents
encircled by the
loop are used in
Ampere's law.
(29-14)
.
:
product B ! ds: is to be
p. The current ienc is the
Amperian
loop
i1
its integral, let us first
θ
i3
B
The figure shows cross
ds
i2, and i3 either directly
i2
op lying in the plane of
Direction of
. The counterclockwise
integration
sen direction of integraFig. 29-11 Ampere’s law applied to an
1
That is, the current
thatAmperian
flows through
anyencircles
surface two
bounded by the loop.
arbitrary
loop that
772
Ampère’s Law
CHAPTER 29 MAGNETIC FIELDS DUE T
This is how to assign a
sign to a current used in
Ampere's law.
+i1
–i2
Direction of
integration
direction of
Fig. 29-11, th
are perpend
current is in
be in that pl
In Fig. 29-11
The sca
Thus, Ampe
We can now
A right-hand rule for
of the Ampe
A current through
thelaw,
loop
the general
Ampere’s
to in
determine
thedirection
signs forof your we can inte
currents
byaanplus
Amperian
outstretched
thumbencircled
is assigned
sign, andloop.
a current generally
around the e
The situation
of Fig.
in the opposite
direction isisthat
assigned
a 29-11.
minus sign.
Fig. 29-12
When w
CHECKPOINT 2
Question
The figure here shows three equal currents
i (two
parallel
one antiparalThe
figure here
shows
three
equaland
currents
i (two parallel and one
(29-19)
lel)
and
four
Amperian
loops.
Rank
antiparallel) and four
Amperian
loops.
Rank
thethe
loops according to
H according to the magnitude
loops
of
the
magnitude
of
B
·
ds
along
each,
greatest
first.
:
’s law gives us
! B ! ds: along each, greatest first.
i
i
a
(29-20)
b
c
portional to r ,
hat Eqs. 29-17
d
i
A a, b, c, d
B d, b, c, a
C (a and b), d, c
D d, (a and c), b
1
Halliday, Resnick, Walker, page 773.
CHECKPOINT 2
Question
The figure here shows three equal currents
i (two
parallel
one antiparalThe
figure here
shows
three
equaland
currents
i (two parallel and one
(29-19)
lel)
and
four
Amperian
loops.
Rank
antiparallel) and four
Amperian
loops.
Rank
thethe
loops according to
H according to the magnitude
loops
of
the
magnitude
of
B
·
ds
along
each,
greatest
first.
:
’s law gives us
! B ! ds: along each, greatest first.
i
i
a
(29-20)
b
c
portional to r ,
hat Eqs. 29-17
d
i
A a, b, c, d
B d, b, c, a
C (a and b), d, c
D d, (a and c), b
1
←
Halliday, Resnick, Walker, page 773.
3
Ampère’s Law and the Magnetic Field from a
Current Outside a wire
You might
All of the current is
Suppose we want
to know and
the magnitude
at a B o
encircled
thus all of the magnetic fieldnitude
distance r outside
a wire.in Using
Ampère’s
is that the
is used
Ampere's
law. Law?
Wire
surface
Amperian
loop
r
i
B
ds
(θ = 0)
Using Ampere’s law to find
the magnetic field that a current i produces
outside a long straight wire of circular cross
Fig. 29-13
integration
of an encir
We can
the situatio
the integra
m0(i1 $ i2),
We sh
allow us to
Magnetic
Figure 29-
(Current i3 is not encircled by the loop.) We can then rewrite
B cos " ds a
! # (i $ i ).
Ampère’s Law and the Magnetic Field!from
might wonder why, since current i contributes to the
All of the
current is
Current
Outside
a wire You
nitude B on the left side of Eq. 29-16, it is not needed on the
0
1
2
3
encircled and thus all
is used in Ampere's law.
Wire
surface
Amperian
loop
r
i
B
ds
(θ = 0)
is that the contributions of current i3 to the magnetic field ca
integration in Eq. 29-16 is made around the full loop. In contr
of an encircled current to the magnetic field do not cancel ou
We cannot solve Eq. 29-16 for the magnitude B of the mag
Ampère’s
Law:we do not have enough information
the situation
of Fig. 29-11
the integral. However,
I we do know the outcome of the integrat
m0(i1 $ i2), the value of which is set by the net current passing th
B ·Ampere’s
ds = µ0law
Ienc
We shall now apply
to two situations in w
allow us to simplify and solve the integral, hence to find the m
Magnetic Field Outside a Long Straight Wire with Cu
29-13the
shows
a long straight
that carries
To find the B-field at a distanceFigure
r from
wire’s
centerwire
choose
a : curren
page. Equation 29-4 tells us that the magnetic field B produc
circular path of radius r .
the same magnitude at all points that are the same dista
Using Ampere’s law to find
the magnetic field that a current i produces
outside a long straight wire of circular cross
section. The Amperian loop is a concentric
circle that lies outside the wire.
Fig. 29-13
By cylindrical symmetry, everywhere along the circle B · ds is
constant.
(Current i3 is not encircled by the loop.) We can then rewrite
B cos " ds a
! # (i $ i ).
Ampère’s Law and the Magnetic Field!from
might wonder why, since current i contributes to the
All of the
current is
Current
Outside
a wire You
nitude B on the left side of Eq. 29-16, it is not needed on the
0
1
2
3
encircled and thus all
is used in Ampere's law.
Wire
surface
Amperian
loop
r
i
B
ds
(θ = 0)
is that the contributions of current i3 to the magnetic field ca
integration in Eq. 29-16 is made around the full loop. In contr
of an encircled current to the magnetic field do not cancel ou
We cannot solve Eq. 29-16 for the magnitude B of the mag
Ampère’s
Law:we do not have enough information
the situation
of Fig. 29-11
the integral. However,
I we do know the outcome of the integrat
m0(i1 $ i2), the value of which is set by the net current passing th
B ·Ampere’s
ds = µ0law
Ienc
We shall now apply
to two situations in w
allow us to simplify and solve the integral, hence to find the m
Magnetic Field Outside a Long Straight Wire with Cu
29-13the
shows
a long straight
that carries
To find the B-field at a distanceFigure
r from
wire’s
centerwire
choose
a : curren
page. Equation 29-4 tells us that the magnetic field B produc
circular path of radius r .
the same magnitude at all points that are the same dista
Using Ampere’s law to find
the magnetic field that a current i produces
outside a long straight wire of circular cross
section. The Amperian loop is a concentric
circle that lies outside the wire.
Fig. 29-13
By cylindrical symmetry, everywhere along the circle B · ds is
constant.
The magnetic field lines must form a closed loop ⇒ B · ds = B ds.
Ampère’s Law and
Magnetic
Field from
a wonder w
You might
All ofthe
the current
is
nitude
B on the left si
Current Outside encircled
a wireand thus all
is that the contributio
is used in Ampere's law.
Wire
surface
Amperian
loop
r
i
B
ds
(θ = 0)
Using Ampere’s law to find
I
the magnetic field that a current i produces
outside a long
circular cross
B straight
ds =wire
µ0 Iof
enc
section. The Amperian loop is a concentric
circle that lies outside the wire.
Fig. 29-13
B(2πr ) = µ0 I
And again we get
B=
µ0 I
2πr
integration in Eq. 29of an encircled curren
We cannot solve E
the situation of Fig. 29
the integral. However,
m0(i1 $ i2), the value of
We shall now app
allow us to simplify a
Magnetic Field Ou
Figure 29-13 shows a
page. Equation 29-4 t
the same magnitude
HALLIDAY REVISED
Ampère’s Law
and the Magnetic Field from a
Current Inside a wire
PA R T 3
We can also use Ampère’s Law in another context, where using the
773
29-4 AMPERE’S LAW
Biot-Savart Law is harder.
the wire. We can take advanAmpere’s law (Eqs. 29-14 and
ular Amperian loop of radius
e same magnitude B at every
wise, so that ds: has the direc-
Only the current encircled
by the loop is used in
Ampere's law.
B
:
Eq. 29-15 by noting that B is
:
as is ds:. Thus, B and ds: are
loop, and we shall arbitrarily
:
u between ds: and B is 0°, so
comes
ds
Wire
surface
r
i
R
Amperian
loop
s ! B(2% r).
Fig. 29-14
Using Ampere’s law to find
gment lengths ds around the
the magnetic field that a current i produces
we place the Amperian
loop inside the wire.
ce 2pr of Now
the loop.
inside a long straight wire of circular cross
H
current of Fig. 29-13. The right
section. The current is uniformly distribWe still have B · uted
ds =
2πrB, but now the current
ave
over the cross section of the wire and
through the loop
from the page. An Amperian loop
isemerges
reduced.
is drawn inside the wire.
that flows
Ampère’s Law and the Magnetic Field from a
Current Inside a wire
LLIDAY REVISED
annd
ius
ery
ec-
H
We still have B · ds = 2πrB, but now the current that flow
through the loop is reduced. PA R T 3
773
Assuming the wire has uniform
29-4 AMPERE’S LAW
resistivity, Ienc :
Only the current encircled
by the loop is used in
Ampere's law.
B
:
B is
are
ily
so
he
ds
Wire
surface
r
i
R
Amperian
loop
Using Ampere’s law to find
the magnetic field that a current i produces
inside a long straight wire of circular cross
Fig. 29-14
Ampère’s Law and the Magnetic Field from a
Current Inside a wire
LLIDAY REVISED
annd
ius
ery
ec-
H
We still have B · ds = 2πrB, but now the current that flow
through the loop is reduced. PA R T 3
773
Assuming the wire has uniform
29-4 AMPERE’S LAW
resistivity, Ienc :
Only the current encircled
by the loop is used in
Ampere's law.
B
:
B is
are
ily
so
he
ds
Ienc =
Wire
surface
r
i
R
Ampére’s Law
I
r2
B · ds = 2πrB = µ0 2 I
R
So,
Amperian
loop
Using Ampere’s law to find
the magnetic field that a current i produces
inside a long straight wire of circular cross
Fig. 29-14
πr 2
r2
I
=
I
πR 2
R2
B=
µ0 I r µ0 I r
=
2πR 2
2πR R
Ampère’s Law
For constant currents (magnetostatics):
I
B · ds = µ0 Ienc
The line integral of the magnetic field around a closed loop is
proportional to the current that flows through the loop.
Later we will extend this law to deal with the situation where the
fields / currents are changing.
Additional examples, video
Solenoids
solenoid
29-5 Solenoi
A helical coil of tightly wound wire that can carry a current.
Magnetic Field o
i
i
Fig. 29-16
A solenoid carrying current i.
We now turn our
useful. It concerns
wound helical coil
that the length of t
Figure 29-17 sh
The solenoid’s ma
turn
A single complete loop of wire in a solenoid. “This solenoid has 10
turns,” means it has 10 complete loops.
nuniform, we start with a thin ring and then sum (via integration) the
circled area.
Magnetic Field inside and around a solenoid
P
he central axis of a
ve turns are shown, as are
the solenoid. Each turn proar the solenoid’s axis, the
is directed along the axis.
ong magnetic field. Outside
e field there is very weak.
Each turn of wire locally has a circular magnetic field around it.
The fields from all the wires add together to create very dense field
lines inside the solenoid.
Magnetic Field of a solenoid
FIELDS DUE TO CURRENTS
P2
P1
Magnetic field lines for a real solenoid of finite length. The field is strong
and uniform at interior points such as P1 but relatively weak at external points such as P2.
Fig. 29-18
The wires on opposite sides (top and bottom in the picture) have
currents in opposite directions. The fields add up between them,
vidual turns (windings) that make up the solenoid. For points very close to a turn,
but cancel out outside of them.
:
the wire behaves magnetically almost like a long straight wire, and the lines of B
there are almost concentric circles. Figure 29-17 suggests that the field tends to
cancel between adjacent turns. It also suggests that, at points inside the solenoid
:
"
:
Magnetic Field
B ! ds !of" 0an
ienc , ideal solenoid
(29-21)
:
In an ideal solenoid (with infinite length) the field outside is
:
andwhere
insideB
isisuniform.
to solenoid
a capacitor!)
lenoid negligible
of Fig. 29-19,
uniform (Similar
within the
and
:
using the rectangular Amperian loop abcda. We write ! B ! ds: as
Can use an Amperian loop to find the B-field inside:
d
h
c
i
I
B · ds = µ0 Ienc
B
a
b
pplication of Ampere’s law to a section of a long ideal solenoid carrying
Amperian loop is the rectangle abcda.
"
:
Magnetic Field
B ! ds !of" 0an
ienc , ideal solenoid
(29-21)
:
In an ideal solenoid (with infinite length) the field outside is
:
andwhere
insideB
isisuniform.
to solenoid
a capacitor!)
lenoid negligible
of Fig. 29-19,
uniform (Similar
within the
and
:
using the rectangular Amperian loop abcda. We write ! B ! ds: as
Can use an Amperian loop to find the B-field inside:
d
h
c
i
I
B · ds = µ0 Ienc
B
a
b
Here,
supposelaw
there
are n of
turns
per
unit
lengthcarrying
in the
pplication
of Ampere’s
to a section
a long
ideal
solenoid
then
Inh abcda.
Amperian
loopIisenc
the=rectangle
Bh = µ0 Inh
Inside an ideal solenoid:
B = µ0 In
solenoid,
Summary
• Forces on parallel wires
• Gauss’s Law for magnetism
• Ampère’s Law
• (Solenoids)
Homework
• Collected homework 3, posted online, due on Monday, Nov 16.
Serway & Jewett:
• PREVIOUS: Ch 30, Problems: 3, 5, 9, 13, 19
• NEW: Ch 30, Problems: 21, 25, 31, 33, 34, 47