Electricity and Magnetism
Magnetic Field from Moving Charges
Lana Sheridan
De Anza College
Nov 17, 2015
Last time
• force on a wire with a current in a B-field
• torque on a wire loop in a B-field
• motors
• relating a current loop to a magnet
• magnetic dipole moment
• torque and potential energy of magnetic dipole
• magnetism of matter
Overview
• magnetic fields from moving charges
• magnetic fields around current-carrying wires
• forces between parallel wires
• Gauss’s law
Magnetic fields from moving charges and currents
We are now moving into chapter 29.
Anything with a magnet moment creates a magnetic field.
This is a logical consequence of Newton’s third law.
Magnetic fields from moving charges
A moving charge will interact with other magnetic poles.
This is because it has a magnetic field of its own.
The field for a moving charge is given by the Biot-Savart Law:
B=
µ0 q v × r̂
4π r 2
Magnetic fields from moving charges
B=
1
Figure from rakeshkapoor.us.
µ0 q v × r̂
4π r 2
Magnetic fields from currents
B=
µ0 q v × r̂
4π r 2
We can deduce from this what the magnetic field do to the current
in a small piece of wire is.
Current is made up of moving charges!
qv = q
q
∆s
=
∆s = I∆s
∆t
∆t
We can replace q v in the equation above.
This element of current creates a
Magnetic fields from currents
magnetic field at P, into the page.
ids
ds
A current-length element
a differential magnetic
int P. The green ! (the
w) at the dot for point P
:
dB is directed into the
i
θ
ˆr
r
P
d B (into
page)
Current
distribution
This is another version of the Biot-Savart Law:
Bseg =
µ0 I ∆s × r̂
4π r 2
where Bseg is the magnetic field from a small segment of wire, of
length ∆s.
ummation
Magnetic fields
fromfield
currents
The magnetic
vector
because of
at any point is tangent to
Magnetic field
around a wire segment, viewed end-on:
is a scalar,
a circle.
being the
Wire with current
into the page
a currentB
(29-1)
that points
y constant,
B
(29-2)
f the cross
The magnetic field lines produced by a current in a long straight wire
Fig. 29-2
HALLIDAY REVISED
Magnetic fields from currents
to determine
direction of the field lines (right-hand rule):
GNETICHow
FIELDS
DUE TOthe
CURRENTS
es the dio a curFig. 29-2,
:
eld B at
perpennd dion of the
b) If the
to the
hed rathe page,
i
B
B
i
(a )
The thumb is
current's dire
The fingers re
the field vecto
direction, whi
tangent to a c
(b )
Here is a simple right-hand rule for finding the direction of the mag
set up by a current-length element, such as a section of a long wire:
Right-hand rule: Grasp the element in your right hand with your extended th
Magnetic field from a long straight wire
The Biot-Savart Law,
Bseg =
µ0 I ∆s × r̂
4π r 2
implies what the magnetic field is at a perpendicular distance R
from an infinitely long straight wire:
B=
µ0 I
2πR
(The proof requires some calculus.)
Force between 2 wires
With two current carrying wires, each creates its own magnetic
field:
µ0 I
B=
2πR
The result is that the wires interact, much like two bar magnets
producing magnetic fields would.
Force between 2 wires
Currents in opposite directions repel, currents in the same direction
attract.
1
Figure from salisbury.edu.
Force between 2 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
=
`
2πd
Force between 2 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
=
`
2πd
Where does this come from?
The force on a current carrying wire is:
F = IL × B
Force between 2 wires
µ 0 Ia
Suppose that wire a770
produces
a field:
a = 2πd FIELDS DUE TO CURRENTS
CHAPTER
29 B
MAGNETIC
The field due to a
at the position of b
creates a force on b.
29-3 Force Between Tw
a
d
Fba
b
L L
ia
ib
Ba (due to ia )
Two parallel wires carrying
currents in the same direction attract each
:
bother.
is: Ba is the magnetic field at:wire b produced by the current in wire a. Fba is the resulting force acting on wire b because it
:
0 Ia
◦
carries current in Bµ
a.
Fig. 29-9
The force on wire
F = Ib L
2πd
sin(90 )
FB
µ0 Ia Ib
=
L
2πd
Two long parallel wires carrying
shows two such wires, separated
Let us analyze the forces on thes
We seek first the force on w
current produces a magnetic fie
causes the force we seek. To fi
:
direction of the field Ba at the sit
wire b is, from Eq. 29-4,
The curled – straight right-hand
down, as Fig. 29-9 shows.
Now that we have the field
Equation 28-26 tells us that the
:
nal magnetic field Ba is
:
where L is the length vector of t
dicular to each other, and so with
Force between 2 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.
Magnetic Permeability
A constant we will need is:
µ0 = 4π × 10−7 Tm/A
µ0 is called the magnetic permeability of free space.
It arises when we look at magnetic fields because of our choice of
SI units.
Whenever we use µ0 we assume we are considering the magnetic
field to be in a vacuum or air.
µ0 is not the magnetic dipole moment µ !
Another notation coincidence.
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.
Gauss’s Law for Magnetic Fields
There is more we can say about magnetic fields.
When studying electric fields we used Gauss’s law to understand
the how the electric field looked around a point charge.
There is also Gauss’s law for magnetic fields, but it tells us
something different about magnetic fields.
Reminder about Gauss’s law for electric fields...
sics in attaining this
:
ectric
field E ofLaw
the for Electric Fields basic idea
Gauss’s
1, we considered the
. Then we simplified
ndicularGauss’s
components
law relates the electric field across a closed surface (eg. a
sphere) to the amount of net charge enclosed by the surface.
n save far more work
ematician and physi:
ing the fields dE of
nsiders a hypothetical
his Gaussian surface,
es our calculations of
e distribution. For exose the sphere with a
hen, as we discuss in
ct that
Spherical
Gaussian
surface
?
sian surface to the
c field on a Gaussian
As a limited example,
ally outward from the
mediately tells us that
E
A spherical Gaussian
surface. If the electric field vectors
are of uniform magnitude and point
radially outward at all surface points,
Fig. 23-1
726
Chapter 24
Gauss’s Law
Flux
The number of field lines that
go through the area A› is the
same as the number that go
through area A.
Normal
A
u
w› u
A›
S
w
,
E
From th
coulomb
lines pe
If the
through
the norm
that the
cross th
lar to th
A 5 ,w.
by w ! 5
areas ar
Reminder: Gauss’s Law for Electric 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
S
B
(30.19)
a, then u 5 908 and the
the plane as in Figure
maximum value).
a weber (Wb); 1 Wb 5
Magnetic flux
The magnetic flux
e plane is a
magnetic
r to the plane.
S
dA
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
the surface. X
ΦB =
a surface A is
B · ∆A
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. (This is like a
sum over the flux through many small areas.)
We can interpret it as an assertion that magnetic monopoles do
not exist.
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
a wire, or wires carrying current: the flux through any closed
surface will be zero.
Another law can: Ampère’s Law.
Summary
• field from a moving charge
• field from a current
• force between two parallel wires
• Guass’s law
Homework Halliday, Resnick, Walker:
• PREVIOUS: Ch 28, Problems: 54, 55, 57, 61, 65
• NEW: Ch 29, onward from page 783. Questions: 3; Problems:
1, 11, 21, 23