Transcript Chapter 30

Chapter 30
Sources of the Magnetic Field
A whole picture helps
Charge q as source
Current I as source
Gauss’s Law
Ampere’s Law
Faraday’s Law
Electric field E
FE  qE
Magnetic field B
Ampere-Maxwell Law
Force on q in the field
FB  qv  B
Force on qv or I in the filed
Summarized in
Maxwell equations
Lorentz force
F  qE  q v  B
Math -- review
C  AB
Vector cross product: C  A  B
Magnitude of the vector C :
C  C  AB sin  θ 
B
θ
B
A
θ
Vector cross product: D  B  A  C
Determine the direction. If
D  AB
A
FB  v  B
The right-hand rule:
1. Four fingers follow
the first vector.
2. Bend towards the
second vector.
3. Thumb points to the
resultant vector.
Sources of Electric field, magnetic field
From Coulomb's Law, a point charge dq
generates electric field distance r away
from the source:
dq
1 dq
dE 
rˆ
2
4 0 r
r
dE
From Biot-Savart's Law, a point current Ids
generates magnetic field distance r away
from the source:
dB
0 Ids
dB 
 rˆ
2
4 r
Difference:
1. A current segment Ids , not a point charge dq,
hence a vector.
2. Cross product of two vectors, dB is determined
by the right-hand rule.
P
Ids
Total Magnetic Field


dB is the field created by the current in the length
segment ds , a vector that takes the direction of the
current.
To find the total field, sum up the contributions from
all the current elements Ids
μo
B
4π


Ids  ˆr
 r2
The integral is over the entire current distribution
o = 4 x 10-7 T. m / A , is the constant o is called
the permeability of free space
Example: B from a Long, Straight
Conductor with current I




The thin, straight wire is carrying a
constant current I
Constructing the coordinate
system and place the wire along
the x-axis, and point P in the X-Y
plane.
ˆ  dx cos θ k
ˆ
ds ˆr  dx sin  π2  θ  k
Integrating over all the current
elements gives
μo I θ2
B
cos θ dθ

θ
1
4πa
μ I
ˆ
 o  sin θ1  sin θ2  , or B  Bk
4πa
The math part
ˆ  dx cos θ k
ˆ
ds ˆr  dx sin  π2  θ  k
μo Ids ˆ μoI cos 3 θ ˆ
μo I
ˆ
dB 

r

dx
k


cos
θdθ
k
4π r 2
4π a 2
4πa
a
adθ
 cos θ , x  a tan θ , dx  
r
cos 2 θ
μo I θ2
B
cos θ dθ

θ
1
4πa
μo I

 sin θ1  sin θ2 
4πa
ˆ
B  Bk
Now make the wire infinitely long


The statement “the conductor is
an infinitely long, straight wire” is
translated into:
q1 = /2 and q2 = -/2
Then the magnitude of the field
becomes
μo I
B
2πa

The direction of the field is
determined by the right-hand rule.
Also see the next slide.
For a long, straight conductor the
magnetic field B goes in circles




The magnetic field lines are
circles concentric with the wire
The field lines lie in planes
perpendicular to to wire
The magnitude of the field is
constant on any circle of
radius a
A different and more
convenient right-hand rule for
determining the direction of the
field is shown
Example: B at the center of a
circular loop of wire with current I

From Biot-Savart Law, the
field at O from Ids is
0 Ids
0 Ids ˆ
dB 
 rˆ 
k
2
2
4 r
4 r
O
μo I
μo I
μo I
B
ds 
2πr 
2
2

4πr full circle
4πr
2r

Y
This is the field at the center
of the loop
μo I
ˆ
B
or B  Bk
Off center points
2r
are not so easy
to calculate.
X
r
Ids
How about a stack of loops?
Along the axis, yes, the formula is
μo I
BN
2r
N is the number of turns.
When the loop is sufficiently long,
what can we say about the field
inside the body of this electric
magnet?
We need Gauss’s Law Oops, typo.
We need Ampere’s Law. Sorry, Mr. Ampere.
Ampere’s Law
Connects B with I

Ampere’s law states that the
line integral of B ds around
any closed path equals oI
where I is the total steady
current passing through any
surface bounded by the
closed path:
 B  ds  μoI
This is a line integral
over a vector field
B from a long, straight conductor
re-calculated using Ampere’s Law
Choose the Gauss’s Surface,
oops, not again!
Choose the Ampere’s loop, as a
circle with radius a.
Ampere’s Law says
 B  ds  μ I
o
B is parallel with ds , so
B  ds  B 2π a  μoI
μoI
B
2π a
When the wire has a size: with radius
R

Outside of the wire, r > R
 B  ds  B( 2πr )  μ
o

I
μo I
 B
2πr
Inside the wire, we need I’,
the current inside the
ampere’s circle
r2
 B  ds  B( 2πr )  μo I '  I '  R 2 I
 μ I 
B   o 2 r
 2πR 
Plot the results



The field is proportional
to r inside the wire
The field varies as 1/r
outside the wire
Both equations are
equal at r = R
Magnetic Field of a Toroid


The toroid has N turns of
wire
Find the field at a point at
distance r from the center
of the toroid (loop 1)
 B  ds  B( 2πr )  μ N I
o

μo N I
B
2πr
There is no field outside
the coil (see loop 2)
Magnetic Field of a Solenoid



A solenoid is a long wire
wound in the form of a helix
A reasonably uniform magnetic
field can be produced in the
space surrounded by the turns
of the wire
The field lines in the interior are



nearly parallel to each other
uniformly distributed
The interior of the solenoid
field people use.
Magnetic Field of a Tightly
Wound Solenoid


The field distribution is
similar to that of a bar
magnet
As the length of the
solenoid increases


the interior field becomes
more uniform
the exterior field
becomes weaker
Ideal (infinitely long) Solenoid

An ideal solenoid is
approached when:



the turns are closely spaced
the length is much greater than
the radius of the turns
Apply Ampere’s Law to loop 2:
 B  ds  
B  ds  B
 B  ds  B
 o NI
path 1
B  μo
N

ds  B
path 1
I  μo n I
n = N / ℓ is the number of turns per unit length
Magnetic Force Between Two Parallel
Conductors



Two parallel wires each carry
steady currents
The field B 2 due to the current in
wire 2 exerts a force on wire 1 of
F 1 = I1 ℓ B 2
Substituting the equation for
gives
μo I1 I 2
F1 

2π a
Check with right-hand rule:


same direction currents attract
each other
opposite directions currents
repel each other
The force per unit length on the wire is
FB
μo I1 I 2

2π a
And this formula defines the current unit Ampere.
PLAY
ACTIVE FIGURE
Definition of the Ampere and
the Coulomb




The force between two parallel wires is used to
define the ampere
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
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 second is 1 C
Magnetic Flux, defined the same way as
any other vector flux, like the electric flux,
the water (velocity) flux

The magnetic flux over a
surface area associated with a
magnetic field is defined as
 B   B  dA
unit of magnetic flux is T.m2
= Wb (weber)
The
Magnetic Flux Through a Plane



A special case is when a plane
of area A, its direction dA
makes an angle q with B
The magnetic flux is
B = BA cos q
When the field is perpendicular
to the plane,  = 0 (figure a)
When the field is parallel to the
plane,  = BA (figure b)
PLAY
ACTIVE FIGURE
Gauss’ Law in Magnetism


Yes, that is Gauss’s Law, but in Magnetism.
Magnetic fields do not begin or end at any
point


The number of lines entering a surface equals the
number of lines leaving the surface
Gauss’ law in magnetism says the
magnetic flux through any closed surface is
always zero:
 B  dA  0
Example problems
A long, straight wire carries current I. A right-angle bend is made in
the middle of the wire. The bend forms an arc of a circle of radius r.
Determine the magnetic filed at the center of the arc.
Formula to use: Biot-Savart’s Law, or more
specifically the results from the discussed two
examples:
π
μo I
μo I
2
sin θ 0 
For the straight section B 
4πr
4πr
For the arc
The final answer:
magnitude B 
direction
B
μo I
4πr 2

1 circle
4
ds 
μo I 2πr μo I

2
4πr 4
8r
μo I μo I μo I μo I  2 1 



  
4πr 8r
4πr
4r  π 2 
pointing into the page.