Chapter 30 Sources of the Magnetic Field. A whole picture helps Charge q as source Current I as...
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Transcript of Chapter 30 Sources of the Magnetic Field. A whole picture helps Charge q as source Current I as...
Chapter 30
Sources of the Magnetic Field
A whole picture helps
Charge q as source Current I as source
Electric field E Magnetic field B
Gauss’s Law Ampere’s Law
Force on q in the fieldForce on or I in the filed
qv
Ampere-Maxwell Law
Faraday’s Law
Summarized in
Maxwell equations
E qF E
B qF v B
q qF E v B
Lorentz force
Math -- review
C A BVector cross product:
θ
A
B
C A B
Vector cross product:
D B A C
θ
A
B
D A B
sin
C AB θC
Determine the direction. If
Magnitude of the vector :C
BF 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 away from the source:
r
2
0
1
4
dq ˆdr
E rr
dE
dq
P
From Biot-Savart's Law, a point current generates magnetic field distance away from the source:
Ids
r dB
Ids
024
Id ˆdr
sB r
Difference:1. A current segment , not a point charge dq,
hence a vector.2. Cross product of two vectors, is determined
by the right-hand rule.
Ids
dB
Total Magnetic Field
is the field created by the current in the length segment , a vector that takes the direction of the current.
To find the total field, sum up the contributions from all the current elements
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
dB
24
oμ Id
π r
s rB
Ids
ds
Example: 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.
Integrating over all the current
elements gives
Icos
I ˆsin sin
2
1
1 2
4
, or 4
θo
θ
o
μB θ dθ
πaμ
θ θ Bπa
B k
ˆ ˆˆ sin cos
2πd dx θ dx θs r k k
B
The math part
Icos ˆ ˆˆ cos
cos , tan ,cos
Icos
Isin sin
ˆ
2
1
3
2 2
1 2
4 4 4
2
4
4
o o o
θo
θ
o
μ μ I μId θd dx θdθ
π r π a πaa adθ
θ x a θ dxr θ
μB θ dθ
πaμ
θ θπa
B
sB r k k
B k
ˆ ˆˆ sin cos
2πd dx θ dx θs r k k
Now make the wire infinitely long
The statement “the conductor is an infinitely long, straight wire” is translated into:
= /2 and = -/2 Then the magnitude of the field
becomes
The direction of the field is determined by the right-hand rule. Also see the next slide.
2
IoμB
πa
For a long, straight conductor the magnetic field 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
B
Example: at the center of a circular loop of wire with current I
From Biot-Savart Law, the field at O from is
B
I I I 2 2
24 4 2
o o o
full circle
μ μ μB ds πr
πr πr r
0 02 24 4
Id Ids ˆˆdr r
sB r k
Ids
Ids
O
Y
Xr
Off center points are not so easy to calculate.
I ˆ
or 2oμ
B Br
B k
This is the field at the center of the loop
How about a stack of loops?
Along the axis, yes, the formula isI
2oμ
B Nr
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 LawConnects B with I
Ampere’s law states that the line integral of around any closed path equals oI where I is the total steady current passing through any surface bounded by the closed path:
s od μ IB
dB s
This is a line integral over a vector field
from a long, straight conductorre-calculated using Ampere’s LawB
Choose the Gauss’s Surface, oops, not again!
Choose the Ampere’s loop, as a circle with radius a.
Ampere’s Law says
s od μ IB
a
a
2
2
o
o
B ds B π μ I
μ IB
π
is parallel with , soB
ds
When the wire has a size: with radius R
Outside of the wire, r > R
Inside the wire, we need I’, the current inside the ampere’s circle
22
oo
μd B πr μ B
πr B s
I( ) I
2
2
2
2
2
o
o
rd B πr μ
Rμ
B rπR
B s
( ) I ' I ' I
I
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)
There is no field outside the coil (see loop 2)
2
2
o
o
d B πr μ N
μ NB
πr
B s
( ) I
I
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:
1 1path path
d d B ds BB s B s
od B NI B s
I Io o
NB μ μ n
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 due to the current in wire 2 exerts a force on wire 1 of F1 = I1ℓ B2
Substituting the equation for gives
Check with right-hand rule: same direction currents attract
each other opposite directions currents
repel each other
2B
PLAYACTIVE FIGURE
I I
a 1 2
1 2oμ
Fπ
The force per unit length on the wire isI I
a
1 2
2oB μF
πAnd this formula defines the current unit Ampere.
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 d B A
The unit of magnetic flux is T.m2 = Wb (weber)
Magnetic Flux Through a Plane
A special case is when a plane of area A, its direction makes an angle with
The magnetic flux is
B = BA cos When the field is perpendicular
to the plane, = 0 (figure a) When the field is parallel to the
plane, = BA (figure b)
B
PLAYACTIVE FIGURE
dA
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:
0d B A
Example problemsA 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:
I Isin 2
04 4
πo oμ μ
B θπr πr
For the straight section
For the arcI I I
14
2 2
2
4 4 4 8o o o
circle
μ μ μπrB ds
πr πr r
The final answer:I I I I
2 1
4 8 4 4 2o o o oμ μ μ μ
Bπr r πr r π
magnitude
direction pointing into the page.