Post on 04-Jan-2016
Magnetic Fields Due to Magnetic Fields Due to Currents Currents
Chapter 29Chapter 29
Remember the wire?
Try to remember…
30
20 4
1
4
1
r
dq
r
dq
rd
rrE
VECTOR UNITr
r
The “Coulomb’s Law” of Magnetism
A Vector Equation
For the Magnetic Field,current “elements” create the field.
TmATm
typermeabilir
id
r
id unit
770
30
20
1026.1/104
44
:field fashion tosimilar aIn
rsrs
B
E
This is the Law ofBiot-Savart
Magnetic Field of a Straight Wire
We intimated via magnets that the Magnetic field associated with a straight wire seemed to vary with 1/d.
We can now PROVE this!
From the Past
Using Magnets
Right-hand rule: Grasp the element in your right hand with your extended thumb pointing in the direction of the current. Your fingers will then naturally curl around in the direction of the magnetic field lines due to that element.
Let’s Calculate the FIELD
Note:
For ALL current elementsin the wire:
ds X r
is into the page
The Details
02
0
20
)sin(
2B
it. DOUBLE and to0 from integrate
wesoamount equalan scontribute
wire theofportion Negative
)sin(
4
r
dsi
r
idsdB
Moving right along
R
i
Rs
RdsiB
SoRs
R
Rsr
22
)sin(sin
0
02/322
0
22
22
1/d
Center of a Circular Arc of a Wire carrying current
More arc…
Cpoint at 4
44
44
0
0
0
02
0
20
20
R
iB
dR
i
R
iRddBB
R
iRd
R
idsdB
Rdds ds
The overall field from a circular current loop
Iron
Howya Do Dat??
0rsd
0rsd
No Field at C
Cpoint at 4
)2/(0
R
iB
Force Between Two Current Carrying Straight Parallel Conductors
Wire “a” createsa field at wire “b”
Current in wire “b” sees aforce because it is movingin the magnetic field of “a”.
The Calculation
d
iLi
iFd
iB
ba
b
a
2F
angles...right at are and Since
2
:calculatedjust what weis a"" wire
todue b"" at wire FIELD The
0
b""on
0b""at
BL
BL
Invisible Summary
Biot-Savart Law (Field produced by wires)Centre of a wire loop radius RCentre of a tight Wire Coil with N turnsDistance a from long straight wire
Force between two wires
a
II
l
F
2
210
a
IB
2
0
R
IB
20
R
NIB
20
20 ˆ
4 r
rdsB
id
Ampere’s Law The return of Gauss
Remember GAUSS’S LAW??
0enclosedq
d AESurfaceIntegral
Gauss’s Law Made calculations easier than
integration over a charge distribution.
Applied to situations of HIGH SYMMETRY.
Gaussian SURFACE had to be defined which was consistent with the geometry.
AMPERE’S Law is compared to Gauss’ Law for Magnetism!
AMPERE’S LAWby SUPERPOSITION:
We will do a LINE INTEGRATIONAround a closed path or LOOP.
Ampere’s Law
enclosedid 0 sB
USE THE RIGHT HAND RULE IN THESE CALCULATIONS
The Right Hand Rule .. AGAIN
Another Right Hand Rule
COMPARE
enclosedid 0 sB
0enclosedq
d AE
Line Integral
Surface Integral
Simple Example
Field Around a Long Straight Wire
enclosedid 0 sB
r
iB
irB
2
2
0
0
Field INSIDE a WireCarrying UNIFORM Current
The Calculation
rR
iB
andR
rii
irBdsBd
enclosed
enclosed
20
2
2
0
2
2
sB
R r
B
R
i
2
0
Procedure
Apply Ampere’s law only to highly symmetrical situations.
Superposition works. Two wires can be treated separately
and the results added (VECTORIALLY!) The individual parts of the calculation
can be handled (usually) without the use of vector calculations because of the symmetry.
THIS IS SORT OF LIKE GAUSS’s LAW
#79 The figure below shows a cross section of an infinite conducting sheet carrying a current per unit x-length of l; the current emerges perpendicularly out of the page. (a) Use the Biot–Savart law and symmetry to show that for all points P above the sheet, and all points P´ below it, the magnetic field B is parallel to the sheet and directed as shown. (b) Use Ampere's law to find B at all points P and P´.
FIRST PART
Vertical ComponentsCancel
Apply Ampere to Circuit
Infinite Extent
B
B
Li
: thereforeis loop theinsideCurrent
lengthunit per current
L
The “Math”
Infinite Extent
B
B
20
0
0
B
LBLBL
id enclosedsB
Bds=0
A Physical Solenoid
Inside the Solenoid
For an “INFINITE” (long) solenoid the previous problem and SUPERPOSITION suggests that the field OUTSIDE this
solenoid is ZERO!
More on Long Solenoid
Field is ZERO!
Field is ZERO
Field looks UNIFORM
The real thing…..
Weak Field
Stronger
Fairly Uniform field
Finite Length
Another Way
0
:
0
0
0
niB
nihBhh
id
Ampere
enclosed
sB
Application
Creation of Uniform Magnetic Field Region
Minimal field outsideexcept at the ends!
Two Coils
“Real” Helmholtz Coils
Used for experiments.
Can be aligned to cancelout the Earth’s magneticfield for critical measurements.
The Toroid
Slightly lessdense than
inner portion
The Toroid
r
NiB
so
totalNirBd
Ampere
2
turns)# (N 2
:nintegratio ofpath the
in contained coil INNER about the
only worry need Weagain.
0
0
sB
15. A wire with current i=3.00 A is shown in Fig.29-46. Two semi-infinite straight sections, both tangent to the same circle, are connected by a circular arc that has a central angle θ and runs along the circumference of the circle. The arc and the two straight sections all lie in the same plane. If B=0 at the circle's center, what is θ?
38. In Fig. 29-64, five long parallel wires in an xy plane are separated by distance d=8.00 cm , have lengths of 10.0 m, and carry identical currents of 3.00 A out of the page. Each wire experiences a magnetic force due to the other wires. In unit-vector notation, what is the net magnetic force on (a) wire 1, (b) wire 2, (c) wire 3, (d) wire 4, and (e) wire 5?