Chapter 27 Sources of the magnetic field
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Transcript of Chapter 27 Sources of the magnetic field
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Chapter 27 Sources of the magnetic field
EF qE GF mg
BF qv B
IL B
1 2E 2
kq q ˆF rr
1 2G 2
Gm m ˆF rr
2
Gm ˆg rr
2
kq ˆE rr
Force Equation
Point Object Force
Point Object Field
BF ?
B ?
2
Gdm ˆdg rr
2
kdq ˆdE rr
Differential Field
dB ?
Are B, dB radial?Do B, dB have 1/r2 dependence?
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Magnetic field of point charges
2
kqEr
m
2max
k q vB
r
o
1k4
omk
4
2
kq ˆE rr
m
2
ˆk qv rB
r
Magnitude ~ 1/r2
Constants
o = permeability of free space = 4x 10-7 N/A2
Direction
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Right Hand Rule
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Cross product review
C A B
A
C A B sin
B
x y z
x y z
ˆ ˆ ˆi j kC A A A
B B B
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• What is the direction of the Magnetic Field at the origin?• What is the magnitude of the magnetic field at the origin?• What if the charge was orbiting the origin?
q = 4.5 C
-3.2 x 10-14 T
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Biot-Savart Law – Set-Up
• The magnetic field is dB at some point P
• The length element is dl
• The wire is carrying a steady current of I
o2
ˆId rdB4 r
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Biot-Savart Law
o2
ˆId rdB4 r
For a long Wire:
ˆ ˆ ˆi j kˆˆd r 0 dy 0 k sin dy
sin cos 0
7om
T mk 1x104 A
ds
Permeability of free space
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Magnetic field of a long wireo
2
ˆId rdB4 r
o2
IsinˆdB k dy4 r
o IB2 R
Rtany
o2
I sinˆB k dy4 r
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Magnetic field due to a straight wire
IBr
o IB2 r
o
2k 1Er 2 r
Compare with:
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Force between two parallel wires
MF I B
M2 1
FI B
o 11
IB2 R
M o 1 2F I I
2 R
R
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Force between two parallel wires
BF I B
M o 1 2F I I
2 a
If the currents are in the same direction, the force is attractive.
If the currents are in the opposite direction, the force is repulsive.
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Historical definition of the Ampere
Historical definition of the Coulomb
If 2 long parallel wires 1.0 m apart have the same current in them and the force per unit length on each wire is 2.0 x 10-7 N/m, the current is 1.0 Ampere
If the current is 1.0 Ampere, then 1.0 Coulomb is the amount of charge passing through a cross section in 1 second.
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B for a Curved Wire Segment
• Find the field at point O due to the wire segment
• I and R are constants
will be in radians4
IoμB θπR
d
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5. Determine the magnetic field at a point P located a distance x from the corner of an infinitely long wire bent at a right angle, as shown in Figure P30.5. The wire carries a steady current I.
7. The segment of wire in Figure P30.7 carries a current of I = 5.00 A, where the radius of the circular arc is R = 3.00 cm. Determine the magnitude and direction of the magnetic field at the origin.
10. A very long straight wire carries current I. In the middle of the wire a right-angle bend is made. The bend forms an arc of a circle of radius r, as shown in Figure P30.10. Determine the magnetic field at the center of the arc.
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Magnetic field due to a current loopo
2
ˆId rdB4 r
o2
IddB4 r
The perpendicular components cancel by symmetry.
o// 2
IddB cos4 r
2o o
// 322 2 2
IRIB cos d4 r 2 R x
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Helmholtz Coils
2 2
o o3 3
2 2 222 2
N IR N IRB2 R x 2 R R x
R
R
One Coil:
Two Coils:
2o
32 2 2
N IRB2 R x
o8N IB5 5R
At x~R/2
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Magnetic field due to a current loop
2o o
32 2 2
x 0
IR IB(x 0)2R2 R x
At the center:
ˆIA IAn Magnetic dipole moment vector
for a single loop:
2o o
3 32 2 2 22 2
IRB
2 R x 2 R x
When x>>R: o3B
2 x
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Comparison to an electric dipole
3 3o
p kpE4 x x
o m3 3
2kB2 x x
p 2a +
32 2 2
kpEx a
-q
q
o
32 2 2
B2 R x
om2 2
ˆ ˆId r Id rdB k4 r r
Biot-Savart Law:
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16. Two long, parallel conductors, separated by 10.0 cm, carry currents in the same direction. The first wire carries current I1 = 5.00 A and the second carries I2 = 8.00 A. (a) What is the magnitude of the magnetic field created by I1 at the location of I2? (b) What is the force per unit length exerted by I1 on I2? (c) What is the magnitude of the magnetic field created by I2 at the location of I1? (d) What is the force per length exerted by I2 on I1?
18. Two long, parallel wires are attracted to each other by a force per unit length of 320 μN/m when they are separated by a vertical distance of 0.500 m. The current in the upper wire is 20.0 A to the right. Determine the location of the line in the plane of the two wires along which the total magnetic field is zero.
63. Two long, parallel conductors carry currents in the same direction as shown in Figure P30.63. Conductor A carries a current of 150 A and is held firmly in position. Conductor B carries a current IB and is allowed to slide freely up and down (parallel to A) between a set of nonconducting guides. If the mass per unit length of conductor B is 0.100 g/cm, what value of current IB will result in equilibrium when the distance between the two conductors is 2.50 cm?
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Introduction to Ampere’s Law
Recall the definition of electric potential: b
baa
V E d
What is the value of the integral over a closed path for any electric field?
a
aaa
V E d E d ?
Let’s try the same thing for a magnetic field around a current carrying wire.
o oo
I IB d d 2 r I2 r 2 r
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Ampere’s Law
• The integral is around any closed path• The current is that passing through the surface bounded by the
path• Like Gauss’s Law, useful in finding fields for highly
symmetric problems
o enclosedB d I
This result has been shown experimentally to be true in general
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Applying Ampere’s Law
• Select a surface– Try to imagine a surface where
the electric field is constant everywhere. This is accomplished if the surface is equidistant from the charge.
– Try to find a surface such that the electric field and the normal to the surface are either perpendicular or parallel.
• Determine the charge inside the surface
• If necessary, break the integral up into pieces and sum the results.
• Select a path– Try to imagine a path where the
magnetic field is constant everywhere. This is accomplished if the surface is equidistant from the charge.
– Try to find a path such that the magnetic field and the path are either perpendicular or parallel.
• Determine the current inside the surface
• If necessary, break the integral up into pieces and sum the results.
enclosedE
o
QE dA
o enclosedB d I
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Example: Magnetic field inside a wire
o enclosedB d I
2
encl2
I rI R
2
o 2
rB 2 r IR
o
2
IrB2 R
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Example: Solenoid
o enclosedB d I
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Example: Solenoid
o enclosedB d I
b c d a
oa b c d
B d B d B d B d NI
0 0
small
oB NI o o
NB I nI
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Example: Toroid
o enclosedB d I
oB 2 r NI
oNIB2 r
Inside:
Outside:
B 0
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21. Four long, parallel conductors carry equal currents of I = 5.00 A. Figure P30.21 is an end view of the conductors. The current direction is into the page at points A and B (indicated by the crosses) and out of the page at C and D (indicated by the dots). Calculate the magnitude and direction of the magnetic field at point P, located at the center of the square of edge length 0.200 m.
29. A long cylindrical conductor of radius R carries a current I as shown in Figure P30.29. The current density J, however, is not uniform over the cross section of the conductor but is a function of the radius according to J = br, where b is a constant. Find an expression for the magnetic field B (a) at a distance r1 < R and (b) at a distance r2 > R, measured from the axis.
24. The magnetic field 40.0 cm away from a long straight wire carrying current 2.00 A is 1.00 μT. (a) At what distance is it 0.100 μT? (b) What If? At one instant, the two conductors in a long household extension cord carry equal 2.00-A currents in opposite directions. The two wires are 3.00 mm apart. Find the magnetic field 40.0 cm away from the middle of the straight cord, in the plane of the two wires. (c) At what distance is it one tenth as large? (d) The center wire in a coaxial cable carries current 2.00 A in one direction and the sheath around it carries current 2.00 A in the opposite direction. What magnetic field does the cable create at points outside?
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Magnetic Flux
• The magnetic field in this element is B• dA is a vector that is perpendicular to the surface• dA has a magnitude equal to the area dA• The magnetic flux B is
• The unit of magnetic flux is T.m2 = Wb– Wb is a weber
B d B A
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Gauss’ Law 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:
0B dB A
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Displacement Current• Ampere’s law in the original form
is valid only if any electric fields present are constant in time
• Maxwell added an additional term which includes a factor called the displacement current, Id
• The displacement current is not the current in the conductor– Conduction current will be used to
refer to current carried by a wire or other conductor
o enclosedB d I
I Ed o
dεdt
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Ampere’s Law – General Form
• Also known as the Ampere-Maxwell law
• Magnetic fields are produced both by conduction currents and by time-varying electric fields
I I I E
o d o o odd μ μ μ εdt
B
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Material Magnetism
o oB nI
Empty core magnetic field
m m oB nI
Magnetic field in the solenoid changes due to the tendency of the dipoles to align and strengthen the field
susceptability
total o m o m o m o m oB B B nI nI 1 nI K nI nI
• Ferromagnetic – Fe, Co, Ni - susceptability is large and positive• Paramagnetic – Al, Mg – susceptability is small and positive• Dimagnetic – Cu, Au, Ag, Pb – suceptabiltiy is small and negative
m mK 1 m o m oK 1
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Magnetization Vector• Defined to describe what is happening in the
material as a result of the external applied field.
m m o m oB nI B
m o
o
BM
total m o m oB 1 B K B
total o oB B M
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Dielectrics•A dielectric is a nonconducting material that, when placed between the plates of a capacitor, increases the capacitance
•Materials with Dipoles that can align with an external electric Field. Dielectrics include rubber, plastic, and waxed paper
+ ++
--
-
+ ++
---
oE
indE
oDielectric o ind
EE E E
is the Dielectric ConstantMeasure of the degree of dipole alignment in the material
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Ferromagnetism• Domains• Curie Temperature• Hysteresis
U B ~ kT
Dipoles are randomat higher temperatures
oBM C
T
Curie’s Law