Download Chapter 30 Sources of the Magnetic Field

Chapter 30 Sources of the Magnetic
Field
The earliest known sources of magnetism were permanent magnets. Oersted
discovered that a compass needle was deflected by an electric current --> current
model of magnetism
30.1 The Biot-Savart Law
r
r µ0 Idl × rˆ
dB =
4πr 2
r
r
µ0 Idl × rˆ
B=∫
4πr 2
The sources of magnetic fields are electric currents and moving charges.
r
B
Due to a Current Loop
The magnetic field due to the entire loop at the center
of the loop is:
r
r µ0 I dl × rˆ µ0 I
B=
=
4π ∫ r 2
4π
2π
∫
0
Rdθ ˆ µ0 I ˆ
i=
i
R2
2R
The magnetic field at a point on the axis:
r
r µ0 I dl × rˆ µ 0 I 2π Rdθ
R
µ 0 IR 2
ˆ=
B=
=
i
4π ∫ r 2
4π ∫0 x 2 + R 2 x 2 + R 2
2 x2 + R2
(
)
(
)
3/ 2
iˆ
At great distances from the loop (magnetic dipole):
r
x >> R --> B =
2
2
ˆ ~ µ0 IR iˆ = µ0 IπR iˆ = 2 µ0 µ iˆ
i
3/ 2
2 x3
2πx 3
4πx 3
2 x2 + R2
(
µ 0 IR 2
)
Compare with the electric dipole: E =
1 q
2p
(monopole: E =
)
3
4πε 0 x 2
4πε 0 x
1
1
Example: A circular coil of radius 5.0 cm has 12 turns and lies in the x = 0 plane and
is centered at the origin. It carries a current of 4.0 A so that the direction of the coil is
along the x-axis. Find the magnetic field on the axis at (a) x = 0, (b) x = 15 cm, and (c)
x = 3 m.
r µI
4π × 10−7 (12)(4.0)
B = 0 iˆ =
= 6.03 × 10 −4
2R
2(0.05)
r
µ0 IR
R
B=
iˆ
2
2
2(x + R ) x 2 + R 2
(
)
Example: Find the torque on the magnet.
r µI
r r r
B = 0 iˆ , τ = µ × B
2R
r
B
Due to a Current in a Straight Line
∫ dB =
µ0 I
4π
∫ (a
2
dx
µI
dlˆ × rˆ = 0
2
+x
4π
)
∫ (a
2
dx
+ x2
)
a
a + x2
2
let x = a tan θ
µ0 I a 2 sec 2 θdθ µ 0 I
B=
=
(sin θ 2 − sin θ1 )
4π ∫ a 3 sec3 θ
4πa
For a infinite wire, θ1 = −
--> B =
π
2
& θ2 =
π
2
µ0 I
2πR
Example: Find the magnetic field at the center of a
square loop of edge length L = 50 cm, which carries a
current of 1.5 A.
For one edge, θ1 = −
B=
π
4
& θ2 =
π
4
µ0 I
(sin θ 2 − sin θ1 ) = µ0 I 2
4πR
4πR
Total magnetic field: B = 4
µ0 I
2
4πR
2
Example: The current is 1.7 A. Find the
magnetic field at y = 6 cm on the y-axis.
r µI
B = 0 φˆ
2πr
30.2 The Magnetic Force Between Two
Parallel Conductors
Lorentz force is more fundamental concept.
It’s not good idea to take them as two magnets or to consider
from the view point of magnetic field.
1.
Evaluate the magnetic field due to one of the two conductors.
2. Apply the Lorentz force law to obtain the magnetic force on the other conductor.
r r
r µI
r
r
r r
( B = 0 φˆ ) & ( F = IL × B --> dF = Idl × B )
2πr
dF2 = I 2 dl2
µ 0 I1
--> cannot differentiate I1 from I 2 --> mutual inductance
2πR
Current or Ampere meter:
Example: Two straight rods 50-cm long with
axes 1.5 mm apart in a current balance carry
currents of 15 A each in opposite direction.
What mass must be placed on the upper rod?
F = ILB = IL
µ0 I
= mg
2πR
30.3 Ampere’s Law
I
r
The current is the source of a magnetic field.
r µI
B = 0 φˆ ,
2πr
(
)
2π
r r
µI
µ0 I ˆ
ˆ
∫ B ⋅ dl = ∫ 2πr φ ⋅ rdφφ = ∫0 2π0 r rdφ = µ0 I
3
r r
B
∫ ⋅ dl = µ0 I -->
Ampere’s Law:
r r
B
∫ ⋅ dl = µ0 I enclosed
Example: A long, straight wire of radius R carries a current
I that is uniformly distributed over the circular cross section
of the wire. Find the magnetic field both outside and inside
the wire.
I = πR 2 J --> J =
I
πR 2
r µ0 I
r r
B
⋅
d
l
=
B
2
π
r
=
µ
I
-->
φˆ
B
=
0
∫
2πr
r r
r µ Ir
r2
r < R --> ∫ B ⋅ dl = B 2πr = µ0 I enc = µ0 Jπr 2 = µ 0 I 2 --> B = 0 2 φˆ
R
2πR
r > R -->
Toroid:
r < a -->
r
r r
B
⋅
d
l
=
B
2
π
r
=
0
-->
B
=0
∫
r > b -->
r
r r
B
⋅
d
l
=
B
2
π
r
=
0
-->
B
=0
∫
r µ0 NI
r r
B
⋅
d
l
=
B
2
π
r
=
µ
(
NI
)
-->
B
=
φˆ
0
∫
2πr
b > r > a -->
30.4 The Magnetic Field of a Solenoid
The magnetic field lines of a solenoid are identical to
that of a bar magnet.
R
r
B=
µ 0 IR
(
2 x + R2
2
θ
I
2
)
3/ 2
iˆ
x
r
µ0 R 2 Indx ˆ
dB =
i
3/ 2
2 x2 + R2
(
)
x2
r
µ0 R 2 Indx
µ0 nIR 2
ˆ
ˆ
--> B = i ∫
=i
2
2 3/ 2
2
x1 2 x + R
(
)
x2
∫ (x
x1
dx
2
+ R2
)
3/ 2
let x = R cot θ
4
θ2
r
µ nIR 2 − sin θdθ ˆ µ0 nI
(cosθ 2 − cosθ1 )
=i
B = iˆ 0
2 θ∫1 R 2
2
r
For a infinite lone solenoid: θ1 = π & θ 2 = 0 --> B = iˆµ 0 nI
Apply Ampere’s Law
Example: Find the magnetic field of a very long solenoid,
considering of n closely wound turns per unit length on a
cylinder of radius R and carrying a steady current I.
We know that B does not dependent on the distance from the
surface for the surface current.
r
BL = µ0 nIL
Binside = µ0 nIzˆ
Limitations of Ampere’s Law
Ampere’s law is useful for calculating the magnetic field only when the current is
steady and the geometry has a high degree of symmetry.
30.5 Gauss’s Law for Magnetism
Gauss’s Law Applied to Calculation of Magnetic Flux on an Open
Area:
r r
Φ B = ∫ B ⋅ dA or Φ B = BA cos θ
Example: Magnetic Flux Through a Rectangular Loop
ΦB =
c+ a b
µ0 I
∫ ∫ 2πr dzdr =
c 0
µ 0 Ib  c + a 
ln

2π  c 
Charge or electric monopole is the source for electric field.
q
-q
q -q
E=0
q
You cannot find a magnetic monopole.
r r
φm,net = ∫ B ⋅ da ∝ number _ of _ enclosed _ monopoles = Nqm + N (− qm ) = 0
5
30.6 Magnetism in Matter
Magnetization and Magnetic Susceptibility
r
r
magnetization M : the net magnetic dipole moment ( µ net ) per unit volume of the
material
r
r
r
r m
µ net dµ net
=
M = =
V
V
dV
Current Loop Picture:
Will you be hurt by the large current
flowing in a permanent bar magnet?
dµ = Adi , M =
dµ Adi di
=
=
dV Adl dl
The applied magnetic field is used to align small magnets in
materials so the total magnetic field is higher than the applied
filed.
Atomic Magnetic Moments
Angular momentum of orbital electrons in an atom:
r r r r
r
L = r × p = r × (mv ) = mrv
Is there any relation to magnetic dipole moment?
r
µ = IA = Iπr 2 =
q
1
q
πr 2 = qrv --> µ =
L
2πr / v
2
2m
Magnetism, magnetic moment, and magnetization are in consequences of orbital
motion of charged particles.
The angular momentum and magnetic moment are in the same direction:
r
q r
µ=
L
2m
Bohr’s model: The angular momentum is quantized with a unit of h , so we express
r
r
r qh L
L
(a number). µ =
the magnetic moment in terms of n =
h
2m h
6
r
r
r
qh L − eh L
L
For electrons: µ =
=
= −µB
2m h 2me h
h
r
The unit of magnetization in atomic scale is Bohr magneton µ B =
eh
.
2me
r
The intrinsic spin angular momentum S generates the magnetic moment:
r
r
S
µ s = −2 µ B
h
If all the atoms have the same magnetic moment aligned in the same direction. The
saturated magnetization is:
r
r
r r
r
r
m N r
m = µ = Nµ s or M s = = µ s = µ s
V V
When will the magnetization be saturated?
Paramagnetism
1.
2.
r r
The potential energy of the magnetic moment in a magnetic field is U = − µ ⋅ B .
The magnetic moment of electrons in atoms is quantized to be lµ B and the
potential energy is quantized to be ml µ B B , where n can be positive and negative
integer.
3.
When you turn on the magnetic field, the magnetic moment of atoms could have
different potential energy.
Ferromagnetism
What is the exchange interaction?
Thermal energy
atom
No exchange interaction
What is a magnetic domain?
What is the Curie temperature?
Exchange interaction
A long iron rod placing in a solenoid:
B = Bapp + µ0 M = µ0 nI + µ 0 M
(M =
N
N
N
µ = IA = I = n' I )
V
V
l
7
A solenoid:
A magnet:
l
What is a hysteresis loop?
Diamagnetism
What’s the diamagnetism? --> Lorentz law
What’s the Lenz law?
The magnetic field in the superconductor:
B = (1 + χ m )Bapp = 0 --> χ m = −1
30.7 The Magnetic Field of the Earth
8