Download Lecture 9 - Cornell University

Lecture 9
Magnetoquasistatics
In this lecture you will learn:
• Basic Equations of Magnetoquasistatics
• The Vector Potential
• The Vector Poisson’s Equation
• The Biot-Savart Law
• Magnetic Field of Some Simple Current Carrying Elements
• The Magnetic Current Dipole
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Equations of Magnetoquasistatics
Equations of Electroquasistatics
Equations of Magnetoquasistatics
r
r
∇. ε o E = ρ (r , t )
r
∇×E = 0
r
r r ∂ε E
∇×H = J + o
∂t
r
∇ . µo H = 0
r r r
∇ × H = J (r , t )
r
r
∂ µoH
∇×E = −
∂t
• Electric fields are produced by only
electric charges
• Magnetic fields are produced by only
electric currents
• Once the electric field is determined,
the magnetic field can be found by the
last equation above
• Once the magnetic field is determined,
the electric field can be found by the last
equation above
• Currents in magnetoquasistatics are
solenoidal (i.e. with zero divergence)
(
)
r r
r
∇ . J (r , t ) = ∇ . ∇ × H = 0
In magnetoquasistatics the source of the magnetic field is electrical current
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
1
Ampere’s Law for Magnetoquasistatics
r
r r
r
∫ H . ds = ∫∫ J . da
r r
∇×H = J
electric current
density
A closed contour
Ampere’s Law: The line integral of magnetic field over a closed contour is equal
to the total current flowing through that contour
Right Hand Rule: The positive directions for the surface
normal vector and of the contour are related by the right
hand rule
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Magnetic Field of an Infinite Line-Current
Consider an infinitely long line-current carrying a total current I in the +z-direction,
as shown below
Use ampere’s law on the closed contour shown
by the dashed line:
line current
r r
r
r
∫ H . ds = ∫∫ J . da
⇒ (2π r ) Hφ (r ) = I
⇒
Hφ (r ) =
I
2π r
y
r
ds
r
x
Working in the cylindrical
coordinates
Magnetic field is entirely in the φˆ direction and
falls off as ~1/r from the line-current
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
2
Magnetic Field of a Solenoid
H
Consider a solenoid with N turns per unit
length and carrying a current I
Assumptions:
• The magnetic field inside the solenoid is
uniform and strong
I
• There is a fringing field outside the solenoid
which is very weak and may be neglected
I
Hy
r
r r
r
∫ H . ds = ∫∫ J . da
⇒ L H y = (LN ) I
y
L
⇒
H y = NI
x
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
The Vector Potential - I
A Vector Identity:
r
For any vector F the divergence of the curl is always zero:
(
)
r
∇. ∇×F = 0
The Vector Potential:
In magnetoquasistatics the divergence of the B-field is always zero:
( )
(
)
r
r
∇ . B = ∇ . µo H = 0
So one may represent the B-field as the curl of another vector:
r
r
r
B = µo H = ∇ × A
r
A is called the vector potential
⇒
(
)
r
r
r
∇ . B = ∇ . µo H = ∇ . ∇ × A = 0
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
3
The Vector Potential - II
In electroquasistatics we had:
r
∇×E = 0
r
E = −∇ φ
Therefore we could represent the E-field by the scalar potential:
( )
(
)
r
r
∇ . B = ∇ . µo H = 0
In magnetoquasistatics we have:
Therefore we can represent the B-field by the vector potential:
r
r
r
B = µo H = ∇ × A
A vector field can be specified (up to a constant) by specifying its curl and its
divergence
r
Our definition of the vector potential A is not yet unique – we have only
specified its curl
r
For simplicity we fix the divergence of the vector potential A to be zero:
r
∇.A = 0
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Differential Equation for The Vector Potential
Start from Ampere’s law in differential form:
r
r r
∇×H = J
r
Use: µo H = ∇ × A
(
A Vector Identity:
(
)
r
r
∇ × ∇ × A = µo J
To get:
(
)
(
)
r
r
r
∇ × ∇ × F = ∇ ∇ . F − ∇ 2F
r
)
r
r
r
∇ ∇ . A − ∇ 2 A = µo J
⇒
r
r
∇ 2 A = − µo J
Use: ∇ . A = 0
The differential equation for the
vector potential (also called the
Vector Poisson’s Equation)
r
This is in fact 3 different equations (one for each component of A )
∇ 2 Ax = − µo J x
∇ 2 Ay = − µo J y
∇ 2 Az = − µo J z
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
4
The Superposition Principle for The Vector Potential
r r
• Supposerforr the current distribution J1(r ) we have found the vector
potential A1(r )
r r
• Supposerforrsome other current distribution J 2 (r ) we have also found the vector
potential A2 (r )
(
r r
r r
)
• Then the vector
r rpotential
r r A1(r ) + A2 (r ) is the solution for the current
distribution J1(r ) + J 2 (r )
(
)
Proof:
r r
r r
∇ 2 A1(r ) = − µo J1(r )
+
r r
r r
∇ 2 A2 (r ) = − µo J 2 (r )
=
(
)
(
)
r r
r r
r r
r r
∇ A1(r ) + A2 (r ) = − µo J1(r ) + J 2 (r )
2
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Recall the Superposition Integral for the Potential
In the most general scenario, one has to solve the Poisson equation:
r
r
ρ (r )
∇ 2 φ (r ) = −
r r
r − r'
εo
r
r'
We know that the solution for a point charge
sitting at the origin:
r
φ (r ) =
r
r
q
4π ε o r
r
ρ (r ')
To find the potential at any point one can sum up the contributions from
different portions of a charge distribution treating each as a point charge
r
φ (r ) = ∫∫∫
r
ρ (r ')
r r dV '
4π ε o r − r '
dV ' = dx ' dy ' dz '
A formal solution of the vector differential equation is the vector superposition
integral:
r
r
∇ 2 A = − µo J
r r
r r
µo J (r ')
A(r ) = ∫∫∫
r r dV '
4π r − r '
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
5
The Biot-Savart Law - I
Start from the superposition integral for the vector potential:
r r
r r
µo J (r ')
A(r ) = ∫∫∫
r r dV '
4π r − r '
Now find the magnetic field:
r
r ∇×A
H=
µo
r r
r r
⎡
⎤
J (r ' )
H (r ) = ∇ × ⎢ ∫∫∫
r r dV '⎥
4
r
−
r
'
π
⎣
⎦
r r
⎛ 1 ⎞ r r
1
∇ ⎜⎜ r r ⎟⎟ × J (r ') dV '
H (r ) = ∫∫∫
4π
⎝ r − r' ⎠
r r
r r
J (r ') × nˆ rr '→ rr
H (r ) = ∫∫∫
r r 2 dV '
4π r − r '
A vector Identity:
( )
(
)
r
r
r
∇ × ϕ F = ϕ ∇ × F + (∇ϕ ) × F
rˆ
⎛ 1⎞
r
⎝ ⎠
r
r r
(
1 ⎞
r − r ')
nr r
r r ⎟ = − r r 3 = − r r '→r r2
r − r ' ⎟⎠
r − r'
r − r'
Recall that: ∇ ⎜ ⎟ = − r 2
⎛
∇ ⎜⎜
⎝
Biot-Savart Law
r
nˆ rr '→ rr is a unit vector directed from point r ' on the current source
r
to the observation point r
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
The Biot-Savart Law - II
nˆ rr '→ rr
Biot-Savart Law:
r r
r r
J (r ') × nˆ rr '→ rr
H (r ) = ∫∫∫
r r 2 dV '
4π r − r '
r r
r − r'
P
r
r
r
r'
r r
J (r ')
r
nˆ rr '→ rr is a unit vector directed from point r ' on the current source
r
to the observation point r
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
6
Biot-Savart Law for Line-Currents
Need to find a formula that gives the total
magnetic field at a point due to a current
r
carrying wire as a superposition of magnetic ds '
field contributions from all the small pieces
I
of the wire
y
nˆ sr '→ rr
r
r
r
s'
x
z
line-current
carrying a current I
Start from the Biot-Savart law:
r r
r r
J (r ' ) × nˆ rr '→ rr
H (r ) = ∫∫∫
r r 2 dV '
4π r − r '
r r
r r
I (r ') × nˆ rr '→ rr
J (r ') × nˆ rr '→ rr
= ∫∫∫
r r 2 da' ds ' = ∫∫∫
r r 2 ds '
4π r − r '
4π r − r '
r
I ds ' × nˆ sr '→ rr
=
∫ r r2
4π
r − s'
Integrate over the
cross-section area
of the wire to get
the total current
carried by the wire
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Magnetic Field of a Finite Line-Current
y
Consider a line-current of length L with
current I in the +x-direction.
nˆ sr '→ rr
Find the magnetic field at the point P on
−L 2
the y-axis (as shown)
dx '
Use the Biot-Savart law:
z
r
r
I ds ' × nˆ sr '→ rr
H (0, y ,0 ) =
∫ r r2
4π
r − s'
P
y
x'
r
ds ' = xˆ dx '
I
r
ds '×nˆ sr '→ rr = zˆ
L2
x
y dx '
x '2 + y 2
r r 2
r − s ' = x '2 + y 2
= zˆ
I
4π
L2
∫
−L 2
y
dx '
(x '2 + y 2 )3 2
r
As L → ∞, H (0, y ,0 ) → zˆ
= zˆ
⎡
⎤
L2
⎥
⎢
2π y ⎢ y 2 + (L 2 )2 ⎥
⎣
⎦
I
2π y
I
…..recover the previous result
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
7
Magnetic Field of a Current Loop – Near Field
y
Consider a line-current in the form of a
circular loop of radius a and carrying a
current I, as shown in the figure
r
ds '
I
P
Find the magnetic field at the point P in
the center of the loop
x
a
Use the Biot-Savart law:
r
r
I ds ' × nˆ sr '→ rr
H (0,0,0 ) =
∫ r r 2
4π
r − s'
= zˆ
r
ds ' = φˆ' a dφ '
r
ds '×nˆ sr '→ rr = zˆ a dφ '
r r 2
r − s' = a2
I 2 π a dφ '
I
= zˆ
∫
4π 0 a 2
2a
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Magnetic Field of a Current Loop – Far Field (Magnetic Dipole)
P (r , θ , φ = 0 )
Consider a line-current in the form of a
circular loop of radius a and carrying a
current I, as shown in the figure (r >> a)
Find the magnetic field at the point P far
far away from the loop
A small current loop such as this is a
magnetic dipole
z
r
θ
r
ds '
a
y
φ'
x
I
Use the superposition integral for the A-field:
r r
r
r
ds '
µ J (r ' )
µ I
A(r ,θ , φ = 0 ) = ∫∫∫ o r r dV ' = o ∫ r r
4π r − s '
4π r − s '
r
ds ' = φˆ' a dφ '
φˆ' = cos(φ ') yˆ − sin(φ ') xˆ
r r
r − s ' ≈ r − a sin(θ ) cos(φ ' )
Integrate over the
cross-section of
the wire
Be careful – tricky
integral – the unit
vector is changing
directions within
the integral
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
8
Magnetic Field of a Current Loop – Far Field (Magnetic Dipole)
r r
r
r
µ J (r ' )
µ I
ds '
A(r ,θ , φ = 0 ) = ∫∫∫ o r r dV ' = o ∫ r r
4π r − s '
4π r − s '
P (r , θ , φ = 0 )
z
µo I 2π [cos(φ ')yˆ − sin(φ ')xˆ ] a dφ '
≈
∫
r − a sin(θ ) cos(φ ')
4π 0
≈
r
y
θ
( )
µo I π a 2 sin(θ )
yˆ
4π
r2
φ'
r
ds '
a
x
I
More generally:
(
)
z
r
µo I π a 2 sin(θ )
φˆ
A(r ,θ , φ ) ≈
4π
r2
And:
r
r
∇×A
H (r , θ , φ ) =
µo
≈
( )[2 cos(θ )rˆ + sin(θ )θˆ ]
4π µ r 3
µo I π a 2
o
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Electric and Magnetic Dipoles and Dipole Moments
z
z
H
E
+q
I
−q
r
r r
∇×A
H ( r >> a ) =
r r
r
E ( r >> d ) = −∇ φ (r )
≈
qd
4πε o r
3
µo
( 2 cos(θ ) rˆ + sin(θ )θˆ )
≈
( )[2 cos(θ )rˆ + sin(θ )θˆ]
4π µ r 3
µo I π a 2
o
Electric dipole moment:
Magnetic dipole moment:
( )
r
r
p = qd
r
m = I π a 2 nˆ
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
9
Magnetic Flux and Vector Potential Line Integral
The magnetic flux λ through a surface is the surface integral of the B-field
through the surface
r r
B-field
λ = ∫∫ B . da
r r
= µo ∫∫ H . da
Since:
r
r
r
B = µo H = ∇ × A
We get:
r
r
λ = ∫∫ B . da
(
)
r
r
= ∫∫ ∇ × A . da
r
r
= ∫ A . ds
r
ds
Stoke’s Theorem
A closed contour
The magnetic flux through a surface is equal to the line-integral of the vector
potential along a closed contour bounding that surface
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
10