Magnetic Field of Steady Currents
(Magnetostatics)
EE 141 Lecture Notes
Topic 18
Professor K. E. Oughstun
School of Engineering
College of Engineering & Mathematical Sciences
University of Vermont
2014
Motivation
Magnetic Induction Field B(r)
The force on a charged particle q moving with the velocity v is given
by the Lorentz force relation
F(r) = Fe (r) + qv(r) × B(r),
(1)
where Fe (r) = qE(r) is the electrostatic force. The magnetic part of
the force
Fm (r) = qv(r) × B(r)
(2)
then serves to define the magnetic induction vector B(r).
In MKSA Units, the magnetic induction field strength is in
Webers/m2 (Wb/m2 ) ≡ Tesla (T ), where [from Eq. (2)]
T ≡ Wb/m2 = N/(C · m/s) = N/(A · m)
so that
Wb = N · m/A = J/A.
Biot-Savart Law
If d ~ℓ is a differential element of length oriented along the direction of
flow of a current filament I and if r is the position vector from the
current element Id ~ℓ to a fixed observation point P, then the
differential element of magnetic flux density d B produced at the
point P is given by the Biot and Savart law as
I
dl
dB
r
P
dB =
µ0 d ~ℓ × r̂
µ0 d ~ℓ × r
I
=
I
4π
r3
4π
r2
(3)
Biot-Savart Law
As an illustration, consider determining the magnetic induction field
B of an infinitely long straight filament of current I . With the z-axis
of a polar cylindrical coordinate system (R, ϕ, z) chosen along the
filament in the direction of positive current flow,
d ~ℓ × r = (1̂z d ℓ) × (1̂z z + 1̂R R) = 1̂ϕ Rd ℓ,
so that B = 1̂ϕ Bϕ with magnitude
Z ∞
µ0 I
µ0
R
I
.
d
ℓ
=
Bϕ =
4π −∞ (R 2 + ℓ2 )3/2
2πR
(4)
This is the experimental result first found by Biot and Savart and is
properly referred to as the Biot-Savart law.
Ampére’s Law
Ampére’s experiments were concerned with the force that one current
carrying wire experiences in the presence of another.
The differential element of force d F experienced by a current element
I1 d ~ℓ1 due to its interaction with the magnetic induction field B is
given by the Lorentz force relation
d F = I1 d ~ℓ1 × B.
(5)
For example, the total magnetic force on a curved filament of current
I extending from P1 to P2 inR a uniform magnetic field is then given
P
by Fm = I ~ℓ × B where ~ℓ = P12 d ~ℓ is the distance vector from P1 to
P2 ; if the filament forms a closed loop, then the total magnetic force
vanishes.
Ampére’s Law
If the magnetic field B(r12 ) is due to a closed circuit C2 carrying
current I2 , then the total force F12 that the closed circuit C1 with
current I1 experiences is obtained from Eqs. (3) and (5) as
C1
dl2
C2
r12
dl1
I1
I2
F12
µ0
=
I1 I2
4π
I I
C1
C2
d ~ℓ1 × (d ~ℓ2 × r12 )
.
3
r12
(6)
Ampére’s Law
Because
d ~ℓ1 × (d ~ℓ2 × r12 )
d ~ℓ1 · r12 ~
d ~ℓ1 · d ~ℓ2
=
d
ℓ
−
r12 ,
2
3
3
3
r12
r12
r12
where the first term on the right is a perfect differential in ~ℓ1 and so
integrates to zero over the closed contour C1 , one then obtains
Ampére’s law for the force between two current loops
F12
µ0
= − I1 I2
4π
I I
C1
C2
r12 ~
d ℓ1 · d ~ℓ2
3
r12
which exhibits the inherent symmetry of the problem.
(7)
Ampére’s Law: Example
As an elementary application of Ampére’s law, consider two infinitely
long, parallel, straight wires a distance d apart, carrying currents I1
and I2 . Each of these wires then experiences a force per unit length
that is directed perpendicularly towards the other wire with
magnitude
Z ∞
µ0 I 1 I 2
d
µ0
I1 I2
.
dℓ =
f =
3/2
4π
2π d
−∞ (ℓ2 + d 2 )
This force is attractive if the two currents flow in the same direction
and repulsive if they are oppositely directed.
Comparison of Electric & Magnetic Forces
Consider comparing the electric and magnetic forces between two
moving point charges: charge q with velocity v and charge q1 with
velocity v1 .
From the Lorentz force relation (5), the magnetic force on q is given
by Fm = qv × B where B = (µ0 /4π)(q1/r 2 )v1 × (r/r ) is the
magnetic field produced by q1 [see Eq. (3)]. Because µ0 = 1/(ǫ0 c 2 ),
this magnetic field may be written as
B=
v1 E
× ,
c
c
where E = (1/4πǫ0 )(q1 /r 2 )(r/r ) is the electric field produced by q1
with force Fe = qE exerted on q.
Comparison of Electric & Magnetic Forces
The magnetic force exerted on q is then given by
1 qq1 v v1 r ,
×
×
Fm =
4πǫ0 r 2 c
c
r
which may be rewritten as
Fm =
v v1
×
× Fe
c
c
(8)
This then shows that Fm ≪ Fe for non-relativistic velocities v ≪ c,
so that the magnetic force can usually be neglected in comparison to
the electric force except in exceptional circumstances.
One such exceptional circumstance is that of a conduction current
where both positive and negative charges are present with equal
densities. The macroscopic electric field is then zero while the
magnetic field produced by the moving charges is not.
The Differential Relations of Magnetostatics
The differential relation for the magnetic induction field given in Eq.
(3) can be expressed in terms of the current density J(r) as
ZZZ
µ0
(r − r′) 3 ′
B(r) =
J(r′) ×
d r,
(9)
4π
|r − r′ |3
the integration taken over the region containing the current density.
Because (r − r′ )/|r − r′ |3 = −∇|r − r′ |−1 , this expression may be
rewritten as
ZZZ
1
µ0
J(r′ ) × ∇
d 3r ′
B(r) = −
4π
|r − r′ |
ZZZ
J(r′ ) 3 ′
µ0
∇×
d r,
(10)
=
4π
|r − r′ |
so that
∇ · B(r) = 0
(11)
and the B-field is solenoidal. This is the differential form of Gauss’
law for the magnetic induction field.
The Differential Relations of Magnetostatics
For the curl of the magnetic induction filed vector, Eq. (10) yields
ZZZ
J(r′ ) 3 ′
µ0
∇×∇×
d r
∇ × B(r) =
4π
|r − r′ |
" ZZZ
µ0
1
=
∇
J(r′ ) · ∇
d 3r ′
4π
|r − r′ |
#
ZZZ
1
−
J(r′ )∇2
d 3r ′ .
|r − r′ |
As ∇(|r − r′ |−1 ) = −∇′ (|r − r′ |−1 ) & ∇2 (|r − r′ |−1 ) = −4πδ(r − r′ ),
the above result becomes
ZZZ
1
µ0
J(r′ ) · ∇′
d 3 r ′ + µ0 J(r).
∇ × B(r) = − ∇
4π
|r − r′ |
The Differential Relations of Magnetostatics
Integration by parts with the surface term at infinity vanishing then
results in
ZZZ
∇′ · J(r′ ) 3 ′
µ0
∇
d r.
(12)
∇ × B(r) = µ0 J(r) +
4π
|r − r′ |
Because ∇ · J(r) = 0 for steady-state magnetic phenomena, one
finally obtains the differential form of Ampére’s law
∇ × B(r) = µ0 J(r)
(13)
The Integral Relations of Magnetostatics
Integration of Eq. (11) over a simply-connected region bounded by a
closed surface S yields, after application of the divergence theorem
(n̂ denoting the outward unit normal vector to S),
I
(14)
B · n̂d 2 r = 0
S
which is the integral form of Gauss’ law for the magnetic induction
field.
Application of Stokes’ theorem to the surface integral of the normal
component of ∇ × B given in Eq. (13) over an open surface So
bounded by a simple closed contour C yields the integral form of
Ampére’s law
I
B · d ~ℓ = µ0 I
(15)
C
where I =
R
So
J · n̂d 2 r is the current passing through C.
Vector Potential of the Magnetostatic Field
Because ∇ · B = 0 everywhere, then the magnetic induction field
vector B(r) may always be expressed as the curl of a vector potential
A(r) as
B(r) = ∇ × A(r).
(16)
From Eq. (10) it is seen that the general form of the vector potential
is
ZZZ
J(r′ ) 3 ′
µ0
d r + ∇Ψ(r).
(17)
A(r) =
4π
|r − r′ |
The added gradient of a scalar function Ψ(r) shows that the vector
potential can be transformed as A(r) → A′ (r) = A(r) + ∇Ψ(r)
without effecting the magnetic field vector B(r). This is referred to
as the gauge transformation for the magnetostatic vector potential.1
1
An analogous gauge transformation for the scalar potential of the electrostatic field did not arise because it is
just a trivial constant potential.
Vector Potential of the Magnetostatic Field
Substitution of (16) into the differential form (13) of Ampére’s law
results in
∇(∇ · A) − ∇2 A = µ0 J.
The freedom of choice implied by the gauge transformation for the
vector potential allows one to set
∇ · A = 0,
(18)
so that A(r) is solenoidal. This choice is known as the Coulomb
gauge. The vector potential in the Coulomb gauge then satisfies
Poisson’s equation
∇2 A(r) = −µ0 J(r)
(19)
Vector Potential of the Magnetostatic Field
The choice of gauge given in Eq. (18) implies that the gauge
function Ψ satisfies Laplace’s equation ∇2 Ψ = 0 once one is in the
Coulomb gauge, in which case Ψ is a constant provided that there
are no sources at infinity. The expression (17) for the vector potential
then becomes
ZZZ
µ0
J(r′ ) 3 ′
A(r) =
(Wb/m)
(20)
d r
4π
|r − r′ |
Comparison of this expression with that given in Eq. (1.26) for the
electrostatic scalar potential reveals their similarity in mathematical
form, the principal difference being the scalar and vector forms of the
two potentials.
Magnetic Field of a Localized Current Distribution
Consider the magnetic field structure due to a general steady-state
current distribution J(r) that is localized in a small region of space τ ′
with vector potential given by Eq. (20). By choosing the origin of
coordinates O in close proximity to the localized current distribution
region τ ′ , the denominator in the integrand of Eq. (20) may then be
expanded in inverse powers of the distance r to the field point P as
|r − r′|−1 = 1/r + (r · r′ )/r 3 + [3(r · r′ )2 − r 2 r ′2 ]/2r 5 + · · · . The
vector potential then has the expansion
ZZZ
ZZZ
µ0
µ0
′
3 ′
J(r )d r +
J(r′ )(r · r′ )d 3 r ′
A(r) =
3
4πr
4πr
′
′
τZ Z Z
τ
µ0
+
J(r′) 3(r · r′ )2 − r 2 r ′2 d 3 r ′ + · · · , (21)
5
8πr
τ′
where the integration may be extended to all of space because J
vanishes outside of τ ′ .
Magnetic Field of a Localized Current Distribution
By the divergence theorem, if each of the functions f (r), g (r), and
J(r) is continuous throughout all space, then
ZZZ
I
3 ′
′
′
′
′
∇ · f (r )g (r )J(r ) d r = f (r′ )g (r′)J(r′ ) · n̂d 2 r ′ → 0
as the surface recedes to infinity, so that [because ∇′ · J(r′) = 0]
ZZZ
′
f (r )J(r′) · ∇′ g (r′ ) + g (r′)J(r′ ) · ∇′ f (r′ ) d 3 r ′ = 0.
(22)
With f = 1 and g (r′) = rj′ , this general result gives
ZZZ
J(r′)d 3 r ′ = 0,
(23)
so that the first term (the monopole term) in the expansion (21) of
the vector potential vanishes.
Magnetic Field of a Localized Current Distribution
For the second term in this expansion, let f (r′ ) = ri′ and g (r′) = rj′ ,
RRR ′
ri Ji (r′ ) + rj′ Ji (r′ ) d 3 r ′ = 0, so
in which case Eq. (22) becomes
that
r·
ZZZ
′
′
3 ′
r Ji (r )d r
ZZZ
3
X
=
rj
rj′ Ji (r′ )d 3 r ′
j=1
3
1X
= −
rj
2 j=1
3
ZZZ
3
1 XX
ǫijk rj
= −
2 j=1 k=1
ri′ Jj (r′ ) − rj′ Ji (r′ ) d 3 r ′
ZZZ
(r′ × J(r′))k d 3 r ′ ,
where ǫijk is the permutation symbol defined to be 0 if two or more
indices are equal, to be +1 if ijk is a permutation of the indices 123
(or xyz) and to be −1 if it is not.
Magnetic Field of a Localized Current Distribution
The final result is then given by
ZZZ
ZZZ
1
′
′
3 ′
r·
r J(r )d r = − r ×
2
r′ × J(r′ ) d 3 r ′ .
(24)
The magnetic moment density or magnetization of the localized
current distribution is defined as
1
~
M(r)
≡ r × J(r)
2
(A/m)
(25)
and its integral is defined as the magnetic moment
1
m≡
2
ZZZ
τ′
r′ × J(r′)d 3 r ′
(A · m2 )
(26)
Magnetic Field of a Localized Current Distribution
If the current is confined to a plane loop Hwith arbitrary shape, then
its magnetic moment is given by m = 21 I C r × d ~ℓ.
Since d a = 21 r × d ~ℓ is the oriented differential element of area swept
out by d ~ℓ, then the magnitude of the magnetic moment is given by
the current times the area of the loop (m = I a), regardless of the
shape of the circuit C.
Magnetic Field of a Localized Current Distribution
The first nonvanishing term in the expansion (21) is the magnetic
dipole vector potential
µ0 m × r
.
(27)
A(r) =
4π r 3
The associated magnetic induction vector B(r) = ∇ × A(r) is then
given by
µ0
r
B(r) =
∇× m× 3
4π h
r r
r i
µ0
m∇ · 3 − (m · ∇) 3 .
=
4π
r
r
m
r
r
8π
n̂, where
Because ∇ · r 3 = 3 δ(r) and (m · ∇) r 3 = r 3 − 3 n̂·m
r3
n̂ ≡ r/r is a unit vector in the direction of the position vector r, then
µ0 (3n̂ · m)n̂ − m
B(r) =
,
(28)
4π
r3
where the delta function term is not included because r is bounded
away from zero.
Magnetic Field of a Localized Current Distribution
The magnetic dipole field then has a component along the dipole axis
as well as along the direction of observation from the origin O. With
the dipole axis taken along the polar axis of a spherical coordinate
system r , θ, ϕ with origin O at the center of a circular current loop,
the above expression becomes
B(r) =
µ0 m
1̂
2
cos
θ
+
1̂
sin
θ
,
r
θ
4πr 3
(29)
where m = |m| is the magnitude of the magnetic dipole moment.
Comparison of this result with that given in Eq. (9.7) for the electric
dipole field with dipole axis oriented along the polar axis reveals that
they have the same spatial behavior in the far-zone r ≫ a, where a is
a measure of the spatial extent of the dipole source (e.g., the linear
separation of the charges for the electric dipole and the radius of the
current loop for the magnetic dipole).
Magnetic Field of a Localized Current Distribution
Finally, notice that the magnetic dipole field given in Eq. (29) with r
bounded away from zero can be expressed as the negative gradient of
a scalar function as
B(r) = −∇φm (r)
(30)
with
µ0 m · r
4π r 3
viewed as a scalar potential for the magnetic dipole field.
φm (r) =
(31)
Scalar Potential for the Magnetostatic Field
In general, a scalar potential for the magnetostatic field can be
introduced wherever the current density vanishes. In that case,
Ampére’s law (13) becomes
∇ × B(r) = 0 =⇒ B(r) = −∇φm (r).
(32)
With Gauss’ law (11), this scalar potential then satisfies Laplace’s
equation
∇2 φm (r) = 0
(33)
everywhere that J(r) = 0. Eq. (32) may be integrated to yield an
expression for this scalar potential as
Z r
~ · d ξ~
φm (r) = −
B(ξ)
(34)
Pref
where Pref is some fixed reference point. It is important that the
path of integration taken in Eq. (34) not pass through any region
where J 6= 0.