Download ∫ ) r '

Magnetostatics
Biot-Savart Law
B
(r
)
J
The magnetic field
at point r due to a current density (r ) is
1 r − r'
B ( r ) = ∫ dr ' J ( r ' ) × 3
c
| r − r '|
Here c is the speed of light. We note that the magnetic field has the same dimensions as the
electric field.
Exercise: Find the magnetic field due to an infinite straight wire carrying a constant current I.
Solution: In cylindrical coordinates, in which the wire is put along z, the current density is
J = zˆIδ ( ρ ' 2 )
Since the current is along the z-direction, then
zˆ × (r − r ' ) = zˆ × ρˆ ( ρ − ρ ' ) = ϕˆρ . We hence
have
∞
I
1
ˆ
B ( r ) = ρϕ ∫ dz '
c
( z − z' )2 + ρ 2
−∞
(
and the magnetic field depends only on
∞
)
3/ 2
I
1
= ρϕˆ ∫ dz '
c
z '2 + ρ 2
−∞
(
)
3/ 2
=
2I
ϕˆ
cρ
ρ.
Exercise: Find the magnetic field due to a circular ring (of radius a) carrying a constant current,
on the symmetry axis.
Solution: We put the ring in the x-y place, and then the symmetry axis is the z-direction. In
cylindrical coordinates, the current is along the
r − r ' → zzˆ − aρ̂
ϕ̂
direction. Then,
ϕˆ × (r − r ' ) → zρˆ + azˆ
1
Biot-Savart law gives
2π
I
azˆ + zρˆ
I
a2
ˆ
B ( r ) = ∫ adϕ
= z
3/ 2
c 0
c a2 + z2
a2 + z2
(
)
(
)
3/ 2
2π
and the magnetic field depends only on z.
Exercise: Find the magnetic field at the center of a wire square of side a carrying a current I.
Solution: Let us calculate the contribution from the side along the x-direction. Putting the
coordinate origin at the center of the square, we have
π /4
I a/2
xˆ × (− xxˆ + ( a / 2) yˆ )
I 2
I 2
ˆ
ˆ
B=
dx
=
z
d
α
cos
α
=
z
2
3/ 2
c −a∫/ 2
c a −π∫/ 4
ca
x 2 + ( a / 2) 2
(
)
Calculating the contributions of the other three sides in the same way, the final result is
I 2
B = zˆ
4 2
ca
Let us now calculate the magnetic field due to this square current loop on the z axis. Using
I d × ( zˆz − r ' )
B( z ) = ∫
c
| zˆz − r ' |3
We have the contributions of the four segments of the wire as follow:
a/2
a/2
I
xˆ × ( zˆz − xˆx + yˆ (a / 2)) I
yˆ × ( zˆz − xˆ (a / 2) − yˆ y )
B( z ) =
dx
+
dy
c −a∫/ 2
( z 2 + (a / 2) 2 + x 2 ) 3 / 2 c −a∫/ 2
( z 2 + ( a / 2) 2 + y 2 ) 3 / 2
I
+
c
−a / 2
−a / 2
xˆ × ( zˆz − xˆx − yˆ ( a / 2)) I
yˆ × ( zˆz + xˆ ( a / 2) − yˆ y )
dx
+
dy
∫ ( z 2 + (a / 2) 2 + x 2 ) 3 / 2 c a∫/ 2 ( z 2 + (a / 2) 2 + y 2 ) 3 / 2
a/2
(note the integration limits!). Collecting terms, this expression becomes
2
2I
B( z ) =
c
xˆ × yˆ ( a / 2)
2I
dx
+
∫ ( z 2 + ( a / 2) 2 + x 2 ) 3 / 2 c
−a / 2
a/2
a/2
∫
−a / 2
dy
yˆ × (− xˆ (a / 2))
( z 2 + ( a / 2) 2 + y 2 ) 3 / 2
Now we see that the resulting magnetic field is along the z-direction. Moreover, the two integrals
are identical, and therefore
a/2
4I a
1
2 Ia
1
a/2
B( z ) =
zˆ ∫ dx 2
=
zˆ 2
2
2
2 3/ 2
2
c 2 − a / 2 ( z + ( a / 2) + x )
c
z + ( a / 2)
z 2 + 2 ( a / 2) 2
I 2a 2
B ( z ) = zˆ
c z3
In particular, for z>>a, this becomes
Forces between current carrying conductors
The force exerted by a magnetic field
B on a charge q moving with velocity v is
v×B
F =q
c
Hence the force experienced by a current element
dI1 ≡ I1d 1` is
I dF = 1 d 1 × B
c
When the magnetic field is due to another (closed) current loop we have
II
F12 = 1 2 2
c
(
d 1 × d 2 × r12
∫ ∫ | r12 |3
)
3
Exercise: Find the force acting between two circular current loops, of radii a and b respectively,
placed one on the top of the other such that the z-axis passes through both centers. The
distance between the two centers is d.
Solution: Let us use polar coordinates, such that for the lower ring
r1 = xˆa cos ϕ1 + yˆ a sin ϕ1 ≡ arˆ1
and
d 1 = ( − xˆa sin ϕ1 + yˆ a cos ϕ1 )dϕ1 ≡ aϕˆ1dϕ1
Then
r2 = xˆb cos ϕ 2 + yˆ b sin ϕ 2 + zˆd ≡ b(cos(ϕ 2 − ϕ1 )rˆ1 + sin(ϕ 2 − ϕ1 )ϕˆ1 ) + dzˆ
and
d 2 = (− xˆb sin ϕ 2 + yˆ b cos ϕ 2 )dϕ 2 ≡ bdϕ 2 (− sin(ϕ 2 − ϕ1 )rˆ1 + cos(ϕ 2 − ϕ1 )ϕˆ1 )
Thus,
(d 2 × r12 ) = (bdϕ 2 )[(b − a cos(ϕ 2 − ϕ1 ))rˆ1 − d cos(ϕ 2 − ϕ1 ) zˆ ]
and
d 1 × (d 2 × r12 ) = (ab )(dϕ1dϕ 2 )(− zˆd cos(ϕ1 − ϕ 2 ) + rˆ1 (b − a cos(ϕ1 − ϕ 2 )))
But since
r122 = d 2 + a 2 + b 2 − 2ab cos(ϕ1 − ϕ 2 )
we can do the integrations separately on
ϕ1
and
ϕ1 − ϕ 2 .
We are left with the contribution of
the z-direction alone. Hence,
2π
I1 I 2
cos ϕ
F12 = − zˆ 2 dab 2π ∫ dϕ
c
(d 2 + a 2 + b 2 − 2ab cos ϕ )3 / 2
0
4
Magnetostatics and Ampere’s law
The Biot-Savart law is
1 J (r ' ) × (r − r ' )
B ( r ) = ∫ dr '
c
| r − r ' |3
ρ (r ' )( r − r ' )
E ( r ) = ∫ dr '
.)
| r − r ' |3
(Compare with
We can write the above equation in the form
J (r ' )
1
B ( r ) = ∇ × ∫ dr ' c
| r − r '|
and then
∇ ⋅ B(r ) = 0
Next we consider the rotor of the magnetic field:
J (r ' )
1 ∇ × B ( r ) = ∇ × ∇ × ∫ dr ' c
| r − r '|
Using
(
) (
)
∇ × ∇ × V = ∇ ∇ ⋅ V − ∇ 2V
which holds for any vector, we find
 1 
1 1
1 ∇ × B ( r ) = ∇ ∫ dr ' J (r ' ) ⋅ ∇ − ∫ dr ' J ( r ' )∇ 2  
c
| r − r '|
c
 | r − r '| 
1 1
4π = − ∇ ∫ dr ' J ( r ' ) ⋅ ∇ ' +
J (r )
c
| r − r '|
c
By integrating by parts the first term we have
5
4π 1 ∇ '⋅ J ( r ' )
∇ × B(r ) =
J ( r ) + ∇ ∫ dr ' c
c
| r − r '|
Now we use the continuity equation, by which
∂ρ (r ) + ∇ ⋅ J (r ) = 0
∂t
to finally obtain
1 ∂E 4π J
∇× B −
=
c ∂t
c
(For a stationary situation, the time derivative does not appear. Also, compare this equation with
∇ ⋅ E = 4πρ ).
Applying Stokes theorem, the surface integral
4π
ˆ
∇
×
B
⋅
n
da
=
∫S
c
J
∫ ⋅ nˆda
S
becomes
4π
B
∫C ⋅ d = c
4π
J
∫S ⋅ nˆda = c I
which is, again, Ampere's law.
Exercise: Find the magnetic field due to a surface current K density on a plane perpendicular to
the z-axis.
Solution: Let us denote the direction of the current density on the plane as the x-direction.
Applying Ampere's Law
6
4π
B
∫ ⋅ d = c
C
4π
ˆ
J
⋅
n
da
=
I
∫S
c
for a rectangular loop of a length L along y and a tiny (goes to zero) length c along z, we obtain
B y ( z < 0) L − B y ( z > 0) L =
4π
KL
c
Therefore,
2π

K, z > 0
− yˆ
c
B=
2π
 yˆ
K, z < 0
c

Conclusion: The tangential component of the field has a discontinuity at a current-carrying
surface.
Exercise: Find the magnetic field due to a conducting infinite slab perpendicular to the zdirection, located in the region
− a ≤ z ≤ a , in which there is a current density J = xˆJ .
Solution: Firstly, the field cannot depend on x or on y. Then by
∇ ⋅ B = 0 we find that
dB z / dz = 0 and hence Bz = 0 . (It cannot be simply a constant, being the same close and far
away from the slab.) Secondly, by ∇ × B = ( 4π / c ) J we find that
dB y
dz
= −( 4π / c ) J
Then,
By ( z) = −
4π
Jz − a ≤ z ≤ a
c
Now we use continuity (there is no discontinuity in the tangential component of the field, because
there is no finite surface current density) to find
7
By ( z) = −
4π
c
 Ja

(− Ja )
z≥a
z ≤ −a
Note that a constant magnetic field does penetrate into a conductor. The same result can be
obtained from Ampere's law.
Exercise: Find the magnetic field produced by a very long solenoid, of a circular cross-section,
carrying a current I.
Solution: We use cylindrical coordinates, and take the solenoid to be along the z-direction. The
wire winds around it densely, such that there are n loops per unit length. The surface current
density is then
K = nIϕ̂
For an infinitely long solenoid, the field cannot depend on z and by symmetry it cannot depend on
ϕ ; However, it may depend on ρ . By ∇ ⋅ B = 0 we find that there is no field component along
ρ̂ By ∇ × B = (4π / c ) J , the field is directed along z. So we write
B ( ρ ) = zˆB ( ρ )
B
If we choose to integrate ⋅ d along a loop which lies outside the solenoid, we will obtain zero.
Hence, the magnetic field outside the solenoid is zero. Now let us integrate along a loop with one
arm (of length L) outside the solenoid, the other inside. Then we find
4π
B = zˆ
In
c
(inside).
8
Exercise: Find the magnetic field produced by an infinite cylindrical wire, of radius a, carrying a
current density which is a function of
ρ , J ( ρ ) = zˆJ ( ρ ) .
Solution: By the similar considerations as in the previous example, we know that
B ( ρ ) = ϕˆB ( ρ )
Integrating along a circular loop we find
ρ
4π
B 2πρ =
c
∫ J ( ρ ' )2πρ ' dρ '
0
so that
ρ
2
B( ρ ) =
J ( ρ ' )2πρ ' dρ '
cρ ∫0
When
ρ>a
the integral gives the total current in the wire, I. Hence outside the wire
B( ρ ) =
2I
cρ
as has been also found from the Biot-Savart law. When
J is independent of ρ , the field
increases within the wire, and decreases outside.
The vector potential
Let first summarize our results so far, and compare them to electrostatics.
4π ∇× B =
J
c
∇⋅B = 0
∇ ⋅ E = 4πρ
∇× E = 0
9
B curls around I E diverges from q
We now introduce the vector potential A such that
B = ∇× A
with the (arbitrarily) chosen gauge
∇⋅ A = 0
(Coulomb gauge)
For example, to have a constant magnetic field along the z-direction, we may take
1 1
A = B × r = B (− xˆy + yˆ x )
2
2
Ampere’s law now takes the form of a vectorial Poisson equation
4π J = ∇ × (∇ × A) = −∇ 2 A + ∇ (∇ ⋅ A) = −∇ 2 A
c
and then
1 J (r ' )
A( r ) = ∫ dr ' c
| r − r '|
(note that ∇ ⋅ A = 0 because by the continuity equation for a stationary situation ∇ '⋅ J = 0 .)
From the definition of the vector potential, and using Stokes theorem we find that the magnetic
flux through an area S is given by the contour integral around the loop closing that area
(
)
Φ = ∫ B ⋅ nˆ da = ∫ ∇ × A ⋅ nˆ da = ∫ A ⋅ d This yields that the tangential components of the vector potential are continuous.
10
Exercise: Find the vector potential of a wire (of zero thickness) of length L carrying a current I.
Solution: We take the current along the z-direction, and use cylindrical coordinates. From the
1 J (r ' )
general formula, A( r ) = ∫ dr ' , we know that the vector potential is along the zc
| r − r '|
direction as well. In this special case, the integral becomes
L/2
I
Az ( ρ ) =
dz
c − L∫/ 2
1
2I
=
c
z2 + ρ 2
L/2
∫
dz
0
1
2I
=
ln( z + z 2 + ρ 2
c
z2 + ρ 2
2 I L / 2 + ( L / 2) + ρ
ln
c
ρ
2
=
For
L
2ρ
L/2
0
2
→∞,
Az ( ρ ) = −
2I
ln ρ
c
(up to an unimportant constant).
Exercise: Find the vector potential of a wire of length L and radius a carrying a current I.
Solution: We take the current along the z-direction, and use cylindrical coordinates. Here again
the vector potential is along the z-direction. Therefore, Poisson equation gives
4π
∇ Az = −
J,
c
2
Since
 I
 2
where J =  πa
 0
ρ≤a
ρ>a
Az ≡ Az ( ρ ) (it cannot have an angular dependence) we have
11
1 d
ρ dρ
 I

 dAz 
4π
J where J =  πa 2
 ρ
 = −
c
 dρ 
 0
ρ≤a
ρ>a
.
This equation can be integrated in the two regimes, to give
2
I ρ
Az = −   + C1 ln ρ + C
ca
A = D1 ln ρ + D
where
ρ≤a
ρ>a
C1 , C , D1 , and D are constants. The magnetic field is given by
 2 Iρ C1
− ca 2 + ρ
dAz
= −ϕˆ 
B = ∇ × A = −ϕˆ
D1
dρ


ρ
We therefore find that
ρ≤a
ρ>a
C1 = 0 . The normal component of the magnetic field is continuous and
therefore
D1 = −
2I
c
The tangential component of the vector potential is continuous and therefore
−
I
+ C = D1 ln a + D
c
giving
D=−
I 2I
+
ln a + C
c c
12
The final result, up to the constant C (in the vector potential, but not in the magnetic field!) is
then
2
I ρ
Az = −   + C
ca
2I ρ
B=
ϕˆ
c a2
ρ≤a
I
ρ
Az = − 1 + 2 ln  + C
c
a
2I 1
B=
ϕˆ
c ρ
ρ >a
The magnetic dipole
We look for the asymptotic behavior of the vector potential created by a current density
(confined in space). We start from the expression
1 J (r ' )
A( r ) = ∫ dr ' c
| r − r '|
and introduce there
1
1 rˆ ⋅ r '
= + 2 + .....
| r − r '| r
r
,
r >> r '
The first term of the expansion gives
1
A( r ) ≈ ∫ dr ' J (r ' )
rc
But this is zero. To see this property, we write
(
)
(
J i ( r ' ) = ∇ ⋅ r 'i J ( r ' ) − r 'i ∇ ⋅ J ( r ' ) = ∇ ⋅ r 'i J ( r ' )
)
13
where we used the continuity equation, by which
(
)
(
(
)
∇ ⋅ J (r ' ) = 0 . Now we apply Gauss's law
)
ˆ
d
r
'
∇
⋅
r
'
J
(
r
'
)
=
da
'
n
⋅
r
'
J
(r ' ) = 0
i
i
∫
∫
(since the current density does not go out of the region where it is confined).
In other words, the volume integral of the steady current density vanishes:
d
r
∫ J (r ) = 0
Thus the lowest order in the expansion the vector potential is
1
A(r ) = 2 ∫ dr ' J ( r ' )(rˆ ⋅ r ')
cr
Let us now define the magnetic dipole,
1
m=
d
r
' r '× J ( r ' )
2c ∫
We will now prove that the lowest order of the expansion of the vector potential can be written
in the form
m × rˆ
A( r ) = 2
r
First, we write the result obtained above for the vector potential explicitly,
1
1
A j ( r ) = 2 ∫ dr ' J j ( r ' )(rˆ ⋅ r ') = 2
cr
cr
∑ rˆ ∫ dr ' J
i
j
( r ' ) r 'i
i
14
Secondly, we show that
∫ dr ' r '
i
J j ( r ' ) = − ∫ dr ' r ' j J i ( r ' )
This is accomplished as follows. By Gauss's theorem,
(
)
(
)
ˆ
d
r
'
∇
⋅
r
'
r
'
J
(
r
'
)
=
da
'
n
⋅
r
'
r
'
J
(r ' ) = 0
i
j
i
j
∫
∫
Writing explicitly the left-hand-side (remembering the continuity equation!)
(
)
(
d
r
'
∇
⋅
r
'
r
'
J
(
r
'
)
=
d
r
'
r
'
J
(
r
'
)
+
r
'
J
(
r
' )) = 0
i
j
i
j
j
i
∫
∫
So, we find that
1
1
A j ( r ) = 2 ∫ dr ' J j ( r ' )(rˆ ⋅ r ') = 2
cr
cr
1
=
2cr 2
where
ε ijk
∑
i
∑ rˆ ∫ dr ' J
i
j
( r ' ) r 'i
i
∫ dr ' ( J j (r ' )r 'i rˆi − J i (r ' )r ' j rˆi ) =
(
1
rˆiε ijk ∫ dr ' r '× J ( r ' )
2cr 2
)
k
is the Levi-Civita tensor (it is 1 for i=1, j=2, and k=3, and any other cyclic permutation,
-1 for all other combinations, and 0 if two or more of the indices are equal). We can now write
this result in the form
1
A j (r ) = 2
r
∑ε
rˆ mk
ijk i
i
and in vector notations, this becomes
m × rˆ
A( r ) = 2
r
Remember that the magnetic dipole is
1
m=
d
r
' r '× J ( r ' ) .
2c ∫
15
The magnetic field in this approximation is
3rˆ(m ⋅ rˆ) − m
B ( r ) = ∇ × A(r ) =
r3
Exercise: Find the magnetic dipole moment of a planar current loop, carrying the current I.
Solution: In this case
J (r ' ) dr ' = Id , and the magnetic dipole is
1
1 1
m=
d
r
'
r
'
×
J
(
r
'
)
=
I
r '×d = ISnˆ
2c ∫
2c ∫
c
where S is the area enclosed by the current loop and n̂ is a unit vector normal to it.
This is general. For example, for a circular current loop of radius a, we have
r ' = xˆa cos ϕ + yˆ a sin ϕ , and d = (− xˆa sin ϕ + yˆ cos ϕ )dϕ ,
giving
r '×d = zˆa 2 (cos 2 ϕ + sin 2 ϕ ) dϕ
Exercise: A current loop, carrying a current I, is made of a wire of the shape of square with side
a. Find the magnetic field on the axis perpendicular to the current loop, and passing through its
center.
Solution: We have already solved this problem, and found that for z>>a (the loop lies in the x-y
I 2a 2
. We now obtain the same result from the magnetic dipole.
plane), the result is B ( z ) = zˆ
c z3
We now know that the magnetic moment of this current loop is directed along z, and has the
m = Ia 2 / c . Indeed, since m is along z and the magnetic field is required on the z axis,
3rˆ(m ⋅ rˆ) − m
3
B
(
r
)
=
the result
gives just B ( z ) = 2m / z which is our result above.
3
r
magnitude
16
Exercise: Find the magnetic dipole of a thin disk of radius a carrying surface charge σ and
rotating with angular velocity ω around the z axis which passes through the center of the disk.
Solution: We use the formula
1
m=
d
r
r × J ( r ) . In our case the current is confined to the
2c ∫
disk, and is given by
J (r ) = ϕˆσωrδ ( z )
Since
rˆ × ϕˆ = zˆ , we find
1
m = zˆ
2c
2π
a
(
)
π
a4
∫0 dϕ ∫0 drr r σω = zˆ c σω 4
2
Magnetic dipole in a (constant) magnetic field
A magnetic dipole m is placed in a magnetic field B (which is not created by the dipole). The
magnetic dipole is created by a current loop carrying a current I. The force on each small element
of the wire carrying the current is
I dF = dr × B
c
This force creates a torque,
N , given by
(
I N = ∫ r × dF = ∫ r × dr × B
c
)
Let us write the integrand here in a different form, using (remember that the magnetic field is
constant)
17
d [ r × ( r × B )] = r × ( dr × B ) + dr × (r × B )
Then,
(
)
I
I
N=
r
×
(
d
r
×
B
)
−
d
r
×
(
r
×
B
)
+
d
[
r
× ( r × B )]
2c ∫
2c ∫
The last integral vanishes (it is a complete derivative). The other terms become
(
)
I
I (
)
N=
r
×
(
d
r
×
B
)
+
d
r
×
(
B
×
r
)
=
−
B
×
r
×
d
r
=
−
B
×m = m× B
2c ∫
2c ∫
(For any three vectors,
a × (b × c ) + b × (c × a ) + c × ( a × b ) = 0 )
Forces and potential energy
When a current density, J (r ) , is placed in an external magnetic field,
B (r ) , the total force on
the current density is given by
1 F = ∫ dr J ( r ) × B ( r )
c
In case the magnetic field varies slowly enough over the region where the current is flowing, we
may expand it to obtain the dominant terms:
Bi (r ) ≈ Bi (0) + (r ⋅ ∇ ) B i +...
and thus
18
Fi ≈
(
)
1
ε ijk Bk (0) ∫ dr ' J j (r ' ) + ∫ dr ' J j (r ' )(r '⋅∇) Bk ( r ) + ...
∑
c jk
As we have found above, the volume integral of the current density vanishes. By the same vector
algebra as above, the second term gives
Fi = ∑ ε ijk (m × ∇ ) j Bk
jk
so that
F = ( m × ∇) × B = ∇( m ⋅ B ) − m(∇ ⋅ B ) = ∇( m ⋅ B )
It follows that the potential energy of a dipole moment in an external magnetic field is
U = −m ⋅ B
Magnetic fields in media
m × rˆ
We have found above that the vector potential due to a magnetic dipole is A( r ) =
. Let us
r2
now consider a material ("medium") placed in a magnetic field. The moving bound charges in the
material are affected by that magnetic field, and as a result we may imagine that each atom or
molecule acquires a tiny magnetic dipole moment. Let us denote that magnetic dipole distribution
M
(r ) . This is the magnetization of the material. If in
(magnetic dipole per unit volume) by
addition there is "true" current density J (r ) flowing in the material, the total vector potential
created will be
19
1  J (r ' )
M (r ' ) × (r − r ' ) 
,
A(r ) = ∫ dr '  + c
3

c
|
r
r
'
|
|
r
r
'
|
−
−


The second term here can be written in the form
 1 
 M (r ' ) × (r − r ' ) 
1 

d
r
'
d
r
'
M
(
r
'
)
'
d
r
'
=
×
∇


≈
∇ '×M (r ' )
 | r − r '|  ∫
∫  | r − r ' |3  ∫
| r − r '|


using integration-by-parts, and assuming that the surface contribution may be neglected. Then
1  J (r ' ) + c∇ '×M ( r ' ) 

A( r ) = ∫ dr ' 

c
|
r
−
r
'
|


J
(
r
)
=
c
∇
× M (r ) .
This means that we may define now an effective medium current density, m
As a result,
4π ∇× B =
J + 4π∇ × M ,
c
which is put in the form
4π ∇× H =
J with
c
(Compare with
H = B − 4πM
D = E + 4πP ).
As was the case in electrostatics, we also have here
B = µH
where
µ
M = χH
is the magnetic permeability, and
χ
is the magnetic susceptibility, and both are
properties of the medium.
20
The boundary conditions on the magnetic fields at the interface between two media are: The
B are continuous, the tangential components of H have a discontinuity if
the interface carries a finite surface current density K ,
normal components of
4π nˆ × ( H 2 − H 1 ) =
K
c
21