The Biot-Savart Law
This semester we want to look at ¤elds that are constant in time. That
means that we should not attempt to consider the magnetic ¤eld due to a
single charge, because only moving charges produce magnetic ¤elds (just as only
moving charges experience magnetic force). The simplest source of magnetic
¤eld is steady current, current that does not change in time, and the closest we
can come to a point source is a di¤erential piece. The magnetic ¤eld produced
by a wire segment of length d` carrying current I at a point P is
~ =
dB
¹0 d~` £ ^r
I
4¼
r2
This law is actually a lot like Coulombzs law. Itzs an inverse square law, and it
depends on the vector ~r that points from the wire to P: The new complication
is that the source is a vector, and so the Biot-Savart law involves the cross
product. That means that the magnetic ¤eld at P is perpendicular to both d~`
and r^: The ¤eld produced by a long straight wire forms circles centered on the
wire.
~ we have to integrate over the whole source:
To get the total B
Z
^
d`~ £ R
~ (~r) = ¹0
B
I
2
4¼
R
where now we describe the source point by the vector r~0 ; as we did for electric
~ points from the source to P :
¤elds, and the vector R
~ = ~r ¡ ~r0
R
Letzs begin by ¤nding the magnetic ¤eld due to a straight wire segment.
We put the y¡axis along the wire in the direction of I; and the x¡axis through
P; which is distance s from the wire. Model the wire as a collection of di¤erential
elements d~` = d`^
y at position y 0: Then
~ = s^
R
x ¡ y 0y^
1
and
~ (P ) =
B
=
¹0
4¼
Z
¹0
4¼
Z
y2
y1
y2
y1
dy 0 y^ £ (s^
x ¡ y 0 y^)
I h
i3=2
2
s2 + (y 0 )
dy 0s (¡^
z)
Ih
i3=2
s 2 + (y 0 )2
We integrate from one end of the wire to the other. Now I; s; and the unit
vector z^ are independent of y 0 ; so we can take them out of the integral:
Z
dy 0
~ (P ) = ¡ ¹0 Is z^
B
h
i3=2
4¼
s2 + (y 0 )2
To do the integral, we use a tangent substitution. Let tan µ = y 0 =s: Then sec2 µ
dµ = dy0 =s:
~ (P ) =
B
=
=
=
¡
¹0 Is
z^
4¼
Z
µ2
µ1
Z
s sec2 µdµ
£
¤3=2
s2 + s2 tan2 µ
µ2
¹0 Is2
sec2 µdµ
z^
3=2
4¼
µ1 s3 [sec2 µ]
Z µ2
¹ I
¡ 0 z^
cos µdµ
4¼s µ1
¹ I
¡ 0 z^ (sin µ2 ¡ sin µ 1 )
4¼s
¡
The angles µ1 and µ 2 are shown in the diagram.
We have to be aware of several important facts if we want to use this expression correctly. First, we chose our y axis so that y increases in the direction of
I: That means that y2 is always bigger than y1 ; and also µ 2 > µ1 as a result.
However, there are several possibilities:
1. y1 and y 2 (and hence µ 1 and µ2 ) may both be positive
2. y1 > 0 but y 2 < 0 (and hence µ 1 > 0 and µ 2 < 0)
3. y1 and y 2 (and hence µ 1 and µ2 ) may both be negative
An in¤nitely long wire would be in case 2.
tan µ1 =
ytop e nd
! 1 as y top
s
and thus
µ1 =
2
¼
2
en d
!1
while
tan µ2 =
so
yb o ttom
s
en d
! ¡1 as y b ott om
µ2 = ¡
and so
end
! ¡1
¼
2
¹0 I
¹ I
z^ [1 ¡ (¡1)] = ¡ 0 z^
4¼s
2¼s
~ is indepenCheck the direction and convince yourself it is correct. Note that B
dent of y;and z as it must be due to the translational and rotational symmetry
of the system. Also note that the ¤eld decreases as 1/distance from the wire.
This is exactly the same dependence that we found for the electric ¤eld due to
an in¤nite line charge.
Aside. Now we are ready to look at an important system that is used to
de¤ne the current unit ampere. We have two in¤nitely long wires, each carrying
current I; and separated by distance s The one on the left produces a magnetic
¹ 0I
¤eld ¡ 2¼s
z^ at the position of the one on the right. Thus the second wire
experiences a force
µ
¶
¹ I
¹ I2
dF~ = Id`^
y £ ¡ 0 z^ = ¡ 0 x^ d`
2¼s
2¼s
~ lon g
B
w ire
=¡
on each segment d`: Thus there is a force per unit length
dF~
¹ I2
= ¡ 0 x^
d`
2¼s
The second wire is attracted toward the ¤rst. Go through the analysis to convince yourself that there is an equal and opposite force per unit length on the
wire on the left. The ampere is de¤ned by measuring the force and adjusting
the current until the force/length is 2£10¡ 7 N/m when the wire separation is
s = 1 m. The resulting current is 1 A by de¤nition.
Now Gri¢ths claims that there is no electric force because the wires are
electrically neutral, but thatzs wrong. In a normal lab situation, the system must
have some resistance, so letzs look at an idealized system with the two long wires
being superconductors, but the connectors at in¤nity having a resistance R:
Then we need a potential di¤erence ¢V = IR between the two wires. This can
be accomplished if each carries a uniform line charge density §¸: With origin on
the positively-charged wire and x¡axis pointing toward the negatively-charged
wire, the electric ¤eld at a point between the wires is
µ
¶
¸ x^
1
1
~
E =
+
2¼"0 x
s¡ x
The potential di¤erence is then
¶
Z
Z s¡ a µ
¸
1
1
¸
~
~
¢V =
E ¢ dl =
+
dx =
[ln x ¡ ln (s ¡ x)]
2¼"0 a
x
s¡x
2¼"0
µ
¶
¸
s¡ a
IR =
2 ln
2¼"0
a
3
where a is the radius of each wire. Thus
¸=
¼"0 IR
ln (s=a ¡ 1)
The charge density ¸ ! 0 as a ! 0; but is not zero for a real wire. Then the
electric force on a segment of wire is
dF
=
dF
dl
=
·
¸2
¸2
¼ "0 IR
dl
E ¸dl =
dl =
2¼" 0 s
ln (s=a ¡ 1) 2¼" 0 s
I2
¼ 2 " 0 R2
2¼s [ln (s=a ¡ 1)]2
So
Fe le c
¼ 2 "0 R2
=
2
Fm ag
¹0 [ln (s=a ¡ 1)]
Letzs put in some numbers. Let R = 1 -; s = 10 cm, a = 1 mm. Then we get
¢
2 ¡
Fe le c
(1 -) 8:85 £ 10 ¡12 F/m
2
= ¼ ³
´
Fm ag
4¼ £ 10¡ 7 N/A2 [ln 100 ¡ 1]2
=
=
¼ £ 8:85 £ 10 ¡12
(4 £ 10 ¡7 ) (4:605 2 ¡ 1)
5:3 £ 10¡ 6
N/A
-=
and
F=
-2 F/m
N/A2
=
N/A2
-2 F/m
Letzs check the units.
so
-2 F/m
2
2
V
A
C
V
V2 C
V¢C
J
=
=
N V ¢m
N¢m
J
Thus the ratio is dimensionless, as required.
The ratio is small, but is not zero! A resistance of k- instead of - would
make the ratio > 1.
Amperezs Law
We have shown above that a long wire produces a magnetic ¤eld that curls
around the wire. If we choose a circular path that follows the ¤eld line, we have
I
C
~ ¢ d~l =
B
Z
2¼
0
4
¹0 I
sdµ = ¹0 I
2¼s
The result is independent of s: Now letzs generalize the result to a curve of
arbitrary shape. On such a curve
d~l = ds ^s + sdµ ^µ + dz z^
and
Thus
³
´
~ ¢ d~l = ¹0 I ^µ ¢ ds s^ + sdµ ^µ + dz z^ = ¹0 I dµ
B
2¼s
2¼
I
I
¹0 I
~ ¢ d~l =
B
dµ
2¼
C
C
If the curve surrounds the wire, then the limits are 0 to 2¼; and we get ¹0 I; as
before. But if the wire is outside the curve, the limits go from some minimum
value µ1 to a maximum µ 2 and back to µ1 ; giving zero as the result of the
integral. Now we can add as many wires as we like, to get the general rule
called Amperezs Law:
I
~ ¢ d~l = ¹ (total current through C)
B
0
C
Now letzs use the fact that I is the ¢ux of ~j :
I
Z
~ ¢ d~l = ¹0 ~j ¢ n
B
^ dA
(1)
S
C
which is the integral form of Amperezs law. It is true generally for any ~j; not
only currents in straight wires. As with Gaussz law, this global statement is
~ only in cases with su¢cient symmetry: plane,
useful as a tool for ¤nding B
spherical or cylindrical symmetry, but also toroids (See G Ex 5.10). See LB
page 910 for the method.
Suppose we have current con¤ned to a thin sheet of conductor in the x ¡ y
plane. The current is described by a surface current K (units A/m).
Z
~ = ~j dz = K ^x
K
If we model the plane as a collection of wires, each of thickness dy and carrying
current Kdy; we can see that the magnetic ¤eld above the plane at z > 0 will
be the superposition of ¤elds circulating around the wires. This superposition
gives a ¤eld in the ¡y¡direction, and parallel to the sheet. Below the sheet,
the ¤eld is reversed. We choose an Amperian curve that is a rectangle of length
L and width 2h; lying parallel to the x ¡ z plane. By placing it symmetrically
about the sheet we guarantee that the magnitude of B on the top side equals
the magnitude on the bottom side.
5
Then integrating around this curve, we have
I
Z
~
~
B ¢ dl = 2BL = ¹0 ~j ¢ ^n dA = ¹0K L
C
Thus
K
2
The result is independent of L; as it must be, but it is also independent of h:
We saw a similar result for the electric ¤eld due to a charged sheet. Finally
B = ¹0
~
B
K
y^ above the sheet
2
K
= +¹0 y^ below the sheet
2
= ¡¹0
This result is important in understanding how the Earthzsmagnetic ¤eld can
shield us from charged particles arriving from the sun. Charged particles spiral
around magnetic ¤eld lines. When paricles from the sun reach the Earthzs ¤eld,
the electrons spiral one way and the ions the other. Both signs of charge create
a current sheet of the same sign, and that sheet acts to reduce the ¤eld outside
the sheet and double it inside. The net result is that the Earth;s ¤eld is cut o¤
at the magnetopause. This boundary prevents particles from plunging directly
to the Earthzs surface.
Finally we can use Amperezs law to get a local result. We use Stokesz theorem
to modify the LHS
Z ³
Z
´
~ £B
~ ¢n
r
^ dA = ¹0 ~j ¢ n
^ dA
S
S
Now this is true for absolutely any curve C and any surface S spanning the
curve, so
~ £B
~ = ¹ ~j
r
(2)
0
~ ; this result is true only for static ¤elds.
As with the curl equation for E
6
~ ¢B
~ = 0 because there are no magnetic charges.
We asserted earlier that r
Now letzs verify this starting with the Biot-Savart law.First we re-write the B-S
law in terms of ~j : A general current distribution may be written as a collection
of current loops, each having cross sectional area dA0 :
~ (~r)
B
=
=
~ (~r)
B
=
¹0
4¼
¹0
4¼
¹0
4¼
X
a ll lo op s
X
a ll lo op s
Z
Z
^
d~` £ R
¹
I
= 0
R2
4¼
^
~j £ R
dA0 d`
2
R
X Z
j dA0
all loo ps
Z ~ 0
^
j (~r ) £ R
d¿ 0
R2
V
^
d~` £ R
R2
(3)
Then
~ ¢B
~ (~r ) = r
~ ¢ ¹0
r
4¼
Z ~ 0
^
j (~r ) £ R
d¿ 0
R2
V
Now the grad operator di¤erentiates the unprimed coordinates, so we may move
it inside the integral.
Ã
!
Z
~j (~r 0) £ R
^
¹0
~
~
~
r ¢ B (~r) =
r¢
d¿ 0
4¼ V
R2
Ã
!
#
Z "
^
^
¹0
R
R
~ £
~ £ ~j (~r0 ) d¿ 0
~j (~r 0) ¢ r
=
¡ 2 ¢r
4¼ V
R2
R
~ £ ~j (~r0 ) ´ 0; so
But r
~ ¢B
~ (~r) = ¹0
r
4¼
Ã
!#
Z "
^
R
0
~j (~r ) ¢ r
~ £
d¿ 0
2
R
V
Now
^
R
~ 1
= ¡r
R2
R
~ ¢B
~ = 0; as required.
and the curl of any gradient is zero, so r
Maxwellzs equations for static ¤elds
We now have all four Maxwell equations, limited to the case of static ¤elds.
They are
~ ¢E
~ = ½ ;
~ ¢B
~ =0
r
r
"0
~ £E
~ = 0;
r
~ £B
~ = ¹0~j
r
Here we have grouped them according to the di¤erential operation that appears
(divergence or curl). But we could also group them this way:
7
~ £E
~ = 0;
r
~ ¢E
~ = ½;
r
"0
~ ¢B
~ =0
r
~ £B
~ = ¹0~j
r
where the top two are homogeneous equations (RHS=0) and the bottom two
display the sources ½ and ~j of the ¤elds. To these we add the force law:
³
´
~ + ~v £ B
~
F~ = q E
and charge conservation
@½ ~ ~
+r¢j= 0
@t
~ and B:
~
These equations illustrate the similarities and di¤erences between E
The two source-free equations indicate the lack of magnetic chargew there are
~ has a non-zero divergence
(essentially) no magnetic monopoles. The fact that E
~ ¤eld lines radiate outward from positive charge and into negative
means that E
~ has a non-zero curl, the ¤eld lines wrap around the current
charge. Because B
~ ¢B
~ = 0; magnetic ¤eld lines do not begin or endw they form
sources. Since r
closed loops (which may extend to in¤nity before they close).
A fact that does not emerge obviously from these equations is that magnetic
forces are usually much weaker than electric forces, unless we contrive situations
that make the electric force zero, or almost zero, as is usually the case with
currents in wires, or permanent magnets. This is much more apparent in the
cgs Gaussian unit system, in which the force law is
µ
¶
~v
~
~
~
F = q E+ £B
c
~ and B
~ have the same units, and we can see that for compaIn this unit system E
rable ¤elds wezd need relativistic speeds (v » c) to make the forces comparable.
(See LB pg 914 for an example.)
8