Lecture 16
Ampère’s Law
Sections: 7.2, partially 7.7
Homework: See homework file
LECTURE 16
slide
1
Ampère’s Law in Integral Form – 1
• the field of a straight wire with current (Lecture 15)
I
B  a
,T
2
H  a
I
2
az
I
, A/m
• for this cylindrically symmetric
problem with circular field lines the
result suggests the relation
a
 H  dL  I encl
C
• Ampère’s law can be derived from the Biot-Savart law (section 7.7
in textbook, Hayt/Buck 8th ed.)
LECTURE 16
slide
2
Ampère’s Law in Integral Form – 2
 H  dL  I encl   J  ds
C[ S ]
S
• contour C can be any – does not need to be a field line
• I is total current flowing through area S bounded by C
• positive I relates to the direction of C by the right-hand rule
LECTURE 16
slide
3
Ampère’s Law in Integral Form – 3
examples on the sign of current contributions
I2
I2
I1
I1
C
 H  dL   I1  I 2
C
 H  dL  I1  I 2
C
I1
C
I2
 H  dL  I1  I 2
C
C
LECTURE 16
slide
4
Applications of Ampère’s Law
• problems where the magnetic field is circularly symmetric
• contours exist, along which the magnetic field is constant or zero
– Ampère’s contours
• concept is similar to Gauss law in electrostatics
LECTURE 16
slide
5
Applications of Ampère’s Law: Straight Thin Wire
Example 1: Find the magnetic field of an infinitesimally thin straight
wire with current.
az
 H  dL  Iencl
C
2

 (  d )  I encl
 H a  a


0
 H 2  I  H 
I
dL
I
2
, A/m

C[ S ]
a
H
same result obtained via Biot-Savart law in Lecture 15
LECTURE 16
slide
6
Applications of Ampère’s Law: Straight Thick Wire – 1
Example 2: A straight conductor of circular cross-section and radius a
carries current I. Find the magnetic field inside and outside the
conductor. The conductor has no magnetic properties.
(a) outside – case the same as in Example 1
 P  a  total enclosed current is I
H ( P ) 
I
2 P
, A/m
J
(b) inside  P  a
assume uniform current distribution
I
J  2 a z , A/m 2
a
LECTURE 16
P
P P
P
a
slide
7
Applications of Ampère’s Law: Straight Thick Wire – 2
2

I P
2
P
H   2 P  J   P  I  2  H  ( P ) 
, A/m
2

a
2 a
I encl
H
I
2 a
0
LECTURE 16
 1/  P
 P
a
slide
P
8
Applications of Ampère’s Law: Straight Thick Wire – 3
LECTURE 16
slide
9
Applications of Ampère’s Law: Coaxial Cable – 1
Example 3: Find the magnetic field everywhere in the cross-section
of a coaxial cable.
(a) inside wire, ρP ≤ a – same as in Example 2-b
I c b
I P
H 
, A/m
2
2 a
a
H
(b) between wire and shield, a ≤ ρP ≤ b – same as in Example 2-a
H 
I
2 P
, A/m
(c) inside shield, b ≤ ρP ≤ c
2 P H  I  J   ( 
2
P
 b2 )
I
2  b2 )
I


(

P
 (c 2  b 2 )
LECTURE 16
slide
10
Applications of Ampère’s Law: Coaxial Cable – 2
(c) inside shield, cont.
I
(c 2   P2 )
 H 
 2 2
2 P (c  b )
LECTURE 16
(d) outside cable, ρP > c
H  ?
slide
11
Applications of Ampère’s Law: Toroid
Example 4: Find the magnetic field inside a toroid of N turns carrying
current I.
 H  dL  I encl
I
C
C
H
 H  2  N  I
2
N I
H 
, 1     2
2

I
1
a
What is the field outside the toroid?
LECTURE 16
slide
12
Applications of Ampère’s Law: Surface Current Sheet – 1
Example 5: Find the magnetic field due to an infinite sheet of uniform
current flowing in the y-direction.
H
field does not vary with x and y since the source does not vary with
x and y
LECTURE 16
slide
13
Applications of Ampère’s Law: Surface Current Sheet – 2
Hy = 0 since current is along y (field is always ┴ to current)
Hz = 0 due to cancellation of contributions from two symmetrical
current elements along x
z
y
#j
Hk
P H  H j  H k  H xa x
Hj
x
#k
resultant field has only x-component and does not vary with x
and y
LECTURE 16
slide
14
Applications of Ampère’s Law: Surface Current Sheet – 3
apply Ampère’s law along xz plane contour 1-1'-2'-2-1
 H  dL  I encl
C
1
field does not vary with x
2
2
1
  H x1dx   (  H z )dz   (  H x 2 )dx   H z dz  K y  L
1
1
2 

 2



0
0
zero contribution from segments 1'-2' and 2-1 (Hz = 0)
 H x1L  H x 2 L  K y L  H x1  H x 2  K y
H1
as follows from Biot-Savart’s
law, field is anti-symmetrical
wrt the current sheet
z
y
H x 2   H x1  H x1  K y / 2, H x 2   K y / 2
LECTURE 16
x
IdLa y
H2
slide
15
Applications of Ampère’s Law: Surface Current Sheet – 4
Ampère’s law can be applied to another xz plane contour stretched
in the z-direction, see 3-3'-2'-2-3
3
2
3 H x3dx  2 ( H x 2 )dx  K y  L  H x3  H x 2  K y
 H x 3  H x1  K y / 2
field does not depend on distance from the infinite current sheet
result analogous to D-field of an infinite charged sheet
P1
K  an
, A/m
H
2
D
s
2
K
a n , C/m 2
LECTURE 16
H1
a n1
an 2
H 2 P2
slide
16
Surface current sheet of current density K = 4az A/m lies
in the plane defined by y = 4. Find the magnetic field at
the origin (0,0,0).
4
H
ZAx
is
X-Axis
LECTURE 16
slide
17
Applications of Ampère’s Law: Solenoid – 1
z
field is constant in z and ϕ because the
source is constant in z and ϕ
H  0 because current is along 
Hρ = 0 because the ρ contributions of
two symmetrical turns along z cancel
Hj P
H  H j  Hk  H za z
Hk
a
a
#j K
#k
a z solenoid axis
field has only a z-component
and it is constant in z and ϕ
LECTURE 16
NI
Ka 
d
slide
18
Applications of Ampère’s Law: Solenoid – 2
L
H

d
L

I
encl

3
3
11 221
2 c
d 2
NI
 H z1  L  H z 2  L 
L
az
d
1
1
NI
 H z1  H z 2 
b
a
d
NI
repeat along 11'3'31: H z1  H z 3 
 H z3  H z2
d
outside the solenoid field is constant and zero (see contour a-b-c-d)
H z ,ab  H z ,cd  0
antisymmetry

2H z ,ab  0  H  H z a z  0,   a
 H z3  H z2  0
inside solenoid
NI
 K  K a , A/m
H z1 
d LECTURE 16
slide
19
You have learned:
that the “work” integral of H along a closed path is not zero but is
equal to the enclosed current (Ampère’s law)
how to apply Ampère’s law to symmetric problems
that the field of a current sheet is constant in space, it is tangential
to the sheet but is orthogonal to the surface current (right-hand rule
applies)
what the field is in a toroid and in a solenoid (equal to the surface
current density of the toroid/solenoid)
what the field is inside wires with current (growing linearly
inside the wire, decaying as 1/ρ outside)
LECTURE 16
slide
20