Ampere’s Law
Chapter 5
z
r r
H ◊ dl = I
Biot and Savart’s Law is equivalent to Ampere’s Law, (Chapter 5.4 - 2)
which states that:
z
r r
H ◊ dl = I
I
path
z
“ ” refers to any path, and I is the total current flowing through the area
enclosed by the path. That is, the current must pierce the area within the
path! I is an algebraic sum. For currents in the opposite direction, + and signs indicate directions.
Right Hand Rule
Positive direction for the current is the thumb, if fingers of the right hand
point in the direction of the path.
Demonstration of Ampere’s Law
First, let us calculate the field of an infinitely long, straight current. Then
apply Ampere’s Law.
z
r
r
z=0
x
r
dB
r
R
z
r
dl
Integrate over q , 0 Æ p
Recall Law of Biot and Savart
r
r
m
dB = 2 o I dl ¥ e$R
R 4p
r
r
dl ¥ e$ R = e$f sin q dl = e$f sin q dz
R
2
r I
=F
H sin q K
2
z = - r cotq, , or dz = r csc 2 q dq, ,
or dz =
r
I mo
Putting together: dB =
4p
F sin q I
H r K
2
e$f sin q
p
r
I m o sin q
I mo
- cos q
B = e$f
dq = e$f
4
p
4
p
r
r
0
z
r
I mo
B = e$f
r 2p
r
H = e$f
I
r 2p
p
0
r dq
sin 2 q
r dq
sin 2 q
r
Note that the magnetic field intensity H has units of amp/m, the same as
the surface current.
Demonstration of Ampere’s Law (continued)
We just have shown directly from the Law of Biot and Savart that the field of
an infinite straight wire carrying I in the z direction is:
r
H = e$f
I
r 2p
r
dl
Verify Ampere’s Law!
Choose a path of integration along consant r.
r
Then dI = r d f e$f
and
so
z
z
r r
H ◊ dl =
2p
z
af
H r r df
0
r r
H ◊ dl =
2p
z
I r
df = I
2
p
r
o
Using Ampere’s Law to solve problems
r
If we have a symmetric problem, such that the direction of H is known and
is constant over a chosen path, then:
z
r r
H ◊ dl = H (path length) = I
Therefore: H = I / pathlength
Example:
Consider an infinite current sheet with current density js .
(Practical case: W is width, but can be finite to a reasonable approximation.)
e$R
y
js
H
z
x
b
W
H
Let the current density be js , such that I = js W (sheet current density).
Now consider symmetry.
H will not vary with x (infinite sheet).
Since we take js to be directed along e$z , H will be directed along e$x .
In particular:
r
H = - H e$ x
= H e$x
for y > 0
for y < 0
Then Ampere’s Law says:
a f
- b - H + b H = js b =
Therefore: H =
Ib
W
I
js
, or
.
2W
2
Discussion of why H is as shown:
Vertical components cancel, and horizontal components add.
Example:
Use Law of Biot and Savart to calculate the magnetic field on the axis of
symmetry of a single current loop.
x
radius a
cylindrical coordinates
r
I dl1
e$z
e$r
e$f
y
f
r
I dl 2
a
sinq =
ca
2
+ z2
h
1
2
r
r
dB 2 = dl ¥ e$ R
R
e$R
q
z
z
r
r
m o I dl ¥ e$R
dB1 =
4p R2
r
dB1
r
r m o I dl ¥ e$R
dB =
4p a 2 + z 2
c
dl = a df
h
r
Consider I dl 2 to be 180∞ out of phase.
dBz = dB1 sin q =
=
=
m o I a df sin q
4p a 2 + z 2
m o I a df
4p a 2 + z 2
c
c
a
h ca
m o I a 2 df
c
4p a 2 + z 2
r
m o I a 2 e$z
B=
3
2
2 2
2 a +z
c
h
h
h
3
2
2
+z
2
h
fi Bz =
1
2
mo I a 2 2p
c
4p a 2 + z 2
h
3
2