Today in Physics 217: Ampère’s Law
‰ Magnetic field in a
solenoid, calculated with
the Biot-Savart law
‰ The divergence and curl of
the magnetic field
‰ Ampère’s law
‰ Magnetic field in a
solenoid, calculated with
Ampère’s law
‰ Summary of electrostatics
and magnetostatics so far
6 November 2002
dA
C
A
J
v∫ B ⋅ dA =
C
Physics 217, Fall 2002
4π
c
∫
A
J ⋅ da =
4π
I encl
c
1
Another Biot-Savart law example: the solenoid
Griffiths problem 5.11: find the magnetic field at point P on
the axis of a tightly-wound solenoid (helical coil) consisting
of n turns per unit length wrapped around a cylindrical tube
of radius a and carrying current I. Express your answer in
terms of θ 1 and θ 2 (it’s easiest that way). Consider the turns to
be essentially circular, and use the result of example 5.6.
What is the magnetic field on the axis of an infinite solenoid?
a
θ1
θ2
P
I
6 November 2002
Physics 217, Fall 2002
2
Reminder of the result of Example 5.6
Magnetic field a distance z
along the axis of a circular
loop with radius R and
current I:
2π I
ˆ
B=z
c
(z
R2
2
+R
2 dBz
dB
dBz
dB
dBs
)
2 32
dBs
dBz
r
z
r
dA
R
dA
6 November 2002
Physics 217, Fall 2002
3
The solenoid (continued)
a
θ1
P
θ2
z
I
Suppose that n is so large that we can consider the loops in
the coil to be displaced infinitesimally; then the number of
loops in a length dz is ndz, and
2π Indz
ˆ
dB = z
c
6 November 2002
(z
a2
2
+a
Physics 217, Fall 2002
)
2 32
4
The solenoid (continued)
a
θ1
θ2
P
z
I
Take
so
a
tan θ =
z
2
tan
θ
2
1 + tan θ dθ =
dz
= − 2 dz = −
2
a
cos θ
z
a
⇒ dz = −
sin 2 θ
(
6 November 2002
)
dθ
a
Physics 217, Fall 2002
5
The solenoid (continued)
a
θ1
θ2
I
2π Indz
ˆ
dB = z
c
(
a2
P
z
2π In  adθ  sin 3 θ
ˆ
=
z
−

3
2
2
c
a
2
2
θ
sin


z +a
)
2π In
sin θ dθ ;
c
θ2
2π In
2π In
B = −zˆ
sin θ dθ = zˆ
( cosθ 2 − cosθ 1 ) .
c
c
= −zˆ
∫
θ1
6 November 2002
Physics 217, Fall 2002
6
The solenoid (continued)
a
θ1
θ2
P
z
I
For an infinite solenoid, θ 2 = 0 and θ 1 = π , so
B = zˆ
6 November 2002
2π In
4π In
zˆ = µ0 Inzˆ in MKS .
cos
0
cos
−
π
=
(
)
c
c
Physics 217, Fall 2002
7
The divergence and curl of B
Any vector field is uniquely specified by its divergence and
curl. What are the divergence and curl of B?
Consider a volume V to contain current I, current density J ( r ′ ) :
1 J ( r ′ ) × rˆ
′
B (r ) =
d
τ
c
r2
S
∫
P
Denote gradient with respect to
the components of r and r’ by
— and —′. Now note that
1
1
—   = −—′   (because r = r − r ′),
r
r
rˆ
1
and —   = − 2 .
r
r
6 November 2002
V
Physics 217, Fall 2002
J ( r ′)
r
r
r’
dτ ′
8
The divergence and curl of B (continued)
With these,
1
B (r ) = −
c
rˆ
1
1

—   × J ( r ′ ) dτ ′
× J ( r ′ ) dτ ′ =
2
c
r
r
V
V
J ( r ′)
1
dτ ′ ( remember, J ≠ f ( r ) ) .
= —×
c
r
∫
∫
∫
V
This is a useful form for B, which we will use a lot next
lecture too (the integral turns out to be the magnetic vector
potential, A). Take its divergence:

J ( r ′)
1 
— ⋅ B (r ) = — ⋅ — × ∫
dτ ′  = 0 .

r
c 
V


6 November 2002
Physics 217, Fall 2002
The divergence
of any curl is
zero, remember.
9
The divergence and curl of B (continued)
Integrate this last expression over any volume:
∫ — ⋅ B ( r ) dτ = v∫ B ⋅ da = 0
.
Compare these to the expressions for E in electrostatics, and
we see that magnetostatics involves no counterpart of charge:
there’s no “magnetic charge.”
Now for the curl:
J ( r ′)
1
— × B (r ) = — × — ×
dτ ′ .
r
c
V
Use Product Rule #10:
∫
— × — × A = — ( — ⋅ A) − ∇2 A :
6 November 2002
Physics 217, Fall 2002
10
The divergence and curl of B (continued)
 1
J ( r ′)
J ( r ′)
1 
2
dτ ′  − ∇
dτ ′
— × B (r ) = — — ⋅
 c
c 
r
r
V
 V

∫
∫
 1
J ( r ′)
1 
1
= — —⋅
J ( r ′ ) ∇ 2   dτ ′ .
dτ ′  −
 c
c 
r
r

V
V

Now use your old friend Product Rule #5,
∫
— ⋅ ( fA ) = f ( — ⋅ A ) + A ⋅ ( — f ) ,
to write
—⋅
J ( r ′)
6 November 2002
∫
r
=0 (J independent of r)
1
1
1



dτ ′ =   — ⋅ J ( r ′ ) + J ( r ′ ) ⋅ —   = J ( r ′ ) ⋅ —  
r
r
r
Physics 217, Fall 2002
11
The divergence and curl of B (continued)
 rˆ
1
∇   = — ⋅—  = — ⋅ 2
r
r
r
2 1
Also,

3
=
πδ
4
(r) ,


so — × B ( r ) = 1 — J ( r ′ ) ⋅ —  1  dτ ′ + 4π
 
c
c
r
∫
V
∫
J ( r ′ ) δ 3 ( r − r ′ ) dτ ′
V
1
4π
1
= − — J ( r ′ ) ⋅ —′   dτ ′ +
J (r ) .
c
c
r
∫
V
Use Product Rule #5 again, on the first term:
 J ( r ′)  1
 J ( r ′) 
1

J ( r ′ ) ⋅ —′   = —′ ⋅ 
 − —′ ⋅ J ( r ′ ) = —′ ⋅ 

r
r
r
r
 




=0 in magnetostatics
6 November 2002
Physics 217, Fall 2002
12
The divergence and curl of B (continued)
So,
 J ( r ′ )   4π
1 
J (r )
— × B ( r ) = − — —′ ⋅ 
 dτ ′  +
c 
r   c

V


∫
 4π
1  J ( r ′)
J (r ) .
= − —
⋅ da′  +
 c
r
c 
S

But by definition J = 0 on the surface, so the integral vanishes:
v∫
— × B (r ) =
4π
J (r ) .
c
Ampère’s Law
This can be put into integral form by choosing an area that
some current flows through:
6 November 2002
Physics 217, Fall 2002
13
The divergence and curl of B (continued)
∫
A
4π
( — × B ) ⋅ da =
c
v∫
B ⋅ dA =
C
dA
∫ J ⋅ da
C
A
4π
I enclosed
c
.
A
J
Ampère’s law is to magnetostatics what Gauss’ law is to
electrostatics, except that one
uses an Ampèrean loop to
enclose current, instead of a
Gaussian surface to enclose charge. The same tricks we
learned with Gauss’ law and superposition have analogues in
magnetostatics.
6 November 2002
Physics 217, Fall 2002
14
Example: field in an infinite solenoid
The symmetry of the coil
dictates that the field must
be along z, and must be a
lot stronger inside than out,
so if the number of turns
per unit length is n, and the
current is I,
v∫
C
4π
B ⋅ dA =
c
∫
A
∆z
C
A
4π
J ⋅ da =
I enclosed
c
4π
4π nI
zˆ
B∆z =
In∆z ⇒ B =
c
c
[ = µ0 nI in MKS ] .
Same as before!
6 November 2002
Physics 217, Fall 2002
15
Maxwell’s equations for electrostatics and
magnetostatics
Note the similarities and differences:
— ⋅ Ε = 4πρ
—⋅B = 0
—× E = 0
4π
J
—×B =
c
v∫ E ⋅ da = 4π Qencl v∫ B ⋅ da = 0
v∫ E ⋅ dA = 0
6 November 2002
4π
v∫ B ⋅ dA = c Iencl
Physics 217, Fall 2002
16
Maxwell’s equations for electrostatics and
magnetostatics
Note the similarities and differences (MKS):
ρ
—⋅Ε =
ε0
—× E = 0
6 November 2002
—⋅B = 0
— × B = µ0 J
1
v∫ E ⋅ da = ε 0 Qencl
v∫ B ⋅ da = 0
v∫ E ⋅ dA = 0
v∫ B ⋅ dA = µ0 Iencl
Physics 217, Fall 2002
17