Today in Physics 217: Ampère's Law
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
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
