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PHY 5346
Homework Set 10 Solutions – Kimel
2. 5.2 a) The system is described by
First consider a point at the axis of the solenoid at point z0. Using the results of problem 5.1,
dφm =
μ0
4π
NIdzΩ
From the figure,
Ω = ∫ r̂ ⋅ dA⃗
r2
= ∫ dAcosθ
r2
= 2πz ∫
0
R
ρdρ
ρ2 + z2 3/2
= 2π −
z
R2 + z2
+ 1
φm =
μ0
2
NI ∫
z0
∞
z −
1
R2 + z2
+ 1z
dz =
μ0
2
NI −z0 +
R2 + z0
2
Br = −
μ0
2
NI ∂
∂z0
−z0 +
R2 + z0
2 =
μ0
2
NI
−z0 +
R2 + z0
2
R2 + z0
2
In the limit z0 → 0
Br =
μ0
2
NI
By symmetry, thej loops to the left of z0 give the same contribution, so
B = B l + Br = μ0NI
H = NI
By symmetry, B⃗ is directed along the z axis, so
δφm = −δρ⃗ ⋅ B⃗ = 0
if δρ⃗ is directed ⊥ to the z axis. Thus for a given z, φm is independent of ρ, and consequently
H = NI
everywhere within the solenoid.
If you are on the outside of the solenoid at position z0, by symmetry the magnetic field must be in
the z direction. Thus using the above argument, φm must not depend on ρ. Set us take ρ far away
from the axis of the solenoid, so that we can replace the loops by elementary dipoles m⃗ directed along
the z axis. Thus for any point z0 we will have a contributions
φm
m⃗ ⋅ r⃗1
r1
3
+
m⃗ ⋅ r⃗2
r2
3
where m⃗ ⋅ r⃗1 = −m⃗ ⋅ r⃗2 and r1 = r2. Thus
H = 0
Homework Set 10 Solutions – Kimel
2. 5.2 a) The system is described by
First consider a point at the axis of the solenoid at point z0. Using the results of problem 5.1,
dφm =
μ0
4π
NIdzΩ
From the figure,
Ω = ∫ r̂ ⋅ dA⃗
r2
= ∫ dAcosθ
r2
= 2πz ∫
0
R
ρdρ
ρ2 + z2 3/2
= 2π −
z
R2 + z2
+ 1
φm =
μ0
2
NI ∫
z0
∞
z −
1
R2 + z2
+ 1z
dz =
μ0
2
NI −z0 +
R2 + z0
2
Br = −
μ0
2
NI ∂
∂z0
−z0 +
R2 + z0
2 =
μ0
2
NI
−z0 +
R2 + z0
2
R2 + z0
2
In the limit z0 → 0
Br =
μ0
2
NI
By symmetry, thej loops to the left of z0 give the same contribution, so
B = B l + Br = μ0NI
H = NI
By symmetry, B⃗ is directed along the z axis, so
δφm = −δρ⃗ ⋅ B⃗ = 0
if δρ⃗ is directed ⊥ to the z axis. Thus for a given z, φm is independent of ρ, and consequently
H = NI
everywhere within the solenoid.
If you are on the outside of the solenoid at position z0, by symmetry the magnetic field must be in
the z direction. Thus using the above argument, φm must not depend on ρ. Set us take ρ far away
from the axis of the solenoid, so that we can replace the loops by elementary dipoles m⃗ directed along
the z axis. Thus for any point z0 we will have a contributions
φm
m⃗ ⋅ r⃗1
r1
3
+
m⃗ ⋅ r⃗2
r2
3
where m⃗ ⋅ r⃗1 = −m⃗ ⋅ r⃗2 and r1 = r2. Thus
H = 0