### About Global Documents

**Global Documents provides you with documents from around the globe on a variety of topics for your enjoyment. **

**Global Documents utilizes edocr for all its document needs due to edocr's wonderful content features. Thousands of professionals and businesses around the globe publish marketing, sales, operations, customer service and financial documents making it easier for prospects and customers to find content. **

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