Through the use of Ampere's Circuital Law, find the field in all regions of an infinite length coaxial cable carrying a
Lecture #16 Topics to be covered: (i) Ampere’s Circuital Law Reference: Hayt and Buck, page 218-225
Motivation: To understand the concept of Ampere’s Circuital Law and its application in calculating magnetic field
Ampere’s Circuital Law (ACL) Discovered by Andre-Marie Ampere, a French physicist in 1800s. Analogous to Gauss’s Law.
Special case of charge distribution
Special case of current distribution
Gauss’s Law
Ampere’s Circuital Law
Electric field
Magnetic field
Ampere’s Circuital Law (ACL) Gauss’s Law
Ampere’s Circuital Law
Infinite length of line charge Infinite surface charge Gaussian surface
Infinite length of filament current Infinite surface current Amperian loop
∫ D .ds
Gaussian surface
= Q en
∫ H .dl
Amperian loop
= I en
Ampere’s Circuital Law- Filament current z
I
Step 1: Create an amperian loop (closed loop that surround the filament current)
+∞
dl
Step 2: Identify the loop element, dl Step 3: Use the ACL to find the magnetic field
−∞
∫ H .dl
Amperian loop
= I en
Ampere’s Circuital Law- Filament current From previous slide;
dl = rd φ φˆ
H = H φ φˆ Hence:
∫ H . dl = I ˆ . rd φ φˆ = φ H φ ∫
Ien = I
en
H
φ
=
I
I
2π r I H = φˆ ( A / m ) 2π r
As proved from B-S Law
Ampere’s Circuital Law- Surface Current z y x
dx( xˆ ) J S = J S yˆ
Step 2: Identify the loop element Step 3: Use the ACL
dz (zˆ )
2
dz (− zˆ )
4
dx(− xˆ ) 3
Step 1: Create an amperian loop
1
Ampere’s Circuital Law- Surface Current From previous slide;
⇒ ∫ H .dl = I en 2
3
1
2
⇒ ∫ H x xˆ.dxxˆ + ∫ H x xˆ.dz (− zˆ ) + 4
1
∫ H (− xˆ ).dx(− xˆ ) + ∫ H (− xˆ ).dzzˆ = J L x
3
x
s
4
⇒ 2H x = J s Js ⇒ Hx = 2
1 ∴ H = J S × nˆ ( A / m) 2
Example 1 Plane x=10 carries current 100mA/m along ˆz while line x=1, y=-2 carries filamentary current 20π mA along ˆz . Determine H at (4,3,2).
Example 2 An infinitely long solid conductor of radius a is placed along the z-axis. If the conductor carries current I in the ˆz direction, show that:
Ir ˆ φ H = 2 2π a
Example 3 Through the use of Ampere’s Circuital Law, find the H field in all regions of an infinite length coaxial cable carrying a uniform and equal current I in opposite directions in the inner and outer conductors. Assume the inner conductor to have a radius of a (m) and the outer conductor to have an inner radius of b (m) and an outer radius of c (m). Assume that the cable’s axis is along the z axis.
Example 4 Through the use of Ampere’s Circuital Law, find the H field inside and outside an infinite-length hollow conducting tube whose radius is 0.002 m that carries a current I = 10-7 A, directed in the + ˆz direction. Consider the thickness of the tube very small.
Example 5
Consider two wire transmission line whose cross section is illustrated in figure below. Each wire is of radius 2 cm and the wires are separated by 10 cm. The wire centered at (0,0) carries current 5 A in the direction of + ˆz , while the other centered at (10cm,0) carries the return current. Find H at: (5cm, 0) and (10cm, 5 cm) y
x 10 cm
Next Lecture Please have a preliminary observation on the following topics: (i) Stokes’ Theorem (ii) Magnetic Flux Density (iii) Maxwell’s equation for static field (iv) Vector Magnetic Potential Please refer to Hayt and Buck, page 225-246
