Prof. D. Wilton ECE Dept.

ECE 2317 Applied Electricity and Magnetism Prof. D. Wilton ECE Dept. Notes 25 Notes prepared by the EM group, University of Houston.

Magnetic Field v r N S q This experimental law defines the magnetic flux density vector B. Lorentz Force Law: In general, (with both E and B present): The units of B are Webers/m2 or Tesla [T].

Magnetic Gauss Law z S (closed surface) N N y x S B Magnetic pole (not possible) ! S No net flux out !

Magnetic Gauss Law: Differential Form Apply the divergence theorem: Since this applies for any volume, B field flux lines either close on themselves or at infinity!

Ampere’s Law I y x Iron filings (exact value)

Ampere’s Law (cont.) Note: the definition of one Amp is as follows: 1 [A] current produces: So

Ampere’s Law (cont.) Define: H is called the “magnetic field” The units of H are [A/m]. (for single infinite wire)

Ampere’s Law (cont.) y C I x The current is inside a closed path.

y C I x Ampere’s Law (cont.) The current is outside a closed path.

C Ampere’s Law (cont.) Hence Ampere’s Law Iencl Although this law was “derived” for an infinite wire of current, the assumption is made that it holds for any shape current. This is an experimental law. “Right-Hand Rule”

Amperes’ Law: Differential Form C S (from Stokes’ theorem) S is a small planar surface Since the unit normal is arbitrary,

Maxwell’s Equations (Statics) electric Gauss law magnetic Gauss law Faraday’s law Ampere’s law In contrast to electric fields, the sources for magnetic fields produce a circulation and are vectors! !

Maxwell’s Equations (Dynamics) electric Gauss law magnetic Gauss law Faraday’s law Ampere’s law

Displacement Current Ampere’s law: A “displacement current” insulator h z + + + + + + + + + + + (This term was added by Maxwell.) Q I The current density vector that exists inside the lower plate.

Ampere’s Law: Finding H • 1) The “Amperian path” C must be a closed path. • 2) The sign of Iencl is from the RH rule. • 3) Pick C in the direction of H (as much as possible). • 4) Wherever there is a component of H along path C , it • must be constant. => Symmetry!

z I y  C r x Example Calculate H First solve for H . “Amperian path” An infinite line current along the z axis.

z I y  C r x Example (cont.)

Example (cont.) Hd cancels 2)Hz = 0 Hzdz=0 I h  C  =  3)H = 0 I Magnetic Gauss law: h S

z I y  r x Example (cont.)

Example y coaxial cable b I a c a I r x C b z c The outer jacket of the coax has a thickness of t = c-b  < a Note: the permittivity of the material inside the coax does not matter here. b <  < c

y b a r x C c Example (cont.) The other components are zero, as in the wire example. This formula holds for any radius, as long as we get Ienclcorrect.

y b a r x C c Example (cont.)  < a a < < b b < < c  > c Note: There is no magnetic field outside of the coax (“shielding property”).

Example Solenoid n = N/L= # turns/meter a Calculate H z L I Find Hz  < a h C 

 > a Hz C h  Example (cont.)

Example (cont.) The other components of the magnetic field are zero: 1)H = 0 since C 2)H = 0from S

n = N/L = # turns/meter a z L I Example (cont.) Summary:

Example z Calculate H y - x x + Infinite sheet of current top view y

Hx- x Hx+ y Example (cont.) (superposition of line currents) (magnetic Gauss Law) By symmetry:

Example (cont.) w - 2y x + C y Note: No contribution from the left and right edges (the edges are perpendicular to the field).

Example (cont.) Note: The magnetic field does not depend on y.

y I h I x w z w >> h Example Find H everywhere

Example (cont.) y h x Two parallel sheets of opposite surface current

“bot” “top” y I h -I x magnetic field due to top sheet magnetic field due due to bottom sheet Example (cont.) Infinite current sheet approximation of finite width current sheets

Example (cont.) Hence

Prof. D. Wilton ECE Dept.