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

Magnetic Field rr Lorentz Force Law: N S q In general, (with both E and B present): This experimental law defines the magnetic flux density vector B. The units of B are Webers/m 2 or Tesla [T]. v

Magnetic Gauss Law x y z B S (closed surface) N Magnetic pole (not possible) ! N S 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 y x Iron filings I (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: The units of H are [A/m]. H is called the “magnetic field” (for single infinite wire)

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

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

I encl Hence “Right-Hand Rule” 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. Ampere’s Law Ampere’s Law (cont.) C

Amperes’ Law: Differential Form C S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: “displacement current” (This term was added by Maxwell.) h A insulator I z Q 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 I encl 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!

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

Example (cont.) y x I C  z r

I S 3 ) H  = 0 h Magnetic Gauss law: 2 ) H z = 0 I C h  H z dz = 0 H  d  cancels  = 

Example (cont.) y x I  z r

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

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

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

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

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

Example (cont.)  > a h HzHz C 

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

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

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

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

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

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

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

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

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

Example (cont.) Hence