Prof. David R. Jackson ECE Dept. Fall 2014 Notes 3 ECE 2317 Applied Electricity and Magnetism Notes prepared by the EM Group University of Houston 1.

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Prof. David R. Jackson ECE Dept. Fall 2014 Notes 3 ECE 2317 Applied Electricity and Magnetism Notes prepared by the EM Group University of Houston 1

Current is the flow of charge: the unit is the Ampere (Amp)Current 1 Amp  1 [ C/s ] Convention (Ben Franklin): Current flows in the direction that positive charges move in Current flows from left to right. Velocity Ampere 2

Current (cont.) Definition of I = 1 Amp: I I d Fx2Fx2 x # 1 # 2 Two infinite wires carrying DC currents The Amp is actually defined first, and from this the coulomb is defined. 3 From later in the semester: Note:  0 = permeability of free space.

Postulate: Positive charges moving one way is equivalent to negative charges moving the other way (in terms of measurable physical effects). Current (cont.) 1 [A] ---- Flow rate is 1 [ C ] per second Velocity This is equivalent to: ++++ Flow rate is 1 [ C ] per second Velocity 4 In both cases: Current flows from right to left.

Sign convention: A positive current flowing one way is equivalent to a negative current flowing the other way. Current (cont.) [A] Flow rate is 1 [ C ] per second Velocity - 1 [A] or Note: The blue arrow is called the reference direction arrow. 5 It is very useful in circuit theory to assume reference directions and allow for negative current.

Mathematical definition of current Current (cont.) Reference direction arrow  Q = amount of charge (positive or negative) that crosses the plane in the direction of the reference arrow in time  t. I QQ More generally, 6 Uniform currentNon-uniform current

Example i(t)i(t) QQ Find the charge Q(t) that crosses the dashed line going from left to right in the time interval ( 0, t ) [ S ]. 7

Example (cont.) 8

Current Density Vector J The current-density vector points in the direction of current flow (the direction of positive charge motion). v Consider a general “cloud” of charge density, where the charge density as well as the velocity of the charges may be different at each point. 9

The magnitude of the current-density vector tells us the current density (current per square meter) that is crossing a small surface that is perpendicular to the current-density vector. I = the current crossing the surface  S in the direction of the velocity vector. Current Density Vector (cont.) 10 v v Hence

Current Density Vector (cont.) 11 Consider a small tube of moving charges inside the cloud: v v Tube

Current Density Vector (cont.)  L = distance traveled by charges in time  t. or Hence so I LL QQ v Reference plane Tube

Current Density Vector (cont.) Summary 13

Current Crossing a Surface This is the current crossing the surface  S in the direction of the unit normal. J 14 The surface  S does not have to be perpendicular to the current density vector.

Current Crossing Surface Integrating over a surface, Note: The direction of the unit normal vector determines whether the current is measured going in or out. S J 15

Example Cloud of electrons N e = 8.47  [ electrons / m 3 ] (a) Find: current density vector The cloud of electrons is inside of a copper wire. Hence There are 8.47  atoms / m 3 There is one electron/atom in the conduction band. Notes: z

Example (cont.) (b) Find: current in wire for the given reference direction I Radius a = 1 [ mm ] I Note: The wire is neutral, but the positive nuclei do not move. 17

Example (cont.) Density of Cu: 8.94 [g/cm 3 ] Avogadro’s constant:  atoms/mol 1 mol = amount of material in grams equal to the atomic weight of atom Atomic weight of Cu: (atomic number is 29, but this is not needed) 1 electron per atom in the conduction band ( N e = atoms / m 3 ) Properties of copper (Cu) 18 Using knowledge of chemistry, calculate the value of N e (that is used in this example).

Example x y z (1,0,0) ( 0,1,0) Find the current I crossing the surface S in the upward direction. 19 ( 1,1,0) S

Example (cont.) 20 x y 1 1 Note: The integrand is separable, and the limits of integration are fixed numbers. Hence, we can split this into a product of two one-dimensional integrals.

Example x y z (1,0,0) ( 0,1,0) S Find the current I crossing the surface S in the upward direction. 21

Example (cont.) x y 1 1 y(x) = 1 - x 22 Note: The integrand is separable, but the limits of integration are not fixed numbers. Hence, we cannot split this into a product of two one- dimensional integrals.