Today’s agenda: Electric Current. You must know the definition of current, and be able to use it in solving problems. Current Density. You must understand the difference between current and current density, and be able to use current density in solving problems. Ohm’s Law and Resistance. You must be able to use Ohm’s Law and electrical resistance in solving circuit problems. Resistivity. You must understand the relationship between resistance and resistivity, and be able to use calculate resistivity and associated quantities. Temperature Dependence of Resistivity. You must be able to use the temperature coefficient of resistivity to solve problems involving changing temperatures.

This makes senseThis makes sense: a longer wire or higher-resistivity wire should have a greater resistance. A larger area means more “space” for electrons to get through, hence lower resistance. It is also experimentally observed (and justified by quantum mechanics) that the resistance of a metal wire is well-described by Resistivity where  is a “constant” called the resistivity of the wire material, L is the wire length, and A its cross-sectional area.

L The longer a wire, the “harder” it is to push electrons through it. R =  L / A,  The greater the resistivity, the “harder” it is to push electrons through it. The greater the cross-sectional area, the “easier” it is to push electrons through it. A Resistivity is a useful tool in physics because it depends on the properties of the wire material, and not the geometry. units of  are  m

Resistivities range from roughly  ·m for copper wire to  ·m for hard rubber. That’s an incredible range of 23 orders of magnitude, and doesn’t even include superconductors (we might talk about them some time). R =  L / A A =  L / R A =  (d/2) 2 geometry!  (d/2) 2 =  L / R Example (will not be worked in class): Suppose you want to connect your stereo to remote speakers.will not be worked in class (a) If each wire must be 20 m long, what diameter copper wire should you use to make the resistance 0.10  per wire.

(d/2) 2 =  L /  R d/2= (  L /  R ) ½ don’t skip steps! d = 2 (  L /  R ) ½ d = 2 [ (1.68x10 -8 ) (20) /  (0.1) ] ½ d = m = 2.1 mm V = I R (b) If the current to each speaker is 4.0 A, what is the voltage drop across each wire?

V = (4.0) (0.10) V = 0.4 V

Homework hint you can look up the resistivity of copper in a table in your text.

Ohm’s Law Revisited The equation for resistivity I introduced five slides back is a semi-empirical one. Here’s almost how we define resistivity: Our equation relating R and  follows from the above equation. We define conductivity  as the inverse of the resistivity: NOT an official starting equation!

With the above definitions, The “official” Ohm’s law, valid for non-ohmic materials. Cautions! In this context:  is not volume density!  is not surface density! Think of this as our definition of resistivity. In anisotropic materials,  and  are tensors. A tensor is like a matrix, only worse.

Example: the 12-gauge copper wire in a home has a cross- sectional area of 3.31x10 -6 m 2 and carries a current of 10 A. Calculate the magnitude of the electric field in the wire. Homework hint (not needed in this particular example): in this chapter it is safe to use  V=Ed. Question: are we still assuming the electrostatic case?