Ohm’s Law and Resistance. Resistivity. 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.

Resistivity It is also experimentally observed (and justified by quantum mechanics) that the resistance of a metal wire is well-described by where  is a “constant” called the resistivity of the wire material, L is the wire length, and A its cross-sectional area. This 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.

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

Resistivities range from roughly 10-8 ·m for copper wire to 1015 ·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). Example (will not be worked in class): Suppose you want to connect your stereo to remote speakers. (a) If each wire must be 20 m long, what diameter copper wire should you use to make the resistance 0.10  per wire. R = L / A A = L / R A =  (d/2)2 geometry!  (d/2)2 = L / R

(d/2)2 = L / R d/2= ( L / R )½ d = 2 ( L / R )½ don’t skip steps! d = 2 ( L / R )½ d = 2 [ (1.68x10-8) (20) /  (0.1) ]½ d = 0.0021 m = 2.1 mm (b) If the current to each speaker is 4.0 A, what is the voltage drop across each wire? V = I R

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.

NOT an official starting equation! 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: NOT an official starting equation! Our equation relating R and  follows from the above equation. We define conductivity  as the inverse of the resistivity:

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

Question: are we still assuming the electrostatic case? Example: the 12-gauge copper wire in a home has a cross-sectional area of 3.31x10-6 m2 and carries a current of 10 A. Calculate the magnitude of the electric field in the wire. Question: are we still assuming the electrostatic case? Homework hint (not needed in this particular example): in this chapter it is safe to use V=Ed.