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PHYS-2020: General Physics II

Course Lecture Notes

Section III

Dr. Donald G. Luttermoser East Tennessee State University

Edition 3.

Abstract

These class notes are designed for use of the instructor and students of the course PHYS-2020: General Physics II taught by Dr. Donald Luttermoser at East Tennessee State University. These notes make reference to the College Physics, Enhanced 7th Edition (2006) textbook by Serway, Faughn, and Vuille.

III–2 PHYS-2020: General Physics II

3. Electrons flow in the opposite direction of the E~-field. a) As an electron (or any charged particle) moves through a conductor, it collides with atoms (and/or molecules) in the conductor =⇒ causes a zigzag motion through the conductor.

b) The amount of charge passing through a wire can be de- termined as follows: i) Let A be the cross-sectional area of a wire and ∆x be a small slice along the length of the wire.

ii) The volume of this small segment of the wire is then V = A ∆x (note that V here is volume not voltage).

iii) Let N be the number of charge carriers contained in this volume and q be the charge per carrier. Then, n =

N

V

N

A ∆x

represents the number of carriers per unit volume.

iv) The total charge contained in this volume is thus ∆Q = Nq = (n A ∆x) q. (III-3)

c) Although the electron makes a zigzag path through the wire, on average, it continues to move down the electric field (remember in the opposite sense) at an average speed called the drift speed vd: vd = ∆x ∆t =⇒ ∆x = vd ∆t. We can then substitute this into Eq. (III-3) giving ∆Q = (n A vd ∆t) q. (III-4)

Donald G. Luttermoser, ETSU III–

d) Dividing both sides by ∆t gives ∆Q ∆t = I = n q vd A (III-5) or vd =

I

n q A

Since we normally talk about electrons in a metal wire, we can rewrite this as vd = I n |e| A

. (III-6)

e) If no current exists in a conductor, the electric field is zero inside the conductor. However, if current exists, an electric field exists inside the conductor (due to Maxwell’s laws — see §IX of the notes).

Example III–1. Problem 17.7 (Page 588) from the Serway & Faughn textbook: A 200-km long high-voltage transmission line 2.0 cm in diameter carries a steady current of 1000 A. If the con- ductor is copper with a free charge density of 8. 5 × 1028 electrons per cubic meter, how long (in years) does it take one electron to travel the full length of the cable? Added question: How long would it take a photon to travel the same distance? Solution: The drift speed of electrons in the line is (from Eq. III-6)

vd =

I

n q A

I

n |e| (πD^2 /4)

where I = 1000 A = 1000 C/s is the current, n = 8. 5 × 1028 electrons/m^3 is the charge carrier density, |e| = 1. 60 × 10 −^19 C is the charge per electron, and D = 2.0 cm = 0.020 m is the diameter of the wire. Solving for the drift velocity gives

vd = 4(1000 C/s) (8. 5 × 1028 m−^3 )(1. 60 × 10 −^19 C)π(0.020 m)^2 = 2.^3 ×^10

− (^4) m/s.

Donald G. Luttermoser, ETSU III–

2. Ohm’s Law: Resistance that remains constant over a wide range of applied voltage differences such that the voltage difference is linearly dependent on current:

∆V = I R. (III-9)

a) Materials that obey this law are called ohmic =⇒ con- ductors are ohmic.

b) Materials that do not obey this law are called nonohmic =⇒ semiconductors are nonohmic.

3. A resistor is a simple circuit element that provides a specific resistance in an electric circuit. It’s symbol is (^) AA
   
   
   AAA
   
   
   AAA
   
   
   AAA
   
   
   AA

C. Resistivity.

1. Since resistance is related to the number of collisions an electron has with atoms/molecules of the wire, we can describe resistance in terms of the geometric properties of the conductor and the composition of the conductor (i.e., ohmic material):

R = ρ

L

A

, (III-10)

where L is the length of the conductor, A is the cross-sectional area of the conductor (both of these are the geometric part), and ρ is the resistivity of the material (which is related to the composition of the material — see Table 17.1 in the textbook).

2. Electric conductors have low resistivity, insulators have high re- sistivity.
3. Resistivities can change with temperature (see §III.D).

III–6 PHYS-2020: General Physics II

Example III–2. Problem 17.17 (Page 588) from the Ser- way & Faughn textbook: A wire 50.0 m long and 2.00 mm in di- ameter is connected to a source with a potential difference of 9.11 V, and the current is found to be 36.0 A. Assume a temperature of 20 ◦C and, using Table 17.1, identify the metal of the wire. Solution: From Ohm’s law (Eq. III-9) we can calculate the resistance of the wire to be R = ∆V I

=^9 .11 V

36 .0 A

The diameter of the wire is D = 2.00 mm = 2. 00 × 10 −^3 m, so the resistivity (using Eq. III-10) of the metal is

ρ =

R A

L

R (πD^2 /4) L

(0.253 Ω)π(2. 00 × 10 −^3 m)^2 4(50.0 m) = 1. 59 × 10 −^8 Ω · m. Comparing this value to those listed in Table 17.1 of the textbook, the metal must be silver.

D. Temperature Variation of Resistance.

1. For most metals, resistivity increases in an approximate linear fashion with temperature: ρ = ρ◦ [1 + α (T − T◦)]. (III-11) a) ρ is the resistivity at some temperature T.

b) ρ◦ is the resistivity at some temperature T◦.

c) α is the temperature coefficient of resistivity (see Table 17.1).

III–8 PHYS-2020: General Physics II

E. Superconductors.

1. There is a class of metals and compounds whose resistance vir- tually goes to zero below a certain temperature, Tc, called the critical temperature (see Figure 17.9 in the textbook).
2. These materials are known as superconductors.
3. Once current has been set up in a superconductor, it persists without any applied voltage (since R = 0).

F. Electrical Energy and Power.

1. A battery can deliver power to an electric circuit. Power (as covered in General Physics I) is the amount of work exerted over an interval of time: P =

W

∆t

. (III-13)

2. Since work is equal to the change of potential energy (see Eq. II-3), we can write

W = ∆PE = q ∆V

or P = q

∆V

∆t

. (III-14)

3. If we have a current of charges ∆q across a voltage difference ∆V , we can rewrite Eq. (III-14) as

P = ∆q

∆V

∆t

= ∆V

∆q ∆t

= (∆V ) I

or P = I ∆V. (III-15)

Donald G. Luttermoser, ETSU III–

As we can see from this equation, the unit for power (W = watt) must be equal to amperes times volts, or

1 W = A · V. (III-16)

4. Using Ohm’s law (Eq. III-9), we can also express Eq. (III-15) as

P = I^2 R = (∆V^ )

2 R

. (III-17)

a) The power delivered to a conductor of resistance R is often referred to as an I^2 R loss.

b) Note that Eq. (III-17) applies only to resistors (and other ohmic devices) and not to nonohmic devices like light- bulbs and diodes.

c) Electric companies measure the amount of power you have used over the billing cycle (= total energy used). The units used by the electric company are kilowatt-hour:

1 kWh = (10^3 W)(3600 s) = 3. 60 × 106 J. (III-18)

Example III–4. Problem 17.32 (Page 589) from the Ser- way, Faughn, & Vuille textbook: If the electrical energy costs 12 cents, or $0.12, per kilowatt-hour, how much does it cost to (a) burn a 100-W lightbulb for 24 hr, and (b) operate an electric oven for 5.0 hr if it carries a current of 20.0 A at 220 V? Solution (a): The energy produced by the 100-W lightbulb in one day (24 hr) is

E = P · ∆t = (100 W)(24 hr) = (0.100 kW)(24 hr) = 2.40 kWh ,

where kWh = kilowatt-hour. At a rate of R = 12-cents (= $0.12)