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Announcements Homework for tomorrow… Ch. 30: CQ 11, Probs. 20, 22, & 32 CQ3: The answer lies in your (recently used) garden hose (look inside while turning it on). CQ4: No 30.4: 9.3 x 10 -4 m 30.6: a) 4.6 x 10 21 b) 4.3 x 10 -12 m Office hours… MW 12:30-1:30 pm TR 9-10 am F 1-2 pm Tutorial Learning Center (TLC) hours: MW 8-10 am, 12-6 pm R 8-12 pm, 2-6 pm T 8-1 pm, 2-6 pm F 8-10 am, 2-5 pm
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Announcements

Dec 30, 2015

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Announcements. Homework for tomorrow… (Ch . 30, CQ 11, Probs. 18, 20, 38) 30.3: 2.1 x 10 24 30.6: a) 4.6 x 10 21 b) 4.4 x 10 -12 m 30.7: 1.0 x 10 19 s -1 Office hours… MW 12:30-1:30 pm TR 9-10 am F 10-11 am Tutorial Learning Center (TLC) hours: - PowerPoint PPT Presentation
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Page 1: Announcements

Announcements Homework for tomorrow…

Ch. 30: CQ 11, Probs. 20, 22, & 32CQ3: The answer lies in your (recently used) garden hose (look inside while turning it on).CQ4: No30.4: 9.3 x 10-4 m30.6: a) 4.6 x 1021b) 4.3 x 10-12 m

Office hours…MW 12:30-1:30 pm TR 9-10 am F 1-2 pm

Tutorial Learning Center (TLC) hours:MW 8-10 am, 12-6 pm R 8-12 pm, 2-6 pm T 8-1 pm, 2-6 pm F 8-10 am, 2-5 pm

Page 2: Announcements

Chapter 30

Current & Resistance(Conductivity and Resistivity &

Resistance and Ohm’s Law)

Page 3: Announcements

Current..

Current Density..

Kirchoff’s Junction Rule..

Current Density related to the E-field..

Review…

Page 4: Announcements

The current density in this wire is

1. 4 106 A/m2.

2. 2 106 A/m2.

3. 4 103 A/m2.

4. 2 103 A/m2.

5. Some other value.

Quiz Question 1

Page 5: Announcements

How the current density, J, is related to the E-field driving the current:

where

SI Units?

30.4:Conductivity and Resistivity

Page 6: Announcements

How the current density, J, is related to the E-field driving the current:

Where

SI Units?

30.4:Conductivity and Resistivity

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Both segments of the wire are made of the same metal. Current I1 flows into segment 1 from the left.

How does current I1 in segment 1 compare to current I2 in segment 2?

1. I1 > I2.

2. I1 = I2.

3. I1 < I2.

4. There’s not enough information to compare them.

Quiz Question 2

Page 8: Announcements

Both segments of the wire are made of the same metal. Current I1 flows into segment 1 from the left.

How does current density J1 in segment 1 compare to current density J2 in segment 2?

1. J1 J2.

2. J1 J2.

3. J1 J2.

4. There’s not enough information to compare them.

Quiz Question 3

Page 9: Announcements

Both segments of the wire are made of the same metal. Current I1 flows into segment 1 from the left.

How does the electric field E1 in segment 1 compare to the electric field E2 in segment 2?

1. E1 > E2.

2. E1 = E2 but not zero.

3. E1 < E2.

4. Both are zero because metal is a conductor.

5. There’s not enough information to compare them.

Quiz Question 4

Page 10: Announcements

Define the resistivity…

SI Units?

30.4:Conductivity and Resistivity

Page 11: Announcements

Define the resistivity…

SI Units?

30.4:Conductivity and Resistivity

Page 12: Announcements

Define the resistivity…

SI Units?

30.4:Conductivity and Resistivity

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A 2.0 mm diameter aluminum wire carries a current of 800 mA. What is the E-field strength inside the wire?

i.e. 30.6:The E-field in a wire

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How is the current, I, related to the potential difference, ΔV in a wire?

30.5:Resistance and Ohm’s Law

Page 15: Announcements

How is the current, I, related to the potential difference, ΔV in a wire?

SI Units of R?

30.5:Resistance and Ohm’s Law

- Ohm’s Law

Page 16: Announcements

How is the current, I, related to the potential difference, ΔV in a wire?

SI Units of R?

30.5:Resistance and Ohm’s Law

- Ohm’s Law

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Notice: Resistivity, ρ, describes just the material. Resistance, R, characterizes a specific piece of the

conductor with a specific geometry.

30.5:Resistance and Ohm’s Law

Resistance of a wire

Page 18: Announcements

Wire 2 has twice the length and twice the diameter of wire 1. What is the ratio R2/R1 of their resistances?

1. 1/4.

2. 1/2.

3. 1.

4. 2.

5. 4.

Quiz Question 5

Page 19: Announcements

A battery is a source of potential difference Vbat.

The battery creates a potential difference Vwire = Vbat between the ends of the wire.

The potential difference in the wire Vwire generates an E-field in the wire.

The E-field establishes a current I = JA = σAE in the wire.

The current in the wire is determined jointly by the battery and the wire’s resistance, R to be:

I Vwire/R

Resistance and Ohm’s Law