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cse63146

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## Homework Statement

1. What electric field strength is needed to create a 5.0 A current in a 2.0-mm-diameter iron wire?

2. You need to design a 1.0 A fuse that "blows" if the current exceeds 1.0 A. The fuse material in your stockroom melts at a current density of 500 A/cm^2. What diameter wire of this material will do the job?

3.A car battery is rated at 90 Ahr, meaning that it can supply a 90 A current for 1 hr before being completely discharged. If you leave your headlights on until the battery is completely dead, how much charge leaves the battery? Express your answer in Coulombs to three significant figures

4. The total amount of charge in coulombs that has entered a wire at time t is given by the expression Q = 4t - t^2 , where t is in seconds and t>=0.

Which is an expression for the current in the wire at time t.

http://img100.imageshack.us/img100/5448/currentequationsgd2.jpg

## Homework Equations

E = [tex]\frac{I}{(\sigma)(\pi)r^2}[/tex]

J = [tex]\frac{I}{(\pi)(r^2)}[/tex]

Q = [tex]\frac{I}{t}[/tex]

## The Attempt at a Solution

1. I know [tex]\sigma[/tex] is 10^7 so

E = [tex]\frac{5}{(10^7)(\pi)(0.001^2)}[/tex] = 0.159 N/c

2. 500x10^-2=5 A/m^2

r = [tex]\sqrt{\frac{I}{(\pi)J}}[/tex] = [tex]\sqrt{\frac{1}{(\pi)5}}[/tex] = 0.252 m and to get diameter I multiply it by two to get 0.504. The question asks me to express it in mm so I got 504mm.

3. 1 hour = 3600 seconds

Q = [tex]\frac{I}{t}[/tex] = Q = [tex]\frac{90}{3600}[/tex] = 0.025 C

4. Current is the derivative of charge so it would be the third equation.

Did I make a math (or any other kind) error?

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