+ Jumper Cables and Ohm's Law 3 of 20 > Review | Constants Part B Learning Goal: To learn to apply the concept of current density and microscopic Ohm's law. What electric field E' would have been required to create current of 30.0 A in an aluminum wire of the same diameter? Express your answer in V/m. Use three significant figures. A "gauge 8" jumper cable has a diameter d of 0.326 cm. The cable carries a current I of 30.0 A. The electric field E in the cable is 0.062 V/m. • View Available Hint(s) Below is a list of some common metals with their Dνα ΑΣφ ? resistivities. Resistivity e (N. m) Material E' V/m
+ Jumper Cables and Ohm's Law 3 of 20 > Review | Constants Part B Learning Goal: To learn to apply the concept of current density and microscopic Ohm's law. What electric field E' would have been required to create current of 30.0 A in an aluminum wire of the same diameter? Express your answer in V/m. Use three significant figures. A "gauge 8" jumper cable has a diameter d of 0.326 cm. The cable carries a current I of 30.0 A. The electric field E in the cable is 0.062 V/m. • View Available Hint(s) Below is a list of some common metals with their Dνα ΑΣφ ? resistivities. Resistivity e (N. m) Material E' V/m
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![t Jumper Cables and Ohm's Law
3 of 20
>
Review | Constants
Part B
Learning Goal:
To learn to apply the concept of current density and
microscopic Ohm's law.
What electric field E' would have been required to create
current of 30.0 A in an aluminum wire of the same
diameter?
Express your answer in V/m. Use three significant figures.
A "gauge 8" jumper cable has a diameter d of 0,326
cm. The cable carries a current I of 30.0 A. The
electric field E in the cable is 0.062 V/m
• View Available Hint(s)
Below is a list of some common metals with their
resistivities.
?
Resistivity
p (N. m)
Material
E' =
V/m
copper
1.72 x 10-8
Submit
silver
1.47 x 10-8
aluminum 2.75 x 10 8
Part C Complete previous part(s)
gold
2.44 x 10-8
Part D Complete previous part(s)
steel
20 x 10-8
Next >
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Transcribed Image Text:t Jumper Cables and Ohm's Law
3 of 20
>
Review | Constants
Part B
Learning Goal:
To learn to apply the concept of current density and
microscopic Ohm's law.
What electric field E' would have been required to create
current of 30.0 A in an aluminum wire of the same
diameter?
Express your answer in V/m. Use three significant figures.
A "gauge 8" jumper cable has a diameter d of 0,326
cm. The cable carries a current I of 30.0 A. The
electric field E in the cable is 0.062 V/m
• View Available Hint(s)
Below is a list of some common metals with their
resistivities.
?
Resistivity
p (N. m)
Material
E' =
V/m
copper
1.72 x 10-8
Submit
silver
1.47 x 10-8
aluminum 2.75 x 10 8
Part C Complete previous part(s)
gold
2.44 x 10-8
Part D Complete previous part(s)
steel
20 x 10-8
Next >
Provide Feedback
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