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+ 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 > Provide Feedback
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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