Question

Consider a wire of length L and a circular cross section of radius a. The ends of the wire are maintained at a potential difference of V. Let the current in the wire be I and assume that the metal of the wire obeys Ohm's Law, J = ? E where ? is the metal's conductivity (which is a real constant). Prove that the problem parameters must be related as V = I R and give the expression for the resistance, R, of the wire.

          Consider a wire of length L and a circular cross section of radius a. The ends of the wire are maintained at a potential difference of V. Let the current in the wire be I and assume that the metal of the wire obeys Ohm's Law, J = ? E where ? is the metal's conductivity (which is a real constant). Prove that the problem parameters must be related as V = I R and give the expression for the resistance, R, of the wire.
        
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Consider a wire of length L and a circular cross section of radius a. The ends of the wire are maintained at a potential difference of V. Let the current in the wire be I and assume that the metal of the wire obeys Ohm's Law, J = ? E where ? is the metal's conductivity (which is a real constant). Prove that the problem parameters must be related as V = I R and give the expression for the resistance, R, of the wire.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Consider a wire of length L and circular cross section of radius a. The ends of the wire are maintained at a potential difference of V. Let the current in the wire be I and assume that the metal of the wire obeys Ohm's Law, J = σ E where σ is the metal's conductivity (which is a real constant). Prove that the problem parameters must be related as V = I R and give the expression for the resistance, R, of the wire.
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Transcript

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00:01 All right, so let's say we have a wire that has a length l and a cross -sectional area a, or we'll call it a rate, say it has a radius of a.
00:13 And it obeys sort of a certain version of oamslaw, which says the current density is equal to sigma times the electric field.
00:21 So what we'll do is just like write j over sigma.
00:26 This is our electric field strength across the wire.
00:30 Let's integrate this from the length of the wire.
00:34 So e...
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