Learning Goal: To learn to apply the concept of current density and microscopic

Question

Learning Goal:

To learn to apply the concept of current density and microscopic Ohm's law.

A slab of metal of volume V is made into a rod of length L. The rod carries current I when the electric field inside is E.

Part A

Find the resistivity of the metal ?.

Expess your answer in terms of the given quantities.

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Part B

The rod is now stretched so that its length is doubled. If the electric field remains the same, what is the new current I? in the rod?

Express your answer in terms of some or all of the quantities given in the problem introduction.

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Part C

A piece of copper is made into a rod with a square cross-section. The side of the square is 2.00 centimeters. The resistivity of copper is 1.7?10?8 ??m. An unknown electric field E, directed along the rod, creates a current of 12.0 amperes through the rod. Find the magnitude of E.

Use two significant figures in your answer. Express your answer in newtons per coulomb.

Learning Goal:

To learn to apply the concept of current density and microscopic Ohm's law.

A slab of metal of volume V is made into a rod of length L. The rod carries current I when the electric field inside is E.

Part A

Find the resistivity of the metal ?.

Expess your answer in terms of the given quantities.

SubmitMy AnswersGive Up

Part B

The rod is now stretched so that its length is doubled. If the electric field remains the same, what is the new current I? in the rod?

Express your answer in terms of some or all of the quantities given in the problem introduction.

I? =

SubmitMy AnswersGive Up

Part C

A piece of copper is made into a rod with a square cross-section. The side of the square is 2.00 centimeters. The resistivity of copper is 1.7?10?8 ??m. An unknown electric field E, directed along the rod, creates a current of 12.0 amperes through the rod. Find the magnitude of E.

Use two significant figures in your answer. Express your answer in newtons per coulomb.

E =   N/C

Explanation / Answer

as resisatnce R = rho L/A

resistivity rho = RA/L

where A is area and L is length so

also Currrnt density J = current/area = Electric field /ressitivity

so

J = i/A = E /rho

so ressiitity Rho = Ei/A

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if L is doubled, as from ohms law V = iR

if L dobubles, R also Doubles,

current recuced by half

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part C:

apply Eelctric fieled E = rho i/A

E = 1.7 e-8 * 12/(0.02*0.02)

E = 5.1 *10^-4 N/C

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