What Is the Electric Field Magnitude in the Copper Section of a Composite Wire?

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The discussion focuses on calculating the electric field magnitude in the copper section of a composite wire made of silver and copper. A potential difference of 5.0V is applied across a 2.0m wire, which consists of a 1.2m silver wire and an 0.8m copper wire. Participants are attempting to find the electric field (E) in the copper section using the formula E = ρJ, where ρ is the resistivity of copper and J is the current density calculated from the current and cross-sectional area. There is confusion regarding the correct resistivity values and the calculation of current, with one participant noting discrepancies in the resistivity of silver. The conversation emphasizes the importance of correctly applying formulas and values to achieve the correct electric field magnitude.
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Efield in Composite Wire..Pls help!

Homework Statement


A 2.0m length of wire is made by welding the end of a 1.2m long silver wire to the end of an 0.8m long copper wire. Each piece of wire is 0.60mm in diameter. The wire is at room temperature. A potential difference of 5.0V is maintained between the ends of the 2.0m composite wire. What is the magnitude of E in the copper?



Homework Equations


E='roe'J , 'roe' is resistivity
J=I/A, A is cross sectional area
I= 45amperes
A= 2.827*10^-7 sqm

Resistivity of copper= 1.72*10^-8
Resistivity of silver= 1.47*10^-8

The Attempt at a Solution


I tried to calculate J first by substituting in values of I and A using the formula J=I/A and multiplying by 'roe'_copper but i still can't get the right answer! I would like to ask the concept behind the E of a composite wire too..
 
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First, if you could post what you've done it might be easier to spot a mistake. Secondly what value were you using for I?
 
Mmm.. i found out that I=45amperes and A=2.827*10^-7 sqm and found out that J=1.59*10^8. To find E in copper section i multiplied this J by 'roe'_copper= 1.72*10^-8 to get E=2.74V/m. But the answer is wrong. thanks!
 
When I work out the current I get something different. I take it you found current by:

I = \frac{VA}{\rho l}
 
Last edited:
Yup i found it by using this formula whereby i added the resistance of each component using their respective length and 'roe' since they are in series before dividing 5V with the total resistance to get 45amperes.
i obtained J by I/A, where I=45amperes and A= 2.827*10^-7 sqm to get J=1.59*10^8 A/m^2.
I can't get the idea of finding E using E= 'roe'_copper*J to find E in copper section though.
 
It could just be that you are using the wrong resistivity for silver. It is listed as 1.59x10-8 elsewhere.
 
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