# Efield in Composite WirePls help

#### aeris

Efield in Composite Wire..Pls help!!

1. 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?

2. 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

3. 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 cant get the right answer!! I would like to ask the concept behind the E of a composite wire too..

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#### Kurdt

Staff Emeritus
Gold Member
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?

#### aeris

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!

#### Kurdt

Staff Emeritus
Gold Member
When I work out the current I get something different. I take it you found current by:

$$I = \frac{VA}{\rho l}$$

Last edited:

#### aeris

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 cant get the idea of finding E using E= 'roe'_copper*J to find E in copper section though.

#### Kurdt

Staff Emeritus
Gold Member
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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