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

• aeris
In summary, the content discusses a problem involving a composite wire made of silver and copper. A potential difference of 5.0V is maintained between the ends of the 2.0m wire and the task is to find the magnitude of E in the copper section. The formula E='roe'J is used, where 'roe' is resistivity and J is current density. After calculating the current density, the resistivity of copper is multiplied by J to find E in the copper section. However, the incorrect resistivity for silver may have resulted in an incorrect answer.
aeris
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..

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.

## 1. What is Efield?

Efield, or electric field, is a physical quantity that describes the influence that an electric charge has on other charges in its vicinity. It is a vector quantity, meaning it has both magnitude and direction.

## 2. What is a composite wire?

A composite wire, also known as a composite material or composite structure, is a material made up of two or more different materials that are combined to create a new material with improved properties. In the case of a composite wire, different materials are used to create a wire with enhanced electrical conductivity.

## 3. How is Efield affected by a composite wire?

The Efield in a composite wire is affected by the different materials used in its structure. Each material may have a different dielectric constant, which is a measure of how easily an electric field can polarize the material. This can affect the overall strength and direction of the Efield in the wire.

## 4. Can Efield be measured in a composite wire?

Yes, Efield can be measured in a composite wire using specialized instruments such as an electric field meter or a voltmeter. These instruments can detect the strength and direction of the Efield at different points along the wire.

## 5. How is Efield used in composite wires?

Efield is used in composite wires to control the flow of electricity and to create specific electrical properties in the wire. By carefully selecting and arranging the materials in a composite wire, engineers can create wires with unique properties, such as high conductivity or low resistance, for various applications.

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