# 0.5V across 1.5m tungsten wire w/cross-sect. area 0.6 mm^2 - Find wire current

1. Mar 31, 2009

### student_fun

1. The problem statement, all variables and given/known data
0.5 volts is maintained across a 1.5 meter tungsten wire that has a cross-sectional area of 0.6 mm2. What is the current in the wire?

2. Relevant equations
Resistivity of tungsten: 5.6E-8 ohm meters.

Current Density: $$J = \sigma \cdot \frac{V}{l}, J = \frac{I}{A}$$
Definition of resistivity: $$\rho = \frac{1}{\sigma}$$

Variable meanings in equations used:
I: current in amps, A: cross-sectional area of wire in m2, $$\sigma$$: material conductivity, V: volts, l: length of extruded cross-sectional area (in this case, a wire) in meters, $$\rho$$: resistivity in ohm meters

3. The attempt at a solution

Using the relations and data available it seems to make sense to find the current in the wire by relating the known tungsten resistivity to the two definitions of current density in which most variables are known except for the I (current) that I'm after. Doing this yields:

$$\frac{I}{A} = \sigma\frac{V}{l}$$
$$\sigma = \frac{1}{\rho}$$
$$\frac{I}{A} = \frac{1}{\rho} \cdot \frac{V}{l}$$
$$I = A\left[\frac{1}{\rho} \cdot \frac{V}{l}\right]$$

Then I plug and chug:

$$I = 0.6\left[\frac{1}{5.6E-8} \cdot \frac{0.5}{1.5}\right] = 3571428.571$$ Amps

However this answer is not valid. Can anyone see something I am doing that is outright wrong? Invalid logic somewhere?

Any hints to get me on the right track would be greatly appreciated.

2. Mar 31, 2009

### fball558

Your answer has the right procedure but the area is in mm^2 you need to convert it to m^2. So you are off by a factor of 10^6 meaning 3.57 amps.
Below is an easier way to go about the problem and attain the same answer.

resistance = (resistivity*length)/cross sectional area in m^2

Current = voltage/ resistance

3. Mar 31, 2009

### student_fun

Nevermind, lol, solved my own problem. Found out that I didn't convert millimeters squared to meters squared. Got it right now :)

4. Mar 31, 2009

### student_fun

Thanks for the reply fball558 :) It's nice to get other views while your pourin over pages and pages of physics :)