Electric field in a platinum wire

In summary, the conversation discusses the calculation of the strength of the electric field needed to drive a current of 0.3 amperes through a platinum wire with a cross-sectional area of 1 mm^2 and conductivity of 1.0e+7. The equation I=σAE is used to solve for the electric field, with the only challenge being to correctly handle units. The correct answer is .03.
  • #1
sdevoe
21
0

Homework Statement



Consider a platinum wire (σ = 1.0e+7) with a cross-sectional area of 1 mm^2 (similar to your connecting wires) and carrying 0.3 amperes of current, which is about what you get in a circuit with a round bulb and two batteries in series. Calculate the strength of the very small electric field required to drive this current through the wire.

Homework Equations



σ=qnu
I=qnAuE
Therefore I=σAE
where phi is the conductivity A the cross sectional area and E the electric field

The Attempt at a Solution


This seemed easy but i got an answer of 3e-5 and that is incorrect?
 
Physics news on Phys.org
  • #2
Never mind i figured it out
 
  • #3
How did you do it? I am where you were in the first post.
 
  • #4
Hey I figured it out.
The only problem was dealing with units.
I just multiplied the 3e-5 by 1000 and got .03 as the right answer.
 
  • #5


I would like to clarify that the unit for electric field is not amperes (A), but rather volts per meter (V/m). Therefore, the correct equation to use in this scenario would be E = I/σA, where E is the electric field, I is the current, σ is the conductivity, and A is the cross-sectional area.

Plugging in the given values, we get E = (0.3 A)/(1.0e+7 S/m)(1 mm^2) = 3e-5 V/m. This is the correct answer, so it is possible that there was a miscalculation in the previous attempt. It is also important to note that the electric field in this scenario is very small, which is expected since platinum is a good conductor and therefore requires a low electric field to drive a current through it.
 

What is an electric field?

An electric field is a region in space where electrically charged particles experience a force. It is represented by electric field lines, which show the direction of the force and the strength of the field.

How is an electric field created in a platinum wire?

An electric field in a platinum wire is created by the movement of electrons. When a voltage is applied to the wire, the electrons are forced to move, creating an electric field along the wire.

What is the direction of the electric field in a platinum wire?

The direction of the electric field in a platinum wire is from positive to negative. This means that the electric field lines point towards the negative end of the wire, where the electrons are moving towards.

What factors affect the strength of the electric field in a platinum wire?

The strength of the electric field in a platinum wire is affected by the voltage applied to the wire and the distance between the charges. The greater the voltage and the closer the charges are, the stronger the electric field will be.

How is the electric field in a platinum wire measured?

The electric field in a platinum wire can be measured using a device called an electric field meter. This device measures the strength of the electric field in volts per meter (V/m). Another way to measure the electric field is by using a voltmeter to measure the potential difference between two points along the wire.

Similar threads

  • Introductory Physics Homework Help
Replies
14
Views
4K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
2K
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Back
Top