Quick question on resistance&current in a sheet

[vorcil]

Homework Statement

A sheet of copper 0.6m in total length and 0.2m wide is folded to create a uniform spacing of 1mm.
the sheet is 1mm thick. a potential difference of 1mV is applied at the open end in such a way that the current is distrubuted unformly over the cross sectional area of the metal.
I can neglect the effects of the fold (the vertical bit)

http://yfrog.com/f/5mpicbvwp/
(i also forgot to draw in the battery, connecting the two ends of the open side, it has a potential difference of 1mV)

Find the resistance of the sheet,
and the current.

Homework Equations

https://secure.wikimedia.org/wikipedia/en/wiki/Sheet_resistance

In griffiths, I do not see this formula anywhere, however I'm going to use it anyway.

properties of copper:
electrical resistivity: 1.7*10^-8 ohms per metre
density: 8.9*10^3 kg m^-3
atomic weight = 64amu

The Attempt at a Solution

$$R = \rho \frac{L}{A} \Rightarrow A = 0.2m * 1mm$$

the length of the sheet is 0.6m

plugging in the values

$$R = (1.7*10^-8) * \frac{0.6}{0.2(1*10^-3)} = (5.1*10^-5) =51micro ohms$$

-

finding the current through the sheet, I'm just going to use Ohm's law,
$$\mathbf{J} = \sigma\mathbf{E} \Rightarrow \mathbf{V=IR}$$

plugging in the values, 1mV and 51 micro ohms, I get 19.6 Amps,
however I don't know if I'm right, because that seems superlarge to me?
where have I gone wrong?

-> can I not use the formula on this page? https://secure.wikimedia.org/wikipedia/en/wiki/Sheet_resistance

 

[mark726]

Thank you for your post. I am a scientist and I would like to help you with your problem.

First, let's go over the information that was given in the problem. We have a sheet of copper with dimensions of 0.6m x 0.2m x 1mm (length x width x thickness). The sheet is folded to create a uniform spacing of 1mm, and a potential difference of 1mV is applied at the open end. The current is distributed uniformly over the cross-sectional area of the metal, and we can neglect the effects of the fold.

To find the resistance of the sheet, we can use the formula R = ρL/A, where ρ is the electrical resistivity of copper, L is the length of the sheet, and A is the cross-sectional area. Plugging in the values given, we get:

R = (1.7*10^-8 ohms*m)/((0.6m)(0.2m)) = 141.7 microohms

So, the resistance of the sheet is 141.7 microohms.

To find the current through the sheet, we can use Ohm's law, which states that V = IR, where V is the potential difference, I is the current, and R is the resistance. Plugging in the values given, we get:

1mV = (I)(141.7 microohms)

Solving for I, we get:

I = 1mV/(141.7 microohms) = 7.06 mA

So, the current through the sheet is 7.06 mA.

I noticed that you used a different formula for sheet resistance (R = ρL/A) than the one given on Wikipedia (R = ρ/t), which may have caused some confusion. The formula you used is the correct one to use in this situation.

I hope this helps and let me know if you have any further questions.