PLEASE HELP! REALLY LOST!
The electric field in a 3.0mm×3.0mm square aluminum wire is 2.2×10^−2V/m. What is the current in the wire?

Question

PLEASE HELP!  REALLY LOST!  
The electric field in a 3.0mm×3.0mm square aluminum wire is 2.2×10^−2V/m.  What is the current in the wire?

anonymous · Answer

Hi there. I think we can use  current as the electric flow of charge per second. So let me present it:
First, we take the distance as 3mm or 0.003m. Using the formula E=V/d, where d is distance, v voltage and E electric field strength, we make V the subject, being V=Ed or 2.2*10^-2*0.003=6.6*10^-5V
Are you not given the force or charge?

anonymous · Answer

no, this is all i'm given.  it's supposed to be I=.  so i think i need an answer in amps.  i tried just reversing J=I/A using the 2.2 as my J and the area, but that didn't work.

irishboy123 · Answer

V = E•length, R = p • length/area, I = V/R = p • area / p.  p (rho) for al is 2.65 x 10^-8.

anonymous · Answer

not really sure i follow that.  up until I=V/R=....don't even know what R is supposed to be.  e is electric field.  but other than that....

anonymous · Answer

i'm ok up until I=V/R, then the equation seems all jumpy and weird.  p/p and the p*area doesn't even match up with what you said V was supposed to be....

irishboy123 · Answer

R is resistance.  all you are given is a field, a x-sect area, and that it is Al wire.  you need to calc volts from E • length, and resistance from rho•length/area.  lengths cancel out.

irishboy123 · Answer

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/resis.html

anonymous · Answer

Current density is equal to the electric field times the conductivity.  If the area of the wire is A then the current is J*A = conductivity * E*A  look up the conductivity of Aluminum and you are home.  note:  conductivity is the inverse of resistivity