The resistivity of gold is 2.44×10-8 Ω•m at room temperature. A gold wire that is 1.9 mm in diameter and 33 cm long carries a current of 250 mA. What is the electric field in the wire?

Question

fifciol · Answer

$$E=\frac{ V }{ l}=\frac{ IR }{ l }$$
$$R=\frac{ \rho l }{ A }$$
$$E=\frac{ I \rho l }{Al }=\frac{ 4I \rho  }{\pi d^2 }$$

anonymous · Answer

what does p, stand for

fifciol · Answer

it's rho, resistivity

anonymous · Answer

rho as,p=charge / area,

fifciol · Answer

p=RA/l

anonymous · Answer

so if I want to substitute from your equation, instead of p, what should i put

fifciol · Answer

resistace * area / length

anonymous · Answer

okay let me check, ill get back to you

anonymous · Answer

so the equation will end up to be E=IR/L

fifciol · Answer

yes, but you have given resistivity rho, not resistance R

anonymous · Answer

do I have to change the unit of the current to be A not mA, if it so how will I do that

fifciol · Answer

1 A = 10^3 mA

anonymous · Answer

look this are the possible answers that I have .2×10−3 V/m
	5.4×10−3 V/m
	2.7×10−3 V/m
	5.4×10−4 V/m
	1.7×10−3 V/m

anonymous · Answer

sorry buddy, I have to go there is something that I have to do. but please can you do it for me and when you are done post it for me

fifciol · Answer

I find 2.15 * 10-3