Electric field in the narrow wire

[DottZakapa]

Homework Statement

A narrow copper wire of length L and radius b is attached to a wide copper wire of length L and radius 2b, forming one long wire of length 2L. This long wire is attached to a battery, and a current is flowing through it. If the electric field in the narrow wire is E, the electric field in the wide wire is

Relevant Equations

current

i've started from this I₁=I₂
then
I₁= JA₁=$\frac {E l} R$

I₂= JA₂=$\frac {E_2 l} R$

but can't get anything useful relating them. Am i forgetting any other useful formula?
I get as result E4

 

[kuruman]

Why are you using the same J in the two equations? If the current is the same in the two wires would the current density also be the same? Also the wires don't have the same resistance.

 

[DottZakapa]

Why are you using the same J in the two equations? If the current is the same in the two wires would the current density also be the same? Also the wires don't have the same resistance.

you right
$I_1= J A$ , $A_2 = pi 4b^2$
$I_2=J_2 A_2$
$j_2= \frac j 4$
then ?
$I_2 = \frac j 4 \ A_2 = \frac {E l} {R}\space but\space then \space A_2\space simplify\space with\space the\space 4\space and \space i\space get\space back\space to\space the\space beginning...$
sorry I'm not getting anywhere

 

[kuruman]

There are many equivalent ways to get to the answer, but you need a goal and a strategy. The goal is to find the ratio $E_2/E$. The strategy is the use of Ohm's law and the formula for the resistance of a cylindrical wire $R=\dfrac{\rho L}{A}$. You need to write Ohm's law for each wire separately then write separate expressions for each wire's resistance and then take the appropriate ratio.

 

[DottZakapa]

There are many equivalent ways to get to the answer, but you need a goal and a strategy. The goal is to find the ratio $E_2/E$. The strategy is the use of Ohm's law and the formula for the resistance of a cylindrical wire $R=\dfrac{\rho L}{A}$. You need to write Ohm's law for each wire separately then write separate expressions for each wire's resistance and then take the appropriate ratio.

haven't thought about R
now it works, thanks a lot for your time