Calculating diameter of a different type of wire with same current and e-field

[RichardEpic]

Homework Statement

What diameter should the nichrome wire in the figure be in order for the electric field strength to be the same in both wires?

Homework Equations

σ = 1/ρ
J = σ*E...I/A = σ*E...I = σ*E*A

Then, equate the current equations for each wire.

The Attempt at a Solution

nichrome aluminum
σ*A = σ*A
Since, the E-fields are to be the same, they should just cancel out.

ρ(aluminum)= 2.82*10^-8 Ωm
ρ(nichrome)= 1.1*10^-6 Ωm

pi*r^2 = σ(Al)*A/σ(nichrome)

...& after solving for r, I calculated the diameter to be approximately 6.246 mm, however this was apparently wrong, but I have no idea where I went wrong. Help would be greatly appreciated!

 

[SammyS]

It would help if you stated the problem exactly as it was given to you.

There are lots of things missing.

 

[RichardEpic]

HAHA...sorry, but did you take a look at the thumbnail..? It's all there. Same current is flowing through both wires, and aluminum has a diameter of 1.0 mm. All good?

I'd at least like to know if the equation(s) I ended up with were wrong or not, or if I'm not including something in my calculations that should be included, because everything seems to be done right.

By the way, the thumbnail and the stated question is everything, exactly as it was given to me, copy and pasted.

 

[ehild]

A_NiCRσ_NiCr=A_Alσ_Al.

So the ratio of cross sections is 39, this means that the diameter of the NiCr is 6.2 times the diameter of aluminium. Your result is correct.

ehild

 

[SammyS]

Why is the current the same in each?

 

[ehild]

They look connected in series, both metals. A current is shown to flow in, it has to flow out, charge does not accumulate and so on... I just cannot imagine what can be the problem.

ehild.

 

[SammyS]

Does it flow out the other end, or through the side?

 

[ehild]

As nothing is said, only a straight arrow shows the direction of current, it should flow through both wires and go out the other end. The electric field is the same in both wires, nothing about the distribution of the electric field, or having different directions at different points of the wires. These problems must be very simple, the website solution might be wrong.

ehild