Current in a nichrome tube from resistivity and dimensions

[krugertown]

a 20cm long hollow nichrome tube of inner diameter 2.8mm, outer diameter 3.0mm is connected to a 3.0V battery. What is the current in the tube?



I think this is a I = V/R problem. So i used R=($$\rho$$*L)/A. Where L is the length of the wire and A is the outer area minus the inner area and rho is the resistivity of nichrome.


A =.003^2*$$\pi$$ - .0028^2*$$\pi$$ = 3.64*10^-6
R = ((1.5*10^-6) * (.02))/3.64*10^-6 = .0823
I = 3/.0823 = 36.44Amps

However the answer is 9.1A apparently. Where have i gone wrong?

Thanks!

btw if the greek symbols appear superscript they are not supposed to be I can't quite figure that part out

 

[Rake-MC]

You made a simple error, 3.0mm and 2.8mm are diameters, not radii.

Area is $$\pi r^2$$

 

[krugertown]

haha thanks! spent 45 minutes on that problem!