Calculating Resistance for Cylindrical Conductors

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delsoo
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Homework Statement



can anybody help me with this question? my ans is different from the ans given... i need explanation on this ! here's my attempt by the way.
r=pl/a . r=resistance , p = resistivity l=length

r=(1.69e-8)(2pi )(6e-2 +1.5 e-3 )(300) / (pi )(1.5e-3)^2

what's wrong with my working? p/s e =1 x 10^x

Homework Equations





The Attempt at a Solution

 
Attachments
  • Diagram of a hollow copper cylinder made of closely wound wire coils viewed end-on in cross-section, with a 6.0 cm inner radius marked and wire thickness labelled 1.5 mm.
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  • Printed textbook problem: The bare copper wire has a square cross section of sides 1.5 mm. If there are 300 turns, calculate the resistance between opposite ends of the copper cylinder. (Resistivity of copper = 1.69 × 10⁻⁸ Ω m)
    DSC_0135~3[1].jpg
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You ignored the perfect electrical contact between successive turns. You don't have a single long wire, you have a conducting cylinder.
(6e-2 +1.5 e-3 )
Minor detail: this is the outer radius, not the average radius.