Question

An aluminium wire has diameter 0.5 mm and resistivity 2.8 × 10^-6 ohm m. How much length of this wire is required to make 10 ohm coil​

Answer 1

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Answer 2

Given: diameter, d = 0.5 mm

= $0.5 × 10^{-3}m$

Resistance, R = 10Ω

Area: πr² = 3.14 × $(\frac{0.5 × 10^{-3}}{2})^{2}$

=>$\frac{3.14 × 0.25 × 10^{-6}}{4}$

We know that: $R = ρ \frac{L}{A}$

=> $L = \frac{RA}{ρ}$

=> $L = \frac{10 × 3.14 × 0.25 × 10^{-6}/4}{2.8 × 10^{-6}}$

=> $L = \frac{2.5 × 3.14 × 0.25}{2.8}$

=> L = 0.70089 m