Question

An aluminium wire has diameter 0.5 mm and resistivity 2.8 x10-6Ω-cm. How much length of this wire is required to make a 10 ohm coil?

Answer 1

Given data :

• wire diameter, d = 0.5 mm = 0.5 * 10^( - 3 ) m
• resistivity, ρ = 2.8 * 10^( - 6 ) Ω cm = 2.8 * 10^( - 8 ) Ω m
• Resistance, R = 10 Ω

Solution : Now, by formula;

⟹ R = ρL/A

where,

• R = Resistance
• A = cross sectional area
• L = length
• ρ = resistivity

Here, we know that;

⟹ A = π * r²

⟹ A = π * (d/2)²

⟹ A = π * d²/4 ----{1}

Now,

⟹ R = ρL/A

⟹ L = RA/ρ

Now, from given and eq. {1}

⟹ L = [10 * π * d²/4]/2.8 * 10^( - 8 )

⟹ L = 10 * 3.4 * [0.5 * 10^( - 3 )]²/4 * 2.8 * 10^( - 8 )

⟹ L = 34 * 25 * 10^( - 8)/11.2 * 10^( - 8 )

⟹ L = 850 * 10^( - 8)/11.2 * 10^( - 8 )

⟹ L = 850/11.2

⟹ L = 75.8928 m

Answer : Hence, the length of wire 75.8928 m is required to make a 10 ohm coil.

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