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How long is a tungsten wire if it has a diameter of 0.119×10−3 m and a resistance of 2530 Ω ? The resistivity of tungsten is 5.62×10−8Ω⋅m.

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Final answer:

The length of the tungsten wire with a diameter of 0.119×10⁻³ m and a resistance of 2530 Ω, using a resistivity of 5.62×10⁻⁸ Ω⋅m, is calculated to be 5.03×10⁴ m.

Step-by-step explanation:

To calculate the length of a tungsten wire with a given diameter and resistance, we can use the formula for resistance of a uniform cylindrical conductor:

R = ρ L/ A

where:

R is the resistance,

ρ (rho) is the resistivity of the material,

L is the length of the conductor, and

A is the cross-sectional area of the conductor.

We are given:

R = 2530 Ω (ohms),

ρ (resistivity of tungsten) = 5.62×10⁻⁸ Ω⋅m,

Diameter (d) = 0.119×10⁻³ m.

The cross-sectional area A can be calculated using the area formula for a circle, A = π d²/4.

By plugging in the values and solving for L, we get:

A = π(0.119×10⁻³ m / 2)² = π(0.119×10⁻³ m / 2)² = 1.11×10⁻⁸ m²

R = 2530 Ω

ρ = 5.62×10⁻⁸ Ω⋅m

Solving the formula for L, we get:

L = {R ⋅ A}/{ρ} = {2530 Ω ⋅ 1.11×10⁻⁸ m²}/{5.62×10⁻⁸ Ω⋅m} = 5.03×10⁴ m

So, the length of the tungsten wire is 5.03×10⁴ m.

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