Question

A copper wire of resistivity 1.63 x 〖10〗^(-8) Ω-m has cross-section area of 10.3 x 〖10〗^(-4) cm². Calculate length of wire required to make a 20 Ω coil .

Answer 1

Given :

• Resistivity of Wire, ρ = 1.63 × 10⁻⁸  Ω m
• Area of  cross section of Wire, A = 10.3 × 10⁻⁴ cm² = 10.3 × 10⁻⁴ × 10⁻²  m² = 10.3 × 10⁻⁶  m²
• Resistance required, R = 20 Ω

To find :

• Length of Wire, L = ?

Formula required :

• Relation between Resistance (R), Resistivity (ρ), Area of cross section (A), Length of conductor (L)

       R = ρ L / A

Solution :

Using formula

→ R = ρ L / A

→ 20 = (1.63 × 10⁻⁸) ( L ) / (10.3 × 10⁻⁶)

→ 20 = ( 1.63 / 10.3 ) × ( 10⁻¹⁴ ) × L

→ L = 20 × 10.3 / ( 1.63 × 10⁻¹⁴ )

→ L = 126.38 × 10¹⁴ m

Therefore,

• Length of wire required will be 126.38 × 10¹⁴ metres approximately.

Answer 2

Answer:

✡ Question ✡

⬛ A copper wire of resistivity 1.63 x 10-⁸Ω m has cross-section area of 10.3 x 10-⁴cm². Calculate length of wire required to make a 20 Ω coil .

✡ Given ✡

⬛ A copper wire of resistivity (ρ) = 1.63 × 10-⁸ Ω-m

⬛ Cross-section area (A) = 10.3 x 10-⁴cm².

⬛ Resistance required (R) = 20 Ω

✡ To Find ✡

⬛ Length (L) of the wire = ?

✡ Formula Used ✡

▶ $\bold{\huge{\fbox{\color{blue} {R =ρ L/A }}}}$

( Where,

⭐ R = Resistance

⭐ ρ = Resistivity

⭐ L = Length

⭐ A = Area )

✡ Solution ✡

➡ R = 20 Ω

➡ ρ = 1.63 × 10-⁸ Ω-m

➡ A = 10.3 x 10-⁴cm² = 10.3 x 10-⁶ m²

According to the question by using formula we get,

=> 20 = $\dfrac{(1.63×10-⁸)(L)}{10.8×10-⁶}$

=> 20 = $\dfrac{1.63}{10.3}$ × (10-¹⁴)×L

=> L = $\dfrac{20×10.3}{1.63×10-¹⁴}$

=> L = 126.38×10¹⁴ (approx)

∴ The length of wire required is 126.38×10¹⁴ (approx).

Explanation:

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