The current in the primary circuit of a potentiometer wire is 0.5 A, ρ for the wire is 4 × 10^-7 Ω-m and area of cross- section of wire is 8 × 10^-6 m². The potential gradient in the wire would be

Concept:

Potential Gradient in a Wire:

• The potential gradient ΔV/Δx is given by the formula:
• ΔV/Δx = I × ρ / A, where:
  • I = Current through the wire (A)
  • ρ = Resistivity of the wire (Ω·m)
  • A = Cross-sectional area of the wire (m²)
• Potential gradient represents how much the potential changes per unit length along the wire.

Calculation:

Given:

• Current I = 0.5 A
• Resistivity ρ = 4 × 10⁻⁷ Ω·m
• Area A = 8 × 10⁻⁶ m²

ΔV/Δx = (0.5 A × 4 × 10⁻⁷ Ω·m) / (8 × 10⁻⁶ m²)

ΔV/Δx = 25 × 10⁻³ V/m = 25 mV/m

∴ The potential gradient in the wire is 25 mV/meter, which corresponds to Option 1.