A cylindrical wire of radius 0.5 mm and conductivity 5 × 10⁷ S/m is subjected to an electric field of 10 mV/m. The expected value of current in the wire will be x³π mA. The value of x is _________.

Solution

Physics sir · 3 steps

Ohm's law in its local form, J = σ E, gives current density; multiply by the area to get the current.

1. Current density. Ohm's law for a material says J = σ E. With E = 10 mV m⁻¹ = 10⁻² V m⁻¹, J = 5 × 10⁷ × 10⁻² = 5 × 10⁵ A m⁻².
2. Cross-section area. The radius is 0.5 mm = 5 × 10⁻⁴ m, so A = π r² = 25 × 10⁻⁸ π = 2.5 × 10⁻⁷ π m².

   I = JA = 5 × 10⁵ × 2.5 × 10⁻⁷ π = 0.125 π A = 125 π mA

3. So. Matching 125 π = x³ π gives x³ = 125, so x = 5.

The common slip: 125. Stopping at x³ = 125 and reporting 125 skips the last step. The question asks for x, which is the cube root, 5.

The idea to keep. When a question gives conductivity and field instead of voltage and resistance, use J = σ E and then I = JA. Convert mm and mV to SI before multiplying.

Answer from the NTA final answer key. Solution by PrepWiser.