The current density in a cylindrical wire of radius 4 mm is 4 × 10⁶ Am⁻². The current through the outer portion of the wire between radial distances R/2 and R is ____________ π A.

Solution

Physics sir · 4 steps

Uniform current density means current is just J times area, so find the area of the outer ring.

1. Current from density. The current density J is the same everywhere in the wire. So the current through any part of the cross-section is I = J × A, where A is the area of that part.
2. Area of the ring. The outer portion is a ring between radii R/2 and R. Its area is the full circle minus the inner circle: π R² - π R²/4 = 3/4π R².

   I = J × 3/4π R² = 4 × 10⁶ × 3/4π (4 × 10⁻³)²

3. Put in numbers. R² = 16 × 10⁻⁶ m², so 3/4 R² = 12 × 10⁻⁶ m². Then I = 4 × 10⁶ × 12 × 10⁻⁶ π = 48π A.
4. So. The current through the outer portion is 48π A, so the blank is 48.

The common slip: 64. Using the full cross-section π R² instead of the ring gives 64, the current through the whole wire.

The idea to keep. When J is uniform, current through a region is J times its area. For a ring, subtract the inner circle's area from the outer one. Always convert mm to m first.

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