The current density in a cylindrical wire of radius r = 4.0 mm is 1.0 × 10⁶ A/m². The current through the outer portion of the wire between radial distances r/2 and r is xπ A; where x is __________.

Solution

Physics sir · 4 steps

The current density is uniform, so current is just J times the area of the outer ring.

1. Uniform J. The current density is the same at every point of the cross-section. So the current through any part of it is I = JA, where A is the area of that part.
2. Area of the ring. The outer portion is a ring between radius r/2 and r. Its area is the full circle minus the inner circle.

   A = π r² - π(r/2)² = 3/4π r²

3. Put in numbers. With r = 4.0 × 10⁻³ m, r² = 16 × 10⁻⁶ m², so A = 3/4 × 16 × 10⁻⁶ π = 12 × 10⁻⁶ π m². Then I = 1.0 × 10⁶ × 12 × 10⁻⁶ π = 12π A.
4. So. The current through the outer portion is 12π A, so x = 12.

The common slip: 4. Using the area of the inner circle, π(r/2)², instead of the ring gives 4, not 12. The outer portion is the full circle minus that inner part.

The idea to keep. When current density is uniform, current is proportional to area. For a ring, always subtract the inner circle from the outer one, and convert mm to m before squaring.

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