Current Density

Memory Trick: NEVADA for Current Density

Remember the current density formula using NEVADA:

• NE = \(ne\) (number density × elementary charge)
• V = drift velocity (\(v_d\))
• A = area (related to current density)
• D = density (current density \(J\))
• A = area again (cross-sectional area)

So: \(ne \times v_d = J\) (current density)

And: \(J \times A = I\) (total current)

This helps you remember that \(nev_d = J\), which is the microscopic expression for current density!

Example: Radial Current Density

Problem: A cylindrical conductor has current density that varies with radius as \(J(r) = J_0(1 - r/R)\), where \(R\) is the radius. Find the total current.

Solution:

1. Area element: \(dA = 2\pi r \, dr\)
2. Current element: \(dI = J(r) \, dA = J_0(1 - r/R) \cdot 2\pi r \, dr\)
3. Total current: \(I = \int_0^R J_0(1 - r/R) \cdot 2\pi r \, dr\)
4. Integration: \(I = 2\pi J_0 \int_0^R (r - r^2/R) \, dr\)
5. Result: \(I = 2\pi J_0 [R^2/2 - R^2/3] = \pi J_0 R^2/3\)

Answer: The total current is \(\frac{\pi J_0 R^2}{3}\).

Example 1: Basic Current Density Calculation

Problem: A wire with radius 1.0 mm carries a current of 2.0 A. What is the current density in the wire?

Solution:

1. Cross-sectional area: \(A = \pi r^2 = \pi(1.0 \times 10^{-3} \text{ m})^2 = 3.14 \times 10^{-6} \text{ m}^2\)
2. Current density: \(J = \frac{I}{A} = \frac{2.0 \text{ A}}{3.14 \times 10^{-6} \text{ m}^2}\)
3. Calculation: \(J = 6.37 \times 10^5 \text{ A/m}^2\)

Answer: The current density is 6.37 × 10⁵ A/m².

Example 2: Current Density from Microscopic Properties

Problem: A copper wire has 8.5 × 10²⁸ free electrons per m³. If the drift velocity is 1.0 × 10⁻⁴ m/s, what is the current density?

Solution:

1. Current density formula: \(J = nqv_d\)
2. For electrons: \(q = -e = -1.602 \times 10^{-19} \text{ C}\)
3. Substitution: \(J = (8.5 \times 10^{28} \text{ m}^{-3})(1.602 \times 10^{-19} \text{ C})(1.0 \times 10^{-4} \text{ m/s})\)
4. Calculation: \(J = 1.36 \times 10^6 \text{ A/m}^2\)

Answer: The current density is 1.36 × 10⁶ A/m².

Example 3: Current from Non-Uniform Current Density

Problem: A wire has current density that varies as \(J(r) = J_0 r/R\), where \(R\) is the radius. If \(J_0 = 2.0 \times 10^6 \text{ A/m}^2\) and \(R = 1.0 \text{ mm}\), find the total current.

Solution:

1. Current element: \(dI = J(r) \cdot 2\pi r \, dr = J_0 \frac{r}{R} \cdot 2\pi r \, dr\)
2. Total current: \(I = 2\pi J_0 \int_0^R \frac{r^2}{R} \, dr\)
3. Integration: \(I = 2\pi J_0 \frac{R^2}{3}\)
4. Substitution: \(I = 2\pi(2.0 \times 10^6) \frac{(1.0 \times 10^{-3})^2}{3}\)
5. Result: \(I = 4.19 \text{ A}\)

Answer: The total current is 4.19 A.

Example 4: Electric Field from Current Density

Problem: A copper wire has current density 1.0 × 10⁶ A/m². If copper has resistivity 1.68 × 10⁻⁸ Ω·m, what is the electric field in the wire?

Solution:

1. Microscopic Ohm's law: \(E = \rho J\)
2. Substitution: \(E = (1.68 \times 10^{-8} \text{ Ω·m})(1.0 \times 10^6 \text{ A/m}^2)\)
3. Calculation: \(E = 1.68 \times 10^{-2} \text{ V/m}\)

Answer: The electric field is 1.68 × 10⁻² V/m.

Example 5: Current Density in Parallel Wires

Problem: Two wires are connected in parallel. Wire A has radius 1.0 mm and carries 3.0 A. Wire B has radius 2.0 mm and carries 5.0 A. Which wire has higher current density?

Solution:

1. Wire A area: \(A_A = \pi(1.0 \times 10^{-3})^2 = 3.14 \times 10^{-6} \text{ m}^2\)
2. Wire A current density: \(J_A = \frac{3.0}{3.14 \times 10^{-6}} = 9.55 \times 10^5 \text{ A/m}^2\)
3. Wire B area: \(A_B = \pi(2.0 \times 10^{-3})^2 = 1.26 \times 10^{-5} \text{ m}^2\)
4. Wire B current density: \(J_B = \frac{5.0}{1.26 \times 10^{-5}} = 3.98 \times 10^5 \text{ A/m}^2\)

Answer: Wire A has higher current density (9.55 × 10⁵ A/m² vs 3.98 × 10⁵ A/m²).