30: Problem 30.10

INTRO:
The current in a 100 watt lightbulb is 0.870 A . The filament inside the bulb is 0.230 mm in diameter.
-----------------------------------------------------------------------------------------------------
PART A:
What is the current density in the filament?
Express your answer to three significant figures and include the appropriate units.

SOLUTION:
Current density is simply current divided by area...
The area = A = πr² = π/4⋅d²

⇒ J = (0.87 A)/[π/4⋅(2.3×10^-4 m)²]
⇒ J = 2.09×10⁷ A/m²
-----------------------------------------------------------------------------------------------------
PART B:
What is the electron current in the filament?
Express your answer using three significant figures.

SOLUTION:
current = charge/ time
1 Amp = 1 C/s

using equation (30.11) ... I = e⋅i_e

where e is the charge of an electron

∴ i_e = I/e
and I, in terms of C/s -> I = 0.87 C/s

→ i_e = (0.87 C/s)/(-1.602×10^-19 C/electron)
⇒ i_e = 5.43×10^-19 electron/s