g The current density in the 2.9-mm-diameter wire feeding an incandescent lightbulb is 0.33 MA/m2. Part A What's the current den

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ASHA 777

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g The current density in the 2.9-mm-diameter wire feeding an incandescent lightbulb is 0.33 MA/m2. Part A What's the current den

sity in the lightbulb's filament, whose diameter is 0.055 mm

Physics

2 answers:

Yuliya22 [3.3K] · Answer 1 · 2026-03-25T13:54:45+03:00

Yuliya22 [3.3K]1 month ago

5 0

Answer:

$j_{B} = 917.454\,\frac{mA}{m^{2}}$

Explanation:

Current can be determined by multiplying current density by the cross-sectional area. The new current density can be derived through the subsequent relationship, while assuming that the same current flows through the new wire:

$j_{A}\cdot A_{A} = j_{B}\cdot A_{B}$

$j_{B} = j_{A}\cdot \frac{A_{A}}{A_{B}}$

$j_{B} = j_{A} \cdot \left(\frac{D_{A}}{D_{B}} \right)^{2}$

$j_{B} = \left(0.33\,\frac{mA}{m^{2}} \right)\cdot \left(\frac{2.9\,mm}{0.055\,mm} \right)^{2}$

$j_{B} = 917.454\,\frac{mA}{m^{2}}$

Ostrovityanka [3.2K] · Answer 2 · 2026-03-25T13:57:51+03:00

Answer:

Explanation:

Provided data:

Light bulb:

Diameter, d = 2.9 mm

Current density, J = 0.33 MA/m²

Filament:

Diameter, d = 0.055 mm

Current, I = J × A

Area = pi/4 × d²

= pi/4 × (0.0029)²

= 6.6 × 10^-6 m²

The current of the light bulb, I = 3.3 × 10^5 × 6.6 × 10^-6

= 2.18 A

Given that the current in the lightbulb equals the current in the filament, we find the filament current to be 2.18 A

Area = pi/4 × d²

= pi/4 × (5.5 × 10^-5)²

= 2.38 × 10^-9 m²

Current density, J = 2.18/2.38 × 10^-9

= 9.18 × 10^8 A/m²

= 918 MA/m²