Question

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

Answer 1

Answer:

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

Explanation:

The current can be calculated by mutiplying the current density by cross section area. The new current density is obtained by the following relation and considering that same current flows through the new wire:

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

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

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

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

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

Answer 2

Answer:

Explanation:

Given:

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 × d^2

= pi/4 × (0.0029)^2

= 6.6 × 10^-6 m^2

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

= 2.18 A

Since the current in the lightbulb = current in the filament, current in the filament = 2.18 A

Area = pi/4 × d^2

= pi/4 × (5.5 × 10^-5)^2

= 2.38 × 10^-9 m^2

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

= 9.18 × 10^8 A/m^2

= 918 MA/m^2