The current in a 100wattlightbulb is 0.85A. The filament inside the bulb is 0.25mmin diameter.

a. What is the current density in the filament?

b. What is the electron current in the filament?

Short Answer

Expert verified

a. The current density is 1.7·107A/m2.

b. The electron current in the filament islocalid="1648891656477" 5.3×1018electron/s.

Step by step solution

01

Given Information (Part a)

Lightbulb power=100watt

Current=0.85A

Filament diameter=0.25mm

02

Explanation (Part a) 

Let I=0.85Aand d=0.25mm,

The area of the wire is

A=d24π,

So the current density is

role="math" localid="1648891197210" j=IA=4Id2π

Substitute the values,

j=4(0.85A)(0.25mm)2×3.14

role="math" localid="1648891297578" =1.7·107A/m2

03

Final Answer (Part a)

Hence, the current density is=1.7·107A/m2

04

Given Information (Part b)

Lightbulb power=100watt

Current=0.85A

Filament diameter=0.25mm

05

Explanation (Part b)

The electron current is equal to the current divided by the magnitude of the charge of the electron,

Ie=Ie

Ie=0.85A1.6×10-19C

=5.31018electrons/s

06

Final Answer (Part b)

Hence, the electron current in the filament is5.3×1018electron/s.

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