Question

The current in a 100 watt lightbulb is 0.890 A. The filament inside the bulb is 0.280 mm in diameter. What is the current density in the filament? Express your answer to three significant figures and include the appropriate units. What is the electron current in the filament? Express your answer using three significant figures.

Answer 1

Current density
`j=1.44* 10^7A/m^2`

Electron density
`=5.55* 10^(18)electron/sec`

Step-by-step explanation:

We have given power = 100 watt

Current = 0.89 A

Diameter d = 0.280 mm

So radius
`r=(d)/(2)=(0.28)/(2)=0.14mm=0.14* 10^(-3)m`

Area
`A=\pi r^2=3.14* (0.14* 10^(-3))^2=0.016* 10^(-6)m^2`

We know that current density
`J=(I)/(A)=(0.89)/(0.016* 10^(-6))=1.44* 10^7A/m^2`

Now we have to calculate the electron density

We have current i = 0.89 A = 0.89 J/sec

Charge on 1 electron
`1.6* 10^(-19)C/electron`

So electron density
`=(0.89j/sec)/(1.6* 10^(-19)C/electron)=5.55* 10^(18)electron/sec`