# What turns on a light bulb?

1. Mar 17, 2015

### Imabioperson

I am having some trouble with discerning whether current (electrons flowing through the Tungsten), or the energy transfer as described by the poynting vector, is responsible for lighting the lightbulb.

2. Mar 17, 2015

### Quantum Defect

The current flowing through the filament heats the tungsten to incandescence. Light is given off by things heated to incandescence. Every time a photon leaves the filament, the energy of the filament is reduced. During normal operation, the rate of heating (I^2R) is equal to the rate that energy leaves the filament as photons -- i.e. steady state is achieved, and the temperature reaches a constant value.

3. Mar 17, 2015

### Staff: Mentor

They are essentially the same. The power density throughout the bulk of the filament is given by the current density (squared) and the resistivity. This quantity will be equal to the Poynting flux across the surface of the filament.

4. Mar 17, 2015

### Imabioperson

Hi Dale,

Thank you. Although they may have an equal value, is it that the cross product of the two fields, ExB, is responsible for the energy transfer, with the source of this EM wave being related to the E field induced by the current? I simply want to know what the source of the energy is, that we see lost as the heat and light. Is it the EM wave generated by the two fields, or simply electron flow. A bit confusing to me, and thank you very much for taking the time to answer.

5. Mar 17, 2015

### Staff: Mentor

When you say "energy transfer" are you asking about the energy transfer from the field at one point to the field at another point or are you asking about the energy transfer from the field at one point to the matter at the same point? The Poynting flux is the first, and the joule heating is the second.

6. Mar 17, 2015

### anorlunda

Don't you think that you may be overdoing it. Sometimes, P=I2R is sufficient and you don't need to analyze it with Maxwell's Equations.

7. Mar 17, 2015

### baazaar12

the electric field is responsible for turning the light on. The first electron that reaches the wire leading to the light from the power outlet is not the same electron that reaches the light. Without the electric field, the light wouldn't turn on as fast as they do. (Negating any effects on turn-on-time from the light-bulb type itself)

8. Mar 17, 2015

### ZapperZ

Staff Emeritus
Wow, why are you making this more complicated than it is?

When you close the switch, there is a potential difference across the circuit containing the bulb. Current than flows. The flow of current through the high-resistive wire heats up the wire. The heating causes molecules/atoms in the wire to vibrate even more, until the vibrational energy is so high, the wire noticeably glows. You get light!

Zz.

9. Mar 17, 2015

### Staff: Mentor

It should be noted that current flows throughout the circuit all at once. It does not start at the switch and move through the circuit like water through an empty pipe. In effect, you cannot separate current flow from the field. When you flip the switch a change in the EM field moves through the circuit at near light speed and creates both voltage and current.