# In a simple electric circuit with battery and lamp, how does the lamp light up? [duplicate]

I watched Veritasium's video on big misconceptions of electricity, i remember when we learn how does lamp light up in the circuit, our teacher said it is the kinetic energy of electrons transfer to metal ion of filament lamp, then convert to heat and light. but in the video, it talked about poynting vertor, does that mean the energy is transferred to the lamp in electromagnetic radiation form?

• Nov 28, 2021 at 16:13

No. The energy is not transferred through electromagnetic radiation. Poynting vector is a notion not exclusive for EM radiation, but also works for circuits like battery, wires and lamp.

In the case of the video, what these vectors show is a source of electromagnetic energy (battery) and a sink (lamp). It is a kind of account balance.

does that mean the energy is transferred to the lamp in electromagnetic radiation form?

Electrons in a wire are moving about in two different ways. First, they are moving randomly, due to thermal energy. Second, superimposed upon this random motion there may be a drift, or a net movement of the sum total of electrons moving in some direction. This net movement superimposed upon the random motion is conduction current, usually just called current.

When an electric field acts upon electrons which are free to move, the electrons are accelerated, creating such a current. That is the electric field gives kinetic energy to the electrons. Then through "collisions" with atoms, ions, or other electrons, the motion of the accelerated electrons becomes randomized again. This randomization process converts the extra kinetic energy given to the electrons by the electric field into thermal energy. This is known as Joule heating. As long as the electric field is present, electrons with random motion, will be accelerated in a given direction, collide, have their motion randomized again, and then be accelerated again.

An electric field that imparts energy to electrons may be the electric field component of an electromagnetic wave, i.e. electromagnetic radiation, as in your question. However, the electric field that is present in low frequency circuits (such as power common lights) is generally not thought of as "radiation", just a field. So, in common parlance, the answer to your question is NO, the energy is not transferred to the lamp in electromagnetic radiation form. But it is transferred in the form of an electric field.

EDIT:

i still confuse about whether the current (i.e. electrons) carries the energy towards lamp?

The electrons do carry energy toward the lamp. However, they do not carry that energy very far, or very long. The mean free path for electrons in copper is something in the vicinity of 40 nm. A nanometer (nm) is a billionth of a meter. So, on average, an electron only carries energy forward for 40 nm before it's direction becomes randomized. Unless an electron re-accelerates in a common direction with other electrons every 40 nm or so, the current will simply die out in very short order. The electric field is what re-accelerates these electrons. Without an electric field at every point along the current path, the common motion of the elections at one point (a current) would not result in a common motion further down the current path.

and how to understand the energy is transferred in form of electric field? electric field of what? charge on the wire or other things?

There are net charges along the surface of wires (and also in areas where the conductivity of a conductor changes, or the current density changes). These net charges create an electric field within the wire. As mentioned before, current will soon die out if there is no electric field to be constantly re-accelerating electrons whose forward motion becomes randomized in short order. If there were no electric field, and the electrons entered an area where their motion became randomized, then electrons would start accumulating in that area. A combination of the microscopic version of Ohm's law and Maxwell's (Heaviside's) equations, results in the accumulation only "sticking" on the surface (or other exceptions mentioned above). But it is these surface (+ exceptions) charges that are responsible for the electric field that constantly re-accelerates electrons in a wire.

The electric field created by these surface charges doesn't actually transfer net energy to the electrons. The electric field created by charge density is conservative. It is like gravity. Imagine skiers taking a ski lift to the top of a mountain, and then skiing down. Although gravity accelerates the skier going down, that same gravity must be overcome while going up. It is the ski lift that imparts the energy to the skier that is released when the skier goes down. In the same way, some source of energy external to the wire's electric field is responsible for the energy that is released when the electrons accelerate under the influence of the (conservative) electric field caused by surface charges along the wire. That external source of energy is commonly called an electro-motive force. It could be chemical in nature as in a battery, mechanical as in a van de Graaff generator, or due to relative motion between a magnetic field and a conductor, as in a electric generator.

• thanks for your answer, but i still confuse about whether the current (i.e. electrons) carries the energy towards lamp? and how to understand the energy is transferred in form of electric field? electric field of what? charge on the wire or other things?
– DORA
Nov 30, 2021 at 13:42
• @DORA I have expanded my answer. I hope that expansion helps. Nov 30, 2021 at 16:34