5
$\begingroup$

I admit my knowledge of physics is very little, but I did study topics on EM and stuff in school.

I was reading about computer networks and I read about antennas and how they are just wires or strips of metals that have some circuit attached to them. Then it occurred to me that the water heater I use to heat water which is rated 1000v, 1A is also a coiled wire(http://img6a.flixcart.com/image/immersion-rod/g/a/f/immersion-1500-bajaj-400x400-imadmke6xugdcynk.jpeg). So when current passes through this water heater, it should setup a magnetic field spiraling the coils of the water heater. When I bring my mobile near this immersion heater say about 4m or so close to the wire, the magnetic field should setup up a current in the metal of the antenna of my phone right? Which should fry my mobile! But it doesn't!!

I don't remember the exact equations behind induction but hey, even though my phone runs on like 5v DC and I guess 0.1 A battery, it can send data to a device three rooms across!! So how is it that the water heater can't fry my mobile?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
4
$\begingroup$

The electricity through the coil is probably coming directly from the network, so it oscillates at either 50 or 60 Hz. That would be the frequency your antenna radiates. This is very different to the frequencies your phone works in, around 1 GHz (a thousand million† Hertzs, or twenty million times faster). So, essentially, your wave (weak, as Lemon pointed out), just passes through without even seeing your phone's antenna, it is just too small (from an electromagnetic wave point of view).

This is actually a relief, because we are flooded with electromagnetic waves at 50 Hz and above (a few harmonics: 100, 150, 200... decreasing in intensity). The electric wires inside the walls are also emitting, and by sheer numbers, they are emitting much more than your water heater.

__

† Or billion, but not always. So I prefer to keep it explicit.

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ Apart from the low efficiency of such a small antenna at those wavelengths, the phone also of course filters the signal. It has to be able to get rid of other microwaves pretty close to its own signal bandwidth; then it certainly won't have a problem with some low-frequency mains hum (unless it were so strong to drive the circuitry into nonlinear behaviour or something, but for that you would indeed need a huge antenna). $\endgroup$
    – leftaroundabout
    Commented Feb 22, 2015 at 14:30
  • 1
    $\begingroup$ 1 GHz is 10^9 Hertzs, isn't it? $\endgroup$
    – rkosegi
    Commented Feb 22, 2015 at 19:03
  • $\begingroup$ @rkosegi you are right, I brainfarted there, thanks! $\endgroup$
    – Davidmh
    Commented Feb 22, 2015 at 21:05
3
$\begingroup$

That heating element has 4 turns and looks to be a few cm across. It most likely consists of a copper wire coated in nickel. The relative permeabilities of which are ~1 and ~100, respectively.

It follows that the magnetic field created by your heating element, at its strongest point, will be roughly 1 Gauss (only a few times stronger than the magnetic field of the earth).

Improve this answer
$\endgroup$
3
  • $\begingroup$ What metal do they use in making bluetooth antennas? It must have one hell of a permeability to transmit signal across multiple rooms! $\endgroup$
    – Aditya
    Commented Feb 22, 2015 at 20:53
  • $\begingroup$ @Aditya Bluetooth doesn't work via magnetic fields, it transmits radio waves. As Davidmh points out, if you were to run an AC current with a frequency on the order of 1 GHz through your heating element then you'd produce radio waves that your phone could detect. $\endgroup$
    – lemon
    Commented Feb 22, 2015 at 20:59
  • $\begingroup$ What's equivalent of permeability factor of material for EM waves like radio waves? Is there a material constant that decides the intensity of the emitted EM waves? @lemon $\endgroup$
    – Aditya
    Commented Feb 25, 2015 at 3:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.