[ RadSafe ] WHO: Cell phone use

Fri Jun 3 11:29:52 CDT 2011

No, that is in the absence of an applied field external. You have the electron moving at c. Obviously the radius is going to be huge. The beta particle slows down and stops. There are ions generated along the track. You have to consider what happens as it travels through tissue etc. I think the problem is intractable. We have to do the experiments. 
C

-----Original Message-----
From: radsafe-bounces at agni.phys.iit.edu on behalf of McNaughton, Michael
Sent: Fri 6/3/2011 5:16 PM
To: The International Radiation Protection (Health Physics) Mailing	List
Subject: Re: [ RadSafe ] WHO: Cell phone use

Chris

The pictures you have seen of electrons inside a cloud chamber, curling in a spiral, are taken with magnetic fields that are much stronger than the earth's magnetic field. In contrast, the magnetic fields associated with an electro-magnetic wave are weaker, so the effect is extremely small. 

The relationship between the power of an electromagnetic wave and the magnetic field is calculated using Poynting's equation. I will write it using the notation in Wikipedia:
http://en.wikipedia.org/wiki/Poynting_vector : 
S = EB/(mu0). 
E = Bc, 
so S = cB^2/(mu0)
The speed of light is c = 3.0E8 m/s 
The permeability mu0 = 4E-7*pi Wb/(A.m)
Assume S = 10,000 W/m2. Then B = 6.5E-6 T = 6.5 micro-tesla.

The effect on a beta particle is calculated using standard textbook physics and the Lorentz force.
http://en.wikipedia.org/wiki/Lorentz_force 
R = mc/(eB) 	
Where R is the radius of curvature, 
The electron mass is: m = 9.11E-31 kg, 
c = 3.0E8 m/s, 
the electron charge is e = 1.6E-19 C, 
B is the magnetic field = 6.5E-6 T.

So R = 260 m. 

The radius of curvature is much greater than the range so the effect of the magnetic field is negligible.

Mike McNaughton
LANL and University of New Mexico
mcnaught at lanl.gov or mcnaught at unm.edu 

-----Original Message-----
From: radsafe-bounces at health.phys.iit.edu [mailto:radsafe-bounces at health.phys.iit.edu] On Behalf Of Busby, Chris
Sent: Thursday, June 02, 2011 8:59 AM
To: The International Radiation Protection (Health Physics) Mailing List
Subject: Re: [ RadSafe ] WHO: Cell phone use

I am pleased you are interested and look forward to your calculations on radsdafe or off it.
I think it might be difficult to do, but maybe you are able to do this. 
The target variable is the ionisation density per unit volume. Different electron energies 0-250keV.
EM at some representative cellphone frequency, you decide.
Monte Carlo method? FLUKA?
Agree?

Sincerely
Chris

-----Original Message-----
From: radsafe-bounces at agni.phys.iit.edu on behalf of McNaughton, Michael
Sent: Thu 6/2/2011 3:47 PM
To: The International Radiation Protection (Health Physics) Mailing	List
Subject: Re: [ RadSafe ] WHO: Cell phone use

Chris

Shall we try some calculations? Shall we do them on RadSafe or offline?

mike

-----Original Message-----
From: radsafe-bounces at health.phys.iit.edu [mailto:radsafe-bounces at health.phys.iit.edu] On Behalf Of Busby, Chris
Sent: Thursday, June 02, 2011 8:38 AM
To: The International Radiation Protection (Health Physics) Mailing List
Subject: Re: [ RadSafe ] WHO: Cell phone use

-----Original Message-----
From: radsafe-bounces at agni.phys.iit.edu on behalf of McNaughton, Michael
Sent: Thu 6/2/2011 3:21 PM
To: The International Radiation Protection (Health Physics) Mailing	List
Subject: Re: [ RadSafe ] WHO: Cell phone use

Chris

Dear Mike,
Whats your point?
I said thats the effect of the magnetic field H(0). Thats the earths magnetic field. Which I hope is constant.
I think we can conclude that much more intense alternating fields will therefore have the effects I am talking about. Yes?
Sincerely
Chris 

The magnetic field of an electromagnetic wave does not cause an electron to move in a spiral. A constant magnetic field does so, but not the alternating magnetic field of an electromagnetic wave.

mike

-----Original Message-----
From: radsafe-bounces at health.phys.iit.edu [mailto:radsafe-bounces at health.phys.iit.edu] On Behalf Of Busby, Chris
Sent: Thursday, June 02, 2011 4:18 AM
To: The International Radiation Protection (Health Physics) Mailing List; The International Radiation Protection (Health Physics) MailingList
Subject: Re: [ RadSafe ] WHO: Cell phone use

Well thats a lot of questions.
The energy of an RF field is proportional to the square of field strength. All the enrgy is transferred to electrons. Just like in a cathode ray tube.
The range of the elctrons is equal to their CSDA range and depends on the decay energy inthe case of internal nuclides and equal to the gamma photon energy less the binding energy (which is second order) in the case of photons. For Sr90 the range is about 400 cells. For tritium a fraction of the cell diameter. 
In the case  of natural background the photoelectron energy fllows the gamma energy which goes as E^-3 roughly and as Z^5 in terms of the absorber, which is why U238 is so dangerous. Z=92. The B filed causes the electrons to jump about aand spin in spirals, bunny hops. And so increase their LET. For the external background it is the low energy electrons that cause the greatest harm. 
The RF energy has been measured from mobiles. It is very large, watts per cc, but the belief is that these watts are not dangerous as the energy cannot be absorbed by covalent bonds. This is true, it is a question of quantum resonance, and the energies are beyond even the rotational levels of simple diatomic molecules.After all the watts per cc from ionising is Grays per second per cc. (1 Gray = 1 joule per kg).
The mechanism is well known. It is the same mechanism as a cathode ray tube focusing ring. Put an electrron in a E or B field and it moves. I would have thought that radsafers knew that. If you look at cloud chamber trac ks you see that at the end of the track the electron curls up in a spiral. That is the effects of the magnetic filed H(0).
If I havent answered everything let me know and ill try. But the real answer is toi do experiments with a end point and run ioniosing and non ionising together. That will give the answer.
Chris

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