Ultra high magnetic fields using carbon nanotubes.

As described in this report, the limits on the strength of stable magnetic fields are due to the magnetic forces on the conducting elements that tend to
tear them apart:
Magnetic Radiation Shielding: An Idea Whose Time Has Returned? Geoffrey A. Landis "The limit to the mass required to produce a magnetic field is set by the tensile strength of materials required to withstand the magnetic self-force on the conductors [8]. For the min-imum structure, all the structural elements are in tension, and from the virial theorem, the mass required to withstand magnetic force can be estimated as [9]:
M = (rho/S) (B^2 V)/(2 mu) (1)
where rho is the density of the structural material, S is the tensile strength, B the magnetic field, V the characteristic volume of the field, and mu the permeability of vacuum." http://www.islandone.org/Settlements/MagShield.html
You see the strength/density ratio of the material goes by the square of the magnetic field strength. The conducting wire commonly used for producing the electromagnets is made of copper because of its high conductivity and current carrying capacity. The tensile strength of copper is 220 MPa at a density of 8.92 g/cm³. The highest measured strength of carbon nanotubes has been 160 GPa at a density of 1.3 g/cm³. This is an increase of the strength to density ratio over copper of about 5,000. Then conceivably with this stronger material we could get higher magnetic fields strengths by a factor of the square root of this, 70; so to a magnetic field strength of 70 x 30 T = 2100 T. However, I have seen some references that the square of the magnetic field intensity goes only as the tensile strength itself of the conducting material:
Force on ferromagnetic materials. http://en.wikipedia.org/wiki/Electromagnet#Force_on_ferromagnetic_materials
In that case B^2 would only be larger by 800, so B itself larger by a factor of 28, so to 28 x 30 T = 840 T. Still this would be a major increase in the stable magnetic fields attainable. Anyone have a reference that says whether it's the strength to density ratio or just the tensile strength itself that determines the intensity of the field that can be maintained? The nanotubes are only available so far at centimeter lengths. Still it would be interesting to find out on tests with small fields if their use would allow magnetic field strengths in the thousand tesla range. For the nanotubes to be used for this purpose they would have to carry large amounts of current to generate the electromagnets. It has been shown experimentally that they can carry thousands of times the current of copper:
Reliability and current carrying capacity of carbon nanotubes. APPLIED PHYSICS LETTERS, VOLUME 79, NUMBER 8, 20 AUGUST 2001. "From the experimental results described in this letter we can conclude that multiwalled carbon nanotubes can carry high current densities up to 10^9-10^10 A/cm2 and remain stable for extended periods of time at higher temperature in air. Furthermore, they conduct current without any measurable change in their resistance or morphology, indicating that the sp2 bonds that are dominant in carbon nanotubes provide much higher stability against electromigration than small metallic structures." http://www.rpi.edu/~ajayan/locker/pdfs/reliability.pdf
We can estimate the strength of the magnetic field we can obtain from a given current flow and wire size from the formula B = 2(10^-7)I/r, for B the magnetic field in Tesla, I the current in amps, and r the distance from the center of the wire in meters, as described here:
Magnetic Field of Current. http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html#c2
For a 100 micron thick wire composed of carbon nanotube material, using a 10^10 A/cm2 current capacity, we could get 10^6 A of current through. Then 100 microns away from the center the magnetic field would be 2,000 T. One million amps is *quite* a large current. There are gas turbine electrical generating stations that put out 100 megawatts, enough to power a small town, that at a voltage of 120 volts would put out about a million amps. Imagine a generating station with enough power to run a town with all that power going into a single wire the width of a human hair! However, I'm wondering if these ultra high fields could be something that can be reached by amateurs, if not as sustained fields then at least in pulsed fashion. Perhaps not as much current would be needed if a different arrangement was made to create the magnetic field, such as a solenoid for example. This page gives the formula for the magnetic field of a solenoid:
Solenoid Magnetic Field Calculation. http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html#c3
You see that for a solenoid using an air core, there is a 4Pi factor in front instead of 2 as for the long wire case. So it's larger by a factor of about 6 and you would therefore need this smaller amount of current. You could get a higher field with the same current by using a metal core:
Magnetic Properties of Ferromagnetic Materials. http://hyperphysics.phy-astr.gsu.edu/hbase/tables/magprop.html#c2
The problem with this is that we are attempting to get the highest field possible while sustaining the stresses. Using other metals for the core, then they have less strength than the carbon nanotubes and will fall apart at lower magnetic field strengths. We could use the nanotubes also for the core but they give little in the way of higher permeability. As for creating a short pulse of high current, this amateurs page describes getting a 25,000 amp pulse from a silicone controlled rectifier (SCR):
The PowerLabs Solid State Can Crusher. http://www.powerlabs.org/pssecc.htm
And this page claims 70,000 to 100,000 amps can be reached in a short pulse:
Coin Crusher. http://webpages.charter.net/tesla/crushed_coin.htm
Experiments at very high magnetic fields are very important for theoretical studies. It is likely the nanotubes could withstand the high stresses induced by the magnetic fields at even higher strengths than 2100 T for short times, especially for nanotubes chosen to be low in defects to have the highest strength. Then carbon nanotubes may be the ideal material to use for producing ultra high magnetic fields for theoretical work.
Bob Clark
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Wouldn't work against neutral "particles", like gammas.
Calculate the gyroradius of a typical cosmic ray proton, and discuss the field energy densities you would need to bring that to human dimensions,
--
ciao,
Bruce

drift wave turbulence: http://www.rzg.mpg.de/~bds /
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I was just looking up some info on the energy storage capabilities of flywheels and noticed that the energy density as measured by the mechanical energy stored *per mass* has the same units as tensile strength to density:
(energy stored)/mass = (1/2)*(tensile strength)/density
from:
Flywheel Basics Tutorial. http://rpm2.8k.com/basics.htm
So it may be then that magnetic field energy *per mass* will go according to the ratio of tensile strength to density. This could be useful for space missions in achieving high storage density in the amount of electrons, positrons or ions stored by Penning traps to the mass of the traps because of the limit on the density of the particles stored according to the square of the magnetic field intensity. Admittedly this is a purely heuristic argument.
Bob Clark
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The homopolar generator discovered by Michael Faraday may provide a low cost way to generate the high currents needed to produce the ultra high magnetic fields with the nanotubes:
Homopolar generator. http://en.wikipedia.org/wiki/Homopolar_generator
This article reports on its use for producing pulsed high current:
Pulsed high-voltage and high-current outputs from homopolar energy storage system. Review of Scientific Instruments -- May 1981 -- Volume 52, Issue 5, pp. 694-697 http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=RSINAK000052000005000694000001&idtype=cvips&gifs=yes
This page describes the current that can be produced with this generator:
1.2 Energy storage system. a) Homopolar generator http://pps.coe.kumamoto-u.ac.jp/streaming/PulsedPower/generator/energy%20storage1.htm
The generator produces current by rotating a conducting disk within a magnetic field. The formula for the current generated is given in formula 28. It's dependent on the radius of the disk a, the angular rotational velocity w, the magnetic field strength B, and a resistance R: i = (a^2)wB/2R . Large homopolar generators can produce high currents using a large disk radius. For a more compact, low cost system, we might take a =1, and B = 1 tesla, about the highest field for commonly available permanent magnets. The highest angular velocity w achievable is limited by the strength of materials and radius. We might be able to make w be 200 rad/s for a disk of 1 m radius. Then we would need a resistor from 10^-3 to 10^-4 ohms to get from 100,000 to 1,000,000 amps. The problem is resistors at such low resistance usually have very low current capacity. Since the resistance is so low, we may be able to use conducting wire itself to provide the resistance. Copper wire can carry perhaps 500 amps/cm^2 without requiring expensive cryogenic cooling equipment. Then we would need from 15 to 45 cm wide wire to carry the needed current. The resistivity however of copper, in ohm-meters, is such it would require hundred meter lengths to get the required resistance. We might be able to use instead carbon nanotubes also for the resistive wire. This page gives their resistivity as 10^-4 ohm-cm:
Physical Properties of Carbon Nanotubes. http://www.pa.msu.edu/cmp/csc/ntproperties/electricaltransport.html
Different types of nanotubes have different physical measurements. It's not certain the nanotubes with this resistivity will have the very high current carrying capacity. But using this value of the resistivity we could get a 10^-4 resistance using a 100 micron wide wire, only 1 micron long.
Bob Clark
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Formula 27 on the "Energy storage system" gives the voltage as: V = (a^2)wB/2 . Then this choice of parameters would give a voltage of 100 V and an amperage of 100,000 to 1,000,000 A would result in a power required of 10 to 100 megawatts. This would be beyond a low cost system. Since what we want is the high amperage, we could choose the parameters to give a low voltage. Choose a = .1 m, w = 20 rad/sec, B =1 Tesla, to give V= .1 volts Now select the diameter and length of the wire to serve as the resistor to give a resistance in the range 10^-6 to 10^-7 ohms to produce 100,000 to 1,000,000 amps. For the wire made of copper, to get 10^-6 ohms and 100,000 amps you could make the wire 15 cm wide to carry the current, and in the range of 10 cm long to get that resistance.
Bob Clark
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