What is the energy density of the ELECTRIC field for a wire that is 10mm in diameter and carrying 10A?
114930
I'm assuming that the current is not changing with time other than the initial turn-on transient.\n\nWhen the current has reached a steady state (didn't say reached zero, but some final state), you are in the area of electromagnetics called electro/magnetostatics and the magnetic and electric fields are uncoupled. For the situation that you describe (I'm assuming no stationary charges external to the wire), the Electric Field Density is zero.\n\nYou might be tempted to use Gauss' Law to find the Electric Field from an individual electron as it moved down the wire, but the summation at any particular point in space from all charges moving in the wire will sum to zero.\n\nWhen the current is first turned on you will have a changing magnetic field and therefore an electric field. However, that electric field will die away rapidly.
1028768
Simply put, it's not a conductor. \n\nhttp://jchemed.chem.wisc.edu/JCEsoft/CCA/CCA1/MVHTM/R0540/S567.HTM
910705
Hello, here is what you need:\n\nEnergy = Planck Constant * Frequency\n\nE = h * f\n\nh = 66.26 x 10^-34 Joule second
794988
Energy radiated per second is given by\nE = e*sigma*A*T^4   where \ne = emissivity; \nsigma = Stefan's constant = 5.67 * 10^-8 W/m^2/K^4\nA = surface area\nT = temperature in kelvin.\nSubstituting the various values,\nE = 249 * 10^-8 Watts.
222936
Consider a conductor of uniform area of crossection A, length L. Let resistivity of conductor be S and resistance of material to current be R.\n\n  Then,  R = SL/A\n\n     Multiply and divide by mass of conductor say m and L\n\n          R = SL*L*m/L*A*m\n\n But L * A  = Volume = V\nAnd m / V   = Density = d\n\n  Hence,  R =  S*L*L*d/m\n          S =  (m*R/L*L)/d\n\nThus, Resistvity (S) is inversely proportional to density(d) when other quantities are kept constant.
924475
= 9u-10u + 6v = 6v - u
331292
See the other postings of this question.  The equation is\n\nhttp://members.aol.com/logan20/elecspon.html
471190
Electromagnetic radiation in the 1 to 1000 m wavelength range.
313019
This is a simple application of Coulomb's Law.\nThe Electric Field is simply given by the charge on the nucleus (26e) divided by 4*Pi*EpsilonNought*distance^2\nWhere e is 1.6*10^-19 Coulombs
1045656
An electron passing through a potential difference of 150 kV picks up 150keV of energy.  So:\n150keV=h*c/lambda\nlambda =h*c/150keV.\nh=6.626*10^-34 J*sec\nc=3*10^8m/sec\n1eV=1.6*10^-19 J\n\nSo, lambda=((6.626*10^-34)(3*10^8))/((150000)(1.6*10^-19))\n\nlambda=8.28*10^-12 m, or 8.28 nm
50175
well, the magnetic dipole moment of a loop is given by mu = I*A where I is the current and A is the area of the loop.  so, to maximize the area for a given perimeter, i think you want a circular loop.\n\nhttp://www.faculty.fairfield.edu/jmac/ther/superlatives.htm\n\nas for whether to make a single loop or lots of loops, consider how much area you get for a wire of length L.  the circumference of a circle is C = 2*pi*r, and the area is A = pi*r^2.  solve the first equation for r and plug it into the formula for area, and you get A = C^2/4pi.  here C is simply the length of wire you use.\n\nso, if you make N loops of length L, then you get a total effective area of N*A.  but, if you make a single loop of length N*L, then you get a total area of (N*L)^2/4pi = N^2 * A, which is a factor of N more area than the case with N loops.  \n\nit's because of the r-squared term in the area - as you increase the amount of wire, the area goes up like the square of the length of wire.  whereas if you just add extra loops, it only goes up linearly with the length of wire.\n\ncheers!
285663
The next power of 10 smaller than nano (E-9) in SI units is pico (E-12).\n\nHowever, if your talking in terms of length, there is a unit called the "Ångström" which is 1E-10 meters.  This unit is sometimes seen when describing the size of atoms or chemical bonds.
713269
Faraday's Law!\n\nMass Plated = t * I * MW / (n * F)\n\nt = time\nI = current (Amps)\nMW = mol weight of metal\nn = valence charge of metal\nF = Faraday's Constant (96,500 Coulombs/mol e-)\n\nSo....\n\nt = 2 days * 24 hours/day * 3600s/hour = 172800s\nI = 7.60 A\nMW = 51.94 g/mol\nn = 3 mol e- / mol M\n\nPlug it all in and you should get 235.62g of Cr.
685980
most wire is marked as 600 volt that is the max voltage the insulation can handle before breakdown.\nI think you need the current handling ability of the wire, which is determined by the wire size. 14 AWG is 15 amp., 12 AWG is 20 Amp., 10 AWG is 30 AMP. any more than 30 amps you need a lot more information as it is dangers at any voltage or current.
1006839
if u recollect the equation,\n             resistance (r)= pl/A;\n        where p=specific resistivity\n                   l= length\n                   A= cross sectional area \n               Area = pie(3.414) * r2\n                  \n              but diameter is twic the radius.\nhence insulation thickness is dependent on pipe diameter.
744501
C = Q/V = εA/d\n\nwhere ε is the permittivity of free space or air\n= 8.8541878176 × 10−^12 F/m (C2/Jm)\n\nQ = (εAV)/d = [(8.85×10^−12)(0.4)(200)]/(0.0005)\n= 1.416 x 10^-6 Coulomb\n\nis the charge on the upper plate.\n\nThe charge on the lower plate that is grounded is ZERO, because the charges have moved down into the ground.
801826
E=hf  where h = 6.6210^-34 Js\n\ne = 6.6210^-34*4.31*10^14 = 2.71 10^-19 Joule=1.7 eV