How would I find the vector potential, A, of a uniform magnetic field?
829822
The vector potential is defined only to within the gradient of a scalar. That is, you can add the gradient of any scalar field to the answer below and get an equally valid expression for the vector potential.\n\nIf B is the constant (vector) value of the magnetic field then the vector potential is:\n\nA(r) = -r x (B/2)\n\nwhere r is the the position vector and x means cross product. In Cartesian coordinates in which the magnetic field is pointing in the direction of the z axis, the three components of A are:\n\nA_x = -(B/2)y\nA_y = (B/2)x\nA_z = 0\n\nwhere B here is a constant number.
950044
In high-energy physics, medical treatments, and space detections, quadrupole field distributions and sextupole field distributions are often required over a large spherical volume. Traditionally, these fields can be generated with cylindrical magnets. In comparison with cylindrical magnets, the spherical magnets have the advantages of smaller values of inductances, lower energy storage, and less magnetic flux leakage. The basic method derives from the target field approach that has been used to solve many of the inverse electromagnetic problems. \n\nCalculation:\n\nThe geomagnetic poles or geocentric dipole, can be computed from the first three Gauss coefficients from a main field model, such as the WMM or IGRF. Based on the current WMM model, the 2005 location of the geomagnetic north pole is 79.74°N and 71.78°W and the geomagnetic south pole is 79.74°S and 108.22°E. \n\nThe magnetic poles or dip pole are computed from all the Gauss coefficients using an iterative method. Based on the current WMM model, the 2005 location of the north magnetic pole is 83.21°N and 118.32°W and the south magnetic pole is 64.53°S and 137.86°E.\n\nThe location of the center of the eccentric dipole, sometimes known as the magnetic center, computed using the first eight Gauss coefficients for 2005.0, is at approximately (r, φ´, λ) = (552 km, 22.2°N, 141.6°E). \n\nThe task of locating the principal magnetic pole by instrument is difficult for many reasons; the large area over which the dip or inclination (I) is nearly 90 degrees, the pole areas are not fixed points, but move tens to hundreds of kilometers because of daily variations and magnetic storms, and finally, the polar areas are relatively inaccessible to survey crews (map of North and South polar wander - courtesy of Dr. John Quinn, U.S. Geological Survey retired). The Geological Survey of Canada keeps track of the North Magnetic Pole, which is slowly drifting across the Canadian Arctic, by periodically carrying out magnetic surveys to redetermine the Pole's location. The most recent survey, completed in May, 2001, determined an updated position for the Pole and established that it is moving approximately northwest at 40 km per year.
1092387
The closest magnetic source.
400015
By newron's ii law\nMa=F-f\nf=10-10*0.6\nf=4 N\nf= -4 i (vector notation)
443144
you should try this site http://www.oberlin.edu/physics/dstyer/SolvingProblems.html
786730
Use the right hand rule, also known as the rule of thumb. If you have a wire (insulated!) with a current flowing through it, grab the wire in your right hand, with your thumb pointing in the direction that the current is flowing. Your fingers will curl around the wire in the direction of the magnetic field.\n\nIf the current alternates (periodically reverse direction) then the magnetic field also alternates. Envision it as a cylindrical zone that has the wire down its axis, and that the cylinder expands and contracts as the current of electrons increases or decreases. \n\nNow, suppose we wrap our wire around a coil form to make a coil. There is also a right hand rule, or rule of thumb, for coils. If you hold the coil (again, use insulated wire!) with your fingers wrapped around it pointing in the direction that the electron current flows, your thumb will point in the direction of the north pole of the coil.\n\nHow does that work? Remember, we are imagining the magnetic field as a cylindrical region around the wire. But the surface of the cylinder is rotating. (We'll stick with current in only one direction for the time being. It's easier to follow.) So think of the flow of the magnetic field as rotating around the wire. \n\nNow, if we wrap the wire into a coil, the rotation of the segements of the magnetic field add up in the same direction. They are working together. This is why the field is stronger inside the coil.\n\nWe should also think of the magnetic field as composed of individual lines of force that repel one another. That is why, when they exit one end of the coil and return to the other, they spread out in space. \n\nWe can see this if we hold a magnet under a sheet of paper or aluminum foil and sprinkle iron filings on top. They cluster most around the poles where the field lines are concentrated-- the field is strongest-- and more thinly around the outside of the magnet, where the field lines spread out, because the field is weaker there. This holds true for any magnet, not just electromagnets.\n\nI hope this is clear enough to help. It is a little tough trying to understand things we can't see.\n\n19 NOV 06, 1816 hrs, GMT.
871715
An equipotential surface of an electrostatic field is given by:\n\nE = -∇Φ\n\nwhere φ(x,y,z) is the scalar field representing the electric potential at a given point. The symbol ∇represent the derivative of the scalar field Φ which results in a vector field that is called the gradient.\n\nSo for example, the electrostatic field of a point charge q is given by:\n\nE = (kq/r^2) ř\n\nwhere k is Coulomb's constant, and is = 1/4πε\nε is the Permittivity of free space\nand ř is the unit vector in the radial direction in polar coordinates\n\nHere we see that the equipotential surface of the electrostatic field of a point charge are "spheres" surrounding the charge q, and its strength decreases by 1/r^2. At some distance r from the charge q, the electrostatic field E has the same strength everywhere on the surface of the sphere.
935341
1 sum the forces and set equal to zero. 2 forces, gravity and electric.\n\n2. conservation of energy. initially all kinetic, finally all potential. KE1=U2. U2 is electric potential. solve.\n\ngood luck
170255
It is much easier to just give you a link instead of explaining it:\n\nhttp://www.shef.ac.uk/physics/teaching/phy107/solvegray.html
474388
Okay, first, since the vector given for B is a unit vector, we know that the vector from the origin to B must be given by (some constant)*(2,-2,1)/3.  So, what you want to do is find the value of that constant such that |A-B|=10.  Do this by setting A-B=(6-2x/3, -2+2x/3, -4-x/3).  Then just find the length of that vector in terms of x, set it to 10, and solve for x.
598602
Since Magnetic field (B) = [K I(dl x r)]/r^3\n\nwhere I = current in the conductor\n      r = distance of the pt. from the conductor where u  want to find the magnetic field\n\nHere dl x r = vector product of small part of L i.e dl and r\n\nSee here that Magnetic field is directly proportional to Current....therefore when we increase the current by a factor of 4, magnetic field will also increase by a factor of 4.
839609
Curl, divergence, gradient.
911239
In standard conditions \ndE = -0.0592 V/n log [Fe 2+]/[Cd2+] = - 0.0592 V/2 log 0.19/1.2 = 0.024 V\n\nYou can take a look here for explanations and examples:\n  http://www.science.uwaterloo.ca/~cchieh/cact/c123/nernsteq.html
950482
OK, Drummanmatthew.  It's been a long time since I did this but I think I remember enough to help you out.\n\nFrom what I remember, the potential, phi, is a function of x, y and z defined so that F = -grad(phi)  (I'm not clever enough to do symbols, I'm afraid) where grad is the operator (d/dx, d/dy, d/dz) where the d's are actually partial differential operators, rather than ordinary differential operators.\n\nIn other words, we need a function which when you diff wrt x gives you -2xy, when you diff wrt y gives -x^2-z^2 and when you diff wrtz gives you -2yz.  To me, it is the y part of this that gives it away.  It clearly has to be something plus something.  You need the bit with the z to differentiate away when you diff wrt x.  It seems to me that -yz^2 will do the trick.  Diff wrt x and it goes, diff wrt y and you get -z^2.  Diff wrt z and you get -2yz, as required.  To get the rest of it right you also need a -xy^2 by similar logic.\n\nSo, I reckon that phi = -y(x^2 + z^2) + c\n\nThe constant c should be obtainable from a boundary condition somewhere (I seem to recall that we often want phi to go to 0 as we go off to infinity but this is clearly not possible here).\n\nI'm not absolutely certain about the - but I think this correct apart from that.\n\nHope this helps.
31718
for an ohmic material:\n\nP=VI\nV=IR\nso\nP=I^2 * R\n\ncheers!
506663
That'll be the integral over C of k.curl(F)\ncurl(F) = 0i,-4j,-1k\nk.curl(F) = -1\narea of C = 9pi\ntherefore W = -9pi
345042
Electric field:\n\nE=(1/4*pi*epsilon0)*(q/r^2)--> [spherical shell at r>Radius]\n\nE=0 --> [spherical shell at r<Radius]\n\nE=(1/4*pi*epsilon0)*(q/r^2) --> [spherical distriburion at r<=Radius]\n\nor E=(q/4*pi*e0*R^3)*r [uniform charge]\n\n-----------------------------------------------------\nPotential field:\n\nit doesn't have a formula but u can find it with this integral:\n\nv=(1/4*pi*e0)S(dq/r)-> S is integral!