magnetic field of a spinning sphere?
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If there are charges on the sphere, then the moving charges will produce a magnetic field. This is called the Biot-Savart law.
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It will utilise quantum mechanics. At the subatomic level, matter behaves very differently to classical mechanics that we are familiar with in everyday life. The spin property of particles is not actually the particle spinning like a top, but is a term given to their angular momentum to help explain their interaction with each other.\n\nParticles can only spin in certain allowable directions, electrons have only 2 directions, other particles can have more. The ability to be in 2 spin states at once is due to uncertainty in that the particle could be in either state until it is observed and probability functions are used to describe its state.\n\nSimilarly a particle in a quantum well (a box with walls of electric potential) will move about with its energy in a fashion that can be described by its wave function. At any measurement it will be in a certain point (although with unknown momentum) but it will vary with every measurement.\n\nThese sorts of uncertainties are what are scientists are hoping to use in quantum mechanics. Calculations will actually be performed millions of times simultaneously providing varying results, but favourable probabilities will lead to the correct answer with immeasurably small innaccurary.\n\nPotential methods will utilise the spin magnetic moments of electrons, or the position of valence (outer) or free electrons moving between pairs of ions (charged atoms) accurately positioned in a semiconductor. There is still a lot of work required.
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Please take the time to\nvote for the best answer\n\nThank You!!\n\nRecall that an object of a certain mass moving with particular speed will have an associated kinetic energy  mass x speed2. An object spinning about an axis will also have associated with it a kinetic energy, composed of the kinetic energies of each individual part of the object. These individual contributions may be summed up to give an expression for the total kinetic energy of the spinning object: \n\nLook this up for more
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1. magnetic field B\n\n2. relativistic mass at double light speed, as you see it's imaginary ;)\n\n10. relativistic kinetic energy\n\ncould only guess about the rest. i hate tensors.
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the earth has a molten metallic core that's spinning...creates an em field
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Look up the mecahnical univers 43? on magnetic induction.
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I haven't done any physics problems for ages, but I'll give it a try.\n\nThe centripetal acceleration = V^2/R \n\nBut, because the acceleration due to gravity is about 9.8m/sec^2 and this is important to the answer, you first want to express V in m/sec;\n\nV = 360km/hr = 360,000m/3,600sec = 100m/sec.\n\nSo the centripetal acc. = (100m/sec)^2 /500m = 10000/500 = 20m/sec^2, which is about 20/9.8 = 2.04 g-forces. This force is always directed towards the center of rotation. \n(Note: if you were on a spinning wheel space station where gravitational forces were negligible, and it had a radius of 500m and had a tangential V of 100m/sec, and you were walking around the inside of the station at the 100m extremity, you would have the apparent weight of 2.04 gs)   \n\nAs far as his "weight" is concerned: if the seat of the plane measured his weight at different points around the loop then it would vary. \n\nAt the top of the loop: his recorded weight would be 70kg x (20-9.8)/9.8 = 72.86kg. This is because gravity is reducing the  weight. \n\nAt the bottom: it would be 70x(20+9.8)/9.8 = 212.86 kg\nThis is because gravity is adding to the weight.\n\nAt the mid-points gravity and the centripetal acc. would be perpendicular to each other and the recorded weight would simply be 70x20/9.8 = 142.86kg\n\nI hope that helps.
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This is the part of the wikipedia that I understood.  I can explain it to you if you want to send me a message.  I think it makes alot of sense.\nNow let’s see what happens to our accelerating particles when a torque is applied to the spinning wheel. Assume the axis of rotation created by the torque is through the center of the wheel at 90 degrees to the primary rotation of the wheel. Let’s look at a particle that is on this axis of rotation. Since the particle is on the axis of rotation there is no direct motion applied to the particle at the instant of the applied torque. But let’s look at what will need to happen at the next moment in time. The particle is now going to be forced to curve again. This time in the direction of the curve so as to accommodate the tilt of the wheel. Now we have a particle that is already moving and it wants to keep moving in a straight line. So the particle will exert a force on the wheel. If you look at a particle on the other side of the wheel you will see that the force of the second particle is in the opposite direction of the first particle. That pair of unmatched forced is what causes the precession torque that is 90 degrees to the applied torque
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You're already on a website.  What is it with people these days?  Look in your book.  \nInteresting though: the derivative of the area of a circle is equal to the circumference of it.  Works the same for a sphere too.
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L = μ(0) μ(r) N^2 A/l\n\nwhere\n \nμ(0) is the permeability of free space (4π × 10-7 henrys per metre) \nμ(r) is the relative permeability of the core (dimensionless) \nN is the number of turns. \nA is the cross sectional area of the coil in square metres. \nl is the length of the coil in metres\n\nHere we assume that μ(r) = 1, since the solenoid doesn't enclose anything like a metal core. Then\n\nL = [μ(0) N^2 (πR^2)] / l, where R is the radius of the solenoid\n\n= [(4π × 10-^7)(2920)^2(π(0.005)^2)] / (0.1)\n\n= 8.44 × 10-^3 henries
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Oh dear.  You should be able to do this one.  You simply need to calculate the mass of the core and divide by the total mass.  You have a density and a radius.\n\nMultiply the density of the core by the volume of the core (be sure you use the same units!)  I will leave it to you to look up/remember the equation for the volume of a sphere.\n\nTake the result and divide by the total mass. Are you sure that is the density, btw?  It seems awfully light for iron at the center of the Earth.  Even water has a mass of 1000 kg/m^3.
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The earth turns with respect to the motion of force which was placed upon it with in the vacuum of space, and also with respect to the vacuum of space there is no friction to slow it down, therefor it shall continue to spin forever if it's motion is uninterrupted, or unaltered by another matter energy object of aether large size, or fast speed impacting it with direct respect to the angle of the interaction.\n\nas for your second question the axis is the center point at which the spin turns, and is therefore recognized by us with respect to the earths motion spin as the axis, and if the motion and direction of spin were changed tomorrow to be perpendicular of that which it is today, then the new axis would be perceived tomorrow as existing at some point at which we recognize the equator today.
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The closest magnetic source.
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1 and 2 are easy. I'll help with #3.\n\n3. You need I, the rotational inertia. From a table in my physics book, I = MR^2 for the reel. You need to convert the force to torque, T = F*R. \n\na. Then you can get the angular acceleration, alpha, from \nT = I*alpha\n\nb. In t = 0.25 s, omega will reach alpha*t. The angular displacement, theta, will be (1/2)*omega*t. From that and the circumference you can get the line drawn from the reel.
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Restmasses of charged particles are measured most accurately using Penning traps. Penning traps use a strong homogeneous magnetic field to confine particles radially and a quadrupole electric field to confine the particles axially. The magnetic field causes charged particles to move in a helix, and the electric field prevents particles from spiraling out of the trap along the magnetic field lines.\n\nThe frequency with which the particles rotate in the magnetic field is proportional to q/m where q is the charge of the particle.
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Let L be the initial rope's length and R be the radius of the equator in metres.  Then \n\nL = 2piR\n\nAdding 1 to the length, we get:\n\nL+1 = 2pi(R+x)\n\nwhere x is the extra radius, i.e. the height at which the rope would float.  Combining these two equalities we get\n\n2piR + 1 = 2pi(R+x)\n2piR + 1 = 2piR + 2pi*x\n1 = 2pi*x\nx = 1/(2pi) = 0.159 or about 16 centimetres.\n\nThe cool part about this is, it's independent of the earth's radius!  The result is the same whether you start with the earth, a basketball or a marble.
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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.