Generation of a magnetic field by rotating a charged ring

  1. Let's assume the case for metallic ring that possesses a net negative charge within. To generate a magnetic field, the field generated by negative charges should not be canceled out by the positive charge. There are two ways to do this:

    1) Move only the electrons.
    2) Move the electrons with the protons.

    The first way is the only way in an electrically neutral ring. However, with charged objects having more electrons than protons, the second option is also possible to generate a net magnetic field.

    My concern is the following: The input of energy to spin the charged ring would be tangent to the ring itself, yet the electrical flux that would generate would create a magnetic field at right angles to that input. This suggests that a compact charged ring can be devised that would possess arbitrarily high "effective" current densities that are not limited by ohmic heating. The potential of this would be huge, allowing for arbitrarily strong magnetic fields (in excess of rare earth magnets) to be produced by relatively small circuits.

    What is not clear to me is how much resistance to motion would be generated if spinning the ring on an axis parallel to an external magnetic field. Since spinning the north pole of a cylindrical permanent magnet on its axis produces minimal magnetic flux on a loop oriented flat below it, I doubt that a cogging torque would exist between the charged ring and the external magnetic field.
  2. jcsd
  3. Here is a conceptually similar problem that is easily solvable. Take a 1 meter long aluminum tube, with a 10 cm outer diameter. Outside of this place a 1 meter long, 10.2 cm inner diameter aluminum tube. The gap is 1 mm, and the surface area between the tubes is 0.314 square meters. The capacitance is e0A/d = 2.78 x 10-9 Farads. Charge this cylindrical capacitor to 10,000 volts (in vacuum). The charge on the inner cylinder is 2.78 x 10-5 Coulombs. Spin the inner aluminum tube at 10,000 revolutions per second (neglect the tensile strength of aluminum for this experiment). The spinning current is 0.278 amp-turns per meter. The field inside is B=u0NI = 0.349 microTesla, or 3.49 milliGauss, about 1/200 of the Earth's field.
  4. The potential energy in this system = (1/2) * B^2 * Volume / magnetic constant

    This equals 3.8E-10 joules. That's very small indeed.
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