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Tornados and magnetism

  1. Jan 30, 2004 #1
    I was wondering if someone knows how nature tornados affect different materials magnetic properties? Is there any published litterature on this subject?
  2. jcsd
  3. Jan 30, 2004 #2


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    Why would a tornado have any effects on magnetic properties?
  4. Jan 30, 2004 #3
    Maybe the corelius? effect have something to do with it. However thats more connected to the rotation of earth i think. But does tornados emit EMP or smth that can distrub electronic equipments?
  5. Jan 31, 2004 #4
    When a charged particle (or a bunch of them) move in a circle, it creates a magnetic field that runs through the center of the circle. It points in the direction that your thumb would point if you curled the fingers on your right hand in the direction that the charges are rotating.

    The sorts of things that generally end up being flung around in circles by tornadoes (rain droplets, ice particles, cows) tend to be neutral, and so they wouldn't create a B-field (magnetic field.) But if they somehow managed to get charged, the way that the water droplets in clouds do to form lightning, then I suppose that the tornado would create a B-field.

    Here's a VERY crude estimate of the strength of B-field that the perfect tornado might create.

    A lightning bolt contains 10 Coulombs of electric charge. (All of these numbers come from web pages googled at random and taken out of context.)

    The mass of a cloud is 4 million kilograms (based on a density distribution of 0.1 g/m^3 to 5 g/m^3, and a cloud radius of one kilometer.) So the charge to mass ratio of charged water droplets is on the order of 2.5E-6 Coulombs per kilogram.

    I found that wind speeds in tornados can exceed 200 mph, let's use 100 mph or 44 m/s. They can have a (funnel) diameter of 20 to 100 meters, I'll take the low end at 20 m.

    The volume of a tornado must be a lot more dense than the 5 g per cubic meter of a rain cloud. But I'm going to multiply the mass by the charge to mass ratio, and I don't want to include the masses of all of the big fat raindrops and cows and other things that couldn't be charged to the same degree that clouds are. So I'll just use 5 g/m^3 for that purpose.

    A cylinder of 40 meters in radius and 100 m in height could hold 2500 kilograms of water droplets, which could support a charge of 0.00628 Coulombs. Suppose, for simplicity, that the tornado makes all of this charge rotate around its center at a speed of 40 m/s and a radius of 20 m. This corresponds to a current of 0.002 Amperes.

    The magnetic field at the center of the tornado is given by the Biot-Savart Law, and once you get rid of that danged integral, it says

    B = ('mu-naught')*I / (2 * R)

    where I is the current and R is the radius of 20 meters. mu-naught is (4*pi)*10^-7 and I forget the units.

    So the mag field would have a strength of 6.28 * 10 -11 Teslas.

    Compare this to the mag field of the earth, which is about 50E-6 Teslas at the Earth's surface. So the strength of the tornado's B-field is about one-millionth the strength of the Earth's B-field.

    You may argue that my estimate has no validity at all, since it is based on so many unjustified guesses, estimates, and simplifications.

    I hold that whereas one estimate may lead you to the wrong answer, several estimates will generally not affect your answer that much: as many of them will be overestimates as underestimates, and so your final answer will be pretty close to the actual value.

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