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01 April 2009

Magnetic field due to a current carrying wire

Reader Scott Milczewski, who teaches at Brooklyn Technical High School, asked me about the deflection of a compass due to a current carrying wire. The question was, can I easily get enough current to see a visible deflection? Or do I need a special high-voltage setup?

The picture to the right shows my setup. The power supply is a standard Elenco Precision, and I'm using the variable voltage input. I don't use much voltage at all -- just a few volts gives me enough current in the wire to see a deflection.

The pictures below show the experiment itself. On the right you see the wire carrying no current... it's aligned (approximately) with the earth's magnetic field. The white end of the compass needle points north. In the left-hand picture, the wire is carrying a current from right to left. By the second right hand rule, this current produces a magnetic field beneath the wire pointing west. (Toward the bottom of the picture. Yeah, sorry, I should've aligned the compass markings with the earth's cardinal directions, but I didn't think of that.

So, why doesn't the compass needle point straight to the bottom of the picture, in the direction of the wire's magnetic field? Because the earth's field is still stronger. In fact, this setup could be used to calculate the earth's magnetic field. Use an ammeter to measure the current in the wire. Use the equation

to calculate the magnetic field due to the wire at the position of the compass needle -- that means the variable r will be equal to only a few millimeters, the distance from the wire nearly to the table. Then, use vector analysis -- the wire's field is perpendicular to the earth's field, and we know that the vector sum of the two magnetic fields points about 30 degrees down from the earth's field. That should be sufficient to get close to the 10-5 T that is the typical magnitude of the earth's magnetic field.


  1. Typically, what would the raw results from this experiment be (in order to findf the strength of the earth's magnetic field)

  2. Hmmm... depends on what you mean by "raw." If we're just making a single calculation, we'd use the measured distance r and current I to calculate the magnetic field of the wire. Then vector analysis gives the horizontal component of earth's magnetic field, which should (in Virginia, anyway) be about 2 x 10^-5 T. That's what my students have measured.

    You could go an extra step: use the compass and earth's known magnetic field to measure the magnetic field produced by the wire when the wire is placed at numerous distances from the compass. Graph the field of the wire as a function of r... this should be a hyperbola. Graph magnetic field of the wire as a function of 1/r, you'll get a straight line with slope mu-nought I / 2 (pi).