(Demonstration idea courtesy of Rob Geller)

The photograph above shows the right-hand rule model resting in its stand. It is a copper ring with three steel arrows, equally spaced about it, to indicate the direction of an imaginary current flowing through the ring, to which is attached a plastic tube bearing decals that indicate the direction of the magnetic field associated with this current. The north and south poles are labeled “N” and “S,” respectively, and each side has a set of arrows that get smaller as they get farther from the center. The arrows are not made to any particular scale, but show the attenuation of the magnetic field as one gets farther from the current loop. The tube is attached via an acrylic disc that has been cut out to leave three spokes, one fixed at each of the steel arrows.

The magnetic dipole moment, μ, associated with an electric current flowing in a loop of wire is μ=

NiA, whereNis the number of turns in the loop,iis the current (in amperes) andAis the area of the loop. Current is, by convention, the flow ofpositivecharge, which is, of course, in the opposite direction to electron flow. The right-hand rule states that if you curl the fingers of your right hand in the direction of the current in the loop, your thumb will point in the direction of the dipole moment, μ. You can illustrate this as shown below:

You can, of course, lift the model off its stand, hold it in your left hand to show to the class, and, in whatever position is most comfortable, use your right hand to illustrate the right hand rule. (It is also best not to block the “N” label or “S” label, as in the photograph at above right. How embarrassing!) As the thumb and the direction of the arrows indicate, magnetic field lines exit the north pole of the dipole, and they enter at the south pole.

You can also think of the field around a straight (long) current-carrying wire, whose direction is also given by the right-hand rule, which states that if you point the thumb of your right hand in the direction of the current flowing in the wire, then curl your fingers around the wire, then your fingers will point in the direction of the magnetic field. If you then make the wire into a loop, the field lines in the center will all point in the direction perpendicular to the loop. You can demonstrate this as below:

The magnetic field around a straight, long wire carrying a current,

i, is (1/4πε_{0}c^{2})(2i/r), where ε_{0}is thepermittivity constant(= 8.854187818 × 10^{-12}C^{2}/N·m^{2}) and r is the distance from the wire at which you wish to know the magnitude of the field. Because thepermeability constant, μ_{0}, = 4π × 10^{-7}tesla·meter/ampere (= 1 henry/meter), 1 tesla = 1 N/[C(m/s)] = 1 N/(A·m), and 4πε_{0}c^{2}= 10^{-7}, the magnetic field can also be expressed as (μ_{0}/4π)(2i/r), or (μ_{0}/2π)(i/r), or μ_{0}i/2πr. You can arrive at this via Ampère’s law (∮B ·dl= μ_{0}i), integrating over a circular path, l, around the wire, or via the Biot-Savart law (dB= ((μ_{0}i/4π)(dl×r)/r^{3}), finding the contribution to the field at a given point at some distance from the wire, of the current through each element,dl, of the wire, then integrating. The two methods yield the same result, B = μ_{0}i/2πr (in tesla). (1 tesla also equals 1 weber/meter^{2}. The weber, which also equals 1 V·s (= 1 N·m/A, or 1 kg·m^{2}/A·s^{2}), is the unit ofmagnetic flux.)If we bend the wire into a loop or coil, consider elements along its circumference,

dl, and the magnetic field,dB, due to each of these, and then integrate around the loop, we find that at the center of the loop or coil, the field isB= (μ_{0}/2)(Ni/r), whereris the radius of the loop, andNis the number of turns in the loop (coil). (From the Biot-Savart law,dB= (μ_{0}i/4π)(dlsinθ/r^{2}). Since the current for each elementdlis tangent tor, sinθ= 1. Also,dBis in the same direction for each element.All the elementsdlsum to 2πr. Integrating around the loop thus givesB= (μ_{0}/2)(Ni/r).)For points distant from a current loop,

B= μ_{0}ir^{2}/2x^{3}, whereris the radius of the loop andxis the distance of the point along the axis perpendicular to the loop. Substituting πr^{2}for the area of the loop and including the number of turns givesB= (μ_{0}/2π)(NiA/x^{3}), orB= (μ_{0}/2π)(μ/x^{3}).

References:1) Feynman, Richard P., Leighton, Robert B. and Sands, Matthew.

The Feynman Lectures on Physics(Menlo Park, California: Addison-Wesley Publishing Company, 1963) Volume II, section 13-5.

2) Halliday, David and Resnick, Robert.Physics, Part Two, Third Edition(New York: John Wiley and Sons, 1977), pp. 727, 747-750, 759-762.

3) Sears, Francis Weston and Zemansky, Mark W.College Physics(Reading, Massachussetts: Addison-Wesley Publishing Company, Inc., 1960), pp. 641-8.