Let us consider a pair of parallel, cylidrical metal rails of radius *r*, the centers of which are separated by a distance *d*. A movable metal connector is placed between these, so as to slide freely along the rails.

Now, suppose we run a current *I* through this by connecting a current source to the rails at one end:

What will be the resulting force, if any, on the connector?

Assuming the length of the wires is at least several times *d*, we can use the “infinite wire” formula for the magnetic field resulting from the current in a rail. At a distance *R*>*r* from the center of the wire, the field is , in the direction given by the right-hand rule.

Thus, the field at a point along the connector a distance x from the center of one rail will be

Now, the magnetic force on a current-carrying wire of length *L*, sometimes called the Laplace force, is given by:

*F*=*L**I*×*B*.

Here, our connector, and the current in it, are perpendicular to the magnetic field; the cross product *I*×*B* has magnitude *IB* and is directed away from the end to which the current source is connected.

Thus, for an infinitesimal segment *dx* of our connector, the force on the segment is

To find the total force, we integrate this over *x* from *r* to *d*–*r*:

So the connecter will experience a force seeking to propel it along the rails. This physics is the basic principle of the rail gun. Since *μ*_{0}=4π×10^{-7} N·A^{-2}, the non-geometric term will be . Thus, for *d*/*r*~10, to obtain forces on the order of a newton, the current needs to be on the order of a kiloampere. Thus lies the difficulty of the railgun; massive currents and voltages are required to obtain effective projectile launching.

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Tags: Electricity & Magnetism, Friday Physics, physics, rail gun

This entry was posted on September 12, 2008 at 12:01 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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