Physics Friday 37

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=LI×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 μ₀=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