Force between two current-carrying coils in a coilgun

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Arv
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Hi

I am currently involved in building a linear motor which is based around the coil gun principle. I am having trouble understanding the equation for force used to desribe the coilgun's action.

namely:
Fx = I1*I2*dM/dX

Fy = I1*I2*dM/dX

Here we are considering two coils of wire, with an rapidly
changing current allowed to flow through one of the coils (The active coil)which in turn induces a current in the nearby 'passive' coil (via Lenz Law). So I1 is the active current and I2 is the passive current.

So bascially u end up with two magnetic poles with same polarity and there is a repulsion action. If the vertical movement is constrained, you are left with horizontal thrust which is what the linear motor makes use of.

I can relate this force equation to physics textbooks' description of force betweentwo parallel current carrying conductors which uses the Lorentz Force equation (F = I*L*B) and Magnetic Field near straight wire (B = mu*I/(2*PI*R) to arrive at F = (mu*I1*I2)/(2*PI*D).

I don't how to explain the dM/dX part.

Can someone help me get from the F = I*L*B description to the
F = I*dM/dX description. Thanks

Arvind.
 

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oh..the M in dM/dX is the mutual inductance..(a coefficient to describe the linking flux between the two coils..coupling). I can see how 'good coupling' can help induce large currents however, i don't understand how its rate of change contributes to force..