Force on a conducting ring due to solenoid with AC current

[Spriteling]

Homework Statement

A long circular solenoid of radius a and N turns per unit length has its axis in the z direction. A small highly conducting ring, or area A, resistance zero but self inductance L, is place with its plane horizontal and its centre on the axis and near the top of the solenoid. Show that when an alternating current is passed through the solenoid, a current is induced in the ring, which thus develops a magnetic dipole moment proportional to the instantaneous local value of the solenoid's magnetic field. Hence show that there is a lifting force on the right, which is maximum when the ring is a distance a/√15 below the top of the solenoid.

Homework Equations

I have that the magnetic field near the end of the solenoid is \frac{\mu_0 N I_1(t)}{2}. The flux will be \pi a^2 = \frac{\mu_0 N I_1(t) \pi a^2}{2} + L \frac{dI}{dt}. Also, \epsilon = I_2R = -\frac{d \Phi}{dt}.

Also, F = \nabla{ (m.B)}

The Attempt at a Solution

Well, for the current, I was just going to solve the differential equation given by \epsilon = I_2R = -\frac{d \Phi}{dt} = -\frac{\mu_0 N \pi a^2}{2}\frac{dI_1}{dt} - \frac{L d^2 I_2}{dt^2} because this will be easy to solve since R is 0. However, I then got confused because it's an alternating current, so the sign of dI/dt will always be changing? Will this not mean that the induced current in the conducting loop will also be constantly changing direction? Am I missing something here, or am I on the right track? How do you account for the continuously changing sign?

 

[Spriteling]

I guess I should also say I'm not sure if the magnetic field from the solenoid is correct; for the force to have a maximum at any point along the solenoid, would the magnetic field not have to be changing?