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I took my son to a science museum where they had a solenoid oriented vertically with a plastic cylinder passing through the solenoid. An employee dropped an aluminum ring over the top of the cylinder when there was no current going through the solenoid. Then they turned on the current going through the solenoid and they aluminum ring went flying up and off the top of the solenoid. What law of electro-magnetics causes the force on the aluminum ring?

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An excellent experiment. My undergrad E&M course was centered around this setup, explaining the physics behind it one step at a time until we had covered all of the subject. –  Chris White Oct 20 '12 at 7:45

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I'll start this with Right Hand Grip rule for solenoids...

"The coil (solenoid) is held in the right hand so that the fingers point the direction of current through the windings. Then, the extended thumb points the direction of magnetic field". (which would be along the axis of the coil)

The higher the current, the more the magnetic field would be produced... For your example, let us assume the aluminium ring as a circular coil. When the uniform magnetic field is produced, there is a change in magnetic flux (such as this increase in magnetic field) along the axis of the ring, According to Faraday's law, induced current flows through the ring whose direction is given by Lenz's law. This induced current in the ring flows in a direction such that it opposes the magnetic field in the solenoid (the one which actually produces it). (But, the magnitude of induced magnetic field is always lesser than the field in the solenoid). Anyways, there's a repulsion. With the maximum repulsive force produced, the ring is thrown off from the solenoid. This force always depends on the magnitude of $B$ in the solenoid.

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This sounds like a Thompson's Jumping Ring setup. Turning on the solenoid creates an increasing magnetic field B. Maxwell's (Faraday's Law) tells us $$\frac{-\partial B}{\partial t}=\nabla \times E$$ So the increasing B field produces an azimuthal E field "curling" around the aluminum ring and since the aluminum is a conductor this will produce a current flowing around the ring, $I$.

Say the solenoid produces a B field in the positive vertical direction, z. The electric field and current in the aluminum ring will then be in the -$\theta$ direction of a cylindrical r,$\theta$,z coordinate system, i.e., $$I=I_{\theta }<0$$

Finally the magnetic field produced by the solenoid will not be precisely vertical but will flare outward radially; that is there will be a positive component of B in the r direction. Now we can see that the $I\times B$ force arising from the interaction between the solenoid field and the current in the aluminum ring is in the vertical direction $$F=I\times B=-I_{\theta }B_r\hat{z}$$ and since $I_{\theta }<0$, the force is in the vertical direction.

If the $I\times B$ is unfamiliar, it is just the summation of forces on a charges moving in an magnetic field.

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