# Why does a metal ring, slid onto a solenoid, fly off?

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In summary: This is the same principle at work in the demonstration with the solenoid and metal ring. In summary, our professor demonstrated electromagnetic induction with a solenoid and a metal ring in our physics II class. The resulting force on the ring was explained by the induced EMF and current, causing a magnetic force pushing the ring upward due to the opposing current. This is similar to the principle of repulsion between parallel wires with opposite currents.
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Our professor did a demonstration today for our physics II class. She was demonstrating electromagnetic induction with a solenoid and a metal ring. When she slid the ring down the solenoid and passed alternating current through the solenoid, the metal ring was flung upward off the solenoid.
My professor was attempting to demonstrate magnetic induction but she never explained the net upward force. I get that currents are induced and the resulting field opposes the original field, but why in the upward direction? why not downward?

Let's see. You can take the solenoid as a magnet with a time-dependent magnetic field ##\vec{B}=B_0 \cos(\omega t) \vec{e}_z## (quasistationary approximation, valid for ##\omega/c \ll L##, where ##L## are the typical length scales of the magnet and the ring etc.

Let ##a## be the radius of the ring, with the surface normal vector taken as ##\vec{e}_z##. Then the magnetic flux through the ring is
$$\Phi=\pi a^2 B_0 \cos(\omega t),$$
and the induced EMF
$$\mathcal{E}=-\frac{1}{c}\dot{\Phi}=\frac{\pi a^2 B_0}{c} \sin(\omega t).$$
Now this EMF causes a current, which is subject to the equation
$$L \dot{I}+R I=\frac{\mathcal{E}}{R}=\frac{\pi a^2 B_0}{Rc} \sin(\omega t),$$
where ##R## is the resistance of the ring and and ##L## its self-inductance, and this gives
$$I=\frac{\pi a^2 B_0}{R^2+\omega^2 L^2} \left [L \omega \exp \left (-\frac{R t}{L} \right )+ R \cos(\omega t)-L \omega \sin(\omega t) \right].$$
Now there is a magnetic-field component also in radial direction ##B_{\rho} =B_{0\rho} \cos(\omega t)##, where ##K## is some constant. The ##z##-component of the force on a "current element" ##I \mathrm{d} \vec{\ell} = I a \mathrm{d} \varphi \vec{e}_{\varphi}## is ##\mathrm{d} F_z=-I B_{\rho} a/c \mathrm{d} \varphi##, leading to a total force ##F_z=2 \pi I a B_{\rho}/c##.
Using the above result you get after some algebra that there is indeed a steady positive ##F_z## plus some contributions oscillating with the frequency of the magnetic field from the solenoid:
$$F_z=\frac{\pi^2 a^3 B_0 B_{0\rho} \omega L}{(R^2+\omega^2 L^2) c^2}+\text{time-dependent contributions}.$$
So there's indeed a resulting upward force (don't nail me on all the constants in front of the expression, I'd have to check carefully again ;-)).

Delta2
where does the radial magnetic field comes from? The fact that the solenoid is not infinitely long ?

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Yes, there is both a radial, and azimuthal component, because of the finite length of the solenoid. For a complete treatment (with heavy use of elliptic functions, of course), see the following preprint:

https://archive.org/details/nasa_techdoc_19980227402

vanhees71 said:
valid for ω/c≪L

where does this statement come from? Is it dimensionally correct (i am assuming that c is speed of light

Argh, it must of course be ##c/\omega \gg L##. This comes from the typical length scale over which retardation effects are important, i.e., ##c/\omega=\lambda/2\pi \gg L## means that the typical extensions of your setup is much smaller than the wave length, over which the wave changes significantly, i.e., you can consider the changes over the extensions of the setup as instantaneous along this extension ("near field approximation"="quasitstatic approximation").

"I get that currents are induced and the resulting field opposes the original field, but why in the upward direction? why not downward?"

Think of two coils on the same axis. If the current in the lower coil increases, there will be an increasing flux through the upper coil. By lenz's law,
There will be an induced current in the upper coil that is opposite to the current in the lower coil. The magnetic force between parallel wires with opposite currents is repulsive, pushing the upper coil up.

## 1. Why does the metal ring fly off when slid onto a solenoid?

The metal ring flies off when slid onto a solenoid because of the electromagnetic force produced by the solenoid. When an electric current flows through the solenoid, it creates a magnetic field which interacts with the metal ring, causing it to move.

## 2. Can the metal ring fly off in the opposite direction?

Yes, the metal ring can fly off in the opposite direction if the direction of the current flowing through the solenoid is reversed. This would result in a change in the direction of the magnetic field, causing the metal ring to move in the opposite direction.

## 3. How does the strength of the magnetic field affect the movement of the metal ring?

The strength of the magnetic field directly affects the movement of the metal ring. The stronger the magnetic field, the greater the force it exerts on the metal ring, causing it to move with more velocity. This can be achieved by increasing the current flowing through the solenoid or by adding more coils to the solenoid.

## 4. Why does the metal ring sometimes not fly off when slid onto a solenoid?

If the metal ring does not fly off when slid onto a solenoid, it could be due to a weak magnetic field. This could be caused by a low current flowing through the solenoid or a small number of coils in the solenoid. Additionally, the metal ring may not be made of a material that is easily affected by magnetic fields.

## 5. Can the metal ring be stopped from flying off when slid onto a solenoid?

Yes, the metal ring can be stopped from flying off when slid onto a solenoid by using a non-conductive material, such as plastic, to cover the solenoid. This would prevent the metal ring from being affected by the magnetic field and thus, it would not move. Alternatively, the current flowing through the solenoid can be turned off, which would also stop the metal ring from flying off.

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