# Why does a ring falling through a magnetic field experience an upward force?

The Problem states:

A metallic ring of Mass $M$ and radius $r$ falls freely under the influence of gravity in the direction along the negative Z-axis. A magnetic field $B_z = B_0(1-z\lambda)$ where $z$ is the Z coordinate of the center of the ring, also exists along the positive Z-axis. Initially , the center of the ring is at the origin. The resistance of the ring is $R$ .The plane of the ring is perpendicular to the Z-axis. Find the terminal velocity of the ring.

I figured, since the question is asking for a terminal velocity , there should be a force acting on the ring in the direction opposite to the gravitational force. But the force on a small segment of the ring , by the magnetic field due to the induced current would be radially inwards. Hence , the total force on the ring by the magnetic field would be zero.

My question is where would this force arise from?

The magnetic field through the ring changes as it falls, and whenever you have a conducting loop with a changing magnetic field through it, you get a magnetically induced current in the loop (remember Faraday's law). The current flows in the XY plane, perpendicular to the magnetic field, and a current flowing perpendicular to a magnetic field experiences a magnetic force ($\vec{F} = I\vec{\ell}\times\vec{B}$). That's going to be your upward force.

Now you at this point you "might be" (i.e. you are) wondering how $\vec{F} = I\vec{\ell}\times\vec{B}$ can ever produce a component of force in the Z direction, given that the magnetic field is in the Z direction and $\vec\ell\times\vec B\perp \vec B$. The reason is that $\vec{B}$ is not actually in the Z direction. It can't be, given the conditions of the problem. The magnetic field has to satisfy $\vec\nabla\cdot\vec{B} = 0$, but

$$\vec\nabla\cdot B_0(1 - z\lambda)\hat{z} = -\lambda B_0 \neq 0$$

The most plausible explanation is that whoever designed the question didn't think it through properly, but one could perhaps claim that the question didn't specify the X and Y components of the magnetic field, and thus there are nonzero components in (at least one of) those directions. These components will produce the vertical force on your ring.

If there are not nonzero components of $\vec{B}$ in the X and/or Y directions, then the magnetic field has to be uniform, and in that case there will be no upward force.

The trick to analyzing this kind of system is to realize that once the ring develops a induced current and thus an induced field it can be treated as a magnet.

The specified external field points up, so you may imagine it as the field between a low north pole and a high south pole.

Next determine the orientation of the induced dipole and image it as a short bar magnet. Which way does it point? Should it be pulled down or up by the external field?

Getting the strength of the effect is probably easier in terms of energy. Some of the gravitational energy is being converted into Ohmic heat.

The next question is "Why did my force-based analysis not work?", and the answer is that you were not given the full magnetic field. It can't be strictly vertical with no transverse components, because then the divergence of the field would be non-zero. That means that the transverse component of the force is not the whole story and there must be a upward component as well. When you sum that component around the ring the result is non-zero.

You have to use Faraday's law to find the current induced (by the change in the magnetic field) in the ring, this current is the movement of charges on the ring, which is perpendicular to the original magnetic field and by Lorentz force will create a force on the ring.

• According to the expression of Lorentz force, the force will be perpendicular to the magnetic field. But the magnetic field is in the z direction. so there can be no force because of it in that direction. And to oppose gravity we need a force in that direction – Sai Mar 13 '15 at 15:15

Assuming the ring is falling flat in relation to the magnetic field: Would the ring act as a single turn solenoid, with the induced magnetic field components around the ring canceling everywhere except the axial direction? And, of course the induced field would oppose the applied field to produce a retarding force, because nature is just contrary like that, Ampere, Biot-Savart and Maxwell not withstanding.

I believe the answer to the force part of your question is summed up in Lenz's Law:

If an induced current flows, its direction is always such that it will oppose the change which produced it.

Since the metallic ring is going through the magnetic field due to gravity, the current induced in the magnetic field (which is along the positive z axis) will oppose gravity, or be in the upward direction.

In practice, this can be viewed by observing a magnetic being dropped down a copper rod to demonstrate Lenz's Law, as seen in various demos.

Sources:

Electrical Engineering Student

http://en.wikipedia.org/wiki/Lenz%27s_law

http://www.phy-astr.gsu.edu/hsu/LCh23.pdf

http://video.mit.edu/watch/physics-demo-lenzs-law-with-copper-pipe-10268/