Origin of repulsive force on a magnet approaching conducting ring?

  • #1
zenterix
525
74
Homework Statement
Consider a conducting ring fixed in the ##(x,y)## plane. A bar magnet is moving at constant velocity towards the center of the ring along the ring's axis of symmetry.

At time ##t=t_1## the magnet is very far from the ring and at a later time ##t=t_2## the center of the magnet is at the center of the ring.

Consider the magnetic flux through the circle with ring as boundary. The element of area ##d\vec{A}## is in the ##+\hat{k}## direction.
Relevant Equations
During the time interval ##t_1<t<t_2## is the force exerted by the ring on the magnet repulsive or attractive?
1711583315390.png


Between ##t_1## and ##t_2## the magnetic flux is positive and increasing.

Thus, we have a negative emf and from the point of view of the little stick figure above, the induced current is clockwise.

It is not clear to me where the repulsive force on the approaching magnet comes from.

The induced magnetic field on the axis of symmetry of the ring points in the ##-\hat{k}## direction. But then the cross product between ##\vec{v}## and this field is 0.

I think this is related to another doubt I have about this experiment in which we have a coil, a solid iron bar, and a ring that fits on the iron bar. When we current on in the coil, if the ring is initially at the base of the solid bar, the ring jumps up into the air. If we instead turn on the current and only then place the ring along the iron bar then it levitates.

I don't understand the forces that provoke this behavior. I think it is the same force that is repulsive in the example above. Is this true?
 
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  • #2
Here is the solution to this problem on MIT OCW
Since the bar magnet approaching the ring causes changing magnetic flux, the resulting force will want to slow that approach, i. e. a repulsive force. Thinking in more detail, the clockwise current will produce a magnetic field pointing to the left. That looks like a bar magnet with its North end on the left. So, the North ends of the bar magnet and the “ring magnet" are facing each other and like magnetic poles repel each other.

Perhaps I am missing something very basic that I already learned. But I cannot remember what the technical reason is that when we have two north poles facing each other they repel.
 
  • #3
zenterix said:
It is not clear to me where the repulsive force on the approaching magnet comes from.
Answer the following questions and it will become clear. Don't think in terms of positive or negative emf. Think in terms of induced currents that are formed to oppose any proposed change in the magnetic flux through the ring.
1. In what direction is the induced current in the ring?
2. As you know, a ring of current produces a magnetic field and has a "North" and "South" pole. If you were to model the ring as a bar magnet, which one of its poles would be closer to the approaching bar magnet on the left?
3. Do the same analysis after the magnet has gone through the loop. Is the force between the loop and the magnet attractive or repulsive n this case?
 
  • #4
1. The induced current in the ring is clockwise from the point of view of the stick figure and this generates a magnetic field that opposes the magnetic field from the approaching bar magnet.

2. The induced magnetic field is such that the ring behaves like a bar magnet (ie, like a magnetic dipole). The north pole is on the left and the south pole on the right. The induced north pole is closer to the approaching bar magnet's north pole.

3. After the magnet has gone through the loop magnetic flux decreases, the electromotive force changes sign, the induced current changes sign (flows in the opposite direction as before, counterclockwise viewed from the right side) and the induced magnetic field is that of a bar magnet with the north pole now facing to the right (and nearer to the south pole of the departing bar magnet than to its north pole).

For some reason (practical real-world experience with magnets) I know that this configuration means that the force on the now departing magnet will be attractive.

But my question is what is the origin of this force?
 
  • #5
Here is a sort of qualitative argument as to why the same poles on two bar magnets repel and opposite poles attract.

I don't fully understand all the steps below but here goes.

Consider a current loop in a uniform magnetic field.

Let ##\vec{\mu}=I\vec{A}=IA\hat{n}## be the magnetic dipole moment vector.

The torque felt by the current loop is ##\vec{\tau}=\vec{\mu}\times\vec{B}## and we can define a potential energy function ##U(\theta)=-\vec{\mu}\cdot\vec{B}##.

From a lecture video I saw, we can write the total force on the current loop as

$$\vec{F}=-\nabla U\tag{1}$$

$$=\nabla (\vec{\mu}\cdot \vec{B})\tag{2}$$

If the magnetic field is uniform then its gradient is ##\vec{0}## and so ##\vec{F}=0##.

Again, there is a torque, but no resultant force.

But now consider a non-uniform magnetic field created by a second current loop (which we know is approximately like a bar magnet).

1712188894313.png


In the figure above, the initial current loop is at the purple point and we can see ##\vec{\mu}## and ##\vec{B}## point in the ##\hat{k}## direction.

Note that in this configuration, the "north" of the bottom current loop is above the current loop and this is near the "south" of the initial current loop.

$$\vec{F}=\nabla(\vec{\mu}\cdot\vec{B})\tag{3}$$

$$=\nabla(\mu_zB_z(z))\tag{4}$$

At this point the lecture says to assume that

$$\frac{\partial B_z}{\partial z}=\frac{\partial B_z}{\partial y}=0\tag{5}$$

Then

$$\vec{F}=\mu_z\frac{\partial B_z}{\partial z}\hat{k}\tag{6}$$

We expect ##\frac{\partial B_z}{\partial z}<0##. The dipole is set up such that ##\mu_z>0##.

Therefore

$$F=\mu_z\frac{\partial B_z}{\partial z}<0\tag{7}$$

The dipole thus experiences a force downwards. The current loop attracts the dipole.

If ##\vec{\mu}## were to point in the ##-\hat{k}## direction then we'd have two "north" poles near one another, ##\mu_z<0## and from the equations above we would have ##F>0## so that the top dipole would be repelled.

The thing I don't really understand is (1) and the assumption in (5). And I don't know yet how to formulate a question about it at this moment.
 
Last edited:
  • #6
Here is a second derivation related to the first.

Consider again the situation in my previous post. Let's denote the lower current loop as a bar magnet

1712192351087.png


$$d\vec{F}=Id\vec{s}\times\vec{B}\tag{1}$$

$$=Ird\theta\hat{\theta}\times(B_r\hat{r}+B_z\hat{k})\tag{2}$$

$$=-IrB_rd\theta\hat{k}+IrB_zd\theta\hat{r}\tag{3}$$

By symmetry, the ##\hat{r}## component vanishes when we integrate over the loop. Thus

$$\vec{F}=-IrB_r2\pi\hat{k}\tag{4}$$

So once again we see the bar magnet attracting the magnetic dipole.

Note that the argument is different from the previous post. There we had

$$\vec{F}=\mu_z\frac{\partial B_z}{\partial z}<0\tag{5}$$

How do we obtain this expression from the analysis above?

Consider the following setup

1712193178336.png


Now we use conservation of magnetic flux

$$\oint \vec{B}\cdot d\vec{a}=0\tag{6}$$

$$2\pi r\Delta z B_r+B_z(z+\Delta z)\pi r^2-B_z(z)\pi r^2=0\tag{7}$$

$$\frac{B_z(z+\Delta z)-B_z(z)}{\Delta z}=-B_r\frac{2\pi r}{\pi r^2}\tag{8}$$

$$\frac{r}{2}\lim\limits_{\Delta z\to 0} \frac{B_z(z+\Delta z)-B_z(z)}{\Delta z}=-B_r\tag{9}$$

$$\frac{r}{2}\frac{\partial B_z}{\partial z}=-B_r\tag{10}$$

So now we go back to our expression for force

$$\vec{F}=-IrB_r2\pi\hat{k}\tag{4}$$

and sub in for ##B_r##

$$\vec{F}=Ir2\pi \frac{r}{2}\frac{\partial B_z}{\partial z}\hat{k}\tag{11}$$

$$=I\pi r^2\frac{\partial B_z}{\partial z}\hat{k}\tag{12}$$

$$=\mu_z\frac{\partial B_z}{\partial z}\hat{k}\tag{13}$$
 

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