How to find eddy current (circular) path in this pendulum experiment?

[microdosemishief]

Homework Statement

I am confused as to how we know the circular direction of current (Figure 2 from this link)
https://courses.lumenlearning.com/suny-physics/chapter/23-4-eddy-currents-and-magnetic-damping/

Relevant Equations

Fleming's Right hand rule?

Does the direction my second finger points in indicate the literal direction of magnetic field at that point in space, or do I need to further use right hand rule or something for a circular path around the finger

 

[Herman Trivilino]

I don't know what you mean by your second finger because I don't see any fingers in the attachment.

There's a reference to RHR-1, which I assume involves fingers and is used to find the direction of the force. But without seeing an illustration of it I can only guess that the second finger points in the direction of the force?

The magnets shown in the figure create magnetic field lines directed from the face of the block labelled "N" to the face of the other block labelled "S". We call this collection of field lines the external field, or the applied field.

The circular direction is whichever way it needs to be to induce a magnetic field that opposes the applied field. So you curl the four fingers of your right hand in the circular direction of the current and your thumb points in the direction of this induced field.

You have to use a trial-and-error approach. There are only two possible circular directions for the current. So you try each of them to find the one that makes your thumb point in the desired direction, that is, the direction of the induced field.

 

[kuruman]

Consider figure (A) below middle. It shows two conducting rods moving with velocity $\mathbf v$ to the right. The rod on the right in figure (A) is in a region of magnetic field $\mathbf B$ directed into the screen. The free electrons in it will experience a force $q(\mathbf v\times \mathbf B)$ which is down because although $(\mathbf v\times \mathbf B)$ is up, $q$ for an electron is negative. This means that there will be an accumulation of net negative charges at the bottom of the rod leaving net positive charges at the top. Thus, the top of the rod will be at a higher electric potential than the bottom. No such potential difference exists in the rod outside the field.

Now look at figure (B). It shows the two rods connected with horizontal conducting wires forming a closed circuit. The rod on the left will act as a battery and establish a counterclockwise current, from the positive to the negative terminal. This potential difference will exist as long as one rod is inside the field and the other is outside. When the other rod enters the field, the potential difference between the tops of the two rods will be zero and the current will stop flowing.

Figure (C) shows the current in the rod inside the field. The magnetic force $\mathbf F = I\mathbf L\times \mathbf B$ is to the left, opposing the motion as indicated by the green arrow. The other rod experiences no force because it's not in the field. The segments of the horizontal wires that are inside the field experience equal and opposite forces which cancel in pairs. Thus, the net force on this loop is opposing the motion.

Finally, if you have a rectangular plate entering the field, you can imagine it as an assembly of nested rectangular loops as shown above. The result will be nested counterclockwise eddy currents and a net force opposing the motion.

 

[microdosemishief]

Regarding using the RHR, my thumb is in direction of motion (parallel to page), my index finger pointing along magnetic field (into or out of page), which leaves third finger for induced current pointing down like gravity or up - this doesnt seem right…

 

[Steve4Physics]

Regarding using the RHR, my thumb is in direction of motion (parallel to page), my index finger pointing along magnetic field (into or out of page), which leaves third finger

I'm guessing you mean second finger - sometimes called the middle finger. (Maybe you don't want to appear rude!)

for induced current pointing down like gravity or up - this doesnt seem right…

It sounds right. There are different versions of the right-hand rule (RHR); I’m guessing you mean Fleming's right hand rule (image below from https://upload.wikimedia.org/wikipedia/commons/9/9a/RightHandOutline.png)

It is typically used at an introductory level to find the direction of an induced current (or emf) when a conductor moves through a magnetic field.
_______________

Try these (easier) questions using the RHR. They may clear things up.

1. Conducting loop ABCD moves into a magnetic field as shown:

a) In which section(s) of the loop is/are emf generated?
b) What is the direction of the current in the section(s) where emf is generated (apply RHR to the section)?
c) What is the induced current’s direction around the loop (clockwise or anticlockwise)?

2. Same as question 1, but the loop is leaving the field:

Edit: extra question added.

 

[microdosemishief]

So in 1. is it anticlockwise cuz current induced between points C and B go from C to B
and for 2. it is clockwise

 

[Steve4Physics]

So in 1. is it anticlockwise cuz current induced between points C and B go from C to B
and for 2. it is clockwise

Yes, you got it.

The same thinking can be applied to the Figure 2 diagram from your link. The exact shape of the loop and the exact direction-of-motion don’t change the sense (CW or ACW) of the induced current. You can reflect on why this is so!

Edit - typo'.

 

[microdosemishief]

Then do I use right hand thumb rule to find

Consider figure (A) below middle. It shows two conducting rods moving with velocity $\mathbf v$ to the right. The rod on the right in figure (A) is in a region of magnetic field $\mathbf B$ directed into the screen. The free electrons in it will experience a force $q(\mathbf v\times \mathbf B)$ which is down because although $(\mathbf v\times \mathbf B)$ is up, $q$ for an electron is negative. This means that there will be an accumulation of net negative charges at the bottom of the rod leaving net positive charges at the top. Thus, the top of the rod will be at a higher electric potential than the bottom. No such potential difference exists in the rod outside the field.
View attachment 366289
Now look at figure (B). It shows the two rods connected with horizontal conducting wires forming a closed circuit. The rod on the left will act as a battery and establish a counterclockwise current, from the positive to the negative terminal. This potential difference will exist as long as one rod is inside the field and the other is outside. When the other rod enters the field, the potential difference between the tops of the two rods will be zero and the current will stop flowing.

Figure (C) shows the current in the rod inside the field. The magnetic force $\mathbf F = I\mathbf L\times \mathbf B$ is to the left, opposing the motion as indicated by the green arrow. The other rod experiences no force because it's not in the field. The segments of the horizontal wires that are inside the field experience equal and opposite forces which cancel in pairs. Thus, the net force on this loop is opposing the motion.

Finally, if you have a rectangular plate entering the field, you can imagine it as an assembly of nested rectangular loops as shown above. The result will be nested counterclockwise eddy currents and a net force opposing the motion.

Thanks. Now for a thin plate, why is the eddy current circular? Is it perfectly circular?

 

[microdosemishief]

Yes, you got it.

The same thinking can be applied to the Figure 2 diagram from your link. The exact shape of the loop and the exact direction-of-motion don’t change the sense (CW or ACW) of the induced current. You can reflect on why this is so!

Edit - typo'.

Then for the emf resulting from the eddy current, do i use flemings left hand rule to find that direction?

 

[Steve4Physics]

Then for the emf resulting from the eddy current, do i use flemings left hand rule to find that direction?

I don’t understand the question.

Referring to Post #5, Q1, the current around loop ABCD is the eddy current in this situation.

(If loop ABCD were replaced by a conducting sheet, a similar-ish loop of current would flow around the sheet.)

The emf is generated in section BC of the loop. BC acts like a cell having its positive terminal at B and its negative terminal at C. This is what's driving the eddy current anticlockwise around the loop.

Fleming’s left hand rule (LHR) is used to find the direction of the force on a current-carrying conductor in a magnetic field.. It has nothing to do with the emf.

If you want, you can apply the LHR to section BC (with induced current flowing from C to B).

You will find that the force on BC acts left. This means that the loop is being pushed left. This is consistent with Lenz’s law: the loop is moving right, into the field resulting in a force (acting left) which opposes the motion.

Minor edit.