2 Iron Bars (one magnetized) Question


by kkrizka
Tags: bars, iron, magnetized
kkrizka
kkrizka is offline
#1
Dec20-06, 07:15 PM
P: 85
Hello everyone,
I was studying for my electromagnetism test by doing questions from a text book that I won at the university's open house. This means that I don't have the complete answer key and neighter does my physics teacher.

While reading through it, I came upon an interesting question. Here it is:
"Given two identical iron bars, one of which is a permanent magnet and the other unmagnetized, how could you tell the difference by using only the two bars?"
I can't seem to figure it out and I even asked my physics teacher for help. Some theories that we came up with are that you could hit one bar with the other and split it in half, and then see if both halves of the split bar will attract/repel. The problem with this is that the impact could cause the pernament magnet to demagnetize and it might be quite hard to hit them against each outher so they split.
Another theory we came up with is that you could balance one on the corner of other and see if it turns toward the earth's magnetic poles (this is just like hanging it from the ceiling with a string), but then to balance it perfectly would be very complicated.

Does anyone have any idea how this problem could be solved? Or have the answer key to College Physics by Wilson and Buffa (4th Edition)? It is question #4 from chapter 19 in that book.
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cesiumfrog
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#2
Dec20-06, 08:00 PM
P: 2,050
Will the end of an unmagnetized iron object be attracted to the midpoint of a permanent bar magnet?
ObsessiveMathsFreak
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#3
Dec21-06, 05:53 AM
P: 406
Quote Quote by cesiumfrog View Post
Will the end of an unmagnetized iron object be attracted to the midpoint of a permanent bar magnet?
Actually, yes it will, but not as strongly as a magnetised iron end wiould be attracted to the midpoint of an unmagnetised bar.

browncat
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#4
Dec26-10, 07:52 PM
P: 3

2 Iron Bars (one magnetized) Question


Perhaps this mental experiment will be useful: if you build two physical pendulums using the two bars, and make them oscillate, in the very same conditions, the magnetized one will have a greater rate for slowing down its motion (i.e., its associated damping will stop it faster), because the magnetized one will be producing a changing magnetic field, and hence a changing electric field, both of which are associated to an electromagnetic wave which has an energy associated to it. Hence, the mechanical energy of the magnetized iron bar will be consumed in producing such electromagnetic waves, finally breaking its movement.

I know what you are thinking: "I can't build pendulums, because I don't have a rope, I only have two bars". Well, imagine then any kind of accelerated movement which can report that a part of the energy must be used to produce the electromagnetic waves.

Regards,

D.
Vanadium 50
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#5
Dec26-10, 08:13 PM
Mentor
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First, this is four years old. Second, this effect will be imperceptible.
NEILS BOHR
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#6
Dec26-10, 10:31 PM
P: 79
take any piece ( u dont know which one is magnet) and bring its end towards the middle of another piece

if it is attracted or repelled then the piece which u brought towards the stationary one is a magnet and other is not

similarly vice versa...
cmos
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#7
Dec27-10, 10:05 PM
P: 367
Quote Quote by NEILS BOHR View Post
take any piece ( u dont know which one is magnet) and bring its end towards the middle of another piece

if it is attracted or repelled then the piece which u brought towards the stationary one is a magnet and other is not

similarly vice versa...
As far as I can tell, this won't work. I think that in both cases you would have attraction. Perhaps you can elaborate?
cmos
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#8
Dec27-10, 10:38 PM
P: 367
Here's my attempt at an answer:

Let one bar be stationary (Bar 1) and let the other be movable (Bar 2). Now move Bar 2 so that both bars are end-to-end. Then take Bar 2 away, turn it 180 degrees, and bring both bars end-to-end again. At this point, there are two possibilities....

Possibility 1 - The movable bar (Bar 2) was the unmagnetized one:
Then regardless of which end you bring to Bar 1, you will always have attraction. This is obvious because both ends of Bar 2 were originally unmagnetized.

Possibility 2 - The movable bar (Bar 2) was the magnetized one:
Let's call the first end "north" and the second end "south". Then when you originally bring the bars together, you will magnetize the end of Bar 1 south. Then, when you rotate Bar 2 and again bring both bars end-to-end, you will be bringing the originally magnetized south end of Bar 2 to the newly magnetized south end of Bar 1. As such, you will have repulsion.

Summary:
If your sequence is attraction followed by attraction, the movable bar was the originally unmagnetized one. If you sequence is attraction followed by repulsion, the movable bar was the originally magnetized one.
shoestring
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#9
Dec28-10, 07:20 AM
P: 96
Quote Quote by cmos View Post
As far as I can tell, this won't work. I think that in both cases you would have attraction. Perhaps you can elaborate?
Think about iron filings around a permanent magnet. More will stick to the end surfaces than to the middle, because it's at the ends that the field is stronger, and also less uniform.

The shape of the field matters because in a uniform field the force on a dipole will be rotational only, not translational. Except for diamagnetism, an induced dipole will tend to move toward a stronger field.
NEILS BOHR
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#10
Dec28-10, 07:58 AM
P: 79
when i say no attraction i mean very less or negligible amount of attraction
as explained by shoestring....
cmos
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#11
Dec28-10, 01:52 PM
P: 367
Quote Quote by shoestring View Post
Think about iron filings around a permanent magnet. More will stick to the end surfaces than to the middle, because it's at the ends that the field is stronger, and also less uniform.
Quote Quote by NEILS BOHR View Post
when i say no attraction i mean very less or negligible amount of attraction
as explained by shoestring....
Yes, I agree with this. The point I was trying to make is that there wouldn't be any case of repulsion as NEILS BOHR originally stated; only one case of stronger attraction and one case of weaker attraction.


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