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Moving Charges and Magnetic Moments |
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| Feb11-13, 01:02 PM | #1 |
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Moving Charges and Magnetic Moments
Does any moving charge generate a magnetic moment? I thought so because a moving charge generates a magnetic field.
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| Feb12-13, 03:48 PM | #2 |
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I know that charges moving in a circle generate a magnetic moment, but I was wondering if charges travelling in a straight line also generate a magnetic moment.
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| Feb13-13, 10:36 PM | #3 |
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If a moving charge travelling in a straight line generates a magnetic field, wouldn't it also have a magnetic moment?
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| Feb14-13, 07:17 AM | #4 |
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Moving Charges and Magnetic Moments
Well, consider the dipole moment in electrostatics... its created by placing two charges... So we can say that there will be a magnetic moment if there are two magnetic poles. Now when a charge is moving in a closed path, it creates a magnetic field similar to what would have been created if two poles had been there instead of the charge. What I mean is that when a charge rotates, it creates two poles, North in the anticlockwise sense and South in the clockwise. So we get a magnetic moment in this case. But when a charge is moving in an open path, it doesn't create these to poles and hence we get no magnetic moment.
Thats what I think. Let's wait for others' opinions as well. What do you think? |
| Feb14-13, 10:14 PM | #5 |
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I think since a magnetic field should have a magnetic moment, which you can use to calculate the magnetic field, a moving charge no matter what kind of motion, generating a magnetic field, should have a magnetic moment.
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| Feb15-13, 12:37 AM | #6 |
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| Feb16-13, 02:37 PM | #7 |
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But if I wanted to, would I be able to describe the magnetic field of a moving charge in terms of its magnetic moment?
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| Feb17-13, 01:44 AM | #8 |
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| Feb17-13, 03:53 AM | #9 |
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Taking a charge on circular path, the moment at the centre of rotation is calculated by
[itex]\vec m = I*\vec A[/itex] (A=circular area) And in the case of single charge the current is: [itex]I= \frac{dQ}{dt}[/itex] where [itex]t=\frac{2\Pi r}{v}[/itex] and the moment becomes: [itex]\vec m=\frac{\vec r \times \vec v Q}{2}[/itex] Now if we start incrasing the size of cicular path towards ∞, the moment increases linearly. But the "reference point" of the moment approaches infinite distance from the charge. Note that the moment itself is not a field; all calculations using moments are done assuming a moment on definite point. Another formula for the moment is: [itex]\vec m=\frac{I}{2}\int \vec {r} \times \vec {dr}[/itex] Which can be used to calculate magnetic moment at origin when the conductor is at (a,y) (vertical line having distance a from origin): [itex]\vec m = \frac{I}{2}\int ^{\infty}_{-\infty}(a\vec i+y\vec j)\times(\vec j dy) [/itex] [itex]\vec m = \frac{I}{2}\int ^{\infty}_{-\infty}a\vec i\times\vec j dy[/itex] [itex]\vec m = \frac{aI}{2}\vec k \infty[/itex] Thus the magnetic moment for a straight line, calculated at distance a from the line, is infinite, if a>0. An analogy in mechanics would be calculating the moment of inertia of a disc at point which is NOT the center of mass. Which seems ridiculous. I would say that the magnetic moment is a property of a system, and valid only at the "centre" of the system (how to define the centre is another problem). As it is not a field, it cannot be used as such for calculating field quantities. BR, -Topi |
| Feb17-13, 08:46 AM | #10 |
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And about the magnetic moment, why do say that it is valid only at the "center"? Is it not distributed over the entire surface? I mean any arbitrarily coiled loop carrying charges have well defined magnetic moments! You said it yourself... the product of the current and the area...!! |
| Feb17-13, 02:58 PM | #11 |
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The magnetic moment points from the south pole to the north pole of a magnet, which I believe to imply magnetic dipoles. If a moving point charge generates a magnetic field according to the Biot-Savart law, (which shows the two magnetic poles of the moving point charge) shouldn't the moving charge also generate a magnetic moment regardless of what type of motion it is executing? Please correct me if I am wrong.
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| Feb17-13, 11:24 PM | #12 |
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| Feb18-13, 12:45 AM | #13 |
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Yes, you are right about the field appearing to be circular, but isn't the field zero in the direction of velocity of the charge? This would give our moving charge a kind of "donut" shape of some sort. We can't define which one is south or north, but we can sure see the poles
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| Feb18-13, 01:26 AM | #14 |
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| Feb18-13, 02:06 AM | #15 |
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What I meant by "define" is that it depends on the observer on what he chooses to be the south and north pole. The way you can see that the poles are there is by seeing the two opposite points that have no field at them. I am very sure you are familiar with the Biot-Savart equation. According to the equation, the magnetic field of the moving point charge in both the direction of the velocity and in the direction opposite of the velocity is zero, hence the "donut" shape of the magnetic field. Since it has two endpoints, (which, by the way, i'm assuming implies magnetic dipoles) then the magnetic moment of the moving charge can be defined since it is a constant vector pointing from the south pole to the north pole of the magnetic field. Again, this horrible way of seeing it like this may be wrong, so please correct me. If it makes sense, please let me know
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| Feb18-13, 04:32 AM | #16 |
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I am sorry if my answers are not good, but honestly, I can't visualize what you are trying to say! I just can't see the two poles at two distinct places throughout the spread of the field and hence i can't see any sign of the magnetic moment. I am really sorry that after almost a week, I couldn't provide you with a satisfactory answer! :( |
| Feb18-13, 07:34 AM | #17 |
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Nooooo! I don't think we are on the same page here is all. (: I don't think imagining our charge to be a wire is going to help because a wire and a point charge have different magnetic field shapes. What I meant by the donut shape of our point charge is that it is not completely in the shape of a donut, but it has some characteristics. In a real donut, there is a hole, which I totally understand why you got confused (that was my mistake by not being clear enough), but in the magnetic field of our point charge there is no hole. It has the same curving features of a donut but it doesn't have a whole. It has the same magnetic field shape as a solenoid. Our moving point charge is moving at a constant velocity and through a vacuum (classical vacuum). If it is not accelerating or near any force, how will the direction between north and south change?
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