Force on Magnetic Dipole

  • #1
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Homework Statement


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Homework Equations





The Attempt at a Solution



I found dB/dr to be constant near r = 0, so it's a straight line, implying B increases linearly with r near r = 0.

For the second part, I assume the setup is as such, if not placing the dipole with its axis parallel to the coil would make B = 0, and hence no force at all..Not sure what energy has to do in this case.

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Answers and Replies

  • #2
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Not sure what energy has to do in this case.
There is a relationship between potential energy and force.
 
  • #3
TSny
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I found dB/dr to be constant near r = 0, so it's a straight line, implying B increases linearly with r near r = 0.
Sounds good.

For the second part, I assume the setup is as such, if not placing the dipole with its axis parallel to the coil would make B = 0, and hence no force at all..Not sure what energy has to do in this case.
[Edited to correct a misstatement] Although there is only an r-component of field on the axis of the Maxwell loops (r-axis), off the axis there will be a radially inward field (perpendicular to the r-axis and toward the r-axis). A dipole in the form of a small current loop would experience a force due to this radially inward field.

But, it's easier to find the force by using the idea that force is the negative gradient of potential energy: F = -∇E. See if you can use this to find the force. [I see MisterX posted this suggestion before me. Sorry.]
 
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  • #4
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Sounds good.



[Edited to correct a misstatement] Although there is only an r-component of field on the axis of the Maxwell loops (r-axis), off the axis there will be a radially inward field (perpendicular to the r-axis and toward the r-axis). A dipole in the form of a small current loop would experience a force due to this radially inward field.

But, it's easier to find the force by using the idea that force is the negative gradient of potential energy: F = -∇E. See if you can use this to find the force. [I see MisterX posted this suggestion before me. Sorry.]
Then, wouldn't the force simply be F = - ∂U/∂r = m ∂Bext/∂r at r = 0?

This is simply the earlier part of the question of finding ∂B/∂r at r = 0.
 
  • #5
TSny
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Then, wouldn't the force simply be F = - ∂U/∂r = m ∂Bext/∂r at r = 0?

This is simply the earlier part of the question of finding ∂B/∂r at r = 0.
Yes, that's right.
 

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