What is the force on a magnetic dipole in a coil at r = 0?

In summary, the conversation discusses the relationship between potential energy and force in the context of a dipole in a magnetic field. It is mentioned that the force can be found using the negative gradient of potential energy, and that there is also a radially inward field present off the axis of the Maxwell loops. The conversation also addresses a question about finding the force at r=0.
  • #1
unscientific
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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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  • #2
unscientific said:
Not sure what energy has to do in this case.
There is a relationship between potential energy and force.
 
  • #3
unscientific said:
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
TSny said:
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
unscientific said:
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.
 

What is a magnetic dipole?

A magnetic dipole is a small magnet with a north and south pole that is created by the movement of electrons in an atom or molecule.

What is the force on a magnetic dipole?

The force on a magnetic dipole is the product of the magnetic field strength and the dipole moment, which is a measure of the strength of the dipole.

What factors affect the force on a magnetic dipole?

The force on a magnetic dipole is affected by the strength of the magnetic field, the orientation of the dipole, and the distance between the dipole and the magnetic field source.

How is the force on a magnetic dipole calculated?

The force on a magnetic dipole is calculated using the equation F = m x B, where F is the force, m is the dipole moment, and B is the magnetic field strength.

What are some real-life applications of the force on a magnetic dipole?

The force on a magnetic dipole is used in a variety of technologies, such as electric motors, generators, and magnetic resonance imaging (MRI) machines. It is also important in understanding the behavior of compass needles and the Earth's magnetic field.

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