Ferromagnetic dipole in a uniform magnetic field

In summary, when a ferromagnetic dipole is placed in a uniform magnetic field with the axis of the dipole moment vector not parallel to the field lines, there will be a net force and torque on the dipole. However, when the axis of the dipole moment vector is parallel to the field lines, there will only be a net torque but no net force. In non-uniform fields, the net force and torque will depend on the specific orientation of the dipole moment vector with respect to the field lines. To determine the net force and torque in non-uniform fields, the right-hand rule can be used to calculate the force on each side of a rectangular loop and then summed together.
  • #1
aquabum619
29
0
a- I place a ferromagnetic dipole in a uniform magnetic field so that the axis of the dipole moment vector does not lie parallel to the magnetic field lines. Is there a net force on the dipole? Is there a net force?
b- I place a ferromagnetic dipole in a uniform magnetic field so that the axis of the dipole moment vector does lie parallel to the magnetic field lines. Is there a net force on the dipole? Is there a net torque?
c- I now place a ferromagnetic dipole in a magnetic field that varies in space so that the axis of the dipole moment vector does not lie parallel to the magnetic field lines. Is there a net force on the dipole? Is there a net torque?
d- I now place a ferromagnetic dipole in a magnetic field that varies in space so that the axis of the dipole moment vector does lie parallel to the magnetic field lines. Is there a net force on the dipole? Is there a net torque?




no relevant equations



a: If the dipole is not parallel to the magnetic field lines, the force will be strongest, right? So I think there will be a net force, and there will be a torque
b: Since the dipole is parallel to the magnetic field lines, the net force is zero? And there would not be any torque?
c: I am not sure
d: I am not sure


Thanks in advance for all your help.
 
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  • #2
There is a simple relation that, when derived and then committed to memory, will make these sorts of problems much easier (see http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magmom.html" for a derivation).

Net torque:

[tex]
\bf{\tau} = \bf{\mu} \times \bf{B}
[/tex]

where [itex]\mu[/itex] is the dipole moment vector and [itex]B[/itex] is the magnetic field.

Furthermore, in uniform fields, there can't be a net force on a dipole due to the magnetic field. But in non-uniform fields, you should reason through what's going to happen.
 
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  • #3
so for A and B there is no net force?
 
  • #4
Think of the magnetic dipole as a loop of current (which, by definition, it can be replaced by), oriented properly. Then, using the Biot-Savart Law or even [itex]\vec{F} = I\vec{L} \times \vec{B}[/itex], figure out whether or not there is a net force on the loop of current. The conclusion will apply to the magnetic dipole.
 
  • #5
But with F = L x B, all I know is that I have a constant current "I" and a uniform "B"

How can I derive a force quantity with such little information?
 
  • #6
can anybody help me with this problem?
 
  • #7
I am not sure what math level you are in, or what how complicated an explanation you want. If you want a completely rigorous proof on these things, then you will need to have an arbitrary loop of current (any shape), do some path integrals using some calculus theorems.

If you don't want to do that, then just imagine a rectangular loop with constant current in space, and calculate the force exerted on the loop by magnetic fields. the direction of the dipole is the direction of the area vector of the loop.

Notice that when there is a torque, there does not necessarily be a net force. as Saketh pointed out, there is no force in constant magnetic field (try to reason it out). when magnetic field is not uniform, things don't cancel out nicely. try to reason it out and see what happens.
 
  • #8
dude.. first of all, relax. Stop assuming that I am being arrogant in any way. I don't need a "complicated explanation" or a "rigorous proof" I just need help trying to understand what is going on here, nothing more. If I needed "complicated explanations" and "rigorous proofs" to accommodate the "level of math" that I am in, I would surely understand this concept and probably wouldn't be here asking for help.

As for trying to "reason out" why there is no force in a uniform magnetic field, I don't know. I was hoping perhaps you guys could help me understand that. My reasoning is that there would be a force on the dipole within a uniform magnetic field because when I apply the right hand rule, my fingers would point in the direction of the magnetic field, my thumb would point into the direction of the current and my palm would face the direction of the exerted force.

Forgive me, I simply just don't understand. I am not ungrateful and I appreciate all the help I can get but I am more or less frustrated at this problem that I've been attempting to solve since yesterday afternoon.
 
  • #9
As for trying to "reason out" why there is no force in a uniform magnetic field, I don't know. I was hoping perhaps you guys could help me understand that. My reasoning is that there would be a force on the dipole within a uniform magnetic field because when I apply the right hand rule, my fingers would point in the direction of the magnetic field, my thumb would point into the direction of the current and my palm would face the direction of the exerted force.
As tim_lou said, http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magmom.html" . Put your thumb in the direction of the dipole vector and curl your fingers, and the current flows in that direction.

If you have a loop of current in a uniform magnetic field, what is the net force on it? (Hint: You have to do the right-hand rule four times, once for each side of the rectangle.) If you want to make this completely general, which you don't need to for this problem, you'll make an infinite number of sides (i.e. a curve) and do a path integral (i.e. right hand rule an infinite number of times), as tim_lou was saying.

For parts (c) and (d), do the right hand rule with the rectangular loop in a non-uniform field. You'll have to be more careful in reasoning through this, but the essential argument and method is the same.
 
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  • #10
Saketh, I thought that you had to know te direction of the current before you can apply the right hand rule?

OK, i think I got this one. (thanks for the hint btw!) When the right hand rule is applied to each of the 4 sides, the Fb of each side will cancel itself out and there will be no net force? (If this is right, you have no idea how big of a breakthrough you just helped me reach!)

With parts C and D, The magnetic force would be going into the page, assuming the current is moving clockwise. But I am not entirely sure how force will be calculated here.

Thanks again Saketh for your patience and hlp!
 
  • #11
aquabum619 said:
dude.. first of all, relax. Stop assuming that I am being arrogant in any way. I don't need a "complicated explanation" or a "rigorous proof" I just need help trying to understand what is going on here, nothing more. If I needed "complicated explanations" and "rigorous proofs" to accommodate the "level of math" that I am in, I would surely understand this concept and probably wouldn't be here asking for help.

I apologize for my post. I do not mean to offend anyone . I myself have gone through the lack of rigor when learning dipoles in introductory physics (and struggled with it). the general case is simply much more complicated and require a lot more mathematics.

as for part C and D, do the same thing as in part A and B. but suppose the magnetic field on left is greater than the field in the right. You'll see that the forces do not cancel out nicely.

a nice rule of thumb is that magnetic dipole tends to align itself in the opposite direction of the magnetic field (lowest potential energy). a magnetic dipole is like a compass. also notices that
[tex]\vec{\tau}=\vec{\mu}\times\vec{B}[/tex]
what is the torque when mu is parallel to B? what happens to the cross product?

aquaabum619 said:
The magnetic force would be going into the page, assuming the current is moving clockwise. But I am not entirely sure how force will be calculated here.

for part C, the dipole is not parallel to the field. so this scenario does not work for part C. (the direction of the diopole is the direction of the area vector). for simplicity, you might as well make the dipole perpendicular to the field.
 
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1. What is a ferromagnetic dipole?

A ferromagnetic dipole is a tiny magnetic unit that has a north and south pole, just like a regular magnet. However, unlike other magnets, a ferromagnetic dipole is made up of atoms that have their own individual magnetic fields that align with each other, creating a strong overall magnetic field.

2. How does a ferromagnetic dipole behave in a uniform magnetic field?

In a uniform magnetic field, a ferromagnetic dipole will align itself with the direction of the field. This means that the north pole of the dipole will point towards the south pole of the magnetic field, and vice versa.

3. What factors affect the strength of a ferromagnetic dipole in a uniform magnetic field?

The strength of a ferromagnetic dipole in a uniform magnetic field is affected by the strength of the magnetic field, the size of the dipole, and the type of material the dipole is made of. Materials with higher magnetic permeability, such as iron, will have stronger ferromagnetic dipoles.

4. Can a ferromagnetic dipole be demagnetized in a uniform magnetic field?

Yes, a ferromagnetic dipole can be demagnetized in a uniform magnetic field if the strength of the field is strong enough to overcome the alignment of the dipoles within the material. This can happen if the material is heated above its Curie temperature or if an alternating magnetic field is applied.

5. What are some real-life applications of ferromagnetic dipoles in a uniform magnetic field?

Ferromagnetic dipoles in a uniform magnetic field have several applications, including in motors, generators, magnetic storage devices, and magnetic sensors. They are also used in medical imaging techniques like MRI machines and in the production of electricity from renewable sources like wind turbines.

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