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Force exterted on a ferromagnetic object in a magnetic field 
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#1
May912, 03:33 PM

P: 2

Hello,
I'm building a coilgun and I'm confused. I understand that within a solenoid, the magnetic field is homogeneous. I've also read that a ferromagnetic projectile in a coilgun is sucked to the center of the coil. If the magnetic field is the same everywhere through the coil, what is special about the center of the coil? What is it that determines the force felt by the ferromagnetic projectile? Thanks! 


#2
May912, 03:35 PM

P: 3,014

The magnetic field is not homogeneous in a solenoid with finite length and spacing between the turns.



#3
May912, 03:46 PM

P: 2

Hm, so in an ideal solenoid (of infinite length and with no spacing between the coils), a ferromagnetic object would feel no force, correct?
So I probably want to keep my coil length pretty short in order to avoid wasting energy creating a relatively homogeneous (and therefore relatively worthless) magnetic field, huh... That makes some sense. Thanks! I'm still wondering: what exactly determines the force that is exerted on a ferromagnetic object by a magnetic field? Is there a simple formula that answers that question? 


#4
May912, 03:51 PM

P: 3,014

Force exterted on a ferromagnetic object in a magnetic field
The formula is the following:
[tex] \vec{F} = \int_{V}{d\mathbf{x}' \, \left(\vec{M}(\mathbf{x}') \cdot \nabla' \right) \vec{B}(\mathbf{x}')} [/tex] 


#5
May912, 04:07 PM

P: 349

Can you quote a source for this equation please?
It would be good to have the terms defined for those needing help and guidance. 


#6
May912, 06:15 PM

P: 3,014

Also, there may be a torque on a magnetic material given by:
[tex] \vec{\tau} = \int{d\mathbf{x}' \, \left( \vec{M}(\mathbf{x}') \times \vec{B}(\mathbf{x}') \right)} [/tex] As for the source, of these formulas, these are simply summations of the formulas given here: http://en.wikipedia.org/wiki/Magneti...agnetic_moment but for a continuous distribution of magnetization [itex]\vec{M}(\mathbf{x}')[/itex]. 


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