Magnetic Dipole Moment of a Surface Current Loop

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SUMMARY

The magnetic dipole moment for a surface current loop can be derived from the surface current density 'K'. Specifically, for a rotating disc with surface charge density 'σ' and angular velocity 'ω', the magnetic dipole moment 'm' can be expressed as 'm = i ∫ dA', where 'i' is the current. The discussion highlights the relationship between surface current and magnetic dipole moment, emphasizing the need to consider the velocity of charge elements as they move in a circular path. Additionally, the forum addresses the challenge of generalizing this concept to arbitrary surface current distributions 'K(r')' and volume current distributions 'J(r')'.

PREREQUISITES
  • Understanding of magnetic dipole moment concepts
  • Familiarity with surface current density 'K'
  • Knowledge of multipole expansion in electromagnetism
  • Basic principles of angular momentum and charge motion
NEXT STEPS
  • Research the derivation of magnetic dipole moment from surface current density 'K'
  • Study the relationship between angular velocity 'ω' and magnetic dipole moment in rotating systems
  • Explore the multipole expansion of the magnetic vector potential
  • Investigate the effects of volume current distributions 'J(r')' on magnetic dipole moments
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Physicists, electrical engineers, and students studying electromagnetism, particularly those interested in magnetic properties of current distributions and their applications in theoretical and applied physics.

siddharth
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For a current loop, the magnetic dipole moment is defined as [tex]i \int d\vec{a}[/tex]

Now, if I'm given a surface current 'K', how can I find the magnetic dipole moment? For example, if a disc of surface charge [itex]\sigma [/tex] is rotating with angular velocity [itex]\omega [/tex], what is it's magnetic dipole moment?<br /> <br /> I thought about taking the dipole term from the multipole expansion of the magnetic vector potential, and then trying to separate it into something dependent on current distribution, and something dependent on the point where I'm trying to find the potential, could lead to an expression for m. But I'm getting nowhere that way.<br /> <br /> So, is it possible to get a term for m in terms of K (and maybe r' ?) ,analogous to a closed current loop?[/itex][/itex]
 
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Qv = iL = j HW L . For a disk that's rotating, each dQ moves around a ring
with velocity v = omega x r. just add them up.
 
lightgrav said:
Qv = iL = j HW L . For a disk that's rotating, each dQ moves around a ring
with velocity v = omega x r. just add them up.

Yeah, that would work in this case. Thanks.

But for a general surface current distribution K(r') is there a way to find the magnetic dipole moment 'm'? For that matter, what if there is a volume current distribution J(r') ?
 

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