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# Class_25new - PHYS809 Class 25 Notes Magnetic dipole moment...

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1 PHYS809Class25Notes Magnetic dipole moment The dipole part of the scalar potential for a current loop can be written in the form ( ) 0 3 , 0 , 4 md r r μ π Φ = > m r (25.1) where for a circular loop of radius a and unit normal k , 2 . a I π = m k (25.2) (Note that the potential in equation (25.1) is such that the magnetic induction B is related to the scalar potential by . = -∇Φ B Later we will see that it is more natural to relate the magnetic field H to the scalar potential. In vacuum, 0 , μ = B H and hence the potentials differ by a constant factor of 0 . μ ) The dipole magnetic induction is given by ( ) ( ) 0 3 ˆ ˆ 3 , 0 . 4 md r r μ π - = > m r r m B (25.3) For a general loop, the magnetic dipole moment is , I = m S (25.4) where S is the vector area of the loop. By applying Stokes’ theorem to a vector field ( ) , = × A r k r where k is arbitrary but fixed, it can be shown that 1 . 2 loop d = × S r r (25.5) Hence an alternative expression for the magnetic dipole moment is . 2 loop I d = × m r r (25.6) For a general current density, the equivalent expression is ( ) 3 1 . 2 allspace d x = × m r J r (25.7) The vector potential of magnetic dipole The vector potential of a magnetic dipole is

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2 0 3 . 4 md r μ π × = m r A (25.8) Taking the curl gives the dipole magnetic field ( ) 0 0 3 3 3 . 4 4 md md r r r μ μ π π × = ∇× = ∇× = ∇⋅ - ⋅∇ m r r r B A m m (25.9) The first term gives a delta function contribution and the second term is the same as for the electric dipole. The result is ( ) ( ) ( ) ( ) 0 0 0 3 ˆ ˆ 3 ˆ ˆ . 4 md r μ μ δ μ δ π - = + - m r r m B m r m r r r (25.10) Hence there is a difference between the singular part of the dipole field from a current loop and the dipole field from a pair of oppositely charged particles. This difference is relevant to atomic and
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Class_25new - PHYS809 Class 25 Notes Magnetic dipole moment...

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