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Unformatted text preview: 1 PHYS809 Class 25 Notes Magnetic dipole moment The dipole part of the scalar potential for a current loop can be written in the form ( ) 3 , 0 , 4 md r r = > m r (25.1) where for a circular loop of radius a and unit normal k , 2 . a I = 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 . = - B Later we will see that it is more natural to relate the magnetic field H to the scalar potential. In vacuum, , = B H and hence the potentials differ by a constant factor of . ) The dipole magnetic induction is given by ( ) ( ) 3 3 , 0 . 4 md r r - = > 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 2 3 . 4 md r = m r A (25.8) Taking the curl gives the dipole magnetic field ( ) 3 3 3 . 4 4 md md r r r = = = - 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 ( ) ( ) ( ) ( ) 3 3 ....
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This note was uploaded on 12/02/2011 for the course PHYS 809 taught by Professor Macdonald during the Fall '11 term at University of Delaware.
- Fall '11