No Title
PH317 MAGNETISM
WINTER Quarter 2009-2010
Solution to Homework Assignment 4
Assigned: Thursday 1/07/10
Due: Monday 1/11/10
  1. Text: 5.34
    SOLUTION

      (a) The magnetic moment is
      m
      =
      Ia
      =
      IpR2 
      ^
      z
       

      (b) The magnetic field is, approximately,
      B
      �
      moIpR2
      4pr3
      		 �
      �       		2cosq  
      ^
      r
       
      		 +sinq 
      ^
      q
       
      		 �
      �

      (c) On the z axis, q = 0, r=z (for z > 0), [^(r)]=[^(z)], so B � (moIR2/2z3)[^(z)] (for z < 0, q = p, [^(r)]=-[^(z)], so the field is the same, with |z|3 in place of z3). The exact answer (Eq. 5.38) reduces ( for z >> R) to B � moIR2/2|z|3, so they aggree.
  2. Text: 5.36
    SOLUTION
    Draw a differential ring of charge on the surface of the sphere concentric with the z-axis. The total charge in the ring is
    dq = s(2pRsinq)Rdq
    The time for one revoultion is
    dt = 2p/w
    The current in the ring is
    I
    =
    dq
    dt
    =
    swR2sinqdq
    The area of the ring is A=p(Rsinq)2, so the magnetic moment of the ring is
    dm
    =
    (swR2sinqdq)pR2sin2 q
    the total dipole moment is
    m
    =
    swpR4 �
    �     		p

    0 
    		 sin3qdq
    =
    swpR4 �
    �     		�
    �     		p

    0 
    		 sinqdq - �
    �     		p

    0 
    		 sinqcos2qdq �
    �
    =
    4swpR4/3
    m
    =
    4p
    3
    		 swR4 
    ^
    z
     
    The dipole term in the multipole expansion for A is
    Adip
    =
    moswR4
    3
    		 sinq
    r2
    		  
    ^
    f
     
    which is also the exact potential (Eq. 5.67); evidently, a spinning sphere produces a perfect dipole field, with no higher multipole contributions.
  3. A long, right-circular cylinder of radius R carries a magnetization
    M = ks2 
    ^
    f
     

      (a) Derive the equation for the bound volume current density, Jb.
      SOLUTION

      Jb
      =
      �×M
      =
      Mf 
      ^
      f
       
      =
      1
      s
      		 �
      �       		�
      �s
      		 ( sMf) �
      �
      ^
      z
       
      =
      1
      s
      		 �
      �       		�
      �s
      		 (ks3 ) �
      �
      ^
      z
       
      =
      3ks 
      ^
      z
       

      (b) Derive the equation for the bound surface current density, Kb.
      SOLUTION

      Kb
      =
      M×
      ^
      n
       
      =
      ks2 
      ^
      f
       
      		 ×
      ^
      s
       
      		 �
      �
      s=R 
      =
      -kR2 
      ^
      z
       

      (c) Describe the current flow in the cylinder. To see what is happening, evaluate the following two integrals:
      Ivol
      =
      �
      �       		Jbda
      Isurf
      =
      �
      �       		Kbdl
      SOLUTION
      The current contribution due to Jb is
      IJb
      =
      �
      �       		Jb·da
      =
      �
      �       		R

      0 
      		 (3ks)(2ps)ds
      =
      2pkR3
      The current contribution due to Kb is
      IKb
      =
      �
      �       		Kbdl
      =
      (-kR2)(2pR)
      =
      -2pk R3
      The bound current flows up the cylinder and returns down the surface. Also, the two currents add to zero.
      (d) Determine the equations for the magnetic field both inside and outside the cylinder.
      Since both currents add to zero, the magnetic field outside the cylinder is zero,
      Bout=0
      Applying Ampere's law inside the cylinder in the usual way,
      �
      �       		B·dl
      =
      moIencl
      B(2ps)
      =
      mo �
      �       		s

      0 
      		 Jb(s�(2ps� ds�)
      =
      2pkmos3
      B
      =
      moks2
      The vector field is then
      B = moks2
      ^
      f
       
      		 = moM


File translated from TEX by TTH, version 3.85.
On 16 Jan 2010, 11:40.