No Title
PH317 MAGNETISM
WINTER Quarter 2009-2010
Solution to Homework Assignment 6
Assigned: Tuesday 1/19/10
Due: Tuesday 1/26/10
  1. Text: 6.25(b)
    SOLUTION
    Use the Fon,by notation. Label the three magnets 1, 2, and 3 in vertical ascension. Let all magnets have the same mass md. Do a free body diagram on the middle and top magnets. The corresposnding equilibrium equations are:
    F32-F31-mdg
    =
    0            top magnet
    F21-F23-mdg
    =
    0            middle magnet
    The magnitudes of the three magnetic forces are:
    F21
    =
    3mom2
    2pz24
    F31
    =
    3mom2
    2pz34
    F32
    =
    3mom2
    2p(z3-z2)4
    Distance z2 is measured from the bottom of magnet 1 to the bottom of magnet 2. Distance z3 is measured from the bottom of magnet 1 to the bottom of magnet 3. The the equilibrium equations become
    3mom2
    2p(z3-z2)4
    		 - 3mom2
    2pz34
    		 - mdg
    =
    0
    3mom2
    2pz24
    		 - 3mom2
    2p(z3-z2)4
    		 -mdg
    =
    0
    Cancel as many terms as possible and rewrite the equations in the form
    1
    (z3-z2)4
    		 - 1
    z34
    =
    2pmdg
    3mom2
    1
    z24
    		 - 1
    (z3-z2)4
    =
    2pmdg
    3mom2
    Set the left-hand-sides equal to each other and rearrange.
    1
    (z3-z2)4
    		 - 1
    z34
    =
    1
    z24
    		 - 1
    (z3-z2)4

    2
    (z3-z2)4
    		 - 1
    z34
    		 - 1
    z24
    =
    0
    2
    z34 æ
    è     		1- z2
    z3
    		 ö
    ø     		4

     
    		 - 1
    z34
    		 - 1
    z34
    		 1
    æ
    è     		z2
    z3
    		 ö
    ø     		4

     
    =
    0
    2
    æ
    è     		1- z2
    z3
    		 ö
    ø     		4

     
    		 - 1
    æ
    è     		z2
    z3
    		 ö
    ø     		4

     
    		 -1
    =
    0
    Define u as
    u
    º
    z2
    z3
    Then, set up the function
    f(u)
    =
    0
    f(u)
    =
    2
    (1-u)4
    		 - 1
    u4
    		 - 1
    Using Maple, the solution for u is
    u
    =
    0.45949
    =
    z2
    z3
    In terms of the book's variables x and y,
    x
    y
    =
    z2
    z3-z2
    =
    1
    z3
    z2
    		 -1
    =
    1
    1/0.45949-1
    =
    0.8501
  2. Text: 6.26
    SOLUTION
    Let the region BELOW the interface be labeled region 1 and the region ABOVE the interface be labeled region 2. Take a normal vector to the interface pointing from 1 to 2. The boundary conditions to satisfy are Eqs. (6.25) and (6.26).
    H||above - H||below
    =
    Kf×
    ^
    n
     
    B^above - B^below
    =
    0
    Converting to the new notation and setting Kf=0,
    H||2 - H||1
    =
    0
    B^2 - B^1
    =
    0
    Taking components
    H2sinq2
    =
    H1sinq1
    B2cosq2
    =
    B1cosq1
    Dividing the two equations and using B=mH,
    1
    m2
    		 tanq2
    =
    1
    m1
    		 tanq1
    tanq2
    tanq1
    =
    m2
    m1


File translated from TEX by TTH, version 3.85.
On 26 Jan 2010, 16:54.