36.PERMANENT MAGNETS by nehalwan

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									                                                                      CHAPTER – 36
                                                           PERMANENT MAGNETS


1.   m = 10 A-m,                                   d = 5 cm = 0.05 m
           m      10 7  10    10 2         –4                                                                                                N            S
     B= 0 2 =                  =       = 4 × 10 Tesla
          4 r     5  10    
                          2 2    25          
2.   m1 =m2 = 10 A-m
     r = 2 cm = 0.02 m
     we know
                                                                                               0 m1m 2   4  10 7  10 2           –2
     Force exerted by tow magnetic poles on each other =                                                =              4
                                                                                                                            = 2.5 × 10 N
                                                                                              4 r 2       4  4  10
            dv                            –3               –3
3.   B=–        dv = –B dℓ = – 0.2 × 10 × 0.5 = – 0.1 × 10 T-m
            d
     Since the sigh is –ve therefore potential decreases.
4.   Here                dx = 10 sin 30° cm = 5 cm
     dV     0.1 10 4 T  m
        =B=
     dx        5  10  2 m
     Since B is perpendicular to equipotential surface.
                                                              –4
     Here it is at angle 120° with (+ve) x-axis and B = 2 × 10 T
                 –4
5.   B = 2 × 10 T
     d = 10 cm = 0.1 m
                                                                                                              V                   0.1×10–4 T-m
     (a) if the point at end-on postion.                                                                                                                   0.3×10–4 T-m
                                                          7
                 0 2M         –4 10  2M                                                                                         30°        30°
        B=              2 × 10 =
                4 d3              (10 1 )3                                                                             0.2×10   –4
                                                                                                                                       T-m            –4
                                                                                                                                                                       X     In cm
                                                                                                                                             0.4×10        T-m
                         4             3
             2  10               10                                2
                        7
                               = M  M = 1 Am
                 10  2
     (b) If the point is at broad-on position
        0 M           –4 10 7  M             2
                2 × 10 =             M = 2 Am
        4 d 3
                           (10 1 )3
6.   Given :
               –1                             2
      = tan     2  tan  = 2  2 = tan                                                                                                                                   
                             tan 
      tan  = 2 cot             = cot                                                                                                                                    P
                               2                                                                                                                                       
                tan 
     We know           = tan                                                                                                                                     
                  2
     Comparing we get, tan  = cot                                                                                                          S                                   N
     or, tan  = tan(90 – )               or  = 90 –           or  +  = 90
     Hence magnetic field due to the dipole is r to the magnetic axis.
7.   Magnetic field at the broad side on position :
                  M
     B= 0                                  2ℓ = 8 cm              d = 3 cm
                
           4 d2   2 3 / 2     
                    –6               10 7  m  8  10 2                          –6         10 9  m  8
      4 × 10            =                                            4 × 10            =
                             9  10     4
                                               16  10        
                                                          4 3 / 2
                                                                                             10 
                                                                                               4 3 / 2
                                                                                                           25 3 / 2
                4  10 6  125  10 8                                  –5
     m=                                            = 62.5 × 10               A-m
                             8  10  9

                                                                                36.1
                                                                                                           Permanent Magnets
8.   We know for a magnetic dipole with its north pointing the north, the neutral point in the broadside on
     position.                                                                               N
                            0M
     Again B in this case =
                            4d3
         M
      0 3 = BH due to earth
        4d                                                                                                 P
                                                                                                                        d
         10 7  1.44
                       = 18 T
              d3
         10 7  1.44            –6                                                     S
                       = 18 × 10
             d3
          3          –3
      d = 8 × 10
                    –1
      d = 2 × 10 m = 20 cm
     In the plane bisecting the dipole.
9.   When the magnet is such that its North faces the geographic south of earth. The neutral point lies along
     the axial line of the magnet.
      0 2M               7                          2  0.7  10 7
            = 18 × 10  10  2  0.72 = 18 × 10  d =
                     –6                        –6  3

     4 d3                   d3                         18  10 - 6
                                1/ 3
          8  10  9       
     d=                                        –1
                                       = 2 × 10        m = 20 cm               S                              N
          10  6                                                                                                     BM   BH
                           
                                                        2                              N2              N
10. Magnetic moment = 0.72 2 A-m = M
        M
                                                                                              –1
                                                                                        tan        2                        BH
    B= 0 3           BH = 18 T
       4 d
          4  10 7  0.72 2                          –6                  W       S                               N            E
                       3
                                         = 18 × 10
               4  d
          3   0.72  1.414  10 7                                                                                N1
     d =                          6
                                             = 0.005656                                                S
                18  10
     d ≈ 0.2 m = 20 cm
11. The geomagnetic pole is at the end on position of the earth.
            0 2M   10 7  2  8  10 22          –6                                                                       d
     B=         3
                  =                       ≈ 60 × 10 T = 60 T
           4 d       (6400  10 3 )3
               –5
12. B = 3.4 × 10 T
           M            –5
    Given 0 3 = 3.4 × 10
          4 R
              3.4  10 5  R 3  4                        2   3
     M=                           7
                                             = 3.4 × 10 R
                 4  10
                  2M           –5
    B at Poles = 0 3 = = 6.8 × 10 T
                  4 R
13. (dip) = 60°
    BH = B cos 60°
                  –6
     B = 52 × 10 = 52 T
                                3       –6
     BV = B sin  = 52 × 10       = 44.98 T ≈ 45 T
                              2
14. If 1 and 2 be the apparent dips shown by the dip circle in the 2r positions, the true dip  is given by
         2        2        2
    Cot  = Cot 1 + Cot 2
            2        2         2
     Cot  = Cot 45° + Cot 53°
            2
     Cot  = 1.56   = 38.6 ≈ 39°

                                                                    36.2
                                                                                               Permanent Magnets
                                              0in
15. We know                         BH =
                                              2r
                                    –5
    Give : BH = 3.6 × 10                 T                     = 45°
                   –2
    i = 10 mA = 10 A                                          tan  = 1
    n=?                                                       r = 10 cm = 0.1 m
         BH tan   2r   3.6  10 5  2  1 10 1               3
    n=                 =                             = 0.5732 × 10 ≈ 573 turns
               0i          4  10  7  10  2
                                                                  –4 2
16. n = 50                        A = 2 cm × 2 cm = 2 × 2 × 10 m
                –3
    i = 20 × 10 A                 B = 0.5 T
            
                                                  –3        –4
     = ni A  B = niAB Sin 90° = 50 × 20 × 10 × 4 × 10 × 0.5 = 2 × 10 N-M
                                                                             –4


17. Given  = 37°                              d = 10 cm = 0.1 m
    We know
     M    4 ( d2   2 ) 2         4  d4
        =                   tan  =       tan  [As the magnet is short]
     BH   0      2d                 0 2d
              4               (0.1)3                                –3   7            4          3   2 –1
    =                 7
                                      tan 37 = 0.5 × 0.75 × 1 × 10 × 10 = 0.375 × 10 = 3.75 ×10 A-m T
     4  10                     2
    M                                            3   2 –1
18.    (found in the previous problem) = 3.75 ×10 A-m T
    BH
     = 37°,             d=?
      M    4 2
         =    (d   2 )3 / 2 tan 
     BH    0
    ℓ << d                          neglecting ℓ w.r.t.d
        M     4 3                               1
          =    d Tan  3.75 × 103 =            7
                                                         3
                                                     × d × 0.75
        BH    0                              10
          33.75  10 3  10 7         –4
    d =                       = 5 × 10
                 0.75
     d = 0.079 m = 7.9 cm
           M            2
19. Given     = 40 A-m /T
          BH
    Since the magnet is short ‘ℓ’ can be neglected
          M   4  d3
    So,     =       = 40
          BH  0   2                                                                                          S

          3  40  4  10 7  2         –6
    d =                         = 8 × 10
                     4
                  –2
     d = 2 × 10 m = 2 cm                                                                                     N
    with the northpole pointing towards south.
20. According to oscillation magnetometer,
                    
    T = 2
                   MBH

                 1.2  10 4
           = 2
         10      M  30  10  6
                  2
       1      1.2  10 4
          =
       20    M  30  10  6
                  1.2  10 4  400                  2    2               2
    M=                           6
                                             = 16 × 10 A-m = 1600 A-m
                       30  10


                                                                   36.3
                                                                                                  Permanent Magnets

                       1 mB H
21. We know :  =
                      2     
    For like poles tied together                                                            N     S          S   N
    M = M 1 – M2
    For unlike poles M = M1 + M2                                                           N     S          N   S
                                          2
     1          M1  M2    10  M  M2        M  M2
        =                   = 1         25 = 1
     2          M1  M2    2   M1  M2       M1  M2
        26   2M1  M    13
          =      1 =
        24   2M2  M2   12
                       –6
22. BH = 24 × 10            T                 T1 = 0.1 
                                     –6       o i          –6 2  10 7  18                –6         –6
    B = BH – Bwire = 2.4 × 10             –        = 24 × 10 –                = (24 –10) × 10 = 14 × 10
                                              2 r                  0 .2
                                             T1       B
    T = 2                                       =
                  MBH                         T2       BH
                                                 2
        0 .1          14  10 6       0 .1  14    2  0.01 14
            =                         T  = 24  T2 =
                                                              T2 = 0.076
        T2            24  10  6      2                 24
                   
23. T = 2                                    Here  = 2
                  MBH
            1
    T1 =      min                             T2 = ?
           40
     T1          
        =
     T2          
          1             1      1        1     2   1
            =                       =    T2 =      T2 = 0.03536 min
        40T2            2   1600 T2 2   2        800
    For 1 oscillation Time taken = 0.03536 min.
    For 40 Oscillation Time = 4 × 0.03536 = 1.414 =                     2 min
24. 1 = 40 oscillations/minute
    BH = 25 T
                                   2
    m of second magnet = 1.6 A-m
    d = 20 cm = 0.2 m
    (a) For north facing north
                  1 MBH                               1 MBH  B 
        1 =                                  2 =
                 2                                 2    
                 0 m      10 7  1.6
        B=                          = 20 T
                 4 d3      8  10  3
        1              B      40                25        40
           =                     =                  2 =    = 17.88 ≈ 18 osci/min
        2            BH  B   2                 5         5
    (b) For north pole facing south
                  1 MBH                               1 MBH  B 
                                        2 =
                 2                                 2    
        1              B      40                25             40
           =                     =                  2 =              = 53.66 ≈ 54 osci/min
        2            BH  B   2                45             25 
                                                                   
                                                                45 

                                                                   
                                                                     36.4

								
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