# 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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