35. MAGNETIC FIELD DUE TO CURRENT by nehalwan

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HC VERMA SOLUTIONS PHYSICS

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									                                                     CHAPTER – 35
                                  MAGNETIC FIELD DUE TO CURRENT


                      F     F          N              N
1.   F = q  B or, B =    =     =                  =
                        q   T   A. sec . / sec .   A m
          0                          2rB     mN     N
     B=                     or, 0 =        =         = 2
          2r                                A m A  A
2.   i = 10 A,     d=1m
                                                                                        Z axis
            i   10 7  4   10             –6
     B= 0 =                         = 20 × 10 T = 2 T
          2r         2  1                                                                                     X axis
     Along +ve Y direction.                                                                         1m

3.   d = 1.6 mm
     So, r = 0.8 mm = 0.0008 m
     i = 20 A
            i   4  10 7  20              –3                              r
      B = 0 =                         = 5 × 10 T = 5 mT
           2r    2    8  10  4
4.   i = 100 A, d = 8 m
                                                                                                            100 A
            i
     B= 0
          2r
                                                                                                                     8m
        4  10 7  100
     =                   = 2.5 T
           28
                   –7
5.   0 = 4 × 10 T-m/A
                                                                  –5
     r = 2 cm = 0.02 m,                 = 1 A,          B = 1 × 10 T
                                                                                        P
                                                                              0
     We know: Magnetic field due to a long straight wire carrying current =                      2 cm
                                                                              2r
                                                                                                             i
                     7
                 4  10  1             –5
     B at P =                  = 1 × 10 T upward                                                        
                   2  0.02                                                               2 cm
                           –7
     net B = 2 × 1 × 10 T = 20 T                                                      Q
                        –5
     B at Q = 1 × 10 T downwards
                   
     Hence net B = 0 
                                                
6.   (a) The maximum magnetic field is B  0 which are along the left keeping the sense along the
                                               2r
     direction of traveling current.
                               
     (b)The minimum B  0                                                                          0i
                              2r                                                                           i
                                                                                                  2r
                 0
         If r =        B net = 0
                 2B                                                                                      r
               0
         r<          B net = 0
               2B
                                 
         r > 0 B net = B  0
               2B                2r
                    –7                                         –4
7.   0 = 4 × 10 T-m/A,              = 30 A,     B = 4.0 × 10 T Parallel to current.        
                                                                                             B =–40 ×–10–4 T
                                                                                                            –
     B due to wore at a pt. 2 cm                                                         –  –
                                                                                            –    –      –        –    –
         0   4  10 7  30         –4
     =       =                 = 3 × 10 T                                                   –    –      –        –    –
         2r     2  0.02
                                                                                            –    –      –        –    –
     net field =   3  10   4  10 
                           4 2           4 2           –4
                                                  = 5 × 10    T                                         30 A


                                                              35.1
                                                                                       Magnetic Field due to Current

8.               ˆ
     i = 10 A. ( K )
     B = 2 × 10
                   –3                      ˆ
                        T South to North ( J )
                                                                                                        ˆ
     To cancel the magnetic field the point should be choosen so that the net magnetic field is along - J
     direction.
      The point is along - ˆ direction or along west of the wire.
                            i
          0
     B=
          2r
                 –3       4  10 7  10
      2 × 10         =
                              2  r
               2  10 7   –3
     r=               = 10 m = 1 mm.
            2  10  3
9.   Let the tow wires be positioned at O & P
                                                         8  10 4 = 2.828 × 10
                                                                              –2
     R = OA, =          (0.02)2  (0.02)2 =                                        m
                                                   7
                               4  10  10                –4                                                             A4
     (a) B due to Q, at A1 =                     = 1 × 10 T (r towards up the line)
                                  2  0.02
                              4  10 7  10                –4
         B due to P, at A1 =                    = 0.33 × 10 T (r towards down the line)
                                  2  0.06                                                                                    2 cm
                                                                                         O
                        –4               –4            –4
         net B = 1 × 10 – 0.33 × 10 = 0.67 × 10 T                                                                                
                                                                                       A1                        A2       A3
                             2  10 7  10           –4
     (b) B due to O at A2 =                   = 2 × 10 T            r down the line
                                   0.01
                            2  10 7  10                –4
         B due to P at A2 =                  = 0.67 × 10 T          r down the line
                                   0.03
                              –4              –4              –4
         net B at A2 = 2 × 10 + 0.67 × 10 = 2.67 × 10 T
                                   –4
     (c) B at A3 due to O = 1 × 10 T                 r towards down the line
                                   –4
         B at A3 due to P = 1 × 10 T                 r towards down the line
                             –4
         Net B at A3 = 2 × 10 T
                            2  10 7  10               –4
     (d) B at A4 due to O =                  = 0.7 × 10 T           towards SE
                             2.828  10  2
                                      –4
         B at A4 due to P = 0.7 × 10 T               towards SW
             
                               2
                                              2
                                                             
         Net B = 0.7  10 - 4  0.7  10 - 4 = 0.989 ×10–4 ≈ 1 × 10–4 T
10. Cos  = ½ ,                              = 60° & AOB = 60°
                           7
         0      10  2  10          –4
                                                                                                                      O
     B=        =                   = 10 T                                                                   2 cm          2 cm
         2r        2  10  2
                   –4 2        –4 2       –8    1/2
     So net is [(10 ) + (10 ) + 2(10 ) Cos 60°]
          –4                      1/2          -4                                                           
     = 10 [1 + 1 + 2 × ½ ]              = 10        ×     3 T = 1.732 × 10–4 T                          A
                                                                                                                  2 cm
                                                                                                                                    B
                 
11. (a) B for X = B for Y
                                                
         Both are oppositely directed hence net B = 0
                      
     (b) B due to X = B due to X both directed along Z–axis                                  (–1, 1)                       (1, 1)
              2  10 7  2  5          –6
         Net B =                  = 2 × 10 T = 2 T
                         1
                      
     (c) B due to X = B due to Y both directed opposite to each other.                       (–1, –1)                          (1, –1)
                    
         Hence Net B = 0
                                         –6
     (d) B due to X = B due to Y = 1 × 10 T both directed along (–) ve Z–axis
                                     –6
         Hence Net B = 2 × 1.0 × 10 = 2 T
                                                    35.2
                                                                                                                 Magnetic Field due to Current
12. (a) For each of the wire                                                                                                       4           3
        Magnitude of magnetic field                                                                                       Q1                       Q2

           i                        0  5     2                                                                              D               A
        = 0 ( Sin45  Sin45) =
          4r                      4  5 / 2 2                                                                         5 2
        For AB  for BC  For CD  and for DA .                                                                                          P             5 cm

        The two  and 2 fields cancel each other. Thus Bnet = 0                                                                       5 2/2
    (b) At point Q1                                                                                                            C                   B
                                                                                                                     Q3                                 Q4
                                     0i                     4  5  2  10 7                   –5
        due to (1) B =                                  =                              = 4 × 10        
                          2  2.5  10  2                      2  5  10  2
                                       0i                           4  5  2  10 7                    –5
        due to (2) B =                                  2
                                                                 =                         = (4/3) × 10 
                          2  (15 / 2)  10                          2  15  10  2
                                            0i                           4  5  2  10 7                    –5
        due to (3) B =                                       2
                                                                     =                         = (4/3) × 10 
                          2  (5  5 / 2)  10                            2  15  10  2
                                     0i                     4  5  2  10 7                   –5
        due to (4) B =                            2
                                                        =                              = 4 × 10        
                          2  2.5  10                          2  5  10  2
                                                            –5       32     –5          –5         –4
        Bnet = [4 + 4 + (4/3) + (4/3)] × 10                      =      × 10 = 10.6 × 10 ≈ 1.1 × 10 T
                                                                     3
    At point Q2
                              oi
        due to (1)                                 
                     2  (2.5)  10  2
                               oi
        due to (2)                                      
                     2  (15 / 2)  10  2
                              oi
        due to (3)                                 
                     2  (2.5)  10  2
                               oi
        due to (4)                                     
                     2  (15 / 2)  10  2
       Bnet = 0
    At point Q3
                        4  10 7  5                                     –5
        due to (1)                                2
                                                        = 4/3 × 10                     
                     2  (15 / 2)  10
                       4  10 7  5                                –5
        due to (2)                           2
                                                       = 4 × 10                        
                     2  (5 / 2)  10
                       4  10 7  5                                –5
        due to (3)                           2
                                                       = 4 × 10                        
                     2  (5 / 2)  10
                        4  10 7  5                                     –5
        due to (4)                                2
                                                        = 4/3 × 10                     
                     2  (15 / 2)  10
                                                            –5       32     –5          –5         –4
        Bnet = [4 + 4 + (4/3) + (4/3)] × 10                      =      × 10 = 10.6 × 10 ≈ 1.1 × 10 T
                                                                     3
    For Q4
                           –5
       due to (1) 4/3 × 10                  
                        –5
       due to (2) 4 × 10                    
                           –5
       due to (3) 4/3 × 10                  
                        –5
       due to (4) 4 × 10                    
       Bnet = 0

                                                                                35.3
                                                                             Magnetic Field due to Current
13. Since all the points lie along a circle with radius = ‘d’                           2             R
    Hence ‘R’ & ‘Q’ both at a distance ‘d’ from the wire.
                           
    So, magnetic field B due to are same in magnitude.                              i       S         d                    Q
    As the wires can be treated as semi infinite straight current carrying
                                                i
    conductors. Hence magnetic field B = 0                                                                                 1
                                               4d                                                        P

    At P                                                                                                  i

        B1 due to 1 is 0
                           i
        B2 due to 2 is 0
                         4d
    At Q
                           i
        B1 due to 1 is 0
                         4d
        B2 due to 2 is 0
    At R
        B1 due to 1 is 0
                           i
        B2 due to 2 is 0
                         4d
    At S
                           i
        B1 due to 1 is 0
                         4d
        B2 due to 2 is 0
           i
14. B = 0 2 Sin 
         4d
        i      2 x                0ix
    = 0                      =                                                                                 
      4d             x 2
                                         x2                                                                       d
           2  d2             4d d2                                                                x/2             i
                       4                  4                                                                   x
    (a) When d >> x
    Neglecting x       w.r.t. d
           0ix      ix
    B=            = 0 2
        d d   2  d
              1
    B
            d2
    (b) When x >> d, neglecting d w.r.t. x
            0ix    2 0i
    B=            =
         4dx / 2   4d
            1
    B
            d
15.  = 10 A, a = 10 cm = 0.1 m                                                                           O
                                                                                                          10 A
                   3                                                                    A                                  B
    r = OP =          0 .1 m
                  2                                                                                  10 cm
                                                                                                                    Q2
    B = 0 (Sin1  Sin 2 )                                                                          30
                                                                                                          30
       4r
                                                                                                Q1
         10 7  10  1       2  10 5             –5                                                    P
     =                    =             = 1.154 × 10 T = 11.54 T
            3                  1.732
               0 .1                                                                                      P
            2


                                                         35.4
                                                                                                                Magnetic Field due to Current
              0i                               0i               i              2                  0i
16. B1 =          ,                    B2 =         (2  Sin) = 0                             =
             2d                               4d              4d                        2                                                     
                                                                                                              2                                d
                                                                             2 d2                 4d d2 
                                                                                           4                   4                                         i
                                                                                                                                                 l
               1        i                             0i                 0i
    B1 – B2 =     B2  0                                               =
              100     2d                                        2       200 d
                                                 4d d2 
                                                                 4
                0i                    0i  1 1 
                                  =            
                              2       d  2 200 
        4d d2 
                            4
                                                                                      2
                                   99                     2            99  4    156816
                              =                                     =          =        = 3.92
                      2           200                          2       200        40000
        4 d2                                         d2 
                      4                                         4
         2                 2       3.92 2
     ℓ = 3.92 d +                     
                                     4
     1  3.92  2        2            2          2                              d2       0.02   d            0.02
               = 3.92 d  0.02 ℓ = 3.92 d                                         =        =   =               = 0.07
         4                                                                     2       3.92               3.92
17. As resistances vary as r & 2r
                                                                                                                                             B
                                     i                                           2
    Hence Current along ABC =          & along ADC =                                                                                  i/3
                                    3                                            3i
    Now,                                                                                                                                                     C         i
                                                                                                                             A    a/2                a
                       i  2  2  2     2 2 0i                                                                                                 2
    B due to ADC = 2 0                   =                                                                                 2i/3a
                       
                             43a       
                                              3 a
                                                                                                                                            D
                      i  2  2                                                                                                                      C         i
                                       2 2 0i
    B due to ABC = 2 0             =
                       43a 
                                        6a                                                                           D

                                                                                                                                                            3a/4
              2 2 0i    2 2 0i      2 0 i
    Now B =            –          =                                                                                a                       
               3a         6a       3a                                                                                         
                                                                                                                                     O         
                                                                                                                                                                 a/4
              a2 a2                    5a 2   a 5                                                               i
18. A0 =            =                      =
              16   4                   16      4                                                                        A   a/2              a/2         B

                       2               2
               3a  a                        9a 2 a 2              13a 2   a 13
    D0 =                              =            =                  =
               4   2                        16    4                16       4
    Magnetic field due to AB
                  i
    BAB = 0           (Sin (90 – i) + Sin (90 – ))
         4 2a / 4 

    =
         0  2i          2i
                 2Cos = 0     2
                                     a / 2 = 2 0i
         4a             4a       a( 5 / 4)    5
    Magnetic field due to DC
                   i
    BDC = 0             2Sin (90° – B)
         4 23a / 4

    =
         0i  4  2
                     Cos =
                              0i
                                  
                                     a / 2 = 2 0i
         4  3a              3a ( 13a / 4) a3 13
    The magnetic field due to AD & BC are equal and appropriate hence cancle each other.
                                                   2 0i             2 0i        2 0i  1   1 
    Hence, net magnetic field is                           –                 =                
                                                    5           a3 13            a  5 3 13 

                                                                              35.5
                                                                  Magnetic Field due to Current
                                                                                                                 D
19. B due t BC &                                                                B
     
    B due to AD at Pt ‘P’ are equal ore Opposite
                
    Hence net B = 0
                                                                            i                                         i
    Similarly, due to AB & CD at P = 0
                                                                                   A                         C
     The net B at the Centre of the square loop = zero.
                                      i
20. For AB      B is along     B = 0 (Sin60  Sin60)                                           2i
                                    4r
                                                                                                    A
                                      i
    For AC      B              B = 0 (Sin60  Sin60)
                                    4r                                                     30°

                                      i
    For BD      B              B = 0 (Sin60)                                  i
                                    4r                                                     60°                   i
                                      i
    For DC      B              B = 0 (Sin60)
                                    4r                                                 i                             C
                                                                            B                             i
     Net B = 0
21. (a) ABC is Equilateral
    AB = BC = CA = ℓ/3
    Current = i                                                                               A

                3     3     
     AO =         a =      =                                                    Q
                2     23    2 3                                                              M
                                                                                                              P

    1 = 2 = 60°
                                                                                        60°       60°
                                                                       B
    So, MO =                as AM : MO = 2 : 1                                                O                       C
                  6 3
     
    B due to BC at <.
         i                         i                 i9
    = 0 (Sin1  Sin 2 ) = 0  i  6 3  3 = 0
       4r                         4                  2
               9 0i        27 0i                                             B                                         C
    net B =            3 =
                2            2
                            i8            8 2 0 i
    (b) B due to AD = 0               2 =
                            4              4                                             45° 45°
                8 2 0 i        8 2 0 i                                                           ℓ/8
    Net B =               ×4=                                                   A                                     D
                   4              
                    r
22. Sin (/2) =
                    x
     r = x Sin (/2)
    Magnetic field B due to AR                                                                                C
      0i
           Sin(180  (90  ( / 2)))  1 
     4r
         i[Sin(90  ( / 2))  1]                                                                 r
     0                            
               4  Sin( / 2)                                              A
                                                                                    
                                                                                      x
            0i(Cos( / 2)  1)
                             
             4   Sin( / 2)
                 0i2Cos 4 ( / 4)        i
                                     0 Cot( / 4) 
           4  2Sin( / 4)Cos( / 4)    4x                                                                  B

     The magnetic field due to both the wire.
     2 0i                  i
           Cot(  / 4 )  0 Cot(  / 4 )
      4 x                 2x

                                                           35.6
                                                                                                                Magnetic Field due to Current
    
23. BAB                                                                                                                         D                      C

       0i  2              iSin
                2Sin = 0
        4b                  b                                                                                                           
                                                                                                                                        
               0i                                                2                 (  / 2)                                       
      =                = BDC                             Sin (ℓ +b) =                                  =
         b  2  b 2                                                            2
                                                                                 /4b /4       2                2
                                                                                                                  b2
                                                                                                                               A       l              B
      BBC
       0i  2                  iSin                                       (b / 2)                       b
                2  2Sin = 0                          Sin  =                                  =
        4                                                             2 / 4  b2 / 4                2  b2
                  0ib            
        =                       = BAD
              2  b 2
                             2 0i               2 0ib          2 0i( 2  b 2 )        2 0i  2  b 2
        Net B =                           +                    =                        =
                         b  2  b 2            2  b 2        b  2  b 2                 b
          2                                 2r
24. 2 =      = ,                     ℓ=
          n            n                       n
                            
    Tan  =        x=
              2x          2Tan 
      r
                                                                                                                                  
    2    n
                                                                                                                                A                  B
             0i                   i2Tan   2Sin                                                                                 l
    BAB =          (Sin  Sin) = 0
           4( x )                      4
         0i2Tan(  / n)2Sin(  / n)n   inTan(  / n)Sin(  / n)
    =                                 = 0
                  42r                         2 2r
                                  0inTan(  / n)Sin(  / n)
    For n sides, Bnet =
                                               2 2r
25. Net current in circuit = 0                                                                                                                         P
    Hence the magnetic field at point P = 0
    [Owing to wheat stone bridge principle]
                                           –5
26. Force acting on 10 cm of wire is 2 ×10 N
     dF     ii
        = 012
     dl     2d
         2  10 5            0  20  20
                        =
        10  10  2               2d
                                                                                                                                               d
              4  10 7  20  20  10  10 2                          -3
    d=                                                     = 400 × 10 = 0.4 m = 40 cm
                             2  2  10  5
27. i = 10 A
    Magnetic force due to two parallel Current Carrying wires.
          
    F= 0 12
           2r
                          
    So, F or 1 = F by 2 + F by 3                                                                                                                       1
          10  10       10  10
    = 0                0                                                                                                5 cm
       2  5  10  2 2  10  10  2                                                                                                                2

        4  10 7  10  10           4  10 7  10  10
    =                                                                                                                                                 3
            2  5  10  2              2  10  10  2
        2  10 3 10 3   3  10 3        –4
    =                  =           = 6 ×10 N towards middle wire
            5      5          5


                                                                       35.7
                                                                                                               Magnetic Field due to Current
       0 10i      0i40
28.           =                                                                                                                        i
       2x      2(10  x )
        10       40          1      4                                           10 A                                                       (10–x)       40 A
            =                =                                                                                             x
         x      10  x       x   10  x
     10 – x = 4x  5x = 10  x = 2 cm
    The third wire should be placed 2 cm from the 10 A wire and 8 cm from 40 A wire.
29. FAB = FCD + FEF                                                                                                               10
                                                                            A                                                                      B
         10  10          10  10
    = 0                 0
       2  1 10  2 2  2  10  2                                A
                                                                            C                                                     10               D
                                                                                                                                                       1 cm
             –3       –3         –3
    = 2× 10 + 10 = 3 × 10                 downward.                                                                               10
                                                                            E
    FCD = FAB + FEF                                                                                                                                F
    As FAB & FEF are equal and oppositely directed hence F = 0
     0i1i2
30.          = mg (For a portion of wire of length 1m)
     2d
                                                                                                                                               F
          0  50  i2           –4
                    3
                         = 1 × 10 × 9.8
        2  5  10
           4  10 7  5  i 2                      –4                                                                                        mg
                             = 9.8 × 10
          2  5  10  3                                                                                                                              50
                     –3            –3         –1
     2 × i2 × 10 = 9.3 × 10 × 10
            9 .8
     i2 =        10 1 = 0.49 A
             2
31. 2 = 6 A
    1 = 10 A
    FPQ                                                                                                             S                              R
                    ii           i i dx            30 dx
    ‘F’ on dx = 0 1 2 dx = 0 1 2               = 0                                                             I1
                    2x           2 x                    x
           0  30 dx                                                                                                  I2                              P
                          
                                                 –7        2
     FPQ =                    = 30 × 4 × 10 × [logx]1                                                      A
               x        1 x
                 –7                                                                                        10
    = 120 × 10 [log 3 – log 1]                                                                                               dx
                                              –7                                                                   P                           Q
    Similarly force of FRS = 120 × 10 [log 3 – log 1]
                                                                                                                  1 cm           x
    So, FPQ = FRS
               0  i1i2          0  i1i 2
     FPS =                2
                             
            2  1 10         2  2  10  2
          2  6  10  10 7         2  10 7  6  6                    –4
      =              2
                                                           = 8.4 × 10         N (Towards right)
                10                         2  10  2
                   0  i1i2                 0  i1i2
      FRQ =                     2
                                     
                2  3  10               2  2  10  2
          4   10 7  6  10           4   10 7  6  6              –4            –5                –4
      =                   2
                                                              = 4 × 10        + 36 × 10     = 7.6 × 10        N
         2  3  10        2  2  10  2
    Net force towards down
    = (8.4 + 7.6) × 10–4 = 16 × 10–4 N
32. B = 0.2 mT,                  i = 5 A,                           n = 1,                   r=?
         n 0 i
    B=
          2r
               n   0i   1  4  10 7  5                –3           –3            –1
      r=               =                    = 3.14 × 5 × 10 m = 15.7× 10 m = 15.7 × 10 cm = 1.57 cm
                 2B        2  0.2  10  3

                                                                           35.8
                                                                                                  Magnetic Field due to Current
          n 0 i
33. B =
           2r
     n = 100, r = 5 cm = 0.05 m
                –5
     B = 6 × 10 T
        2rB     2  0.05  6  10 5    3       –1
     i=      =                  7
                                     =      × 10 = 0.0477 ≈ 48 mA
        n 0     100  4  10         6.28
           5
34. 3 × 10 revolutions in 1 sec.
                         1
    1 revolutions in           sec
                      3  10 5
           q   1.6  10 19
     i=      =              A
           t    1 
                         
                3  10 5 
           0i      4  10 7.16  10 19 3  10 5 2  1.6  3
                                                                  10 11 = 6.028 × 10 ≈ 6 × 10 T
                                                                                      –10      –10
     B=          =                      10
           2r              2  0.5  10                 0 .5
35. l = i/2 in each semicircle                                                                                          B
                    1  (i / 2)                                                                   i/2
    ABC = B =  0                   downwards
                     2      2a                                                                 i   A                        C   i
                    1  0 (i / 2)
    ADC = B =                      upwards                                                        i/2
                     2       2a
                                                                                                                       D
    Net B = 0
36. r1 = 5 cm                 r2 = 10 cm
    n1 = 50                   n2 = 100
    i=2A
               n i n  i
    (a) B = 1 0  2 0
                 2r1      2r2
          50  4  10 7  2       100  4   10 7  2
     =                          
           2  5  10  2        2  10  10  2
                 –4           –4           –4
      = 4 × 10 + 4 × 10 = 8 × 10
              n i n  i
     (b) B = 1 0  2 0 = 0
               2r1        2r2
37. Outer Circle
    n = 100, r = 100m = 0.1 m
    i=2A
         n 0 i   100  4  10 7  2              –4
    B =          =                        = 4 × 10                         horizontally towards West.
           2a              2  0 .1
    Inner Circle
    r = 5 cm = 0.05 m,               n = 50, i = 2 A
         n 0 i   4   10 7  2  50            –4
    B =          =                      = 4 × 10                           downwards
           2r            2  0.05

     Net B =     4  10   4  10 
                           4 2               4 2
                                                     =    32 2  10 8 = 17.7 × 10
                                                                                        –4           –4
                                                                                             ≈ 18 × 10
                                                                                                                   –3
                                                                                                          = 1.8 × 10 = 1.8 mT
                                                                6
38. r = 20 cm,         i = 10 A,                     V = 2 × 10 m/s,                 = 30°
            
    F = e( V  B) = eVB Sin 
                  –19               6    0i
     = 1.6 × 10         × 2 × 10 ×           Sin 30°
                                        2r
         1.6  10 19  2  10 6  4  10 7  10                    –19
     =                                  2
                                                         = 16 × 10         N
                    2  2  20  10

                                                                    35.9
                                                                                                        Magnetic Field due to Current
                          
39. B Large loop = 0
                           2R                                                                                              i                  R
    ‘i’ due to larger loop on the smaller loop                                                                                          r

                                            2  
    = i(A × B) = i AB Sin 90° = i × r × 0                                                                                                        
                                              2r
40. The force acting on the smaller loop
    F = ilB Sin 
                                                                                                                                               R
        i2r o 1     ir                                                                                                        r
    =              = 0                                                                                                        i
         2R  2          2R
41. i = 5 Ampere,                     r = 10 cm = 0.1 m                                                                             
    As the semicircular wire forms half of a circular wire,
                1  0i     1 4  10 7  5
    So, B =              =                                                                                                        10 cm
                 2 2r       2      2  0 .1
                   –6              –6             –5
    = 15.7 × 10 T ≈ 16 × 10 T = 1.6 × 10 T
           i            2      i
42. B = 0           =            0
          2R 2         3  2 2R
         4  10 7  6              –6                                                                              120°
     =               2
                           = 4 × 10
          6  10 t10
                         –6                –6            –5
    = 4 × 3.14 × 10 = 12.56 × 10 = 1.26 × 10 T
                         i
43. B due to loop 0
                          2r
    Let the straight current carrying wire be kept at a distance R from centre. Given  = 4i                       i
                                  4i
     B due to wire = 0 = 0
                            2R      2R
                  
    Now, the B due to both will balance each other                                                                      r
                0i      0 4i          4r
    Hence            =         R=
               2r       2R             
    Hence the straight wire should be kept at a distance 4/r from centre in such a way that the direction of current
                                                                                                   
    in it is opposite to that in the nearest part of circular wire. As a result the direction will B will be oppose.
                                                            –2
44. n = 200, i = 2 A,                   r = 10 cm = 10 × 10 n
               n 0 i   200  4  10 7  2               –4
     (a) B =          =                      = 2 × 4 × 10
                2r        2  10  10  2
                               –4              –4
          = 2 × 4 × 3.14 × 10 = 25.12 × 10 T = 2.512 mT
                    n 0ia 2                n 0 i      n 0ia 2
     (b) B =        2        2 3/2
                                                  =
                2(a  d )                    4a      2(a 2  d2 )3 / 2
        1              a2                    2    2 3/2        3               2      2        3 2/3
            =                          (a +d )          2a              a + d = (2a )
       2a        2(a 2  d2 )3 / 2
        2      2      1/3  2                2     2       2/3 2                    –1 2    2      2/3      –1 2
      a + d = (2 a)                    a + d =2 a                       (10 ) + d = 2 (10 )
         –2      2     2/3   –2             –2   2/3      2                   –2   1/3      2
     10 + d = 2 10                     (10 )(2 – 1) = d                 (10 ) (4 – 1) = d
          –2                       2      2     –2
      10 (1.5874 – 1) = d              d = 10 × 0.5874
            10 2  0.5874 = 10 × 0.766 m = 7.66 × 10
                                       –1                                –2
     d=                                                                      = 7.66 cm.
                
45. At O P the B must be directed downwards                                                                           O
    We Know             B at the axial line at O & P
                                                                                                                                   3 cm = 0.03 m
               0ia 2                                                                                             4 cm M
     =                                           a = 4 cm = 0.04 m
         2(a 2  d2 )3 / 2
                                                                                                                       P
         4  10 7  5  0.0016
     =                                           d = 3 cm = 0.03 m
           2((0.0025 )3 / 2                                                                                                                 3 cm
              –6            –5
     = 40 × 10 = 4 × 10 T                        downwards in both the cases

                                                                   35.10
                                                                                                                                    Magnetic Field due to Current
                          –6
46. q = 3.14 × 10              C,                      r = 20 cm = 0.2 m,
                                                                q   3.14  10 6  60           –5
     w = 60 rad/sec.,                                  i=         =                   = 1.5 × 10
                                                                t       2   0 .2
                                                 xQ
      Electric field
                     =
                       4 0 x  a 2            2
                                                            
                                                            3/2
                                                                         =
                                                                                         xQ
                                                                                                           
                                                                                                                
                                                                                                               2 x 2  a2     
                                                                                                                              3/2


      Magnetic field          0ia 2                                                
                                                                             4 0 x 2  a 2        3/2
                                                                                                                     0ia 2
                                         
                                      2 a2  x 2        
                                                        3/2


          9  10 9  0.05  3.14  10 6  2
     =
          4  10  7  15  10  5  (0.2)2
            9  5  2  10 3                     3
      =                         12
                                         =
          4  13  4  10                        8
                                            
47. (a) For inside the tube                B =0
                                                                                                                                                                    P
    As, B inside the conducting tube = o                                                                                                               r/2
              
    (b) For B outside the tube                                                                                                                                       r
                                                                                                                                                             O
           3r
     d=
            2
             i     i 2     i
      B = 0 = 0             = 0
            2d      23r     2r
48. (a) At a point just inside the tube the current enclosed in the closed surface = 0.
                  o
    Thus B = 0 = 0
                  A
    (b) Taking a cylindrical surface just out side the tube, from ampere’s law.
                                    i
    0 i = B × 2b        B= 0 
                                  2b                                                                                                                        a
49. i is uniformly distributed throughout.                                                                                                                           b

                                                                 i                      ia 2
     So, ‘i’ for the part of radius a =                              2
                                                                          a 2 =               =
                                    b                                                  b2
     Now according to Ampere’s circuital law
                                                                                                                                                   b             a
      B× dℓ = B × 2 ×  × a = 0 
                   ia 2   1       ia
      B = 0                = 0 2 
                 b   2   2a     2b
                                –2
50. (a) r = 10 cm = 10 × 10 m
                –2
    x = 2 × 10 m,                      i=5A
    i in the region of radius 2 cm
            5
                       (2  10  2 )2 = 0.2 A
     (10  10  2 )2
                     –2 2
    B ×  (2 × 10 ) = 0(0-2)
               4  10 7  0.2                      0.2  10 7                         –4
     B=                            4
                                             =              4
                                                                         = 2 × 10
                4  10       10
     (b) 10 cm radius
                    –2 2
     B ×  (10 × 10 ) = 0 × 5
               4  10 7  5         –5
     B=                    = 20 × 10
                  10  2
     (c) x = 20 cm
                       –2 2
     B×  × (20 × 10 ) = 0 × 5
                     0  5                           4  10 7  5                             –5                                                          B
     B=                            2 2
                                                 =                           4
                                                                                  = 5 × 10            
                 (20  10              )              400  10                                                                                                       x

                                                                                               35.11
                                                                                      Magnetic Field due to Current

51. We know,      B  dl =  i. Theoritically B = 0 a t A
                                 0
                                                                                                                    P           Q
    If, a current is passed through the loop PQRS, then                                                          ℓ
             0i                                                                                B
    B=             will exist in its vicinity.
          2(  b)                                                                                     R      S
                                                                                                        b
    Now, As the B at A is zero. So there’ll be no interaction
    However practically this is not true. As a current carrying loop, irrespective of its near about position is
    always affected by an existing magnetic field.                                                     P
52. (a) At point P, i = 0, Thus B = 0                                                               
    (b) At point R, i = 0, B = 0
                                                                                                      
    (c) At point ,                                                                                 
    Applying ampere’s rule to the above rectangle
                        l                                                 A                     B                       B
     B × 2l = 0K0      
                        o
                            dl                                                                 

                                                                                 Bb    Ba                                    
                                  0k
      B ×2l = 0kl  B =
                                  2
                                                                                                                        B
                        l
                                                                                                                    l
     B × 2l = 0K0       dl
                        o

                           k
     B ×2l = 0kl  B = 0                                               Bd                                     B
                           2
                
    Since the B due to the 2 stripes are along the same
    direction, thus.                                                        BC                                                  
             k  k
    Bnet = 0  0 = 0k                                                    
             2     2                                         C              D                               l
53. Charge = q,        mass = m
    We know radius described by a charged particle in a magnetic field B
         m
    r=
         qB
     Bit B = 0K [according to Ampere’s circuital law, where K is a constant]
          m          rq 0k
     r=         =           
         q 0 k         m
                                     –2
54. i = 25 A, B = 3.14 × 10 T,                        n=?
    B = 0ni
               –2           –7
     3.14 × 10 = 4 ×  × 10 n × 5
                10 2        1                     4
     n=                  =  10 4 = 0.5 × 10 = 5000 turns/m
            20  10  7      2
55. r = 0.5 mm,                        i = 5 A,          B = 0ni (for a solenoid)
                                            –3
    Width of each turn = 1 mm = 10 m
                            1           3
    No. of turns ‘n’ =             = 10
                          10  3
                       –7        3                –3
    So, B = 4 × 10 × 10 × 5 = 2 × 10 T
     R
56.     = 0.01  in 1 m,               r = 1.0 cm        Total turns = 400,        ℓ = 20 cm,
      l
              –2                      400
    B = 1× 10 T,            n=                 turns/m
                                   20  10  2
         E                E                            E
    i=      =                             =
        R0     R 0 / l  (2r  400 )       0.01 2    0.01 400
     B = 0ni
                                                         35.12
                                                                                                                              Magnetic Field due to Current

                 2                    –7             400                               E
      10 = 4 × 10                        ×               2
                                                                 
                                               20  10               400  2  0.01 10  2
                     10 2  20  10 2  400  2  10 2 0.01
     E=                                                                                   =1V
                                           4   10  7  400
                                                                                           0        a 2indx
57. Current at ‘0’ due to the circular loop = dB =                                            
                                                                                           4                   3/2
                                                                                                             
                                                                                                             2
                                                                                                  2  l
                                                                                                a    x  
                                                                                                
                                                                                                    2      
                                                                 B
      for the whole solenoid B =                                dB
                                                                0

                              0 a 2nidx
     =            
                                                    3/2
                                 
          0                      2
                          
                 4a 2    x                                                                                                                ni dx
                   
                         2                                                                                                                          ℓ/2–x
                                           2                                                                                       2
          0ni                        a dx                            0ni                   dx                          2x                 ℓ/2
     =
          4         
                     0
                                                   2
                                                          3/2
                                                                 =
                                                                      4 a    
                                                                              0
                                                                                                   2
                                                                                                          3/2
                                                                                                                  = 1  
                                                                                                                       
                                                                                                                               
                                                                                                                            2a 
                                            2x                                                  2x 
                             a 3 1                                             1         
                                          2a                                               2a  
                                                                                                   

                                  8                                                                         –31
58. i = 2 a, f = 10 rev/sec,                               n= ?,          me = 9.1 × 10                           kg,
                                –19                                       B
     qe = 1.6 × 10                    c,                   B = 0ni  n =
                                                                           0i
         qB            f 2m e        B        f 2m e          10 8  9.1 10 31
     f=        B=             n=          =           =                                 = 1421 turns/m
       2m e              qe           0i      qe  0i   1.6  10 19  2  10  7  2 A
59. No. of turns per unit length = n,      radius of circle = r/2,     current in the solenoid = i,
    Charge of Particle = q,                mass of particle = m         B = 0ni
            mV 2                  qBr   q 0nir    niqr
     Again         = qVB  V =        =         = 0        
               r                   m     2m         2m
60. No. of turns per unit length = ℓ
    (a) As the net magnetic field = zero
                
     Bplate  B Solenoid
     
    Bplate  2 = 0kdℓ = 0kℓ
              k                                
    Bplate  0             ...(1)                B Solenoid = 0ni …(2)
                2                                                                                                                          Bc             Ba
                           k
    Equating both i = 0
                           2
    (b) Ba × ℓ = kℓ               Ba = 0k BC = 0k
     B=           Ba 2  Bc 2 =                2 0k 2 =           2 0k                                                                       0ni
                                                                                                                                                           
                           2k                                                                                                          A                    C
     2 0k = 0ni                             i=
                           n
                                     –3
61. C = 100 f,       Q = CV = 2 × 10 C,     t = 2 sec,
                                                –3
    V = 20 V, V = 18 V,     Q = CV = 1.8 × 10 C,
          Q  Q     2  10 4     –4
     i=           =           = 10 A        n = 4000 turns/m.
             t           2
                          –7          –4          –7
      B = 0ni = 4 × 10 × 4000 × 10 = 16  × 10 T

                                                                                           
                                                                                             35.13

								
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