# 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            
  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
–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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