Chapter 11 Moving Charge and Magnetism ( Magnetic effects of Current Part 1) – Physics free study material by TEACHING CARE online tuition and coaching classes

Chapter 11 Moving Charge and Magnetism ( Magnetic effects of Current Part 1) – Physics free study material by TEACHING CARE online tuition and coaching classes

File name : Chapter-11-Moving-Charge-and-Magnetism-Magnetic-effects-of-Current-Part-1.pdf

 

 

Biot-Savart’s law is used to determine the magnetic field at any point due to a current carrying conductors.

This law is although for infinitesimally small conductors yet it can be used for long conductors. In order to understand the Biot-Savart’s law, we need to understand the term current-element.

In the figure shown below, there is a segment of current carrying wire and P is a point where magnetic field is

 

to be calculated.

i dl

is a current element and r is the distance of the point ‘P’ with respect to the current element

 

 

i dl . According to Biot-Savart Law, magnetic field at point ‘P’ due to the current element

i dl

is given by the

 

expression,

dB k i dlsinθ

 

also B =

dB = m i .

 

dl sinq

 

r 2

 

In C.G.S. :        k = 1 Þ dB =

ò

idl sinq

r 2

4p ò   r 2

Gauss

 

 

In S.I.     :

k = m0

Þ dB m0 × idl sinq

Tesla

 

4p                 4p       r 2

 

where

m 0 = Absolute permeability of air or vacuum

= 4p ´ 10 -7               Wb          . It’s other units are

Amp metre

Henry metre

 

or  N     or

Amp 2

Tesla metre Ampere

 

(1)

Different forms of Biot-Savarts law

 

 

 

 

  • Similarities and differences between Biot-Savart law and Coulomb’s Law

(i)  The current element produces a magnetic field, whereas a point charge produces an electric field.

  • The magnitude of magnetic field varies as the inverse square of the distance from the current element, as does the electric field due to a point

 

dB = m 0

idl ´ rˆ

Biot-Savart Law

F    1   q1q2 rˆ

Coulomb’s Law

 

4p     r 2                                               4pe 0    r 2

(iii)   The electric field created by a point charge is radial, but the magnetic field created by a current element is

perpendicular to both the length element dl   and the unit vector rˆ.

 

E

 

 

+q

 

Direction of Magnetic Field.

The direction of magnetic field is determined with the help of the following simple laws :

  • Maxwell’s cork screw rule

According to this rule, if we imagine a right handed screw placed along the current carrying linear conductor, be rotated such that the screw moves in the direction of flow of current, then the direction of rotation of the thumb gives the direction of magnetic lines of force.

(2)  Right hand thumb rule

According to this rule if a current carrying conductor is held in the right hand such that the thumb of the hand represents the direction of current flow, then the direction of folding fingers will represent the direction of magnetic lines of force.

 

(3)  Right hand thumb rule of circular currents

According to this rule if the direction of current in circular conducting coil is in

the direction of folding fingers of right hand, then the direction of magnetic field will be in the direction of stretched thumb.

(4)  Right hand palm rule

If we stretch our right hand such that fingers point towards the point. At which magnetic field is required while thumb is in the direction of current then normal to the palm will show the direction of magnetic field.

 

 

 

Note : @ If magnetic field is directed perpendicular and into the plane of the paper it is represented by Ä (cross) while if magnetic field is directed perpendicular and out of the plane of the paper it is represented by ¤ (dot)

i                                           i

CW                        ACW

 

 

In                        Out

 

In : Magnetic field is away from the observer or perpendicular inwards.

Out : Magnetic field is towards the observer or perpendicular outwards.

 

 

Application of Biot-Savarts Law.

(1)  Magnetic field due to a circular current

If a coil of radius r, carrying current i then magnetic field on it’s axis at a distance x from its centre given by

 

 

Baxis

= μ0 .

4π

2πNir 2

(x 2r 2 )3/2

; where N = number of turns in coil.

 

Different cases

Case 1 : Magnetic field at the centre of the coil

 

(i)  At centre x = 0 Þ B

m0 . 2pNi = m 0 Ni B

 

 

centre          4p      r                2r            max

 

 

(ii)   For single turn coil N = 1 Þ B

m0 . 2pim0i

 

(iii)   In C.G.S.

m0 = 1    Þ B

2pi

 

centre

4p    r          2r

4p                   centre             r

 

Note : @ B

centre µ N

(i, r constant),

Bcentre µ i

(N, r constant),

Bcentre µ r

(N, i constant)

 

1

Case 2 : Ratio of Bcentre and Baxis

B          æ      x 2 ö3/2

 

ø
è

The ratio of magnetic field at the centre of circular coil and on it’s axis is given by  centre = ç1 +                  ÷

 

 

a               5 5

Baxis        ç

a                  æ 3 ö3 / 2

r 2 ÷

 

(i)  If

x = ±a, Bc  = 2    Ba

x = ± 2 , Bc =   8   Ba

x = ±      , Bc = ç   ÷      Ba

2           2

 

 

  • If

B   = Bc

a            n

then

x == ±r

and if Ba   =    c

è   ø

B

then x == ±r

 

Case 3 : Magnetic field at very large/very small distance from the centre

 

(i)     If x >> r (very large distance) Þ B

m0 . 2p Nir 2

m0 . 2NiA

 

where A = pr2 = Area of each turn of the coil.

 

axis           4p       x 3

4p     x 3

 

(ii)           If x << r (very small distance) Þ

Baxis

¹ Bcentre , but by using binomial theorem and neglecting higher

 

x 2                    æ      3 x 2 ö

 

power of

r 2 ; BaxisBcentre ç1 – 2 r 2 ÷

 

è              ø

Case 4 : Bx curve

 

 

 

The variation of magnetic field due to a circular coil as the distance x varies as shown in the figure.

 

B varies non-linearly with distance x as shown in figure and is maximum when at the centre of the coil and it is zero at x = ± ¥.

x 2 = min = 0 , i.e., the point is

 

Point of inflection (A and A¢) : Also known as points of curvature change or pints of zero curvature.

 

 

(i)  At these points B varies linearly with x Þ

dB =

dx

constant Þ

d 2 B = 0 .

dx 2

 

(ii)   They locates at

x = ± r

2

from the centre of the coil.

 

(iii)   Separation between point of inflextion is equal to radius of coil (r)

  • Application of points of inflextion is “Hamholtz coils”

 

Note : @The magnetic field at

(2)  Helmholtz coils

x = r

2

is B = 4m0 Ni

 

(i)  This is the set-up of two coaxial coils of same radius such that distance between their centres is equal to their radius.

  • These coils are used to obtain uniform magnetic field of short range which is obtained between the

 

(iii)   At axial mid point O, magnetic field is given by

B = 8mNi = 0.716 m0 Ni = 1.432 B , where

B m0 Ni

 

5  5R                        R                                                       2R

(iv)  Current direction is same in both coils otherwise this arrangement is not called Helmholtz’s coil arrangement.

  • Number of points of inflextion Þ Three (A, A¢, A¢¢)

 

 

 

 

 

 

 

Note : @The device whose working principle based on this arrangement and in which uniform magnetic field is used called as “Halmholtz galvanometer”.

  • Magnetic field due to current carrying circular arc : Magnetic field at centre O

 

 

B m0 . pim0i

 

B = m0 .q i

 

B m0 . (2pq )i

 

 

4p   r        4r

4p    r

4p         r

 

 

 

Special results

If magnetic field at the centre of circular coil

 

is denoted by B

æ= m0 . 2pi ö

 

0 ç    4p    r ÷

è
ø

Magnetic field at the centre of arc which is

making an angle q at the centre is

 

Barc

= æ B0 ö.q

ç      ÷

2p

 

è      ø

 

 

 

  • Concentric circular loops (N = 1)

(i)  Coplanar and concentric : It means both coils are in same plane with common centre

(a) Current in same direction                              (b) Current in opposite direction

i

 

m

i

 

1

B   =     0

4p

æ

1

2pi ç

è r1

ö

1

+      ÷

r2 ø

B   =      0

m
2

4p

2pié 1

ê r1

ë

1 ù

û

r2 ú

 

 

 

B1       æ r2 + r1 ö

 

Note : @

B    = ç r   r ÷

 

2      è 2       1 ø

(ii)   Non-coplanar and concentric : Plane of both coils are perpendicular to each other

 

Magnetic field at common centre

m

B =                       =    0

2r

 

 

 

(5)  Magnetic field due to a straight current carrying wire

Magnetic field due to a current carrying wire at a point P which lies at a perpendicular distance r from the wire as shown is given as

 

B          0=                   f

m    i

. (sin

  • sinf )

 

4p r            1         2

 

From figure a = (90of1) and

m

b = (90o + f2 )

 

Hence B =    o . i (cosa – cos b )

4p r

 

 

 

Different cases

Note : @ When point P lies on axial position of current carrying conductor then magnetic field at P

B = 0                                                                      i

P

@          The value of magnetic field induction at a point, on the centre of separation of two linear parallel conductors carrying equal currents in the same direction is zero.

  • Zero magnetic field : If in a symmetrical geometry, current enters from one end and exists from the other, then magnetic field at the centre is

i                                                                                                     i

 

 

In all cases at centre B = 0

 

 

 

 

Example: 1         Current flows due north in a horizontal transmission line. Magnetic field at a point P vertically above it directed

 

  • North wards
  • South wards
  • Toward east
  • Towards west

 

Solution : (c)       By using right hand thumb rule or any other rule which helps to determine the direction of magnetic field.

Example: 2         Magnetic field due to a current carrying loop or a coil at a distant axial point P is B1 and at an equal distance

 

in it’s plane is B2

then B1 is

B

 

2

 

(a) 2                                (b) 1                                  (c)

1                           (d) None of these

2

 

Solution : (a)       Current carrying coil behaves as a bar magnet as shown in figure.

We also knows for a bar magnet, if axial and equatorial distance are same then Ba = 2Be

 

Hence, in this equation

B1 = 2

 

B2       1                                                                                      B1

 

 

 

Example: 3         Find the position of point from wire ‘B‘ where net magnetic field is zero due to following current distribution

  • 4 cm

 

 

30 cm

7

12 cm

7

 

  • 2 cm

Solution : (c)       Suppose P is the point between the conductors where net magnetic field is zero.

So at P |Magnetic field due to conductor 1| = |Magnetic field due to conductor 2|

 

 

i.e.

m 0 . 2(5i)m 0 . 2(2i) Þ

5   9

Þ x = 30 cm

 

4p     i

4p (6 – x)

x       6 – x                    7

 

Hence position from B = 6 – 30 = 12 cm

 

7       7

 

cm

 

 

 

Example: 4         Find out the magnitude of the magnetic field at point P due to following current distribution

 

(a)

m 0ia

 

pr 2

 

m ia 2

 

(b)

 

(c)

 

(d)

    0      

pr

m0ia

 

2pr 2

2m 0ia

 

pr 2

 

Solution : (a)       Net magnetic field at P, B          = 2B sinq ; where B = magnetic field due to one wire at P = m0 . 2i

 

net

 

and sinq = a        \ B       = 2 ´ m 0 . 2i ´ a = m 0 ia .

 

 

4p   r

 

r                 net

4p   r      r

pr 2

 

Example: 5         What will be the resultant magnetic field at origin due to four infinite length wires. If each wire produces magnetic field ‘B‘ at origin

  • 4 B
  • B
  • 2 B
  • Zero

 

Solution : (c)       Direction of magnetic field (B1, B2, B3 and B4) at origin due to wires 1, 2, 3 and 4 are shown in the following figure.

m 0 2i                                                                                                   i             1        2

 

B1 = B2 = B3  = B4

= 4p . x

= B . So net magnetic field at origin O

i                                                                i

B4

 

Bnet =

=

= 2 2B

B2

O B3 B1

 

4                   i

3

 

Example: 6         Two parallel, long wires carry currents i1 and i2 with i1 > i2 . When the currents are in the same direction, the

 

magnetic field at a point midway between the wires is 10 mT. If the direction of i2

becomes 30 mT. The ratio i1 / i2 is

is reversed, the field

 

1
2

(a) 4                               (b) 3                                  (c) 2                i                                                                    i

Solution : (c)       Initially when wires carry currents in the same direction as shown.

Magnetic field at mid point O due to wires 1 and 2 are respectively

 

B = m 0 . 2i1 Ä and B

 

 

m0 . 2i2 ¤

 

 

 

1     4p    x                     2     4p    x

 

Hence net magnetic field at O      B

= m 0 ´ 2 (ii )

 

 

Þ   10 ´ 10 6 = m 0    2       – i )

 

net

4p     x   1      2

…..(i)

 

4p . x (i1     2

i1                                            i2

 

If the direction of i2 is reversed then

B = m 0 . 2i1 Ä and B  m 0 . 2i2 Ä

1     4p    x                     2     4p    x

 

 

 

 

 

So B

m 0 2

 

 

  • i )

Þ 30 ´ 10 6 = m 0    2

 

 

+ i )…… (ii)

 

net

4p . x (i1      2

4p . x (i1      2

 

Dividing equation (ii) by (i)

i1 + i2 = 3 Þ i1 = 2

 

 

i1 – i2       1        i2       1

Example: 7 A wire of fixed length is turned to form a coil of one turn. It is again turned to form a coil of three turns. If in both cases same amount of current is passed, then the ratio of the intensities of magnetic field produced at the centre of a coil will be                                                                                                                                [MP PMT 2002]

 

  • 9 times of first case (b)

1 times of first case          (c) 3 times of first case (d)

9

1 times of first case

3

 

Solution : (a)       Magnetic field at the centre of n turn coil carrying current i

B m 0 . 2pni

……(i)

 

4p     r

 

For single turn n =1

B m 0 . 2pi

…..(ii)

 

4p    r

If the same wire is turn again to form a coil of three turns i.e. n = 3 and radius of each turn r‘ = r

3

 

So new magnetic field at centre B‘ = m 0 . 2p (3)

Þ B‘ = 9 ´ m 0 . 2pi

……(iii)

 

4p      r‘                                   4p    r

Comparing equation (ii) and (iii) gives B‘ = 9B .

Example: 8         A wire in the form of a square of side a carries a current i. Then the magnetic induction at the centre of the square wire is (Magnetic permeability of free space = m0)                                                               [EAMCET 2001]

 

(a)

 

 

 

m0 i

2pa

m0 i

pa

2 2m 0i

pa

  m 0 i  

i

 

Solution : (c)       Magnetic field due to one side of the square at centre O

i

 

B  = m 0 .

1     4p

2i sin 45 o        Þ

a / 2

B   = m 0 .

1     4p      a

 

 

Hence magnetic field at centre due to all side B

 

net

= 4 B1

m 0 (2 2 i)

pa

a/2

a

 

Example: 9         The ratio of the magnetic field at the centre of a current carrying circular wire and the magnetic field at the centre of a square coil made from the same length of wire will be

 

  • p2
  • p2

   p                                  (d)  p

 

2 2                              4 2

Solution : (b)       Circular coil                                                                 Square coil         i i

i

 

 

 

 

i

Length L = 2p r                                                                                       Length L = 4a

 

 

 

 

Magnetic field B = m 0 . 2pi

m 0 . 4p 2i

 

B = m 0 . 2

2 i B m 0 . 8 2 i

 

 

Hence

4p    r

Bcircular p 2

Bsquare

4p     r

4p      a

4p      a

 

Example: 10       Find magnetic field at centre O in each of the following figure

  • (ii) (iii)

 

 

 

 

 

 

 

 

m 0i

 

m 0i æ 1      1 ö

 

 

m 0i æ 1      1 ö

 

(a)

r

Ä                                                 (a)

ç

4 è r1

–      ÷ Ä                                                 (a)

r2 ø

ç

4 è r1

–      ÷ Ä

r2 ø

 

m 0i

 

m 0i æ 1      1 ö

 

 

m 0i æ 1      1 ö

 

(b)

  • (b)

2r

ç

4 è r1

+      ÷ Ä                                               (b)

r2 ø

ç

4 è r1

+      ÷ Ä

r2 ø

 

m 0i

 

m 0i æ 1      1 ö

 

 

m 0i æ 1      1 ö

 

(c)

Ä                                                 (c)

4r

ç

4 è r1

–      ÷ ¤                                                  (c)

r2 ø

ç

4 è r1

–      ÷ ¤

r2 ø

 

m 0 i ¤                                                  (d) Zero                                                      (d) Zero

4r

 

Solution : (i)       (c) Magnetic field at O due to parts 1 and 3, B1 = B3 = 0

(2)

 

 

While due to part (2)

B  = m 0 . pi Ä

 

 

2     4p r

 

\ Net magnetic field at centre O,

(1)

O                          (3)

 

m    p i                              m i                                                          2

 

Bnet

= B1 + B2 + B3 =    0 .

p

4      r

Ä     Þ Bnet

=    0 Ä

4r

 

  • (b) B1 = B3 = 0

B = m 0 . p i Ä

2     4p r1

B    = m 0 . p i Ä                                                                                                                                4

 

4     4p r2

m 0      æ 1     1 ö

 

So Bnet

= B2 + B4

=        .p iç

4p     è r1

+      ÷ Ä

r2 ø

 

  • (a) B1 = B3 = 0

B = m 0 . p i Ä

2     4p   r1                                                                                                                           4

 

B  = m 0 . p i Ä

 

 

As | B

| > | B  |

 

4     4p r2

2              4

 

m 0i æ 1      1 ö                                                              O

 

So Bnet

= B2 – B4 Þ Bnet   =

ç

4 è r1

–      ÷ Ä                                          1                               3

r2 ø

 

Example: 11       Find magnetic field at centre O in each of each of the following figure

 

 

 

  • (ii) (iii)

 

 

 

(a)

m 0 i ¤                                                     (a)

2r

m i (p – 2) Ä                               (a)

2p r

m 0 2i (p + 1) Ä 2r r

 

 

(b)

m 0 i Ä                                                       (b)

 

m 0i

 

i p + 2) ¤                          (b)

m 0i

 

2i p – 1) Ä

 

 

(c)

2r

3m 0i Ä

8r

(c)

4p . r (

m 0i Ä

4r

4r . r (

  • Zero

 

(d)

3m 0 i ¤                                                   (d)

8r

m 0 i ¤                                             (d) Infinite

4r

 

Solution : (i)        (d) By using B = m 0 . (2p  q )i

Þ B m 0 . (2p p / 2)i = 3m 0i ¤

 

4p        r                           4p          r                  8r

  • (b) Magnetic field at centre O due to section 1, 2 and 3 are respectively

B = m 0 . i ¤

1     4p r                                                                                                      1 i

m    pi

 

2

B    =    0 .     ¤

4p r

B = m 0 . i ¤

3     4p r

 

2

 

 

m 0 i

 

Þ Bnet

= B1 + B2 + B3 = 4p r (p + 2) ¤

 

  • (b) The given figure is equivalent to following figure, magnetic field at O due to long wire (part 1)

B = m 0 . 2i ¤

 

1     4p   r

m    2p i

 

2

Due to circular coil B   =    0 .         Ä                                                             2

4p     r

Hence net magnetic field at O

 

B       = B

B = m 2i p – 1) Ä

 

net

2        1      4p . r (

1

 

Example: 12       The field B at the centre of a circular coil of radius r is p times that due to a long straight wire at a distance r

from it, for equal currents here shows three cases; in all cases the circular part has radius r and straight ones are infinitely long. For same current the field B is the centre P in cases 1, 2, 3 has the ratio                              [CPMT   1989]

 

 

 

æ
(1)                                             (2)                                            (3)

 

 

(a)

æ- p ö : æ p ö : æ 3p – 1 ö

 

 

(b)

– p + ö æ p +

 

ö æ 3p + 1 ö

 

 

 

ç    2 ÷ ç 2 ÷ ç 4      2 ÷

ç    2    1÷ : ç 2

1÷ : ç 4      2 ÷

 

è        ø è     ø è              ø                                                                   è              ø è           ø è               ø

 

 

(c)

– p p   3p

 

 

(d)

æ- p – 1ö : æ p – 1 ö : æ 3p + 1 ö

 

2 : 2 : 4

ç              ÷ ç

2
2
2
2
4

è              ø è

÷ ç               ÷

ø è               ø

 

Solution : (a)       Case 1 : B     = m 0 . i Ä

 

(A)

 

A       4p r

B = m 0 . i ¤

B       4p r

B     = m 0 . i ¤

 

C       4p r

(C)

 

So net magnetic field at the centre of case 1

 

B   = B    – (B    + B   )

Þ B  = m 0 . pi ¤…………… (i)

 

 

1         B                A              C

1     4p   r

 

Case 2 : As we discussed before magnetic field at the centre O in this case

(B)

 

B = m 0 . pi Ä

 

…..(ii)

 

2     4p r

 

Case 3 : BA = 0

  • O

 

B m 0 . (2p p / 2) Ä = m 0 . 3pi Ä

 

 

B       4p          r

B     = m0 . i ¤

 

4p    2r

 

C        4p r

So net magnetic field at the centre of case 3

 

B  = m 0 . i æ 3p – 1ö Ä

….(iii)

 

r
2
è
ø

3      4p     ç              ÷

 

 

From equation (i), (ii) and (iii)

B   : B   : B

= p ¤ : p ¤ : æ 3p – 1ö Ä = – p

p : æ 3p – 1 ö

 

 

 

2
2

1       2        3

ç              ÷                            ç               ÷

2
2

è              ø                            è 4        ø

 

Example: 13       Two infinite length wires carries currents 8A and 6A respectively and placed along X and Y-axis. Magnetic field at a point P (0, 0, d)m will be

 

 

(a)

7m 0

pd

(b)

10m 0

 

pd

(c)

14 m 0

pd

(d)

5m 0

 

pd

 

 

Solution : (d)       Magnetic field at P

 

 

Due to wire 1,

B  m 0 . 2(8)

 

 

 

1     4p     d

m     2(16)

 

and due to wire 2,

B   =    0 .

2

4p     d

 

m      2           5m

 

\   Bnet   =                        =

=    0 ´

4p

´ 10 =      0

d                 pd

 

Example: 14       An equilateral triangle of side ‘a‘ carries a current i then find out the magnetic field at point P which is vertex of triangle

 

 

 

 

 

(a)

 

(b)

m 0i     Ä

m 0i     ¤

2 3pa

 

 

(c)

2 3m 0i ¤

pa

 

  • Zero

Solution : (b)       As shown in the following figure magnetic field at P due to side 1 and side 2 is zero.

Magnetic field at P is only due to side 3,

 

which is B1

m 0 . 2i sin 30 o

 

¤
a

4p         3a                                                                                                                                       

2                                                                                                 2

m 0 . 2 ¤ =    m 0i      ¤                                                                           3

4p     3a                                                                                                        a

Example: 15 A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle q at the centre. The value of, the magnetic induction at the centre due to the current in the ring is                                                                                          [IIT-JEE 1995]

 

(a) Proportional to 2(180 oq )

  • Zero, only if q = 180 o

(b) Inversely proportional to r

(d) Zero for all values of q

 

Solution : (d)       Directions of currents in two parts are different, so directions of magnetic fields due to these currents are different.

 

Also applying Ohm’s law across AB

i1 R1 = i2 R2 Þ i1l1 = i2l 2 …..(i)

 

Also B

m0 ´ i1l1

 

 

and B

= m0 ´ i2l2 ; \

 

 

B2 = i1l1 = 1 [Using (i)]

 

1    4p     r 2

2     4p     r 2

B1      i2l2

 

Hence, two field are equal but of opposite direction. So, resultant magnetic induction at the centre is zero and is independent of q .

Example: 16       The earth’s magnetic induction at a certain point is 7 ´ 105 Wb / m2 . This is to be annulled by the magnetic

induction at the centre of a circular conducting loop of radius 5 cm. The required current in the loop is

[MP PET 1999; AIIMS 2000]

(a) 0.56 A                                (b) 5.6 A                                      (c) 0.28 A                            (d) 2.8 A

 

Solution : (b)       According to the question, at centre of coil B = B

Þ m 0 . 2p i = B

 

 

 

 

Þ 10 -7 ´

2p i

 

(5 ´ 1-2 )

= 7 ´ 10 -5

Þ i = 5.6

H

 

amp.

4p     r            H

 

Example: 17       A particle carrying a charge equal to 100 times the charge on an electron is rotating per second in a circular

 

path of radius 0.8 metre. The value of the magnetic field produced at the centre will be (m0

permeability for

 

vacuum)                                                                                                                              [CPMT 1986]

 

(a)

107

m0

(b)

1017 m0

(c)

106 m0

(d)

107 m0

 

Solution : (b)       Magnetic field at the centre of orbit due to revolution of charge.

B = m 0 . 2p (qn ) ; where n = frequency of revolution of charge

4p       r

 

So,

B m 0 ´ 2p ´ (100e ´ 1)

 

Þ B = 10 17 m .

 

4p            0.8                                0

Example: 18       Ratio of magnetic field at the centre of a current carrying coil of radius R and at a distance of 3R on its axis is

 

 

 

 

(a)

10                             (b)

20                                (c)

2                            (d)

 

ø

B               æ        x 2 ö 3 / 2                                                                B

 

Solution : (a)       By using

     centre   = ç1 +

÷       ; where x = 3R and r = R   Þ    centre = (10)3 / 2 = 10

10 .

 

è

Baxis

ç        r 2 ÷

Baxis

 

Example: 19       A circular current carrying coil has a radius R. The distance from the centre of the coil on the axis where the

 

magnetic induction will be

  • th to its value at the centre of the coil, is [MP PMT 1997]

8

 

(a)

R                                       (b)

R                                             (c)

  • 3 R
  • R

 

 

ø

B               æ        x 2 ö 3 / 2                                                       1

 

Solution : (b)       By using

     centre   = ç1 +

÷       , given r = R and Baxis   =

Bcentre

 

è

Baxis

ç        r 2 ÷                                                           8

 

ï
ø
ø
R 2
è

æ           2 ö 3 / 2

ìæ           2 ö1 / 2 ü3

æ           2 ö1 / 2                             2

 

Þ 8 = ç1 + x    ÷

 

Þ (2)2 = ïç1 + x  ÷         ï     Þ 2 = ç1 + x    ÷

 

Þ 4 = 1 + x

Þ x =     3R

 

ø
è

ç        R 2 ÷

íç               ÷      ý

îè                       ïþ

ç        R 2 ÷                             R 2

 

Example: 20       An infinitely long conductor PQR is bent to form a right angle as shown. A current I flows through PQR. The

 

magnetic field due to this current at the point M is

H1. Now, another infinitely long straight conductor QS is

 

connected at Q so that the current is

1 in QR as well as in QS, the current in PQ remaining unchanged. The

2

 

magnetic field at M is now H2. The ratio H1 / H2 is given by                                      [IIT-JEE (Screening) 1999]

 

  • 12

 

 

– ¥         i

M

90o

Q                          + ¥

 

  • 1

2

P                               90o            S

 

3                                                                                                                       R

  • 2

Solution : (c)       Magnetic field at any point lying on the current carrying conductor is zero.

Here H1 = magnetic field at M due to current in PQ

 

H  = magnetic field at M due to R + due to QS + due to PQ = 0 + H1 + H

2                                                                                                                                 2        1

= 3 H1

2

 

\      H1 = 2

 

H 2        3

Example: 21       Figure shows a square loop ABCD with edge length a. The resistance of the wire ABC is r and that of ADC is 2r. The value of magnetic field at the centre of the loop assuming uniform wire is

  • 3pa   ¤
  • 3pa   Ä

(c)      p a     ¤

(d)     p a     Ä

Solution : (b)       According to question resistance of wire ADC is twice that of wire ABC. Hence current flows through ADC is

 

half that of ABC i.e.

i2 = 1 . Also i   + i

 

= i Þ i   = 2i and i   = i

 

 

 

i1       2           1      2

1       3          2      3

 

 

 

 

 

Magnetic field at centre O due to wire AB and BC (part 1 and 2)

B  = m 0 .

1     4p

2i1 sin 45 o Ä

a / 2

= m 0 . 4p

2 2 i1 Ä

a

 

 

and magnetic field at centre O due to wires AD and DC (i.e. part 3 and 4) B   = B

m 0 2 2 i2 ¤

 

 

Also i1 = 2i2. So (B1 = B2) > (B3 = B4)

Hence net magnetic field at centre O

3       4     4p      a

 

B

 

Bnet

= (B1 + B2 ) – (B3 + B4 )

 

2 2 ´ æ 2 i ö                 2 2 æ i ö ´ 2

 

 

m              ç 3 ÷      m

ç 3 ÷             m

2 m i

 

= 2 ´    0 .              è      ø –    0 .           è   ø       =    0 .

(2 – 1) Ä =          0 Ä

 

4p          a

4p           a

4p     3a

3p a

 

 

 

Tricky example: 1

Figure shows a straight wire of length l current i. The magnitude of magnetic field produced by the

current at point P is                                       l

P

i

l

 

 

 

(a)

   2m 0i

pl

(b)

m 0 i

4pl

   2m 0 i

8pl

m 0 i

2 2pl

 

Solution: (c)      The given situation can be redrawn as follow.

As we know the general formula for finding the magnetic field due to a finite length wire

 

B         0=                        f

m    i

. (sin

  • sin f )

 

4p r           1           2                                                                                  l

P

 

Here f1

= 0o, f = 45o

45o

i

 

\ B = m 0 . i (sin 0 o + sin 45 o ) = m 0 . i 

Þ B    2m 0i                               l

 

4p r

4p     2l

8pl

 

 

 

Tricky example: 2

 

A cell is connected between the points A and C of a circular conductor ABCD of centre ‘O‘ with angle

AOC = 60°, If B1 and B2 are the magnitudes of the magnetic fields at O due to the currents in ABC

 

and ADC respectively, the ratio B1 is

B2

(a) 0.2

  • 6

i1     B

O

 

A

 

300o

 

60o

C

[KCET (Engg./ Med.) 1999]

 

 

 

 

Solution: (c)

  • 1
  • 5

B m 0 . q i

i2   D

 

1A

 

4p    r

Þ B µ q i

Þ B1 = q 1 ´ i1

B2       q 2     i2

 

Also i1

l 2

q 2

Hence

B1 = 1

 

i2       l1      q1                B2       1

Tags:

Leave a Reply

Your email address will not be published. Required fields are marked *