Chapter 12 Moving Charge and Magnetism ( Magnetic effects of Current Part 2) – Physics free study material by TEACHING CARE online tuition and coaching classes

Chapter 12 Moving Charge and Magnetism ( Magnetic effects of Current Part 2) – Physics free study material by TEACHING CARE online tuition and coaching classes

File name : Chapter-12-Moving-Charge-and-Magnetism-Magnetic-effects-of-Current-Part-2.pdf

 

 

Amperes Law.

Amperes law gives another method to calculate the magnetic field due to a given current distribution.

 

Line integral of the magnetic field B around any closed curve is equal to through the area enclosed by the curve

r

m 0 times the net current i threading

 

i.e. ò Bdl = m0 åi = m0 (i1 + i3 – i2 )

 

Also using

 

B = m0 H

(where H = magnetising field)

 

ò m0 H.dl = m0Si Þ ò H.dl = Si

 

Note : @            Total current crossing the above area is

(i1 + i3 – i2 ) . Any current outside the area is not

 

included in net current. (Outward ¤ ® +ve, Inward Ä ® – ve)

@          When the direction of current is away from the observer then the direction of closed path is clockwise and when the direction of current is towards the observer then the direction of closed path is anticlockwise.

 

 

Application of Amperes law.

(1) Magnetic field due to a cylindrical wire

  • Outside the cylinder

 

P

 

 

 

 

 

 

Solid cylinder

P

 

 

 

 

 

 

Thin hollow cylinder

P

 

 

 

 

 

 

Thick hollow cylinder

 

 

 

r

In all above cases magnetic field outside the wire at P ò B.dl

= m0i Þ

Bò dl = m0i

Þ B ´ 2pr = m 0i

Þ Bout

m0i

2pr

 

 

In all the above cases B

 

surface

m0i

2pR

 

  • Inside the cylinder : Magnetic field inside the hollow cylinder is

 

 

 

Cross sectional view Þ

Solid cylinder

Thin hollow cylinder

Thick hollow cylinder

 

 

 

Note : @For all cylindrical current distributions

Baxis = 0 (min.), Bsurface = max (distance r always from axis of cylinder), Bout µ 1/r.

  • Magnetic field due to an infinite sheet carrying current : The figure shows an infinite sheet of current with linear current density j (A/m). Due to symmetry the field line pattern above and below the sheet is Consider a square loop of side l as shown in the figure.

 

b
c
d

b                       c                       d                      a

 

According to Ampere’s law,

òaB.dl + ò B.dl + ò B.dl + ò B.dl = m0i .

 

 

ò

c

Since B dl along the path b ® c and d ® a, therefore,        B.dl = 0 ;

b

a

ò

B.dl = 0

d

 

 

b                      a

d

Also, B || dl along the path a ® b and c ® d, thus  òaB.dl + ò B.dl = 2Bl

The current enclosed by the loop is i = jl

 

 

Therefore, according to Ampere’s law

2Bl = m0

( jl) or B = m0 j

2

 

 

(3) Solenoid

A cylinderical coil of many tightly wound turns of insulated wire with generally diameter of the coil smaller than its length is called a solenoid.

One end of the solenoid behaves like the north pole and opposite end behaves like the south pole. As the length of the solenoid increases, the interior field becomes more uniform and the external field becomes weaker.

A magnetic field is produced around and within the solenoid. The magnetic field within the solenoid is uniform and parallel to the axis of solenoid.

  • Finite length solenoid : If N = total number of turns,

l = length of the solenoid

n = number of turns per unit length = N

l

 

 

Magnetic field inside the solenoid at point P is given by

B = m0 (2p ni)[sina + sin b ] 4p

 

  • Infinite length solenoid : If the solenoid is of infinite length and the point is well inside the solenoid e.

a = b = (p / 2) .

So                         Bin = μ0 ni

(ii) If the solenoid is of infinite length and the point is near one end i.e. a = 0 and b = (p / 2)

 

 

So                         B

 

end

= 1 (μ

2

0ni)

 

Note : @Magnetic field outside the solenoid is zero.

@           B        = 1 B

end           2   in

  • Toroid : A toroid can be considered as a ring shaped closed Hence it is like an endless cylindrical solenoid.

Consider a toroid having n turns per unit length

 

 

Let i be the current flowing through the toroid (figure). The magnetic lines of force mainly remain in the core of toroid and are in the form of concentric circles. Consider such a circle of mean radius r. The circular closed path

 

surrounds N loops of wire, each of which carries a current i therefore from ò B.dl

m 0 inet

 

Þ B ´ (2pr) = m0 Ni

Þ B = m0 Ni = m ni

2pr           o

where n = N  

2pr

 

For any point inside the empty space surrounded by toroid and outside the toroid, magnetic field B is zero because the net current enclosed in these spaces is zero.

 

 

Example: 22         A long solenoid has 200 turns per cm and carries a current of 2.5 A. The magnetic field at its centre is

[m0 = 4p ´ 10–7 Wb/m2]                                                                                                        [MP PET 2000]

(a) 3.14 ´ 10–2 Wb/m2        (b) 6.28 ´ 10–2 Wb/m2           (c) 9.42 ´ 10–2 Wb/m2     (d) 12.56 ´ 10–2 Wb/m2

 

 

Solution : (b)

B = m 0ni = 4p ´ 10 7 ´ 200 ´ 2.5 = 6.28 ´ 10 2 Wb / m 2 .

10 2

 

Example: 23         A long solenoid is formed by winding 20 turns/cm. The current necessary to produce a magnetic field of 20

 

millitesla inside the solenoid will be approximately æ m 0 = 10 7 Tesla metre / ampere ö

[MP PMT 1994]

 

è
ø

ç 4p                                           ÷

(a) 8.0 A                                  (b) 4.0 A                                      (c) 2.0 A                              (d) 1.0 A

 

 

Solution : (a)

B = m 0 ni ; where n = 20 turn = 2000 turn . So,  20 ´ 10 5

= 4p ´ 2000 ´ i Þ

i = 8 A.

 

10 cm                    m

Example: 24         Two solenoids having lengths L and 2L and the number of loops N and 4N, both have the same current, then the ratio of the magnetic field will be                                                                                                              [CPMT 1994]

 

(a)

1 : 2

(b)

2 : 1

(c)

1 : 4

(d)

4 : 1

 

 

Solution : (a)

B = m

 N i Þ B µ N

Þ B1   = N1 ´ L2   =

 

 

N ´ 2l = 1 .

 

 

 

0 L                       L

B2       N 2      L1

4 N      L       2

 

Example: 25         The average radius of a toroid made on a ring of non-magnetic material is 0.1 m and it has 500 turns. If it carries 0.5 ampere current, then the magnetic field produced along its circular axis inside the toroid will be

 

(a)

25 ´ 102 Tesla               (b)

5 ´ 102 Tesla                              (c)

25 ´ 104 Tesla (d)

5 ´ 104 Tesla

 

 

Solution : (d)

B = m ni ; where n =   N

0

2pR

\   B = 4p ´ 10 7 ´    500  ´ 0.5

2p ´ 0.1

 

= 5 ´ 10 4 T.

 

Example: 26         For the solenoid shown in figure. The magnetic field at point P is

 

 

 

 

(a)

 

(b)

m 0ni ( 4

   3m 0ni

4

+ 1)

 

m ni

 

(c)

 

(d)

    0  (

2

m 0ni ( 4

+ 1)

 

– 1)

 

 

Solution : (a)

B = m 0 .2p ni (sin a + sin b ) . From figure a = (90o – 30o) = 60o and b = (90o – 60o) = 30o

4p

 

\ B = m 0 ni (sin 60 o + sin 30 o ) = m 0 ni (

+ 1) .

 

2                                    4

Example: 27         Figure shows the cress sectional view of the hollow cylindrical conductor with inner radius ‘R‘ and outer radius ‘2R‘, cylinder carrying uniformly distributed current along it’s axis. The magnetic induction at point ‘P’ at a

 

distance

  • Zero

3R from the axis of the cylinder will be

2

 

 

 

 5m 0 i  

72p R

 7m 0 i  

18p R

5m 0 i

 

36p R

 

 

 

ìæ 3R ö – R 2 ü

 

 

m i æ r 2a 2 ö                       3R

m i         ïç 2 ÷            ï      5.m i

 

Solution : (d)       By using

B =     0 ç                  ÷ here r =

, a = R,

ab = 2R Þ

B =        0       ´ íè         ø           ý =           .

 

2p r ç b 2 – a 2 ÷                         2

æ 3R ö

 

ï (R 2 ) – R 2 ï

36p r

 

è                 ø                                                                                2p ç        ÷

è  2  ø     ïî                       ïþ

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