# 37.MAGNETIC PROPERTIES OF MATTER

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```					                                                           CHAPTER – 37
MAGNETIC PROPERTIES OF MATTER

B
1.   B = 0ni,                H=
0
 H = ni
 1500 A/m = n× 2
 n = 750 turns/meter
 n = 7.5 turns/cm
2.   (a) H = 1500 A/m
As the solenoid and the rod are long and we are interested in the magnetic intensity at the centre,
the end effects may be neglected. There is no effect of the rod on the magnetic intensity at the
centre.
(b)  = 0.12 A/m
      
We know  = XH             X = Susceptibility
      0.12                     –5
X=         =        = 0.00008 = 8 × 10
H      1500
(c) The material is paramagnetic
–3
3.   B1 = 2.5 × 10 ,                B2 = 2.5
–4   2
A = 4 × 10 m ,                 n = 50 turns/cm = 5000 turns/m
(a) B = 0ni,
–3          –7
 2.5 × 10 = 4 × 10 × 5000 × i
2.5  10 3
i=                                = 0.398 A ≈ 0.4 A
4  10  7  5000
B2         2 .5                        2 .5                        6         6
(b)  =      H =              (B 2  B1 ) =              2.497 = 1.99 × 10 ≈ 2 × 10
0      4  10  7                 4  10  7
M         m      m
(c)  =  =         =
V         A      A
6         –4
 m = A = 2 × 10 × 4 × 10 = 800 A-m
4.   (a) Given d = 15 cm = 0.15 m
ℓ = 1 cm = 0.01 m
2         –4 2
A = 1.0 cm = 1 × 10 m
–4
B = 1.5 × 10 T
M=?
            2Md
We Know B = 0  2
4  ( d   2 )2

–4     10 7  2  M  0.15             3  10 8 M
 1.5 × 10 =                                       =
(0.0225  0.0001)     2
5.01 10  4
1.5  10 4  5.01 10 4
M=                                       = 2.5 A
3  10  8
M         2 .5                6
(b) Magnetisation  =         =                 = 2.5 × 10 A/m
V   10  4  10  2
m           M                      2 .5
(c) H =           =            =
4d2         4d2         4  3.14  0.01 (0.15)2
2
net H = HN + H = 2 × 884.6 = 8.846 × 10
                       –7       6
B = 0 (–H + ) = 4 × 10 (2.5×10 – 2 × 884.6) ≈ 3,14 T

37.1
Magnetic Properties of Matter
5.   Permiability () = 0(1 + x)
Given susceptibility = 5500
–7
 = 4 × 10 (1 + 5500)
–7                 –7         –3
= 4 × 3.14 × 10 × 5501 6909.56 × 10 ≈ 6.9 × 10
6.   B = 1.6 T,           H = 1000 A/m
 = Permeability of material
B    1 .6            –3
=       =       = 1.6 × 10
H   1000
    1.6  10 3             4          3
r =      =             = 0.127 × 10 ≈ 1.3 × 10
0   4  10  7
 = 0 (1 + x)

x=         1
0
3                           3
= r – 1 = 1.3 × 10 – 1 = 1300 – 1 = 1299 ≈ 1.3 × 10 
C    x   T
7.   x=       = 1 = 2
T    x2  T1

1.2  10 5          T2
               5
=
1.8  10            300
12
 T2 =      300 = 200 K.
18
28         3
8.    = 8.52 × 10 atoms/m
3
For maximum ‘’, Let us consider the no. of atoms present in 1 m of volume.
–24     2
Given: m per atom = 2 × 9.27 × 10 A–m
net m                 –24             28        6
=         = 2 × 9.27 × 10 × 8.52 × 10 ≈ 1.58 × 10 A/m
V
B = 0 (H + ) = 0         [ H = 0 in this case]
–7            6            –1
= 4 × 10 × 1.58 × 10 = 1.98 × 10 ≈ 2.0 T
B
9.   B = 0ni,       H=
0
Given n = 40 turn/cm = 4000 turns/m
 H = ni
H = 4 × 104 A/m
H   4  10 4
i=        =          = 10 A.
n    4000



37.2

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