37.MAGNETIC PROPERTIES OF MATTER

CHAPTER – 37

MAGNETIC PROPERTIES OF MATTER

1.

2.

3.

B = 0ni,

H=

B 0

 H = ni  1500 A/m = n× 2  n = 750 turns/meter  n = 7.5 turns/cm (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   X = Susceptibility We know  = XH  0.12 –5 X= = = 0.00008 = 8 × 10 H 1500 (c) The material is paramagnetic –3 B2 = 2.5 B1 = 2.5 × 10 , –4 2 n = 50 turns/cm = 5000 turns/m A = 4 × 10 m , (a) B = 0ni, –3 –7  2.5 × 10 = 4 × 10 × 5000 × i i= (b)  =

2.5  10 3 4  10  7  5000

= 0.398 A ≈ 0.4 A

B2 2 .5 2 .5 6 6 H =  (B 2  B1 ) =  2.497 = 1.99 × 10 ≈ 2 × 10 0 4  10  7 4  10  7

M m m = = V A A 6 –4  m = A = 2 × 10 × 4 × 10 = 800 A-m (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 (c)  =

4.

–4

 1.5 × 10 = M=

10 7  2  M  0.15 (0.0225  0.0001)

1.5  10 4  5.01 10 4 3  10  8

(b) Magnetisation  = (c) H =

m 4d2

=

2

=

3  10 8 M 5.01 10  4

= 2.5 A

2 .5 M 6 = = 2.5 × 10 A/m V 10  4  10  2

M 4d2

=

2 .5 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.

6.

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 B = 1.6 T, H = 1000 A/m  = Permeability of material

B 1 .6 –3 = = 1.6 × 10 H 1000

= r =

 1.6  10 3 4 3 = = 0.127 × 10 ≈ 1.3 × 10 0 4  10  7

 = 0 (1 + x)

 1 0

x=

3

3

= r – 1 = 1.3 × 10 – 1 = 1300 – 1 = 1299 ≈ 1.3 × 10  7.

x= 

x T C = 1 = 2 x2 T1 T

1.2  10 5 1.8  10

5

=

T2 300

12  300 = 200 K. 18 28 3  = 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  T2 =

8.

net m –24 28 6 = 2 × 9.27 × 10 × 8.52 × 10 ≈ 1.58 × 10 A/m V [ H = 0 in this case] B = 0 (H + ) = 0 –7 6 –1 = 4 × 10 × 1.58 × 10 = 1.98 × 10 ≈ 2.0 T =

9.

B = 0ni,

H=

B 0

Given n = 40 turn/cm = 4000 turns/m  H = ni H = 4 × 104 A/m i=

4  10 4 H = = 10 A. n 4000

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37.2