Assignment 4 − with Outline Answers

PHYSICS 208 – 2 nd TERM
Introduction to Materials Science
Assignment 4 − with Outline Answers
Date of distribution: Tuesday, March 27, 2007
Date for solutions to be handed in: Thursday, April 5, 2007
The questions are all taken from the recommended textbook by Callister (6 th edition).
1. [Question 20.7] 4 marks
(a) We calculate the magnetic susceptibility within a bar of metal alloy when M = 1.2 x 10 6 A/m
and H =200 A/m. Solving for χ m from Eq. (20.6) gives
6
M 1.2 × 10 A/
m
χm
= = = 6000
H 200 A/
m
(b) In order to calculate the permeability we employ Eqs. (20.4) and (20.7) as follows:
µ = µ µ = µ 1+
χ
( )
( )
r o o m
= 1.257 × 10 -6 H/m (6000 + 1) = 7.54 × 10 -3 H/m
(c) The magnetic flux density may then be found from Eq. (20.2) as
-3
B = µ H = 7.54 × 10 H/m (200 A/m) = 1.51 tesla
( )
(d) This metal alloy would exhibit ferromagnetic behavior on the basis of the value of its χ m
(6000), which is considerably larger than typical χ m values for diamagnetic and paramagnetic
materials. Also M is nonzero and large.
2. [Question 20.9] 4 marks
We show that there are 2.2 Bohr magnetons associated with each Fe atom. Let n B
'
be the number
of Bohr magnetons per atom, which we will calculate. By definition of magnetization
n B
'
= M s
µ B
N

where N is just the number of atoms per cubic meter. This found from the number of atoms per
unit cell (two for BCC) divided by the unit cell volume, so
2 2
N = =
3
V a
with a being the BCC unit cell edge length (i.e., 0.2866 nm). Thus
=
n B
'
C
= M s
µ B
N = M s a3
2µ B
3
6 −9
( 1.70 × 10 / )
⎡( 0.2866 × 10 ) /
⎢⎣
−24 2
(2 atoms / unit cell) ( 9.27 × 10 A- m / BM )
A m m unit cell⎤
⎥⎦
= 2.16 BM/atom
3. [Question 20.10] 4 marks
We want the number of Bohr magnetons per atom of a hypothetical metal that has a SC crystal
structure, an atomic radius of 0.125 nm, and a saturation flux density of 0.85 tesla. From Eqs.
(20.8) and (20.11) it follows that
M s = B s
µ o
= n B µ B
V C
For a SC crystal structure the cell volume V C
= (2r) 3 , where r is the atomic radius. Substituting
this relationship into the above equation and solving for n B
yields
n =
B
3
s ( 8r
)
µ µ
−9
3
(0.85 tesla)(8) ( 0.125 × 10 m)
−6 −24 2
( 1.257 × 10 H / m)( 9.27 × 10 A- m / BM )
B
= = 1.14 Bohr magnetons/atom
o
B
4. [Question 20.25] 5 marks
(a) The saturation flux density for the steel (using the B-H behavior shown in Fig. 20.25) is 1.30
tesla, the maximum B value shown on the plot.
(b) The saturation magnetization is found from Eq. (20.8) as
=
M s = B s
µ o
1.30 tesla
1.257 x 10 −6 H /m = 1.03 x 106 A/m
2

(c) The remanence B r
is read from this plot (Fig. 20.25), giving the value is 0.80 tesla.
(d) The coercivity H c
is read from this plot as 80 A/m.
(e) On the basis of Tables 20.5 and 20.6, this is most likely a soft magnetic material. The
saturation flux density (1.30 tesla) lies within the range of values cited for soft materials.
Although the remanence (0.80 tesla) is close to the values given in Table 20.6 for hard
magnetic materials, the H c is significantly lower than for hard magnetic materials. Finally an
estimate the area of the hysteresis loop gives ~ 250 J/m 3 , which is in line with the hysteresis
loss per cycle for soft magnetic materials.
5. [Question 20.28] 3 marks
To demagnetize a magnet having a coercivity of 4000 A/m, an H field of 4000 A/m must be
applied in a direction opposite to that of magnetization. According to Eq. (20.1)
I = H l
N
(4000 A / m) (0.15 m)
= = 6.0 A
100 turns
6. [Question 20.33] 5 marks
We are asked to find which of the SC elements in Table 20.7 are superconducting at 2 K and in a
magnetic field of 40,000 A/m. First of all, in order to be SC at 2 K within any magnetic field, the
critical temperature must be greater than 2 K. Thus, aluminum, titanium, and tungsten cannot be
SC at this temperature.
Then, for each of lead, mercury, and tin, it is necessary, using Eq. (20.12), to find the value of
H c (T). If this value is greater than 40,000 A/m then the element will be SC.
Hence, for Pb
2
B
C(0)
⎡ T ⎤
H
C( T) = ⎢1
−
2 ⎥
µ
0 ⎣ TC
⎦
0.0803 tesla ⎡ (2.0 K)
⎤
= ⎢ − ⎥
1.257 × 10 H / m ⎣ (7.19 K)
⎦
2
4
1 = 5.89 x 10 A/m
−6 2
3

Since this value is greater than 40,000 A/m, lead will be SC.
For Hg
2
0.0411 tesla ⎡ (2.0 K)
⎤
4
H
C( T ) = 1 = 2.51 x 10 A/m
−6 ⎢ −
2 ⎥
1.257 × 10 H / m ⎣ (4.15 K)
⎦
This value is less than 40,000 A/m, so mercury will not be SC.
For Sn
2
0.0305 tesla ⎡ (2.0 K)
⎤
3
H
C( T ) = 1 = 1.73 x 10 A/m
−6 ⎢ −
2 ⎥
1.257 × 10 H / m ⎣ (3.72 K)
⎦
Therefore, tin is not SC.
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