Question 19.12

When a metal is heated its density decreases. This is for two reasons: thermal expansion of the solid; and the formation of vacancies. Consider a specimen of copper at 20 C with a density of 8940 kg/m³.

1. Determine its density on heating to 1000 C if only thermal expansion is concerned.
2. Repeat the calculation including the formation of vacancies. Assume that the energy of vacancy formation is 0.9 eV/atom, and that the volume coefficient of thermal expansion, alpha_v, is three times the linear coefficient, alpha_l.

Answer 19.12

In Part 1, the linear coefficient of expansion for copper is 17 * 10^-6 /C. (From Table 19.1, sixth edition). So the volume coefficient of expansion is three times this, or 51 * 10^-6 /C. So if we originally have a cubic metre of copper, at 1000 C we will have 1 + 51 * 980 * 10^-6 cubic metres, and the density will be 8940/(1+ 51 * 980 * 10^-6) = 8514 kg/m³.

In Part 2, the number of vacancies is given by Equation 4.1,

N_v = N exp(-Q_v/kT)

It follows that the density at 1273 K will be related to the density at room temperature by:

Density₁₂₇₃= density₂₇₃ * (1 - exp(-Q_v/k * 1273))

if we take no account of the thermal expansion. (Note that we don't need to know anything about the size or number of atoms in a unit cell.)

Evaluating this expression, we find that the density is reduced due to hole formation by a factor of 0.9997; so if we include both expansion and hole formation, the final density is 0.9997 * 8514 kg/m³, or 8512 kg/m³. (We note that hole formation is a relatively minor factor in density reduction.)

Question 19.26

Derive Equation 19.8

sigma = E * alpha_l * (T₀-T_f)

from Equation 19.3:

(l_f - l₀)/l₀ = alpha_l(T₀-T_f)

Answer 19.26

Equation 19.8 says that the stress resulting from a temperature change in a rod clamped at both ends increases with Young's modulus, the linear expansion coefficient, and the temperature change.

We could think of the rod as being allowed to expand freely during the temperature change, then being compressed back to its original length by an imposed force. If the rod is originally of length l, then it expands by Delta l = l alpha_l Delta T. So after compression, the strain is Delta l/l = alpha_l Delta T, and the stress is therefore E * strain = E alpha_l Delta T

Question 19.D2

The ends of a cylindrical rod 6.4 mm in diameter and 250 mm long are mounted between rigid supports. The rod is stress-free at 20 C, and can withstand a tensile stress of up to 138 MPa. If the rod has to survive being cooled to - 60 C, which of the following materials could it be made from: aluminium, copper, brass, 1025 steel, or tungsten?

Answer 19.D2

This involves using the equation we've just derived. If we substitute in the material values for these five materials, we find that the thermally induced tensile stresses are, in MPa, 130.2, 149.6, 155.2, 198.72 and 146.52 respectively. Only the first of these, corresponding to aluminium, is below the limiting stress for the application.

Question 20.2

A coil of wire 0.1m long and having 15 turns carries a current of 1.0A.

1. Compute the flux density if the coil is in a vacuum
2. A bar of an iron-silicon alloy, the B-H behaviour of which is shown in Figure 20.24, is placed in the coil. What is the flux density in the bar?
3. If the bar of iron-silicon is replaced by a bar of molybdenum, what current must be used to produce the same field in the molybdenum as was produced in the iron-silicon alloy by the 1.0-A current?

Answer 20.2

(This was marked out of 3.)

1. We calculate the flux density from Equations 20.1 and 20.2:

   B = mu H = mu N I / l = 4 pi * 10 ^-7 * 15 * 1 / 0.1 = 4 pi * 10 ^-7 * 150 = 1.88 * 10^-4 Tesla

2. If we put a bar of iron-silicon alloy in the coil, the value of H stays the same, namely, 150 A/m. From Figure 20.24, a field of 150 A/m induces a flux of 1.65 tesla in the alloy.

3. Molybdenum is paramagnetic, so the flux density in the molybdenum bar is

   B = mu₀H + mu₀M = (mu₀ + mu₀chi_m))H = (4 pi * 10^-7 * (1 + 1.19 * 10^-4))H

   which is approximately 4 pi * 10 ^-7 H. So to produce a flux of 1.65 tesla, we would require an H of:

   H = 1.65 / (4 pi * 10 ^-7) = 1.313 * 10⁶ A-m.

   So NI/l = 1.313 *10⁶, from which we get that I = 1.313 * 10⁶ * 0.1 / 15 = 8.73 kA

Question 20.7

The magnetization within a bar of some metal alloy is 1.2 * 10⁶ A/m at an H field of 200 A/m. Compute the magnetic susceptibility, permeability and magnetic flux density within the material. What types of magnetism are displayed by the material? What is the evidence for this?

Answer 20.7

(Marked out of 4)

The relevant Equation is 20.6:

M = chi_mH

So chi_m = M/H = 1,200,000/200 = 6,000 (unitless)

The relative permeability of the material can be calculated from Equation 20.7:

mu_r = chi_m + 1 = 6,001

So the actual permeability is, from Equation 20.4,

mu = mu_r mu₀ = 6,001 * 4 pi * 10^-7 = 0.0075

And the magnetic flux density is (Equation 20.2):

B = mu H = 0.0075 * 200 = 1.5082 Tesla.

We can conclude that the material is ferromagnetic, because of its high relative permeability.

Question 20.10

Consider a hypothetical metal having ferro-magnetic behaviour, a simple cubic crystal structure, an atomic radius of 0.125 nm, and a saturation flux density of 0.85 tesla. Find the number of Bohr magnetons per atom for this material.

Answer 20.10

We note that (p. 681) the saturation magnetization is the product of the net magnetic moment per atom and the number of atoms present.

M_s=n mu_B N

where n is the number of magnetons per atom, B is a Bohr magneton, and N is the number of atoms per cubic metre.

For the simple cubic structure, a unit cell is of side length 2r, and contains one atom. So the density of the material is 1 atom per (0.25 * 10^-9)³ cubic metres or 6.4 * 10²⁸ atoms per cubic metre.

So n = M_s/(mu_BN) = 1.14 Bohr magnetons per atom.

Question 20.21

Schematically sketch on a single plot the B-H behaviour for a ferromagnetic material at 0 K, and at temperatures just below and just above its Curie temperature. Briefly explain why these curves have different shapes.

Answer 20.21

The most helpful figures from the text are Figure 20.6 and Figure 20.10. Note that above the Curie temperature, ferromagnetic materials become paramagnetic. So the figure should look something like this:

(See q2021.jpg)

(The saturation magnetization drops as we approach the Curie temperature, and becomes very small beyond it.)

It's a mistake to sketch the above-Curie curve as a straight line going up to the same height as the hysteresis loop for 0 K; above the Curie temperature, the magnetization is typically several thousand times less than it is at absolute zero, so the above-Curie line should lie very close to the x-axis.

Question 20.28

An iron bar magnet having a coercivity of 4,000 A/m is to be demagnetised. If the bar is inserted in a cylindrical wire coil 0.15 m long and having 100 turns, what electric current is required to generate the necessary magnetic field?

Answer 20.28

This is an extremely easy question: to demagnetise the magnet, we must apply a field of strength H_c=4,000 A/m.

So H = NI/l implies that I = Hl/N = 4000 * 0.15 / 100 = 6 amps.