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IZhO 2014, Теория — Задача 3. Metamaterials

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X International Zhautykov Olympiad/Theoretical Competition — THEORETICAL COMPETITION, January 14, 2014 · 14 януари 2014 г. · 10 т.

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The time available for the theoretical competition is 4 hours. There are three questions.
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Условие

Metamaterials are composite materials whose properties are due not so much to the properties of its constituent elements but due to artificially tailored periodic structures. Metamaterials are synthesized in modern nanolaboratories by implantation different periodic structures with a variety of geometric shapes into the original natural material, which substantially modifies its physical properties. In a very rough approximation, those implants can be treated as artificially made atoms of extremely large size immersed into the original material. While synthesizing the metamaterial Ddveloper has the opportunity of varying various free parameters (structure sizes and constant or varying period between them, etc.).

Photograph of a metamaterial: a gold-coloured plate with a regular square grid of holes; inset: a grey rectangular block with a gold-coloured end face.

In one nanolaboratory the metamaterial has been manufactured in the form of a wire of the length $L = 5.00\mathrm{cm}$ and radius $R = 1.00\mathrm{mm}$ , whose conductivity depends on the distance from its axis according to the law $\sigma_0 = \beta r$ . Physical properties of the wire have been experimentally measured and are presented in the following table:

PHYSICAL PROPERTY	NUMERICAL VALUE
Conductivity $\sigma_0 = \beta r$	$\beta = 1.00\times 10^9\,\mathrm{S/m^2}$
Heat transfer coefficient	$\alpha = 20\mathrm{W/(m^2\cdot K)}$
Thermal conductivity coefficient	$\kappa = 0{,}01\mathrm{W/(m\cdot K)}$
Young's modulus	$E = 1.00\times 10^7\,\mathrm{Pa}$
Linear expansion coefficient	$\gamma = 1.00\times 10^{-6}\,\mathrm{K^{-1}}$

1. [1.0 points] Find an analytic formula for the total resistance $R_0$ of the wire, and calculate its numerical value.

An electric current $I = 1A$ is made to pass through the wire. It is known that the heat exchange with the environment obeys the Newton-Richman law,

P_{ext} = \alpha(T_s - T_0),

where $P_{ext}$ stands for the power loss per unit surface of the wire with the surface temperature $T_s$ , $T_0 = 293K$ denotes the ambient temperature and $\alpha$ is a constant, called the heat transfer coefficient.

2. [1.0 points] Find an analytic formula for the surface temperature $T_s$ of the wire and calculate its numerical value.

The wire temperature varies with the depth due to the phenomenon known as thermal conductivity, which is described by the Fourier law

P = -\kappa S\frac{\Delta T}{\Delta x},

where $P$ designates the power of the heat flow between the opposite faces of the parallelepiped with the square $S$ , $\Delta T$ is the temperature difference between the faces of the parallelepiped situated at a distance $\Delta x$ from each other, and $\kappa$ is called the heat transfer coefficient.

3. [2.5 points] Find an analytic formula for the temperature $T_{\max}$ in the center of the wire, and calculate its numerical value.

4. [0.5 points] Find an analytic formula for the change $\delta R_T$ of the wire radius due to its thermal expansion and calculate its numerical value.

Attention! In all further calculations assume that the wire is infinitely long.

5. [0.5 points] Find an analytic formula for the magnetic induction inside the wire as a function of the distance $r$ from its axis.

6. [1.0 points] Find an analytic formula for the energy of the magnetic field inside the wire, and calculate its numerical value.

7. [1.0 points] The electric current causes an appearance of mechanical stress in the wire. Find an analytic formula for the pressure $p(r)$ inside the wire as a function of the distance $r$ from its axis.

8. [1.0 points] Find an analytic formula for the mechanical stress energy $W_{\sigma}$ of the wire, and calculate its numerical value.

9. [1.0 points] Find an analytic formula for the change $\delta R_{\sigma}$ of the wire radius due to its mechanical stress, and calculate its numerical value.

10. [0.5 points] Find the value of the thermal expansion coefficient $\gamma$ such that the total change of the wire radius would be zero when an electric current was passing through it.

Help! The value of the magnetic constant is $\mu_0 = 4\pi\cdot 10^{-7}\,\text{Гн}/\text{м}$ .

Отговори

Решение

Consider the conducting layer disposed radially at the interval $[r, r + dr]$ . Its conductivity $d\rho$ is

d\rho = \sigma_0\frac{dS}{L} = \beta r\frac{2\pi rdr}{L}, \qquad (1)

and, hence, the total conductivity is given by

\rho = \int_0^R d\rho = \frac{2\pi\beta R^3}{3L}. \qquad (2)

Thus, the resistance of the wire is found as

R_0 = \frac{1}{\rho} = \frac{3L}{2\pi\beta R^3} = 2.39\times 10^{-2}\mathrm{Ohm}. \qquad (3)

The amount of heat generated in the wire per unit time is determined by Joule law

P_I = I^2R_0. \qquad (4)

In steady regime, the same amount of heat must be removed through the surface of the wire into the environment, therefore, according to the Newton-Richman law

P_I = 2\pi RLP_{ext} = 2\pi\alpha RL(T_s - T_0), \qquad (5)

whence

T_s = T_0 + \frac{3I^2}{4\pi^2\alpha\beta R^4} = 297\mathrm{K}. \qquad (6)

Consider a cylinder of radius $r$ . Let us find an amount of heat generated per unit time inside that cylinder. To do this, let us find the electric field strength in the wire. According to Ohm's law, the current density is

j = \sigma_0E, \qquad (7)

therefore, the total current can be written as

I = \int_0^r j2\pi rdr = E\int_0^r \sigma_02\pi rdr = \frac{2\pi R^3\beta E}{3}. \qquad (8)

Hence

E = \frac{3I}{2\pi\beta R^3}. \qquad (9)

The electric power generated in the cylinder is determined by the Joule law in differential form

P_r = \int_0^r \sigma_0E^22\pi rLdr = \frac{3I^2Lr^3}{2\pi\beta R^6}. \qquad (10)

It is evident that the power dissipated inside the cylinder must be taken away through the surface of the cylinder, thus,

P_r = P = -\kappa 2\pi rL\frac{dT}{dr}. \qquad (11)

Solving differential equation (11), using (10) together with the initial condition

T(R) = T_s, \qquad (12)

the following solution is obtained in the form

T(r) = T_0 + \frac{I^2(\alpha R^3 + 3\kappa R^2 - \alpha r^3)}{4\pi^2\alpha\beta\kappa R^6}. \qquad (13)

Thus, the temperature in the center of the wire is

T_{\max} = T_0 + \frac{I^2(\alpha R^3 + 3\kappa R^2)}{4\pi^2\alpha\beta\kappa R^6} = 299\mathrm{K}. \qquad (14)

The radius change of the wire is determined by the law of thermal expansion of solids and can be written as

\delta R_T = \int_0^R \gamma\left[T(r) - T_0\right]dr = \frac{3\gamma(\alpha R + 4\kappa)I^2}{16\pi^2\alpha\beta\kappa R^3} = 5.70\times 10^{-9}\,\mathrm{m}. \qquad (15)

The magnetic field induction is determined by the circulation theorem, which, in this case, is written as

B2\pi r = \int_0^r j2\pi rdr = E\int_0^r \sigma_02\pi rdr. \qquad (16)

Using expression (9), we finally obtain

B(r) = \frac{\mu_0Ir^2}{2\pi R^3}. \qquad (17)

The energy density of the magnetic field is given by

w_B(r) = \frac{B^2(r)}{2\mu_0}, \qquad (18)

therefore, the energy of the magnetic field inside the wire

W_B = \int_0^R w_B(r)2\pi rLdr = \frac{\mu_0I^2L}{24\pi} = 8.33\times 10^{-10}\,\mathrm{J}. \qquad (19)

Let us write the equilibrium condition for the wire layer of small width $l$ and length $L$ , disposed at the interval $r, r + dr$ . The total Ampere force acting on this layer is written as

dF_A = jB(r)Lldr. \qquad (20)

Hence, the pressure difference is obtained as

dp(r) = \frac{dF_A}{lL} = \frac{3\mu_0I^2r^3}{4\pi^2R^6}dr. \qquad (21)

Taking into consideration that the pressure at the wire pressure is zero, one gets

p(r) = \frac{3\mu_0I^2(R^4 - r^4)}{16\pi^2R^6}. \qquad (22)

As a result of the mechanical pressure the mechanical stress appears in the crystal lattice whose energy density is determined by the expression

w_{\sigma} = \frac{\sigma^2}{2E} = \frac{p^2(r)}{2E}, \qquad (23)

thus, the total energy of mechanical deformations is found as

W_{\sigma} = \int_0^R w_{\sigma}2\pi rLdr = \frac{3\mu_0^2I^4L}{320E\pi^3R^2} = 2.39\times 10^{-18}\,\mathrm{J}. \qquad (24)

The radius change of the wire is determined by Hooke's law, which, in this case, can be written in the form

\varepsilon = \frac{\sigma}{E} = \frac{p(r)}{E}, \qquad (25)

where $\varepsilon$ is the relative change in radius.

Thus, the radius change due to mechanical stress is found as

\delta R_{\sigma} = \int_0^R \varepsilon dr = \frac{1}{E}\int_a^R p(r)dr = \frac{3\mu_0I^2}{20\pi^2ER} = 1.91\times 10^{-12}\,\mathrm{m}. \qquad (26)

Comparing expressions (15) and (25) we obtain

\gamma = \frac{4\mu_0\alpha\beta\kappa R^2}{5E(\alpha R + 4\kappa)} = 3.35\times 10^{-10}\,\mathrm{K^{-1}}. \qquad (27)

Grading scheme

№	Content	Points
1	Formula (1)	0.25	1.0
	Formula (2)	0.25
	Formula (3)	0.25
	Correct numerical numerical value in (3)	0.25
2	Formula (4)	0.25	1,0
	Formula (5)	0.25
	Formula (6)	0.25
	Correct numerical numerical value in (6)	0.25
3	Formula (7)	0.25	2.5
	Formula (8)	0.25
	Formula (9)	0.25
	Formula (10)	0.25
	Formula (11)	0.25
	Formula (12)	0.25
	Formula (13)	0.5
	Formula (14)	0.25
	Correct numerical numerical value in (14)	0.25
4	Formula (15)	0.25	0.5
	Correct numerical numerical value in (15)	0.25
5	Formula (16)	0.25	0.5
	Formula (17)	0.25
6	Formula (18)	0.5	1.0
	Formula (19)	0.25
	Correct numerical numerical value in (19)	0.25
7	Formula (20)	0.25	1.0
	Formula (21)	0.25
	Formula (22)	0.5
8	Formula (23)	0.5	1.0
	Formula (24)	0.25
	Correct numerical numerical value in (24)	0.25
9	Formula (25)	0.5	1.0
	Formula (26)	0.25
	Correct numerical numerical value in (26)	0.25
10	Formula (27)	0.25	0.5
	Correct numerical numerical value in (27)	0.25
Total			10,0

Оригинал в Архива: IZhO-2014-Theory_eng.pdf · официални решения: IZhO-2014-Theory_eng_sol.pdf