Question 11.6.2: A magnetised needle fixed on a vertical rod passing through ...

1. In a stationary state, the time derivatives vanish, which implies that Fick’s equation (11.53) for the conduction electrons is reduced to,

∂_t n_A = −∇· j_A.

∇· j_e = 0.

Using definition j_q = q_e j_e, taking into account the fact that ∇ q_e = 0, we obtain the stationary condition imposed on the electric current density j_q,

∇· j_q = 0.

At the junctions between metals A and B, the electric current densities are equal, i.e.   j_{qA} = j_{qB}, because there is no electric charge accumulation.

2. In the limit where the chemical potential gradients ∇μ_A and ∇μ_B of the electrons in materials A and B are negligible with respect to the electrostatic potential gradients ∇(q_e \varphi _A) and ∇(q_e \varphi _B), the second empirical relation (11.95) is reduced to,

\begin{cases}j _Q=- \kappa \nabla T +T\varepsilon j _q\\j _q = – \sigma \varepsilon \nabla T -\sigma \nabla \varphi \end{cases}.

j_{qA} = σ_A ε_A (−∇ T_A) + σ_A (−∇ \varphi_A).

j_{qB} = −σ_B ε_B ∇ T_B − σ_B ∇\varphi_B.

The integrals of the electric charge transport equations over the volume are the product of integrals over the cross-section area A times integrals over the length \ell of the thermoelectric materials,

\int_{S}{j _{qA} .dS}\int_{0}^{\ell }{dr .\hat{x} } =\sigma _A \varepsilon _A \int_{0}^{\ell }{dr.(-\nabla T_A )}\int_{S}{dS .\hat{x} } +\sigma _A\int_{0}^{\ell }{dr .(-\nabla \varphi _A )} \int_{S}{dS .\hat{x} }.

\int_{S}{j _{qB} .dS}\int_{0}^{\ell }{dr .\hat{x} } =-\sigma _B \varepsilon _B \int_{0}^{\ell }{dr.(\nabla T_B )}\int_{S}{dS .\hat{x} } -\sigma _B\int_{0}^{\ell }{dr .(\nabla \varphi _B )} \int_{S}{dS .\hat{x} }.

The electric charge transport equations integrated over the volume reduce to,

I = \sigma _A \varepsilon _A \frac{A}{\ell } \Delta T +\sigma _A \Delta \varphi _A.

I = -\sigma _B \varepsilon _B \frac{A}{\ell } \Delta T -\sigma _B \Delta \varphi _B.

Since the electric current is the same in each thermoelectric material, it can be recast as,

I = \frac{A}{2\ell }\Bigl((\sigma _A \varepsilon _A – \sigma _B \varepsilon _B)\Delta T +\sigma _A\Delta \varphi _A -\sigma _B\Delta \varphi _B\Bigr).

Thus, if metals A and B are identical, the electric current vanishes, i.e. I = 0.
3. At junctions (1) and (2), the electrostatic potentials of metals A and B are equal. Thus, the electrostatic potential difference\Delta \varphi between the junctions is the same in both metals,

\Delta \varphi = \Delta \varphi_A = \Delta \varphi_B .

Thus, the electric current in each material is recast as,

I = \frac{A}{\ell } \Bigl(\sigma _A \varepsilon _A \Delta T + \sigma _A \Delta \varphi \Bigr).

I = -\frac{A}{\ell } \Bigl(\sigma _B \varepsilon _B \Delta T + \sigma _B \Delta \varphi \Bigr).

Multiplying the first equation by σ_B and the second by σ_A and then adding them together, we obtain,

(σ_A + σ_B) I = \frac{A}{\ell } σ_A σ_B (ε_A − ε_B)ΔT.

This implies that the intensity of the electric current density j_q can be expressed in terms of the empirical parameters σ_A, σ_B, ε_A, ε_B and the temperature difference ΔT as,

I = \frac{A}{\ell } \Bigl(\frac{1}{\sigma _A}+\frac{1}{\sigma _B} \Bigr) ^{-1}(\varepsilon _A-\varepsilon _B)\Delta T.

The terms in the first bracket correspond to the equivalent resistivity of the Seebeck loop, taking into account the fact that the metals A and B are connected in series. The terms in the second bracket represent an effective Seebeck coefficient of the loop.