Formula: Ettingshausen Effect Temperature gradient Electric Current Density Magnetic flux density (B-field) Ettingshausen coefficient
$$\frac{\text{d} T}{\text{d} x} ~=~ C_{\text E} \, \class{red}{j_{\text y}} \, \class{violet}{B_{\text z}}$$
$$\frac{\text{d} T}{\text{d} x} ~=~ C_{\text E} \, \class{red}{j_{\text y}} \, \class{violet}{B_{\text z}}$$
$$\class{red}{j_{\text y}} ~=~ \frac{ 1 }{ C_{\text E} \, \class{violet}{B_{\text z}} } \, \frac{\text{d} T}{\text{d} x}$$
$$\class{violet}{B_{\text z}} ~=~ \frac{ 1 }{ C_{\text E} \, \class{red}{j_{\text y}} } \, \frac{\text{d} T}{\text{d} x}$$
$$C_{\text E} ~=~ \frac{ 1 }{ \class{violet}{B_{\text z}} \, \class{red}{j_{\text y}} } \, \frac{\text{d} T}{\text{d} x}$$
Temperature gradient
$$ \frac{\text{d} T}{\text{d} x} $$
Temperature difference in a current-carrying conductor which is in a magnetic field \(B_{\text z}\). The temperature difference is caused by the deflected electrons in the magnetic field due to the Lorentz force. Because the slow electrons are deflected more than fast ones, one side of the conductor (e.g. a Hall plate) becomes cooler than the other.
The temperature gradient in this case arises in the \(x\) direction.
Electric Current Density
$$ \class{red}{j_{\text y}} $$ Unit $$ \frac{ \mathrm A }{ \mathrm{m}^2 } $$
Electric current through the conductor, which in this case is directed in the \(y\) direction. This current is deflected in the magnetic field due to the Lorentz force, which is why a temperature gradient is created.
Magnetic flux density (B-field)
$$ \class{violet}{B_{\text z}} $$ Unit $$ \mathrm{T} = \frac{\mathrm{kg}}{\mathrm{A} \, \mathrm{s}^2} $$
The magnetic field penetrates the current-carrying conductor perpendicularly and in this case points in the \(z\) direction.
Ettingshausen coefficient
$$ C_{\text E} $$
Ettingshausen coefficient is a material-specific quantity. This coefficient determines how well the temperature gradient can be formed due to the external magnetic field.