Question

IfE= electric field andB= magnetic field, is $\mathit{E}\mathbf{\times }\mathit{B}$ a vector or a pseudovector? Comment: is called the Poynting vector; it points in the direction of transfer of energy. Does that tell you from the physics whether it is a vector or a pseudovector?

Short answer

$E\times B$ is a vector

Step-by-step solution

Given Information

E=electric field, B= magnetic field.

Definition of a cartesian tensor

The first rank tensor is just a vector. A tensor of the second rank has nine components (in three dimensions) in every rectangular coordinate system.

Find the value

The electric field vector is a vector while the magnetic field is pseudo-vector.

$$\begin{gathered} \frac{\partial E}{\partial t}=E \\ \frac{\partial B}{\partial t}=-B \end{gathered}$$

The units of the two coordinates are related.

$$\begin{gathered} \frac{\partial \left(-e_i\right)}{\partial t}=-e_i \\ \frac{\partial \left(-E_i\right)}{\partial t}=-E_i \\ \frac{\partial B_i}{\partial t}=B_i \end{gathered}$$

Calculate the cross product in two coordinate systems are given below.

$$\begin{gathered} \frac{\partial \left(E\right)}{\partial t}\times \frac{\partial \left(B\right)}{\partial t}=\left|\begin{array}{ccc}\frac{\partial \left(-e_1\right)}{\partial t}& \frac{\partial \left(-e_2\right)}{\partial t}& \frac{\partial \left(-e_3\right)}{\partial t}\\ \frac{\partial \left(E_1\right)}{\partial t}& \frac{\partial \left(E_2\right)}{\partial t}& \frac{\partial \left(E_3\right)}{\partial t}\\ \frac{\partial \left(B_1\right)}{\partial t}& \frac{\partial \left(B_2\right)}{\partial t}& \frac{\partial \left(B_3\right)}{\partial t}\end{array}\right| \\ \frac{\partial \left(E\right)}{\partial t}\times \frac{\partial \left(B\right)}{\partial t}=\left|\begin{array}{ccc}-e_1& -e_2& -e_3\\ -E_1& -E_2& -E_3\\ B_1& B_2& B_3\end{array}\right| \\ \frac{\partial \left(E\right)}{\partial t}\times \frac{\partial \left(B\right)}{\partial t}=\left|\begin{array}{ccc}{e}_1& e_2& e_3\\ E_1& E_2& E_3\\ B_1& B_2& B_3\end{array}\right| \end{gathered}$$

Hence, $E\times B$is a vector.