Question

$\mathit{F}\mathbf{=}\mathit{q}\mathbf{(}\mathit{E}\mathbf{+}\mathit{v}\mathbf{\times }\mathit{B}\mathbf{)}$

Short answer

B is an axial vector.

Step-by-step solution

Given information.

Physics definitions are given.

Definition of Biot-Savart law.

Define the Biot Savart law.

$\mathit{B}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}{\mathbf{\int }}_{\mathbf{\Omega }}\mathbf{}\frac{\mathbf{J}\mathbf{\left(}{\mathbf{r}}^{\mathbf{\text{'}}}\mathbf{\right)}\mathbf{\times }\mathbf{(}\mathbf{r}\mathbf{-}{\mathbf{r}}^{\mathbf{\text{'}}}\mathbf{)}}{\mathbf{|}\mathbf{r}\mathbf{-}{\mathbf{r}}^{\mathbf{\text{'}}}{\mathbf{|}}^{\mathbf{3}}}{\mathit{d}}^{\mathbf{3}}{\mathit{r}}^{\mathbf{\text{'}}}$

Write about the nature of various vectors.

Write about the nature of various vectors.

F, E are polar vectors. Therefore,$V\times B$is also a polar vector.

It is known that the vector product of a polar vector and an axial vector results in a polar vector. It is known that velocity is a polar vector. This implies that B is an axial vector.

Another method. Begin with the definition of the Biot Savart Law.

Define the Biot Savart Law.

$B\left(r\right)=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}{\int }_{\mathrm{\Omega }}\frac{\mathrm{J}\left({\mathrm{r}}^{\text{'}}\right)\times (\mathrm{r}-{\mathrm{r}}^{\text{'}})}{|\mathrm{r}-{\mathrm{r}}^{\text{'}}{|}^3}{d}^3{r}^{\text{'}}$

Since J is a polar vector in this equation, write the conclusion.

$J(r\text{'})\times (r-r\text{'})$

Due to the reasoning explained above, the equation above is an axial vector.

Write the final conclusion.

Since, the above vector is an axial vector, B is an axial vector.