Background : I am Fresher Mathematics Student in University. We are currently learning cross products. I am struggling with this question.
Questions
- If a charged particle of charge $q$ is travelling with a velocity $\bf v$ in a magnetic field $\bf B$, then the force that the charged particle feels is given by $$\textbf{F}=q\bf{v\times B}$$ In this case, the force $\bf F$ is also a vector quantity, since it has both the magnitude and the direction. So, the cross product plays an important role in physics and engineering.
Now suppose that a proton with some positive charge $q$ is travelling in the $xy$-plane with a velocity in the direction of the vector $$\textbf v = \begin{pmatrix}3\\-1\\0\end{pmatrix}$$ and that the magnetic field $\bf B$ is a uniform field pointing straight up in the $z-$direction, perpendicular to the $xy-$plane. Then, Find the direction of the force that the moving proton feels. - This
My Attempt: I am aware that the result of the cross product of two vectors is such a vector that if I dot multiply with the either of the original vectors I get zero, Which means the cross product is perpendicular. I don't know how using that example how I can solve the problem.
I would appreciate any help in the questions. Thanks!
Best Answer
Take $\vec{B} = \begin{pmatrix}0 \\ 0\\ k\end{pmatrix} k>0$
Now, $\vec{F} = q\vec{v} \times \vec{B} = q\begin{pmatrix}3 \\-1\\ 0\end{pmatrix}\times\begin{pmatrix}0 \\ 0\\ k\end{pmatrix} = -q\begin{pmatrix}k \\ 3k\\ 0\end{pmatrix}$
Try taking dot products $\vec{F}\cdot\vec{v}$ and $\vec{F}\cdot\vec{B}$