# Is there a more clever way to apply the cross product of two vectors to magnetism?

I am just beginning to learn magnetism and my book used two ways to define the force caused by the magnetic field, brushing over the latter. The first:

$$F = q v B \sin (\theta).$$

And:

$$\vec{F} = q(\vec{v} \times \vec{B})$$

where the $\times$ is the cross product. After looking up the cross product, I found that it was defined as:

$\vec{a} \times \vec{b} = |\vec{a}||\vec{b}| \sin (\theta) \vec{n}$ where $\vec{n}$ is the unit vector found via the "right hand rule."

While I have no doubt the right hand rule is a useful tool, I wonder if there is a more "on paper" way to find the direction of the force a magnetic field applies to a charged particle. Basically, shouldn't there be a way to avoid the right-hand rule and to do it all via equations?

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-1: This is trivially answered by the wikipedia article on cross product –  user2963 Mar 7 '12 at 2:16
@zephyr i'd rather sugges the wiki page on oriented coordinate systems: en.wikipedia.org/wiki/… –  luksen Mar 7 '12 at 3:15
@zephyr I guess I thought that there might be a quicker way than the jumble of variables wikipedia had. I guess the jumble is inherent in matrices. (I am pre-calculus). –  Eric Thoma Mar 7 '12 at 3:21

Any equations you set up will be uglier than the cross product (they will basically by expansions of $\times$)

Remember, vectors only can define a direction. The issue with the magnetic field is that it is heavily related to planes, and is a perpendicular force. We can't assign a magnetic field analogous to the electric field (as in: a field which shows the direction of force), as it depends upon the direction of velocity.

The cross product enables us to define a plane with a vector perpendicular to it. Aside from that, it gives perpendicular stuff. This clearly points to the use of the cross product.

Besides, the cross product isn't that dirty. It makes stuff simpler, if anything.

If you want to do it all via equations, use this:

$$\vec{a}\times\vec{b}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \\ \end{vmatrix}$$ These rules may help you avoid the determinant form for simpler cross products: $$\hat{i}\times\hat{j}=\hat{k};\:\hat{j}\times\hat{k}=\hat{i};\:\hat{k}\times\hat{i}=\hat{j};\:\vec{a}\times\vec{a}=0;\:\vec{a}\times\vec{b}=-\vec{b}\times\vec{a}$$

Over here, the right hand rule is implied by our choice of a right handed system of $\hat{i} , \hat{j} , \hat{k}$ or $x,y,z$. So you don't need to touch it, it is already inherent in the coordinate system.

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Thanks for the great answer. –  Eric Thoma Mar 7 '12 at 3:20