TITLE: Is there an intuitive explanation for why Lorentz force is perpendicular to a particle's velocity and the magnetic field?
QUESTION [53 upvotes]: The Lorentz force on a charged particle is perpendicular to the particle's velocity and the magnetic field it's moving through. This is obvious from the equation:
$$ \mathbf{F} = q\mathbf{v} \times \mathbf{B} $$
Is there an intuitive explanation for this behavior? Every explanation I've seen simply points at the equation and leaves it at that.
I can accept mathematically why $\mathbf{F}$ will be perpendicular to $\mathbf{v}$ and $\mathbf{B}$ (assuming the equation is correct, which it is of course). But that doesn't help me picture what's fundamentally going on.
Trying to create an analogy with common experiences seems useless; if I were running north through a west-flowing "field" of some sort, I wouldn't expect to suddenly go flying into the sky.
I'm hoping there's a way to visualize the reason for this behavior without a deep understanding of advanced theory. Unfortunately, my searching for an explanation makes it seem like something one just has to accept as bizarre until several more years of study.

REPLY [1 votes]: For intuition perhaps you can think about it from an experimental perspective.
If you look at a charged particle moving in a uniform magnetic field the motion is circular. For example looking at the path of a particle in a cloud chamber.

You can then do things like measure the radius and since it moving in a circle, you realize that there needs to be an inward force that is at right angles to the direction of the velocity motion.

Since there is no other charge present at the center of the circle for the particle to orbit around, or mass or gravitation force for the particle to orbit around something must be generating the force (we call it the Lorentz force).
Thinking back to your classical mechanics you know the
You can do other experiments like change the mass of the charge particle and see that radius of the circle changes. When you do that you find the radius is
$\rho=\frac{mv}{qB}$
or thinking about the acceleration for a particle of mass in a circle is $\frac{mv^2}{\rho}$ you come up with the force
$f=qvB = \frac{mv^2}{\rho}$
At that point you f= qvB is the same as qvBsin(90) since the angle is 90 degrees and that the definition of the cross product can be written as
$\vec{v} \times \vec{B} = |v||B|\sin\theta\hat{n}$
And you do some more experiments and find out that the angle between the velocity of the particle  and magnetic field matters and writing the vector equations in terms of a cross product is a nice short hand for that.
So $F=q(\vec{v} \times \vec{B})$
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