I assume you're interested in the expected number of picks after the first.  In that case the answer is

$${1\over6}\cdot{6\over5}+{2\over6}\cdot{6\over4}+{3\over6}\cdot{6\over3}={1\over5}+{1\over2}+1={17\over10}$$

The reasoning is simple:  In general, if an event happens with probability $p$, the expected number of trials it takes for it to occur is $1/p$.  With probability $1/6$, the event of interest is picking a non-white ball, which occurs with probability $5/6$; with probability $2/6$, the event is picking a non-green ball, which occurs with probability $4/6$; and with probability $3/6$, the event is picking a non-red ball, which occurs with probability $3/6$.

If you are counting the first pick as well, the answer is simply $1+{17\over10}={27\over10}$.

To prove the general fact that $E(p)=1/p$, you can either set up and sum the infinite series $\sum_{k=1}^\infty kp(1-p)^{k-1}$ or use the fact that after the first pick you're either done in one round, which happens with probability $p$, or, with probability $1-p$, you're back where you started, but with one round under you're belt, i.e.,

$$E(p)=p\cdot1+(1-p)(1+E(p))$$

Expectation of drawing the second color from an urn
So recently in physics class, we learned about the magnetism right hand rules. 

One of them states that the index finger points in the direction of the velocity of a particle, the middle finger points in the direction of the magnetic field, and the thumb points in the direction of the magnetic force.

I'm curious why the magnetic field vector is perpendicular to the magnetic force vector.
