A proof of magnetic forces doing no work? The proof of magnetic forces do no work is given in Introduction to Electrodynamics by David J. Griffiths like this 

Consider a charged particle $q$ moving with velocity $\mathbf{v}$ in a magnetic field $\mathbf{B}$ then the force on it is $\mathbf{F}= q(\mathbf{v}\times {B})$ and if $q$ moves an amount $d\mathbf{l}=\mathbf{v}dt$, the work done is $$ dW= \mathbf{F} \cdot d\mathbf{l} = q(\mathbf{v}\times \mathbf{B}) \cdot \mathbf{v}dt =0$$ hence, magnetic forces do no work.  

My problem is why he has replaced $d\mathbf{l}$ with $\mathbf{v}dt$? This substitution implies that the charged particle was moving with $\mathbf{v}$ only and no force acted on it because when that magnetic force will act on it will cause displacement in a direction perpendicular to $\mathbf{v}$ and force is also in the same direction there the expression for work should not become zero. When he made that substitution he didn't take into account that the direction of velocity will get changed as soon as force comes into the action. I think that particle should move in the direction of the force and we should take $dl$ in the direction of the force and not in the direction of the previous velocity. My main problem is that substitution, $d\mathbf{l} = \mathbf{v}dt$, beaucse what I think is that as force is applied the direction of velocity will get changed and the new velocity $\mathbf{v'}$ will have some direction in the direction of $\mathbf{F}$ and $d\mathbf{l}= \mathbf{v'}dt$. If we assume that $v' = v$ isn't the same thing that we want to prove? However, v' and v and can't have the same direction, the magnetic force would at least change the direction of the velocity.  
Thank you. Any help will be much appreciated.  
 A: When $\mathbf{v}$ changes to $\mathbf{v'}$, $\mathbf{F}$ will change to $\mathbf{F'} = \mathbf{v'} \times \mathbf{B}$. As the velocity changes, the force changes as well. This way, the force is always perpendicular to the velocity, resulting in no work done.
For the purpose of calculus, the symbol $\textrm{d}\mathbf{l}$ should not be taken to mean a small but finite step, but as a limit as the time internal goes to zero. In less precise language, we only consider a time interval small enough that the magnetic force doesn't have enough time to change the velocity by any significant amount. This is why $\textrm{d}\mathbf{l} = \mathbf{v}\textrm{d}t$.
A: For magnetic forces, force is always perpendicular to displacement, hence they do no work. The more pertinent question is why is dl always $\parallel$ dv?.
For any reasonable* function$f(t)$, $df(t)$ is
$$df=f'(t)dt+f''(t)\frac{dt^2}{2}+...$$
$\therefore$ the displacement dl at time t can be calculated via 


*

*$d\textbf{l}=\textbf{v}(t) dt+\textbf{F}(t)\frac{dt^2}{2}+...$
($2^{nd}$ order derivative from Newton's law)


Similarly,  


*$d\textbf{v}=\frac{\textbf{F}(t)}{m}dt+...$
($1^{st}$ order derivative from Newton's law)
Clearly, 1. shows that dl need not necessarily be $\parallel$ to v. This is unlike 2. where Newton's law dictates that we terminate the series at the first term.   
For 1. however,is where the "trickery" of calculus comes in. We assume that second order changes don't contribute in the limit $dt\to 0$(the first term would be dominant anyway). This is reasonable as long as v doesn't vary too rapidly**. This in turn depends on F being well behaved. I am guessing that smoothness of F is enough to get away with this. Only for pathological functions would this assumption break down.
Physically, this means that all changes from F(t) over dt are considered only at the epoch $t+dt$. As a result, the infinitesimal displacement is always $\parallel$ v. 
You are right in thinking that the displacement caused due to external F isn't parallel to v. But this "actual" displacement(that which would be expected from the enacting force) doesn't contribute to the net displacement (and therefore work) till $t+dt$. At $t+dt$ we will update the v (by dv) and therfore l (by dl) and restart from this new $t'=t+dt$. Thats just how we calculate infinitesimals.
Then what about work done by F? Shouldn't that be calculated using only the "actual" displacement from F? No. The definition of work is only concerned with the displacement of the particle-doesn't matter where it comes from. At every instant, one can calculate this entirely in terms of v.
Does this mean that the "actual" work done by F is being ignored? Not at all. Remember that the effect of F over $dt$ shows up in v at $t+dt$. So while calculating displacement for $t+dt$ to $t+2dt$, the displacement would have a contribution $\mathbf{F}dt^2$

*: the function must be smooth and its taylor series must exist.
**: if it did, the $dt^2$ term would contribute even when $dt\to0$
