Is dimensional analysis always sufficient to establish equivalence of quantities? In dealing with the Biot-Savart law, it was argued that 
$$
q\frac{d\vec{s}}{dt}\equiv Id\vec{s}
$$
using the fact that the units are equal. Does this kind of argument always work? It seems too simple to be true.
 A: No, it doesn't always work. Sometimes there are different quantities with the same dimensions that could go in a formula, and sometimes there are numerical constants that dimensional analysis won't give you. But in a situation where you're not dealing with many variables, dimensional analysis does help drastically narrow down the set of possible relationships between them, so it can give you useful starting points for further experimentation.
A: One should recall that things like multiplication in DA can come from dirrerent natural multiplies.  For example, div, grad zand curl all reduce to 1/L.  The dot and cross product of force and length give energy and force, but these are different.  Likewise, pressure and energy density are both M/LT^2, but are different.
The values are not equal if one applies alternate dimensions.  The lhs has d/d, while the rhs has only d.  This cdoes not show up in standard dimensions, but the standard open algebra, but a differen set of folds shows it clearly.
It's not true in cgs units either.
On the other hand, one has q.dl/dt, is dq.l/dt, at qv =il, is based on imcrements in two of the three factors.
A: No not always.By dimensional analysis you usually get the rough idea what it is but not always.There are some exceptions.For example:-In uniform motion we have distance covered in $\mathrm {n}^{th}$ second will be  $\mathrm{S}_{nth}$=$u +\frac{a}{2}[2n-1]$.Where u is initial velocity,a is constant acceleration and s_nth is distance covered in 
$\mathrm {n}^{th}$ second. Here we must see that $\mathrm{s}_{nth}$ has unit of metre[$M^0L^1T^0$] while u has unit metre/second
[$M^0L^1 \mathrm{T}^{-1}$] and acceleration has [$M^0L^1 \mathrm{T}^{-2}$].Hence it appears dimensionally incorrect but on solving we get this result.Here u is multiplied by $1(second)$ and the term $\frac{a}{2}[2n-1]$ by $1\mathrm{(second)}^{2}$. Hence we see that we cannot analyse by justing observing the formula.
