# The need for the Ising model in Mean field theory?


My thought is that it is the latter since $\mean{\vec S_j}$ is not necessarily zero when $\vec B$ is.

You are correct when saying that $\vec m_j = \langle\vec S_j \rangle$ is not necessarily aligned with $\vec B$.
We can write your second equation as $$H_i = -JS_i \cdot (N\vec m + \vec B)$$ where $\vec m$ is the average of the $\vec m_j$s over $j$. It should be intuitively clear that if -- for any reason -- a majority of spins happens to be aligned along an arbitrary axis, then $|\vec m|$ is large, and the $i$-th spin described by the Hamiltonian above will align with $\vec m$, if $\vec B$ is small (even if nonzero). This means that this "arbitrary axis" solution is stable.
Your intuition did tell you that there must be something wrong, considering the system is rotationally symmetric. Indeed, in phase transitions, systems do undergo spontaneous symmetry breakings. In the case $B=0$, one could think that the total magnetisation must be zero because of symmetry, but that is not necessarily the case.