Kalb-Ramond vs electromagnetic field As I understand it string theory says that charged particles like electrons and quarks are actually open strings whose endpoints interact with the electromagnetic field $A_\mu$.
String theory also says that the bulk of these open strings interact with the Kalb-Ramond field $B_{\mu\nu}$.
So is a charged particle a source for both fields $A_\mu$ and $B_{\mu\nu}$?
Could we detect the $B_{\mu\nu}$ field using charged particles?
 A: While I cannot answer your first question, I may provide a partial answer to your second question.
A charged particle, represented by the worldline $x^\mu(s)$ for  $s\in \mathbb{R}$, usually couples to the electromagnetic field in the Lagrangian via a term
$$ 
A_\mu \dot{x}^\mu(s),
$$
where $\dot{x}^\mu(s) = \frac{\mathrm{d}}{\mathrm{d}s} x^\mu(s)$.
By the antisymmetry of $B$, we immediately see that a similar term involving $B$ cannot appear in the Lagrangian of a particle:
$$
B_{\mu \nu} \dot{x}^\mu(s) \dot{x}^\nu(s) \equiv 0
$$
You can also see that, if you start with the usual coupling to a string via a Weiss-Zumino term
$$
\varepsilon^{\alpha \beta}B_{\alpha \beta},
$$
where $\alpha, \beta$ are coordinates on the string worldsheet.
Assuming a point-like string means that we have in adapted coordinates
$$
(x^{\hat{\mu}}(\tau, \sigma), x^z(\tau, \sigma)) = (x^{\hat{\mu}}(\tau), R \sigma),
$$
i.e. we perform a kind of KK dimensional reduction (${\hat{\mu}}=0,...D-2$).
Plugging this ansatz in the Weiss-Zumino term, we find that
$$
\varepsilon^{\alpha \beta}B_{\alpha \beta} = -2R\ B_{{\hat{\mu}} z} \frac{\mathrm{d}}{\mathrm{d}\tau} x^{\hat{\mu}}(\tau).
$$
So the particle only couples to part of the KR field, the part that transforms exactly as a U(1) field and which may or may not be the same as the field $A$ in your question.
