Toroidal magnetic field of EAST tokamak

Using Ampere's circuital law

$$\oint \mathbf{B} \cdot d\mathbf{l}= \mu_0 I,$$ (514)

along the toroidal direction and assuming perfect toroidal symmetry, Eq. (514) is written

$$2 \pi R B_{\phi} = \mu_0 I,$$ (515)

which gives

$$B_{\phi} = \frac{\mu_0 I}{2 \pi R} .$$ (516)

Neglecting the poloidal current contributed by the plasma, the poloidal current is determined solely by the current in the TF (Toroidal Field) coils. The EAST tokamak has 16 groups of TF coils, with each coils having 130 turns of wire (I got to know the number of turns from Dr. Sun Youwen). Denote the current in a single turn by $I_s$, then Eq. (516) is written

$$B_{\phi} = \frac{\mu_0 \times 16 \times 130 \times I_s}{2 \pi R} = 4.160 \times 10^{- 4} \frac{I_s}{R}$$ (517)

Using this formula, the strength of the toroidal magnetic field at $R = 1.7 m$ for $I_s = 10^4 A$ is calculated to be $B_{\phi} = 2.447 T$. This field ( $I_s = 10^4 A$, and thus $B_t = 2.447$ at $R = 1.7 m$) was chosen as one of the two scenarios of EAST experiments during 2014/6-2014/9 (another scenario is $I_s = 8 \times 10^3 A$). (The limit of the current in a single turn of the TF coils is $14.5 \ensuremath{\operatorname{kA}}$ (from B. J. Xiao's paper [13]).

Note that the exact equilibrium toroidal magnetic field $B_{\phi}$ is given by $B_{\phi} = g (\Psi) / R$. Compare this with Eq. (516), we know that the approximation made to obtain Eq. (517) is equivalent to $g (\Psi) \approx \mu_0 I_{\ensuremath{\operatorname{TF}}} / 2 \pi$, i.e. assuming $g$ is a constant function of $\Psi$. The poloidal plasma current density is related to $g$ by $g' (\Psi) B_p / \mu_0$. The constant $g$ corresponds to zero plasma poloidal current, which is consistent to the assumption used to obtain Eq. (517).