$\mathrm{I}=\frac{\varepsilon}{\mathrm{R}+\mathrm{G}} \quad \mathrm{G}=\frac{1}{9}\, \mathrm{K}\, \Omega$
$\frac{1}{2}=\frac{\varepsilon}{R+\frac{G S}{G+\bar{S}}} \times \frac{S}{S+G} \Rightarrow \frac{1}{2}=\frac{\varepsilon S}{R(S+G)+G S}$
$S=\frac{R G \times \frac{1}{2}}{\varepsilon-\frac{(R+G) I}{2}}$
$S=\frac{11 \times 10^{3} \times \frac{1}{2} \times 10^{2} \times 270 \times 10^{-6}}{6-\left(\frac{6}{2}\right)}=110\, \Omega$


$(A)$ If $I_1=I_2$, then B' cannot be equal to zero at the origin $(0,0,0)$
$(B)$ If $\mathrm{I}_1>0$ and $\mathrm{I}_2<0$, then $\mathrm{B}$ can be equal to zero at the origin $(0,0,0)$
$(C)$ If $\mathrm{I}_1<0$ and $\mathrm{I}_2>0$, then $\mathrm{B}$ can be equal to zero at the origin $(0,0,0)$
$(D)$ If $\mathrm{I}_1=\mathrm{I}_2$, then the $\mathrm{z}$-component of the magnetic field at the centre of the loop is $\left(-\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}}\right)$
$(A)$ $\vec{B}(x, y)$ is perpendicular to the $x y$-plane at any point in the plane
$(B)$ $|\vec{B}(x, y)|$ depends on $x$ and $y$ only through the radial distance $r=\sqrt{x^2+y^2}$
$(C)$ $|\vec{B}(x, y)|$ is non-zero at all points for $r$
$(D)$ $\vec{B}(x, y)$ points normally outward from the $x y$-plane for all the points between the two loops
