$f=\frac{q B}{2 \pi m} \quad\left(\because T =\frac{2 \pi m}{q B}\right)$
$\Rightarrow 10 \times 10^6=\frac{1.6 \times 10^{-19} \times B}{2 \pi\left(1.6 \times 10^{-27}\right)}$
$B=0.656 \,T$


$(A)$ $\vec{B}(x, y)$ 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


